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CHEMISTRY Vol. XXII
No. 5
May 2013
he system of Ayurveda is as old as the ancient history of India. Luckily for us, in spite of repeated invasions, we had retained the purity of our ancient knowledge. We are also absorbing modern systems of knowledge followed by research of the highest order, at par with any nation.
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Mahabir Singh Anil Ahlawat (BE, MBA)
Contents AIIMS Special: Assertion and Reason
Research on Superspecialities beckons you
T
Corporate Office:
Managing Editor Editor
edit Q rial
6
India has a large population because the area of the continent is as much as Europe and Britain, excluding the older Russia. A large population with many states and universities and research institutions is our asset. We also need a large quantity of allopathic medicines for our people. Many of them are patented. These medicines are also having forbidding costs. However, many of their patents also lapse. Are we ready to substitute the generic medicines to replace imported ones?
NEET Special (Scrapbook)
13
NCERT Xtract
21
Practice Paper: AIIMS
29
Practice Paper: JEE Advanced
38
attacking the parasites in blood, their mode of action and applicability have not been tried before for our systems of medicine.
Quick Revision of
46
It is a life-time research for many scientists and for generations this will be continued.
Inorganic Chemistry (Part-ll)
We have to completely map the medicines, study their structure and reproduce the generic one of every available medicine including those with patents so that the day the patents lapse for them, one can start making them for our use. We know that there are very powerful and useful medicines produced by not only Ayurvedic laboratories but also Unani and Homeopathy. We know their composition. But w e do not know which molecule is
Solved Paper: JEE Main
60
Every single instrument ranging from X-ray spectroscopy, diffraction and scanning, U.V., I.R. etc. should be used to study them. Then experiments on their action on various molecules of the polluted blood has to be continued.
Practice Paper: Target PMTs
64
Once one can establish their utility for controlling the disease is proved, one has taken a wicket. The game continues. Don't you like to play super-games?
Concept Booster
72
Anil Ahlawat Editor Subscribe online a t W W W . m t g . i n
Brainstorming Problems
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2 CHEMISTRY TODAY | MAY '13
ASSERTION
REASON
Directions : Mark the correct choice as:
AIIMS Special
(a)
If both assertion and reason are true and reason is the correct explanation of assertion.
(b)
If both assertion and reason are true but reason is not the correct explanation of assertion.
(c)
If assertion is true but reason is false.
(d)
If both assertion and reason are false.
PHYSICAL CHEMISTRY 1.
2.
3.
Assertion : The volume of a gas is directly proportional to the number of moles of the gas. Reason : The ratio of the volumes of the reactants to products are in agreement with their molar ratio. Assertion : The decomposition of NH 3 on finely divided platinum surface is first order when the concentration is low, however at higher concentration, the reaction becomes zero order. Reason : In first order reaction, the rate of reaction is proportional to the first power of the concentration of the reactant.
Reason
6.
Reason
so2fe) + Cl2fe),
the reaction shifts to backward direction on increasing pressure. : The increase in pressure at equilibrium shifts the reaction towards lesser number of moles.
Reason
8.
4. , Assertion : Zinc blende and wurtzite both have fee arrangement of sulphide ions. Reason : There are four formula units of ZnS in both. Assertion : Enthalpy changes are positive w h e n Na2S04 -10H20, CUS04 -5H20
3 CHEMISTRY TODAY | MAY '13
methane, methanol etc. as fuels are called fuel cells. : They are designed to convert the energy of combustion of fuels directly into electrical energy.
Assertion : BF3 is a weaker Lewis acid than
and
salt like NaCl, KC1 etc. (which do not form hydrates) are dissolved in water. But enthalpy changes are
in
acidic medium is M/5 (M = mol. wt.) while in basic medium, it is equal to M/3. : In acidic medium, 1 mol of M n 0 4 gains 5 electrons while in basic medium it gains 3 electrons.
Assertion : Galvanic cells containing hydrogen,
Reason
9.
structure. : Se in SeCl 4 has two lone pairs of electron.
Assertion : Equivalent weight of K M n 0 4
Assertion : In the reaction,
so2ci2te)
5.
Assertion : SeCl 4 does not have a tetrahedral Reason
7.
negative when anhydrous salts capable of forming hydrates are dissolved in water. : The difference in the behaviour is due to large difference in the molecular weights of hydrated and anhydrous salts. The substance with larger molecular weight usually shows positive enthalpy changes on dissolution.
Reason
BC13. : The planar BF3 molecule is stabilized to a greater extent than BC13 by B - X n -bonding.
10. A s s e r t i o n : l n
Reason
Sommerfeld
model,
circular
orbit is possible for the electron in K-shell. : Electron present in JC-shell c o r r e s p o n d s to principal q u a n t u m n u m b e r n = 1. ORGANIC CHEMISTRY
11. A s s e r t i o n : Cyclohexanone Reason
exhibits
keto-enol
tautomerism. : In cyclohexanone, one form contains the keto g r o u p (C = 0 ) while other contains enolic g r o u p (-C= C-OH) •
12. A s s e r t i o n : Oxiranes are cleaved u n d e r acidic Reason
a n d basic conditions. : The oxirane ring is strained a n d the tetrahedral b o n d angle (109.5°) is distorted to 61.5
13. Assertion : Greater the stability of carbanion, greater is its ease of f o r m a t i o n a n d hence, m o r e reactive is the aryl halide. Reason Chlorobenzene is less reactive t h a n
reaction
Reason
19. A s s e r t i o n : Fructose does n o t contain an a l d e h y d e g r o u p b u t still reduces Tollen's reagent. Reason : In presence of a base, fructose u n d e r g o e s r e a r r a n g e m e n t to f o r m a molecule containing an a l d e h y d e group. O 20. A s s e r t i o n : H — C — O H
: S e 0 2 oxidizes the m e t h y l e n e g r o u p or ketones to carbonyl g r o u p .
15. A s s e r t i o n : m-Chlorobenzoic acid is a stronger acid t h a n p-chlorobenzoic acid. : In wi-chlorobenzoic acid both - /-effect a n d + R- effect of CI operate b u t in p-chlorobenzoic acid only +R- effect of CI operates.
16. Assertion : [10]-Annulene
Reason
is
O II 17. A s s e r t i o n : ( C H 3 ) 2 C H — C — N H 2
8
not
aromatic
t h o u g h it contains H u c k e l n u m b e r of Tt-electrons. : Steric interaction b e t w e e n internal h y d r o g e n s m a k e s it non-planar.
faster
CHEMISTRY TODAY | MA Y '13
is
less
acidic
than
O II HO-C-OH. : - H is electron releasing while - O H is electron w i t h d r a w i n g g r o u p . INORGANIC CHEMISTRY
adjacent to C = 0 g r o u p in a l d e h y d e s
Reason
: (CH 3 ) 2 CH— migrates faster t h a n C H 3 — C H 2 — at electron-deficient n i t r o g e n atom.
directly
into glyoxal. Reason
to
18. A s s e r t i o n : M e 3 N reacts w i t h BF 3 w h e r e a s P h 3 N d o e s not. Reason : The electron pair o n nitrogen a t o m in P h 3 N is delocalised in the b e n z e n e ring a n d is n o t available to b o r o n in BF 3 .
Reason dioxide can be converted
comparison
O II CH3-CH2- C-NH2-
p-chloroanisole t o w a r d s nucleophilic substitution reactions. 14. Assertion : Ethanal o n t r e a t m e n t w i t h selenium
in
undergoes
Hofmann-bromamide
21. A s s e r t i o n : N i t r o g e n is unreactive at r o o m t e m p e r a t u r e b u t b e c o m e s reactive at elevated t e m p e r a t u r e s (on heating) or in the presence of catalyst. Reason : In nitrogen molecules, there is extensive derealization of electrons. 22. A s s e r t i o n : F - a t o m h a s less negative electron affinity t h a n CI-atom. Reason : A d d i t i o n a l electrons are repelled m o r e effectively b y 3p- electrons in C I - a t o m t h a n b y 2p- electrons in F-atom. 23. A s s e r t i o n : A1 is obtained b y h i g h t e m p e r a t u r e r e d u c t i o n of a l u m i n a w i t h carbon. Reason : A l u m i n a reacts w i t h carbon to f o r m a l u m i n i u m carbide w h i c h d e c o m p o s e s at h i g h t e m p e r a t u r e to
24. Assertion :
Reason
:
25. Assertion : Reason
:
form Al, while carbon is oxidized to CO. The degree of complex formation in actinides decreases in the order M4+>MO22+>M3+>MO2. Actinides form complexes with Ji-bonding ligands such as alkyl phosphines and thioethers. In water, orthoboric acid behaves as a weak monobasic acid. In water, orthoboric acid acts as a proton donor.
However at higher concentration, the surface of the catalyst is fully adsorbed, so any amount of ammonia a d d e d does not change the adsorption process so the reaction rate becomes independent of the concentration and it becomes zero order reaction. 3.
(a): As the pressure of the gaseous system is increased the volume occupied by the system decreases so that total number of moles per unit volume increases. To u n d o this stress, the reaction proceeds towards lesser number of moles of gaseous substances so that the increased pressure is reduced. This is in accordance to Le Chatelier's principle.
4.
(d) : Wurtzite is built from two interpenetrating hep lattices as opposed to the fee lattices in zinc blende. Wurtzite has two ZnS while zinc blende has four ZnS in unit cell.
5.
(c): The exothermic or endothermic behaviour of dissolution depends only u p o n the nature of the salt and not on its molecular weight.
6.
(c): SeCl 4 has distorted trigonal bipyramidal structure. Se in SeCl 4 has one lone pair of electron.
7.
(a): Mn04" + 8H + + 5e~ • Mn 2+ + 4H z O (In acidic medium) M n 0 4 + 2 H 2 0 + 3e~ • M n 0 2 + 4 0 H (In basic medium)
8.
(a)
9.
(a): There is greater pn-pn
26. Assertion : Ozone is a more powerful oxidizing agent than O z . Reason
: Ozone is diamagnetic but 0 2 paramagnetic.
27. Assertion : [Ti(H z O) 6 ] 3+ is coloured [SC(HZO)6]3+ is colourless. Reason
is
while
: d-d transition is not possible in [SC(H 2 0) 6 ] 3+ .
28. Assertion : Lanthanoids show a limited number of oxidation states whereas actinoids show a large number of oxidation states. Reason : Energy gap between 4/, 5d and 6s-subshells is large, whereas that between 5/, 6d and 7s-subshells is small. 29. Assertion : In fused state, calcium chloride cannot be used to dry alcohol or NH 3 . Reason
: CaCl 2 is not a good dessicant.
30. Assertion : Wolframite impurities are separated from cassiterite by electromagnetic separation. being magnetic is Reason : Cassiterite attracted by the magnet and forms a separate heap. SOLUTIONS 1.
(b): V =
,i.e.,V<*n- Volumes of gases are in
the ratio of their moles. 2.
10
(b): In a heterogeneous system, the reactant is absorbed on the surface of a solid catalyst. The fraction of the surface of the catalyst covered by the reactant is proportional to its concentration at low values and the rate of reaction is first order. CHEMISTRY TODAY | M A Y ' 1 3
back bonding in BF3.
10. (b): Sommerfeld model is an extension of Bohr's model. Electrons in an atom revolve around the nuclei in elliptical orbit. The circular path is a special case of elliptical orbit. 11. (a): In cyclohexanone, keto-enol tautomerism is as follows. O
Keto form
12. (b) 13. (c): On comparing the relative stabilities of carbanion of chlorobenzene and p-chloroanisole, CL^H
OCH3
The electron d o n a t i n g g r o u p ( - O C H 3 ) in anisole t e n d s to intensify the negative charge relative to carbanion in chlorobenzene. Thus, p-chloroanisole is less reactive t h a n chlorobenzene.
23. (c) 24. (b): H i g h e r the charge o n the metal ion, smaller is the ionic size a n d m o r e is the complex f o r m i n g ability. Thus, the d e g r e e of complex f o r m a t i o n
14. (a)
decreases in the o r d e r M 4 + > M O f + > M 3 + > M 0 2 .
15. (c): m-chlorobenzoic s h o w only - I effect b u t p-chlorobenzoic acid s h o w b o t h - I a n d +I< effects, hence m-chlorobenzoic acid is a stronger acid.
MO|+
16. (a): D u e to severe n o n - b o n d e d interactions b e t w e e n the internal h y d r o g e n s (as s h o w n in figure), the ring a s s u m e s n o n - p l a n a r geometry.
17. (a)
18.
(a)
19. (a): U n d e r basic conditions, the fructose molecules have the location of the carbonyl b o n d switched to convert t h e m into a glucose molecule. 20. (d): Peroxycarboxylic acids are a b o u t 1000 times weaker t h a n the p a r e n t carboxylic acid, o w i n g to the absence of resonance stabilization of the anion. 21. (c): At elevated t e m p e r a t u r e or in the presence of catalyst, strong b o n d s b e t w e e n t w o N - a t o m s r u p t u r e a n d m a k e s N 2 molecule reactive. 22. (c): Additional electrons are repelled m o r e effectively by 2p-electrons in F - a t o m t h a n b y 3p- electrons in Cl-atom.
The higher t e n d e n c y of complex f o r m a t i o n of as c o m p a r e d
to M 3 + is d u e
to
MO2+.
25. (c): Orthoboric acid (H 3 B0 3 ) is soluble in water a n d b e h a v e s as a m o n o b a s i c acid. It does not d o n a t e protons. Hence, it is not a protic acid b u t is a Lewis acid. H3B03 + 2H20 H 3 0 + + [B(OH) 4 ]" 26. (b): O z o n e can liberate nascent oxygen, so it acts as a p o w e r f u l oxidizing agent. o3 • o 2 + [O] 27. (a): [SC(H Z O) 6 ] 3+ h a s n o u n p a i r e d electron in its d-subshell a n d t h u s d-d transition is not possible w h e r e a s [Ti(H 2 0) 6 ] 3 + h a s one u n p a i r e d electron in its d-subshell w h i c h gives rise to d - d transition to i m p a r t colour. 28. (a) 29. (c): CaCl 2 f o r m s a d d i t i o n p r o d u c t s w i t h alcohol and ammonia. 30. (c): Wolframite ( F e W 0 4 ) b e i n g m a g n e t i c is attracted b y the m a g n e t i c roller a n d f o r m s a h e a p u n d e r it.
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12
CHEMISTRY TODAY | M AY '13
high
concentration of charge o n metal a t o m M in
OO
NEET Special
Mukul C. Ray, Odisha Inorganic Chemistry always provides breathing space in I IT entrance examination as these questions take very little time to answer. Also there is hardly any chance (except a few chapters) where we see twists and turns in questions. So it is more important for you to go through proper texts rather than practicing the questions. These few pages will give you a vivid revision of general discussion on metals. No need to remember the data such as melting points, radii etc. Just remember the trend. They are incorporated for your convenience.
GROUP ! A
O
Members : Li, Na, K, Rb, Cs •
General Properties
O
Size Li
Na
K
Rb
v a n der Waals radii (in p m )
180
231
231
244
Metallic radii (in p m )
157
Covalent radii (in p m )
123
Ionic radii (M + ) (in p m )
60
+
H y d r a t e d salts : Except Li salts, they crystallise f r o m a q u e o u s solutions in the a n h y d r o u s state. (Explanation : The h y d r a t i o n energy is not sufficient to overcome the loss in lattice energy d u e to inclusion of the larger h y d r a t e d ion). Li + f o r m s trihydrate : LiX.3H 2 0, X = C f , Br , I", ClOy C104", M n 0 4 , N O " N o t e s : Li + does not f o r m a l u m s probably because
191
235
the small Li(H z O)g is too small to form a stable
250
lattice r e q u i r e d for a l u m formation. 154
203
—> N a B i O , perchlorates of K, Rb and Cs are
216
w a t e r insoluble.
+
95
133
148
+
Note : K (133 p m ) , Cu (96 p m ) Ag (126 p m ) O Melting points, boiling points a n d h e a t of atomisation all decreases g r a d u a l l y f r o m t o p to bottom. Density: Li Na K Rb Cs (g/mL at 20°C) 0.534 0.968 0.86 1.53 1.87 Note : K h a s lower density t h a n Na, also Li, Na, K h a v e density values less t h a n the density cf water. Li is the lightest a m o n g all metals k n o w n . O Relative h y d r a t e d ionic radii decreases in the o r d e r : Li + > N a + > K + > Rb + > Cs + a n d c o n d u c t i n g p o w e r of ions in aq. solution. Cs T > Rb + > K + > N a + > Li + O S t a n d a r d oxidation p o t e n t i a l : Li Na K Rb Cs E° 0P (in volts) 3.05 2.71 2.93 2.93 2.99 Note : Li is the strongest r e d u c i n g agent in a q u e o u s solution.
O
Complexing ability : It decreases from Li > N a > K.
•
Abnormal behaviour of Lithium
O
It is immiscible w i t h molten K, Rb a n d Cs but miscible w i t h N a only above 380°C.
O
L i 2 C 0 3 a n d L i N 0 3 d e c o m p o s e readily o n heating to oxides like M g salts.
O
Li o n h e a t i n g w i t h N 2 f o r m s Li 3 N; others fail.
O
M a n y simple lithium salts are normally h y d r a t e d a n d the a n h y d r o u s salts are extremely hygroscopic (e.g. LiCl, LiBr). N o t e : As r e g a r d s nitrides, very h i g h lattice energy is r e q u i r e d to overcome the h i g h b o n d energy of n i t r o g e n p l u s its negative electron affinity. This seems to b e attained only in Li 3 N.
O
Reactivity decreases f r o m Li to Cs. If you see it carefully it is exclusively because of kinetic factors. Because Li p r o d u c t s are t h e r m o d y n a m i c a l l y m o r e stable o w i n g to h i g h lattice energy a n d h y d r a t i o n energy. CHEMISTRY TODAY | MAY '13
13
2 -
•
Extraction
O
Lithium is usually extracted f r o m s p o d u m e n e [LiAl(Si0 3 ) 2 ]. It is converted into LiCl a n d Li is extracted f r o m f u s e d m i x t u r e containing 55% LiCl + 45% KC1 at 450°C.
O O
The peroxides contain peroxide ion, 0 2 a n d t h u s liberate H 2 0 2 w i t h w a t e r or dilute acids. M2Oz + 2HzO N a 2 0 2 reacts a s : Na202 + CO
S o d i u m is extracted b y electrolysis of f u s e d CaCl 2 (65%) a n d N a C l m i x t u r e at 505°C. Potassium is isolated b y r e d u c i n g m o l t e n KC1 in a packed fractionating tower b y rising N a - v a p o u r . KC1 + N a NaCl + K
Uses
O
Li metal is u s e d as scavenger in m e t a l l u r g y to remove 0 2 a n d N 2 gases. Li is a g o o d alloying metal a n d f o r m s h i g h strength, low density a l u m i n i u m alloys for aircraft construction.
O
N a metal is u s e d as a r e d u c i n g agent in the extraction of Ti, Zr etc. Na-K alloy is the m o s t f a v o u r e d material for h e a t exchange in fast breeder nuclear reactor. Note : A b r e e d e r reactor is a nuclear reactor capable of generating m o r e fissile material that it consumes. It p r o d u c e s fissile material f r o m Th 2 3 2 a n d U 238 .
•
General Chemical Behaviour
O
Oxides, Superoxides a n d Peroxides W h e n b u r n t in air, Li f o r m s L i 2 0 , N a gives N a 2 0 2 a n d K, Rb, Cs f o r m superoxides. Li z O
Na20
White
White
K20
Rb20
Cs20
Gold
Pale
Orange
white
yellow
These colours are m o s t probably d u e to lattice defects. Li202
Na202
K202
Rb202
Cs202
White
Pale
Orange
Dark
Yellow
yellow
brown
Colours are d u e to defective lattice a n d small a m o u n t s of superoxides. 14
CHEMISTRY TODAY | M A Y 13
> Na2COs
Na202 + C02
> Na2COs + 1 0
N a 2 0 2 + 2Na
> 2NazO
2
N o t e s : Solid d r y h y d r o x i d e s of Na, K, Rb a n d Cs interact w i t h 0 3 to f o r m M 0 3 (orange colour). Na02 K02 Rb02 Cs02
T h o u g h this is contrary to o u r expectation f r o m the reactivity t r e n d (K > Na). But this h a s b e e n possible because of h i g h volatility of K in the t e m p e r a t u r e range 850-880°C. N o t e : Electrolysis of f u s e d KC1 alone is also difficult as its melting p o i n t is 790°C a n d boiling point of p o t a s s i u m metal is 760°C. A t t e m p t s to lower the melting point b y a d d i n g other metal salts have also failed. Because K + ions h a v e a higher discharge potential t h a n the metal a d d e d a n d potassium is very soluble in m o l t e n p o t a s s i u m chloride. •
> 2MOH + H2Oz
Yellow
Orange
Dark brown
Orange
Superoxide ion h a s o n e u n p a i r e d e~; that is w h y s u p e r o x i d e s are coloured a n d paramagnetic. Crystals of Li z O, N a z O , K z O, R b 2 0 h a v e antifluorite structure. Superoxides react vigorously w i t h water. 2K02 + 2HzO > 2KOH + H 2 0 2 + 0 2 Their (KO z , RbO z , C s 0 2 ) t h e r m a l d e c o m p o s i t i o n give sesquioxides w i t h f o r m u l a M 2 0 3 . Solubility of s u p e r o x i d e s a n d oxides increases as the metal ions increases. The alkali metal h y d r o x i d e s are all very strong bases. Boiling point decreases gradually. Vapour mostly consists of dimers. In the gas phase, basic strength increases. In a q u e o u s solutions; they a p p e a r nearly equally strong d u e to levelling effect of the solvent. N a O H is mostly p r o d u c e d b y the electrolysis of brine solution u s i n g g r a p h i t e a n o d e a n d m e r c u r y cathode (brine solution is N a C l solution w i t h concentration r a n g i n g f r o m 3.5% to 26%). O
Halides: Negative value of lattice e n e r g y : LiX > N a X > KX > RbX > CsX (X = F, CI, Br, I) Negative value of e n t h a l p y of f o r m a t i o n : LiF > N a F > KF > RbF > CsF KC1 > CsCl > RbCl > N a C l > LiCl CsBr > KBr > RbBr > NaBr > LiBr Csl > Rbl > K I > N a l > Lil Melting p o i n t s of halides (°C): F
j CI
Br
r
Li
845
610
547
446
Na
995
808
755
661
K
856
772
748
677
Rb
775
717
682
642
Cs
682
645
636
621
In t e r m s of g r a p h ,
O
O
Li N a K Rb Cs N o t e s : D u e to small size of F ion, the lattice e n e r g y term p r e d o m i n a t e s . This leads to a decreasing trend in the e n t h a l p y of f o r m a t i o n f r o m LiF to CsF. For the large halides, lattice energy b e c o m e s less d o m i n a n t . —» Li salts h a v e very l o w melting points because Li + ions are quite small in size that results anion-anion repulsion. The actual Li - X distance is larger t h a n expected. For fluorides, solubility increases greatly f r o m LiF to CsF. (lattice e n e r g y decreases m o r e rapidly). For iodides, solubility decreases (hydration e n e r g y decreases m o r e rapidly) as N a l > Lil > Rbl > KI > Csl. For chlorides a n d bromides, the solubility o r d e r is : LiX > CsX > RbX > N a X > KX (X = C f , Br"). —» All the halides except CsCl, CsBr a n d Csl, a d o p t N a C l t y p e structure. Carbonates, b i c a r b o n a t e s : Thermal stability of carbonates change in the order : Li2C03 < N a 2 C 0 3 < K2C03 < Rb2C03 < Cs2C03 L i 2 C 0 3 is water insoluble. F r o m N a 2 C 0 3 o n w a r d s , solubility increases. N o t e s : H i g h t h e r m a l instability of L i 2 C 0 3 is attributed to t w o m a j o r factors; Li + ion exerts strong polarising action o n C 0 3 ~ ion a n d h i g h lattice e n e r g y of L i 2 0 t h a n L i 2 C 0 3 . Except Li, all other alkali metals f o r m solid bicarbonates. L i H C 0 3 is k n o w n only in solution. Stability increases f r o m L i H C O s to CsHC03. —> Solubility of bicarbonates increases f r o m Li to Cs. Bicarbonates are less soluble t h a n carbonates. —> N a H C O s h a s a n infinite chain of unsymmetrical H-bonds. K H C 0 3 is dimeric. Lower solubility of N a H C 0 3 is probably d u e to its H - b o n d e d structure. Nitrates a n d Sulphates : Except LiNOj, other alkali metal nitrates decompose o n heating, giving off oxygen. L i N 0 3 decomposes directly to the oxide. Thermal stability of nitrates increases with increasing atomic number. Nitrites m a y be p r e p a r e d b y h e a t i n g nitrates w i t h 'C.
2RbN03 + C > 2RbNOz + C 0 2 Nitrites d i s p r o p o r t i o n a t e o n heating in absence of air. 5NaNOz 3 N a N O s + N a 2 0 + N2 All the s u l p h a t e s (except Li 2 S0 4 ) are water soluble. O n f u s i o n w i t h carbon, they are reduced to c o r r e s p o n d i n g sulphides. M 2 S 0 4 + 4C > M2S + 4CO N o t e : LiCl a n d L i N 0 3 are soluble in alcohol. LiCl is deliquescent while NaCl, KBr are not. Hydrides: Stability o r d e r of h y d r i d e s decreases in the o r d e r : LiH > N a H > K H > R b H > C s H This is because of lowering of lattice energy d o w n the g r o u p . Solutions in N H 3 ( ( ) : Alkali metals dissolve in NH 3{() w i t h o u t evolution of H 2 . The solubility increases f r o m Li to N a to K. The solutions are b l u e w h e n dilute b u t become b r o n z e coloured at higher concentration. They are good conductors; conductivity decreases as the solution becomes m o r e a n d more concentrated a n d then again increases sharply after 0.05 M concentration. Saturated solution practically conduct electricity like liquid metals. The dilute solutions are paramagnetic. The paramagnetic susceptibility decreases with increasing concentration. A r o u n d the region of m i n i m u m conductivity, the solution becomes diamagnetic. At still higher concentrations, the solution again b e c o m e weakly paramagnetic. The solutions of alkali metals in liq. N H 3 are not very stable; they readily f o r m amides.
O
O
M + NH3
» M N H 2 + i H2
But in absence of catalytic impurities a n d u n d e r strictly a n h y d r o u s condition, such solutions are fairly stable. N o t e : At h i g h concentration Mam a n d eam exist as (M a m ) 2 (e am ) 2 , a d i m e r that lowers conductivity a n d p a r a m a g n e t i s m . At very high concentration, there are n o liq. N H 3 molecules to solvate e~. GROUP-II A M e m b e r s : Be, Mg, Ca, Sr, Ba •
General Properties
O
Size
Covalent radii (in p m ) Metallic radii (in p m ) Ionic size (M 2+ ) (in p m )
Sr
Ba
Be
Mg
Ca
89
136
174"! 192
198
112
160
197
215
224
31
65
99
113
135
CHEMISTRY TODAY | MAY '13 1 5
Note : Ca 2+ (99 pm), Co 2+ (72 pm), Ni 2 + Zn 2 + (74 p m ) O Melting points (°C): Be Mg Ca Sr 1287 650 839 770 O Density ( g / m L ) : Be Mg Ca Sr 1.84
(69 pm),
Ba
o
Be a n d M g d o not i m p a r t colour to flame.
O
Be a n d M g carbonates are n o t precipitated b y ( N H 4 ) 2 C 0 3 in presence of NH 4 C1. Be 2 C reacts w i t h water to give C H 4 w h e r e a s m a g n e s i u m carbide a n d calcium carbide give p r o p y n e a n d acetylene respectively. Be 2+ ions are extensively h y d r a t e d in aq. solution. D u e to small size of Be 2+ ion, O — H b o n d s get w e a k e n e d a n d the solution b e c o m e s distinctly acidic.
O
727 Ba
O
1.74
1.55 2.63 3.59 N o t e : Ca h a s the lowest density a m o n g s t alkaline earth metals.
N o t e : Be f o r m s very stable [Be(H 2 0) 4 ] 2 + while the other metal ions t e n d to achieve 6 co-ordination. For e x a m p l e [Ca(H 2 0) 6 ] 2 + .
Be, M g - hep, Ca, Sr -fee, Ba - bee O
Standard oxidation p o t e n t i a l : E° op
Be
Mg
Ca
Sr
Ba
1.85
2.37
2.87
2.89
2.91
(in volts) O
Specific heats of alkaline earth metals are greater t h a n those of alkali metals. H y d r a t e d salts : D u e to small size of M 2 + , the alkaline earth metal salts are extensively h y d r a t e d . Examples are C a C l 2 . 6 H 2 0 a n d MgCl 2 .6H z O. Size of h y d r a t e d ion decreases as : Be 2+ > Mg 2 + > Ca 2+ > Sr 2+ > Ba 2+ Ionic mobility : Be 2+ < Mg 2 + < Ca 2+ < Sr 2+ < Ba 2+ General Reactions
O
O •
Be
Mg
Extraction
O
Be is extracted f r o m beryl, 3BeO. A1 2 0 3 . 6Si0 2 . It is first converted to basic carbonate w h i c h is t h e n ignited to BeO. Finally, BeO is h e a t e d w i t h ' C at 600-800°C.
O
Ca-Ba
Air
N o reaction below 600°C but p o w d e r e d metal b u r n s to give BeO + Be 3 N 2
Brilliantly b u r n s to give MgO + Mg3N2
M O + M3N2 Ca —> mainly nitride Ba —» mainly oxide
C
Be 2 C at 1700°C
M g C 2 at red heat
M C 2 at red heat
N o reaction
amalgam reacts to give H 2
M(OH)2 + H2
Cold/ hot water
•
Steam
N o reaction
MgO + H2
M(OH)2 + H2
N2
Be 3 N 2 at 1000°C
M g 3 N 2 at 800 - 850°C
M3N2 on gentle heating
•
Abnormal behaviour of Beryllium
O
Be dissolves in a q u e o u s mineral acids (HC1, H N 0 3 , H 2 S 0 4 ) evolving H 2 , t h o u g h conc. H N 0 3 r e n d e r s it passive. Also Be dissolves in a q u e o u s caustic alkali f o r m i n g s o d i u m beryllate, N a 2 B e 0 2 .
O
BeO is amphoteric in n a t u r e w h e r e a s other oxides are basic.
16
CHEMISTRY TODAY | MAY ' 1 3
O
BeO + C + Cl 2 BeCl 2 + C O A mixture of BeCl 2 a n d N a C l is electrolysed at 800°C to obtain Be. M g is extracted by the electrolysis of f u s e d MgCl 2 (+ CaCl 2 + NaCl) at 750°C. 'C' reduction of M g O (2000°C) also give Mg: k MgO + C Mg + CO Note : M g H 2 d e c o m p o s e s at 250°C a n d low 'P' to give a highly reactive f o r m of Mg. Ca is extracted b y electrolysis of f u s e d CaCl 2 (plus CaF 2 to lower the m.pt. f r o m 774°C to 664°C). Note : R a d i u m is recovered in small quantities f r o m u r a n i u m minerals - less t h a n 100 g a year over the w o r l d .
•
Uses
O
Be is u s e d to i m p r o v e the properties of C u - Ni alloys. M g is largely u s e d as a light w e i g h t construction metal.
•
General Chemical Behaviour
O
Oxides, peroxides a n d h y d r o x i d e s : They f o r m m o n o x i d e s w h e n h e a t e d w i t h 0 2 . BeO is covalent, others exist as N a C l structure. H i g h charges on b o t h cation a n d anion give rise to h i g h lattice energy.
M.pt. O
BeO
MgO
CaO
SrO
BaO
2530
2826
2600
2400
1900
H i g h polarising p o w e r of the smaller cations m a k e s the s u p e r o x i d e s a n d p e r o x i d e s u n s t a b l e to heat. H o w e v e r , stability increases w i t h electropositive character a n d size of metal.
H y d r a t e d peroxides of Mg, Ca, Sr a n d Ba may be obtained by adding H 2 0 2 into aq. solution of these metal salts or hydroxides. Ca(OH 2 ) + H 2 0 2 > Ca02-2H20
Notes : In BeF 2, hydration energy is very high. For BaF2, the large cations touch one another. This lowers the lattice energy considerably. _> Melting points of fluorides are much higher than other halides. CaF 2 has very high melting point (1418°C). For beryllium halides, the melting points show irregular trend. BeF 2 > BeCl 2 < BeBr2 (554°C) (399°C) (508°C) —» Except MgF 2 , which has rutile structure, other fluorides have the characteristics fluorite structure. Magnesium and calcium halides readily absorb water while others are normally anhydrous.
Dilute aqueous acids liberate H 2 0 2 from peroxides and H 2 0 2 + 0 2 from the superoxides. BaO z + H 2 S 0 4 O
- > BaS0 4 + H 2 0 2
> BaS0 4 + H 2 0 2 + O z Ba(0 2 ) 2 + H 2 S 0 4 For hydroxides, solubility in water and basic character increases from Be(OH) 2 to Ba(OH) 2 . Being amphoteric in nature, Be(OH) 2 dissolves in excess alkali. The initial solutions believed to consist of polymeric chain.
O
2 -
\
V O H
HO
O OH /n H Further addition of alkali depolymerises the chain to give the mononuclear beryllate, [Be(OH)4]2~. Halides
Carbonates
200°C
420°C
810°C
1280°C
The solubility in water decreases in the order: Be > Mg > Ca > Sr > Ba. —» The sulphates from Ca onwards being practically insoluble. SrSO^ BaS0 4 and RaS0 4 are anhydrous. —> High solubility of MgSO, is due to loose packing of large sized S 0 4 ions with Mg 2+ ions. —> Nitrates of group II metals can be obtained by treating the oxides, hydroxides or carbonates with H N 0 3 . Crystallisation of the solution gives the hydrated salts. On heating they give oxides but
» Mg(OH)Cl + HC1
600°C
•> MgO + HC1 Mg(OH)Cl A n h y d r o u s MgCl 2 can be prepared by heating the hydrate in vacuum at 175°C or in a current of HC1. BeCl 2 is prepared by heating BeO in CC14 vapour at 800°C. Beryllium halides are covalent and polymeric. Do not conduct electricity in molten state. Notes : Except BeF2, all other beryllium halides dissolve in polar organic solvents. A n h y d r o u s MgX 2 dissolve in alcohols, ethers a n d ketones. BeF2 acts as Lewis acid and exists as glassy solid w h e n 'Be' is 4 co-ordinated. Above 270°C, it adopts crystalline structure. —> A n h y d r o u s BeCl 2 (m.pt. 405°C) is polymeric in solid state. It contains halogen bridges. Be-atoms are in sp 3 -state. In the vapour phase below 1200 K, it exists as dimer and above 1200 K exists as monomer. —» BeF 2 is highly soluble in water. MgF 2 and CaF 2 are almost insoluble. Solubility again increases d o w n . For chlorides, solubility decreases from Mg to Ra.
BeCO, MgCO, CaC0 3 BaC0 3
Decomposition 25°C temp.
These are obtained by reacting the halogen hydracids with the metal, its oxide or carbonate. A n h y d r o u s Be and Mg halides cannot be obtained by simply heating the aq. solution owing to hydrolysis. MgCl 2 + H 2 0 •
Salts of oxo-acids The tendency to form hydrates decreases for sulphates, nitrates and carbonates from Be to Ba. The thermal stability increases. Notes : Strong polarising power of Be2+ ion make BeC0 3 unstable.
O
Be(N0 3 ) 2 1 2 5 ° C >[Be 4 0(N0 3 ) 6 ] This basic beryllium nitrate has four 'Be-atoms a r o u n d the central 'O' and six N 0 3 acting as bridging ligands. Similarly, basic beryllium acetate is also known. Hydrides Mg, Ca, Sr and Ba directly react with H 2 to form hydride. BeH 2 is difficult to prepare. BeH 2 is polymeric containing 3c 2e-bonds. In the gas phase, B 3 H 2 exists in several forms including polymeric chains and rings. Note : Like (BeH 2 ) n , (BeMe 2 ) tl is also polymeric. \ / - C H 3 \ / C H 3 /
\ c h
3
/
\
C
H
3
/
\
Each 'C' has co-ordination n u m b e r 5 and each 'Be' is sp3. CHEMISTRY TODAY | MAY '13
17
O
Complexing tendency : Complexing tendency decreases in the o r d e r : Be > M g > Ca > Sr. Stability of such complexes d e p e n d o n several factors : —» Stability decreases w i t h increasing size of metal ions for small sized a n d h i g h charged ions.
O
O
Density B to Tl.
O
Ionisation e n e r g y : B > Tl > Ga > AI > In. 2nd IP also follow the same trend.
O
S t a n d a r d oxidation p o t e n t i a l : E°op(inV):
2-
O
O
O
—» For larger oxo anions, S 0 4 , N 0 3 , the o r d e r is reversed i.e., M g < Ca < Sr < Ba. —¥ For EDTA, the stability o r d e r is M g < Ca > Sr > Ba. Nitrides : All alkaline earth metals b u r n in N 2 gas to f o r m nitrides (M 3 N 2 ). Be 3 N 2 is rather volatile; while others are not. All nitrides are crystalline solids w h i c h d e c o m p o s e o n heating. Note : Be 3 N 2 is volatile p r o b a b l y because it h a s greater degree of covalence. Carbides : BeO w h e n h e a t e d w i t h 'C' at 1900-2000°C a brick red coloured c o m p o u n d Be 2 C is f o r m e d . Other m e m b e r s f o r m carbides of formula MC 2 . O n heating M g C 2 changes to M g 2 C 3 that contains C 3 ion. Mg 2 C 3 + H 2 0 > Mg(OH)2 + C3H4 Solutions in liquid N H 3 : The metals of g r o u p II Ca, Sr, Ba dissolve in liq. N H 3 . The solutions d e c o m p o s e very slowly releasing amides a n d h y d r o g e n . Evaporation of a m m o n i a solution give group-I metals, b u t similar evaporation of solution of group-II metals give hexa-ammoniates of the metals. They slowly d e c o m p o s e to give amides. M(NH3)6
O
: Density
gradually
increases
AI
Ga
In
Tl
1.66
0.56
0.34
-1.26
M/M3*
O
H y d r a t i o n : All the m e t a l ions exist as [M(H 2 0) 6 ] 3 + in w a t e r solution. Such a q u a ions act as acids d u e to h i g h polarising p o w e r of the cation. [M(H20)6]3+
[M(H20)50H]2+ + H+
Two pK a values are given, [A1(H 2 0) 6 ] 3 + ; pKa = 4.95 [Ga(H z O) 6 ] 3 + ; pKa = 2.6 N o t e : Stability of M 3 + ions decreases f r o m AI to Tl. Also, Tl + is m o r e stable t h a n Tl 3+ . •
General Reactions B With 0 2 at high temperature
B2O3
> M(NH2)2 + 4NH3 + H2
CI2,
In
Ga203
ln203
-
(also
AIX 3
GaX 3
InX 3
With I 2
-
•
With dilute mineral acid
only oxidising acids react
O
Covalent radii (in p m )
B 82
AI 125
Ga 125
In 150
Tl 170
Ionic radii (M 3+ ) (in p m )
B 27
AI 50
Ga 62
In 81
Tl 95
Melting points(°C): B > AI > T1 > In > Ga 2180 660 303 157 30 Boiling points s h o w a regular t r e n d i.e., decrease d o w n the g r o u p . N o t e : Liquid Ga h a s higher density t h a n solid Ga, which is a n u n u s u a l property. Ga exists as liquid in the t e m p e r a t u r e r a n g e 30°C to 2000°C. CHEMISTRY TODAY | M AY ' 1 3 18
TIX3 (T1X is also formed)
M e m b e r s : B, AI, Ga, In, T1
Size:
TI2O3
n 2 o)
Br 2
General Properties
Tl
only AI forms AIN at high temperature
With F2,
AI 2 O 3
Ga
(highly exothermic)
With N 2
Concentrated solutions are b r o n z e coloured d u e to formation of clusters.
AI
GROUP-IN A
O
from
AII3
Gal 3
Inl 3
Tl+i;
gives
gives
gives
AL3+
Ga 3 +
In 3 +
gives +
Tl
N o t e : Cone. H N 0 3 m a k e s AI, Ga, In passive. •
Abnormal behaviour of Boron Boron is a h i g h m e l t i n g solid containing B 12 units. The B 12 units i c o s a h e d r o n structure, a solid figure w i t h 12 vertices a n d 20 equilateral triangles. Crystalline b o r o n is chemically very inert, it is n o t affected b y acids or oxidising agents. Fused N a O H attacks it above 500°C giving NaBO z a n d H2. Oxidising acids like H N O s a n d conc. H 2 S 0 4 dissolve b o r o n b u t b o r o n r e m a i n s inert in its
W h e n concentration of the solution is very high, polymeric m e t a b o r a t e s also exist.
reaction w i t h HC1. 2B + 3 H 2 S 0 4
> 2H3BO3 + 3SO z
O
H y d r i d e s of b o r o n like h y d r i d e s of silicon are s p o n t a n e o u s l y f l a m m a b l e volatile c o m p o u n d s .
O
The c o m m o n halides of b o r o n a n d silicon are mostly gases.
N o t e s : Boric acid can be titrated with strong alkali
O
Boron oxide is acidic in n a t u r e .
u s i n g p h e n o l p h t h a l e i n indicator in presence of
•
Extraction
O
O
3B(OH) 3 1
^ H 3 0 + + [B303(0H)4R + H 2 0
Tl(OH) is water-soluble.
glycerol. The later raises its acidity.
Crystalline b o r o n of h i g h p u r i t y is obtained b y r e d u c t i o n of volatile b o r o n c o m p o u n d s BC13 or BBr 3 w i t h H 2 . O n the other h a n d , w h e n B 2 0 3 is r e d u c e d w i t h M g give a m o r p h o u s boron.
O
H a l i d e s : All b o r o n trihalides except BF^ are h y d r o l y s e d to boric acid. BX 3 + 3 H z O
» B(OH) 3 + 3 H X
H o w e v e r , BF 3 f o r m s a n addition p r o d u c t with
A l u m i n i u m is extracted electrolytically f r o m bauxite. Purified bauxite (melting point 2050°C) is dissolved in cryolite a n d electrolysed at 950°C. Electrolyte contains 80-85% A l 2 O y CaF 2 - 5 to 7%, A1F3 - 5 to 7%. L i 2 C 0 3 is also u s e d to lower the melting point.
H 2 O.
•
Uses
BI 3 > BBr 3 > BC13 > BF 3
O
Boron carbides a n d metal b o r i d e s are extensively u s e d as n e u t r o n shields a n d control rods. C h e a p , easily malleable, light, non-toxic a n d fairly corrosion resistant a l u m i n i u m is w i d e l y used. Ga, A s is largely u s e d in light emitting diodes.
N o t e s : Stability of ether a d d u c t s of BF 3 follow the
Boron halides are Lewis acids.
General Chemical Behaviour
O
Oxides a n d h y d r o x i d e s : All the m e m b e r s of g r o u p 13 f o r m oxides M203 a n d h y d r o x i d e s .
Lewis acidity decreases in the order
order : M e 2 0 > E t 2 0 —> B - F b o n d is longer in BF 3 t h a n as c o m p a r e d to BF~ a n d (CH 3 ) 3 N.BF 3 . 'B' f o r m s BF 4 , w h e r e a s AI, Ga a n d In form The fluorides of AI, Ga, In a n d Tl are ionic and h a v e h i g h melting points. Other halides of AI, Ga,
•
More acidic less basic
AI2O3
In a n d Tl are largely covalent in a n h y d r o u s state
l n 2 0 3 and T1203 1
and
V
Basic :: v •
a n d possess l o w melting point. Vapour density
'
:
of
fi"
Ga203
B(OH) 3 Weakly acidic
Al(OH)3
temperature
I n ( O H ) 3 a n d Tl(OH) 3
and
Basic
dissociates
to
trigonal
planar
o r d e r is MC1 3 > MBr 3 > M I 3 (M = AI, Ga, In) Also, t o w a r d s (CH 3 ) 2 S or (CH 3 CH 2 ) 2 S; the acidity o r d e r is A1C13 > AlBr 3 > A1I3 butMI3>MBr3>MCl3
> CU(B02)2
> Co(B0 2 ) 2
B 2 0 3 also combines w i t h acidic oxides such as P 4 O 1 0 a n d AS 4 O 10 . 2 B 2 0 3 + P4O10
corresponding
N o t e : T o w a r d s (C 2 H 5 ) z O or (C 2 H 5 ) 3 N; the acidity
B 2 0 3 combines w i t h metal oxides to f o r m basic metaborates. CoO + B203
400°C
structure.
Amphoteric
CuO + B203
at
exists in v a p o u r state b e l o w 473 K a n d at higher
Ga(OH)3
O
A1C13 m e a s u r e d
to the f o r m u l a A12C16. This dimeric structure
Amphoteric
/
> R2O + - BF~
r2O + BF 3
A1F
•
B2o3
» H 3 0 + [BF 3 0H]~
BF 3 + 2 H 2 0
> 4BP04
A m o n g h y d r o x i d e s only B(OH) 3 is w a t e r soluble. It's a Lewis acid. Rest all h y d r o x i d e s are insoluble in w a t e r a n d f o r m a gelatinous precipitate. B(OH) 3 + H 2 0 > B(OH)~ + H +
O
(M = G a a n d I n )
Hydrides : Boron h y d r i d e s are m o s t important. Nido-boranes B
H
N « +4
Arachno-boranes B
«HN+6
B2H6
B H
B5H9
B H
4 10
5 11
B 2 H 6 w h e n h e a t e d to 120°C, B 4 H 1 0 is p r o d u c e d . CHEMISTRY TODAY | MAY '13
19
B2H6:
O
\
/IF
H
\
/KCH
1.33 A
^
B - B distance is 1.77 A. B4H10: No. of B - H - B b o n d s = 4 No. of B - B - B b o n d s = 0 No. of B - B b o n d s = 1 No. of BH 2 g r o u p s = 2 In B 5 H 9 , respectively they are 4,1, 2, 0 B 2 H 6 + (CH 3 ) 3 N
> H3B -
O O
N(CH 3 ) 3
B2H6 + 2 N H 3 - 1 2 0 ° C > [ H 2 B ( N H 3 ) 2 r [ B H J Diborane w h e n r e d u c e d w i t h N a H , LiH gives BH~
Transition Elements
• O
V
Cr M n Fe C o N i
Cu Zn O
General Properties Size: Metallic radii (in pm)
Ionic radii (M 2 + ): (Pauling in 6 : 6 co-ordination number)
Sc 164
Sc
Ti 147
Ti
V 135
V
Cr 129
Cr
Mn 137
Mn 80
Fe 126
Fe 75
Co 125
Co 72
Ni 125
Ni 69
Cu 128
Cu
Zn 137
Zn 74
-
-
-
-
-
N o t e : Different m e t h o d s of m e a s u r e m e n t h a v e reported different results. For example, S h a n n o n a n d Prewitt m e t h o d reports Zn 2 + ion r a d i u s in 6 : 6 co-ordination s p h e r e as 88 p m . 20
CHEMISTRY TODAY | MAY ' 1 3
M^/M
T
(inV)
-1.13
V
Cr
Mn
Fe
Co
Ni
Cu
-0.44
-0.28
-0.26
+ 0.34
£ ° values for Mn, Ni a n d Z n are m o r e negative t h a n expected d u e to stability of half-filled Mn 2 + , a n d completely filled Zn 2 + n a t u r e . In Ni, the reason is highest e n t h a l p y of h y d r a t i o n for Ni 2 + ion.
N o t e s : N a B H 4 can be u s e d as r e d u c i n g agent in water solvent b u t LiAlH 4 can not.
Members Sc Ti (3rf-series)
Ionisation potential : 1 s t ionisation potential increases f r o m Sc to Fe a n d t h e n decreases. Zinc h a s the highest 1 s t I.P. while Sc h a s the lowest in 3tf-series. Z n > Fe > Co > C u > Ni 2 n d I.P. is highest for C u followed b y Ni a n d then Zn. A g a i n it is lowest for Sc. Melting p o i n t : W > Re > O s > Cr > Fe > Sc > M n S t a n d a r d electrode potential
O
O
N o t e : Mn 3 + a n d Co 3 + ions are the strongest oxidants in a q u e o u s solution. O n the other h a n d , V 2+ a n d Cr 2+ are strong r e d u c i n g agents. Stabilisation of oxidation states : Stability is a relative term, m e a n i n g f u l only w h e n referred to a given set of conditions. C o m p o u n d s containing Cu(I) are thermally stable, b u t in a q u e o u s solution Cu + readily d i s p r o p o r t i o n a t e s into Cu° a n d Cu 2 + . In presence of cyanide, however, Cu + can be stabilised in a q u e o u s solution. H i g h oxidation states are f a v o u r e d b y alkaline media. L o w oxidation states are stabilised b y r e d u c i n g solvents. N o t e : If the low oxidation state gets oxidised somehow, the r e d u c i n g solvent will again force it back to the original oxidation state. Oxides : V 2 O s , C r 0 3 - acidic CuO, ZnO, Cr2Oy C r 0 2 , V 0 2 , T i 0 2 a n d Fe203-amphoteric. H a l i d e s : CuCl 2 a n d CuBr 2 consists of infinite chlorine or b r o m i n e b r i d g e s w i t h co-ordination n u m b e r 6. Their halides in higher oxidation states are relatively u n s t a b l e a n d u n d e r g o hydrolysis. Strongly r e d u c i n g states p r o b a b l y d o n o t f o r m fluorides while strongly oxidising states f o r m fluorides b u t not iodides. TiF 2 is n o t k n o w n because Ti 2+ is easily oxidised to Ti 4+ a n d is strongly r e d u c i n g agent. Similarly, VF 4 is k n o w n b u t VI 4 is not. OO
•
•
•
NCERTXtract^ ft • •//• • * uestions for NEET Of the following chemical equations. W + 3 L i A l H 4 — » X + 3LiF + 3A1F3 X + 6 H 2 0 — » Y + 6H 2 3X + 3 0 2 B203 + 3 H 2 0 Identify W, X, Y respectively. (a) Na 2 B 4 0 7 , H 3 B0 3 , B 2 0 3 (b) H 3 B0 3 , B 2 H fr BF3 (c) BF,, B 2 H 6/ H 3 B 0 3 (d) H 3 B03, Na 2 B 4 0 7 , B 2 0 3 2.
A hydrocarbon ' F , (C 4 H 8 ) on reaction with HC1 gives a compound 'Q', (C4H9C1), which on eaction with 1 mol of N H 3 gives compound 'R', (C 4 H n N). O n reacting with N a N 0 2 and HC1 followed by treatment with water, c o m p o u n d 'R' yields an optically active alcohol, 'S'. Ozonolysis of 'P' gives 2 mol of acetaldehyde. Identify compound 'S'. CH 2 CH 3
O
(a) CH 3 CH 9 C CHo
C
(c)
H
(b)
H X - C - H I OH
3 \
)CH—CH2OH CH3X
(d) C H 3 — C H , - - C H O I CH, 3.
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano-second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.
(a) 1.65 x 10"10 J (c) 0.5 x 109 J
(b) 8.28 x 1CT0 J (d) 1.65 x 10-19 J
Calculate the mole fraction of ethylene glycol (C 2 H 6 0 2 ) in a solution containing 20% of C 2 H 6 0 2 by mass. (a) 0.932 (b) 0.862 (c) 0.950 (d) 0.779 Table-tennis ball has a mass 10 g and a speed of 90 m/s. If speed can be measured within an accuracy of 4 %. What will be the uncertainty in speed and position? (a) 6.626 x 10"34 m (b) 4 » 10~ 32 m 33 (c) 1.46 x 10" m (d) 0.6 x 0.6 x 10~31 m In which pair most efficient packing is present? (a) hep and bee (b) hep and ccp (c) bee and ccp (d) bee and simple cubic cell. Q u a r t z is extensively u s e d as a piezoelectric material, it contains (a) Pb (b) Si (c) Tl (d) Sn X-ray d i f f r a c t i o n s t u d i e s s h o w t h a t c o p p e r crystallises in an fee unit cell with cell edge of 3.608 x 10 -8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm 3 , calculate the atomic mass of copper, (a) 60.1 u (b) 63.1 u (c) 69.3 u (d) 62.7 u 2P + 3Q > 2R In which of the following reaction mixtures, P is the limiting reagent? (a) 5 mol P + 2.5 mol Q (b) 3 mol P+ 7 mol Q (c) 4 mol P + 1 mol Q (d) 6 mol P + 5 mol Q CHEMISTRY TODAY | MAY '13
21
10. Resistance of a c o n d u c t i v i t y cell filled w i t h 0.1 mol L"1 KC1 solution is 100 Q. If the resistance of the same cell when filled with 0.02 mol L"1 KC1 solution is 520 £2, calculate the conductivity and molar conductivity of 0.02 mol L_1 KC1 solution. The conductivity of 0.1 mol L"1 KC1 solution is 1.29 S/m. (a) 124 x 103 S m 2 mol' 1 (b) 124 x 10~3 S m 2 mol" 1 (c) 124 x 10"4 S m 2 mol - 1 (d) 124 x 104 S m 2 mol" 1
O n reacting w i t h CHC1 3 and alcoholic potash ' P ' p r o d u c e s an o b n o x i o u s smell d u e to the formation of compound 'R'. Reaction of 'P' with benzenesulphonyl chloride gives compound 'S' which is soluble in alkali. With N a N O z and HC1, 'P' forms compound 'T which reacts with phenol in alkaline m e d i u m to give an orange dye 'IT. Identify compound ' T . NC
N-C6H5
11. Which of the following dissolves in hot conc. NaOH? (a) Fe (b) Zn (c) Cu (d) Ag.
NH,
12. p H of a solution of a strong acid is 5.0. What will be the p H of the solution obtained after diluting the given solution 100 times? (a) 6.6 (b) 2.7 (c) 7 (d) 3.01 13. The first order rate constant for the decomptions of ethyl iodide by the reaction C
2
H
5Ife)
C
2
H
4fe) +
H I
fe)
5
at 600 K is 1.60 x 10" s"1. Its energy of activation is 209 kj/mol. Calculate the rate constant of the reaction at 700 K. (a) 6.36 x 10"3 s"1 (b) 6.24 x 10~3 s"1 (c) 6.57 x 10"3 s"1 (d) 6.86 x 10~3 s"1 14. The first ionisation energies of the elements of the first transition series (Ti —» Cu) (a) increase as the atomic number increases (b) decrease as the atomic number increases (c) do not s h o w any change as the a d d i t i o n of electrons takes place in the inner (n - 1) d-orbitals (d) increase from Ti to Mn and then decreases from Mn to Cu. 15. 20 mL of 0.2 M BaCl 2 is mixed with 30 m l of 0.3 M A12(S04)3. The weight of BaS0 4 formed is (a) 0.932 g (b) 0.233 g (c) 0.699 g (d) 0.116 g 16. W h i c h o n e of t h e f o l l o w i n g is a c a s e of adsorption? (a) Silica gel in contact with water vapour. (b) Cotton cloth dipped in a dye solution. (c) A n h y d r o u s C u S 0 4 in contact w i t h w a t e r vapour. (d) Ammonia gas in contact with water. 17. A colourless substance IP' (C 6 H 7 N) is spraingly soluble in w a t e r a n d g i v e s ^ w a t e r s o l u b l e c o m p o u n d ' Q ' on treating w i t h mineral acid.
22
CHEMISTRY TODAY
M A Y 13
N 2 CI
18. For which of the following compounds Lassaigne's test of nitrogen will fail? (a) H 2 N CONH.NH 2 .HCl (b) H 2 NNH 2 .2HC1 (c) H , N C O N H 2 (d) C 6 H 5 - N = N - C 6 H 5 19. The metals present in insulin, haemoglobin and vitamin B12 are respectively (a) Zn, Hg, Cr (b) Co, Fe, Zn (c) Mg, Fe, Co (d) Zn, Fe, Co 20. The most steam volatile compound is OH
OH NO,
NO,
OH
OH
(c)
(d) NO,
NO,
O2N
I NO 2 NO 2
21. W h i c h of the f o l l o w i n g r e a c t i o n s i n c r e a s e s production of dihydrogen from synthesis gas? (a)
CH4(G) + H2OFE)
C OJfS) + J3l lH22 Ife) TO T
(b) C(s) + H 2 O f e ) . 1270 K > C O wfe) + H fe) 673 K
(c) COJ{g) H„ (sr) T+ H 2 0(, Catalyst > C O „ , T+ "2fe) (d) C 2 H 6 + 2H z O
12
^ K > 2CO + 5H 2
Cone. H N 0 3 + conc. H2SO4
Br„ Fe
•(X)
22.
Cone. HNO3
->
(Y) (Major)
then, the p r o d u c t s (X) a n d (Y) will be NO-
NO,
N02
Br
25. Which of the following statements is incorrect? (a) H e l i u m h a s t h e h i g h e s t f i r s t i o n i s a t i o n enthalpy in the periodic table. (b) C h l o r i n e h a s less n e g a t i v e e l e c t r o n gain enthalpy than fluorine. (c) M e r c u r y a n d b r o m i n e are liquids at room temperature. (d) In any period, atomic radius of alkali metal is the highest. 26. Which of the following ore is the best concentrated by froth floatation method? (a) Galena (b) Cassiterite (c) Malachite (d) Magnetite NaN 3 + conc. H 2 S0 4
27.
« C O X
CH3CH2COOH
,IQ.
A
Br,/P
Br
A by reaction
B by reaction R2
NO, Br
Which is correct alternate? A B R,
NO
(a)
(o r o j ,
R,
CH,CHCOOH
o
Schmidt
HVZ
Br
NO, NO,
CH,CH2NH2
24. The correct order of acidic strength is (a) CH 3 CF 2 COOH > CH 3 CCl 2 COOH > CH 3 CBr 2 COOH (b) CH 3 CF 2 COOH > CH 3 CBr 2 COOH > CH 3 CCl 2 COOH (c) CH 3 CBr 2 COOH > CH 3 CCl 2 COOH > J CH3CF|C0OH (d) CH 3 CCl 2 COOH > CH 3 CBr 2 COOH > ! CH3Cf2COOH.
HVZ
Schmidt
(c)
HVZ
Schmidt
CH3CH,NH2
CH 3 CH 2 COBr
(d) N o n e is correct.
NO,
23. The following data are obtained w h e p carbon a n d d i o x y g e n react together to f o r m different compounds. Mass of carbon Mass of oxygen (i) 12 g 16 g (ii) 12 g 32 g Which law of chemical combination is obeyed by the above experimental data? (a) Law of definite proportions (b) Law of multiple proportions (c) Gay Lussac's law of gaseous volumes (d) Avogadro law.
( b ) C H 3 C H 2 C O N H 2 CH 3 CH 2 COBr
28. Hydrolysis of one mole of peroxodisulphuric acid produces (a) two moles of sulphuric acid (b) t w o moles of peroxomonosulphuric acid (c) one mole of sulphuric acid and one mole of peroxomonosulphuric acid (d) o n e m o l e e a c h of s u l p h u r i c a c i d , p e r o x o m o n o s u l p h u r i c acid a n d h y d r o g e n peroxide.
29.
CH 3 NO 2 C 2 H 5 O- > [A]
[B]
HON
H+
°>
W h i c h r e a c t i o n i s / a r e i n v o l v e d in t h e above conversion? (a) Diazotisation and Beckmann rearrangement. (b) Diazotisation and Robinson annulation. (c) Aldol condensation and diazotisation. (d) D i a z o t i s a t i o n a n d p i n a c o l t y p e of rearrangement. 30. Select t h e s p e c i e s w h i c h is r e d u c e d in t h e following reaction. H C H O + 2[Ag(NH 3 ) 2 ];, ) + 3 0 H W — > 2Ag (s) + H C O O L , + 4 N H 3 M + 2H 2 0 ( ( )
CHEMISTRY TODAY | MAY ' 1 3
23
(b) [Ag(NH 3 ) 2 ] + (d) Ag
(a) H C H O (c) N H 3
31. Match list I with list II and select the answer using the codes given below : List I
List II
A
XeF4
1
Distorted octahedral
B
XeF6
2
Tetrahedral
C
XeO s
3
Square planar
D
Xe0 4
4
Trigonal pyramidal
(a) (b) (c) (d)
A - 4, A - 2, A-l, A - 3,
B - 1 , C - 3, B - 3, C - 1, B-4, C-2, B - 1, C - 4,
D - 2 D -4 D-3 D -2
32. Which of the following are examples of Sandmeyer's reaction? CuCl (a) C 6 H 5 N 2 + Cf -*C6H5CI (b) C 6 H 5 C1
L 2(CN)2
"
(c) C6H5N2+C1
> C6H5CH2NH2
_ NaN0 2 /Cu/A
C6H5N2+Cr
»C6H5NO2
Kl/A
(d) ->C6H5I 33. Reaction of glucose with acetic anhydride indicates the presence of (a) — CHO group (b) — CH 2 — group (c) five —OH groups (d) all of these. 34. The m o l e c u l e s of a gas A t r a v e l f o u r t i m e s faster than the molecules of gas B at the same t e m p e r a t u r e . The ratio of molecular w e i g h t s MA/MB
will be
(a) 1/16
(b) 4
(c) 1/4
(d) 16.
35. Which of the f o l l o w i n g h a s a s q u a r e p l a n a r geometry? (a) [PtClJ 2 (b) [C0C14]2" 2 (c) [FeCl4] (d) [NiCl 4 ] 2 36
Thallium shows lower oxidation states because (a) of its high reactivity (b) of inert pair of electrons (c) of its amphoteric nature (d) it is a transition metal.
38. Which of the following statements is false? (a) Artificial silk is derived from cellulose. (b) Nylon-66 is an example of elastomer. (c) The repeat unit in natural rubber is isoprene. (d) Both starch and cellulose are polymers of glucose. 39. A scientist n e e d s a r e f r i g e r a t i o n m a c h i n e to maintain a chemical reaction at a temperature of -13°C. H o w m u c h work not be performed on the system during each cycle of its operation if 3000 J of heat is to be w i t h d r a w n from the -13°C reservoir and discharged to the room at 27°C in each cycle. (Assume machine operates at 100% of its theoretical efficiency) (a) 65000 J (b) 3000 J (c) 154 J (d) 133 J 40. A white crystalline solid A on boiling with caustic soda solution gave a gas B which w h e n passed through an alkaline solution of potassium mercuric iodide gave a b r o w n ppt. The substance A on heating gave a gas C which rekindled a glowing splinter but did not give b r o w n f u m e s with NO. The gas B is (b) N H 3 (c) HC1 (d) C 0 2 (a) H2S
41. X
H-iO
N„ A
->Y-
-> Z (colourless gas)
lc
CuSOi
T (blue colour) The (a) (b) (c) (d)
substances Y and T are Y = Mg 3 N 2 and T = C u S 0 4 - 5 H 2 0 Y = Mg 3 N 2 and T = CuS0 4 -4NH 3 Y = Mg(N0 3 ) 2 and T = CuO Y = MgO and T = CuS0 4 -4NH 3
42. A s a m p l e of air consisting of N 2 and O z was heated at 2500 K until the equilibrium 2NO, N 2fe> + ° 2 f e )
3 7. Which of the following statements is not correct from the view point of the molecular orbital theory? 43. W h i c h of t h e f o l l o w i n g is u s e f u l as w a t e r (a) Be2 is not a stable molecule. softner? (b) He 2 is not stable but He 2 is expected to exist. (a) Thiokol (b) Bakelite (c) Bond strength of N 2 is maximum amongst the (c) Melamine formaldehyde resin homonuclear diatomic molecules belonging (d) Ion exchange resins to the second period. 44. The enthalpy of reaction (d) The order of energies of molecular orbitals in 2H20fe) + 02(?; -*2H2OW N 2 molecule is o2s < cs"2s < a2pz < (jt2px = n2py) is ARH = -572 k j mol" 1 < (t?2Vx = n * 2 p y ) < o * 2 p z CHEMISTRY TODAY | MAY '13 24
What will be standard enthalpy of formation of H2OM? (a) - 5 7 2 k J m o r 1 (b) 286 kj/mol (c) + 572 k j mol - 1 (d) - 286 kj/mol
50. Atmospheric pressures recorded in different cities are as follows:
45. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour? (a) [Ni(NH 3 ) 6 ] 2+ (b) [Zn(NHM3 ) 6 ] 2+
Consider the above data and mark the place at which liquid will boil first, (a) Shimla (b) Bangalore (c) Delhi (d) Mumbai.
(c)
[Cr(NH3)6]3+
(d)
46. W h e n SO z is p a s s e d potassium iodate, the changes from (a) + 5 to 0 (c) - 5 to 0
p in N/m
[CO(NH3)6]3+
1.2x 10
2H 3 B0 3 + 6H 2
B2H6 + 6 H 2 0 (X) 3B 2 H 6 + 3 0 2 (X)
(V)
B2O3- 3 H , 0
2. ( b ) :
CH,
CH 3 CH 2 - - C H — C H 3 I CI
(Q) CI
CH,
2
NH 2
•i^T
(R)
C H 3 — C H 2 — CH-
H3C
-CH 3
- C H 2 — CH - -CH,
(Q)
Br
-CH 3
NH,
C H , — C H , — C H — CH,
49. Which of the following reactions is feasible?
H3C-
HCl
- C H = CH — C H (P)
48. Which of the following elements does not form hydride by direct heating with dihydrogen? (a) Be (b) Mg (c) Sr (d) Ba
NaNH2, NH3 ,
1.21 x 105
1.02 x 10
• 2B 2 H 6 + 3A1F3 + 3LiF (X)
(W)
(b) +5 to - 1 (d) - 7 t o - l
/V-NH
5
SOLUTIONS
(b) gem dihalides (a) vie dihalides (c) A gem dihalide, B vie dihalide (d) A vie dihalide, B gem dihalide
(b)
5
1.01 x 10
Mumbai
Delhi
Bangalore 5
(c) : 4BF3 + 3LiAlH 4
CH 3 CH 2 CHO + PC15 - - > B, A and B are
(a)
2
t h r o u g h a s o l u t i o n of oxidation state of iodine
47. C H 2 - C H C H 3 +PC1 5
CI
Shimla
Cities
NaNH2, NH3 ,
NaNQ2/HCl ^ "
C H
3
NH 2 NH 2
>
iri)
CH2CH2
(R)
H3C
CH 3
H3C
CH 3
CH3—C
-H
OH (Optically active)
(S) CH,
H,C (C)
3.
NaNH2/ NH3 .
NH, H3C
Frequency =
CH,
CI
4.
(d) C H 3 O ^ A ^ C H 2 C H 3 NH 2
C H 3 C K A ^ CH 2 CH 3
o
(b) : Given that period = 2 ns = 2 x 10"9 s, the frequency Period
2 x 10"
= 0.5 x 109 s"1
Given that N = 2.5 x 1015, energy is £ = Nhv = 2.5 x 1015 x 6.626 x 10"34 x 0.5 x 109 = 8.28 x 10~10 J (a) : Assume that we have 100 g of solutions (one can start with and amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water. Molar mass of C 2 H 6 0 2 = 12 x 2 + 1 x 6 + 16 x 2 = 62 g mol" 1 . CHEMISTRY TODAY | MAY '13 2 5
20 g
Moles of C 2 H 6 0 2 =
= 0.322 m o l
Molar conductivity = A = — c
62 g m o l 248 x l O " 3 S m " 1 80 £ Moles of water = = 4.444 m o l 18 g m o l - 1
20 m o l m - 3 11. ( b ) : Z n + 2 N a O H
Na2Zn02 + H2
mole of C 2 H 6 0 2 rglyco1
~ moles of C 2 H 6 0 2 + moles of H z O 0.322 m o l
- = 0.068
0.322 m o l + 4.444 m o l
4.444 m o l
Similarly, x water '
- = 0.932 0.322 m o l + 4.444 m o l Mole fraction of w a t e r can also b e calculated as : 1 - 0.068 = 0.932 5.
(c) : The u n c e r t a i n t y in the s p e e d is 4 %, i. e., 18 = — = 3.6 m s " 1 100 5 Using the f o r m u l a , 90 x
4
h Ax = 4TI mAt>
6.626 x l O " 3 4 J s 4 x 3.14 x 10 x 10
k g x 3.6
Contribution of H 3 0 + ion concentration of acid
8.
( b ) : In case of fee lattice, n u m b e r of a t o m s p e r unit cell, Z = 4 a t o m s ™ , dNAa3 Therefore, M = — — — 8.92 g c m 3 x 6.022 x 10 2 3 a t o m s m o l " 1 x (3.608 x l O " 8 c m ) 3 4 atoms = 63.1 g/mol Atomic m a s s of c o p p e r = 63.1 u (b)
10. (c) : The cell constant is given b y the e q u a t i o n : Cell constant = G* = conductivity x resistance = 1.29 S/m x 100 Q = 129 m" 1 = 1.29 cm" 1 C o n d u c t i v i t y of 0.02 m o l L" 1 KC1 s o l u t i o n = cell constant /resistance =
g: R
=
l » E L l = 0.248 S m " 1 520 D
Concentration = 0.02 m o l L"1 = 1000 x 0.02 m o l m" 3 = 20 m o l m" 3
CHEMISTRY TODAY | MAY ' 1 3
Contribution of H 3 0 + ion concentration of water
p H = 2 x 10"7 = 7 - log 2 = 7 - 0.3010 = 6.6990 13. (a) : W e k n o w that
(b) ( b ) : Q u a r t z is the crystalline f o r m s of silica. It is interconvertible at suitable t e m p e r a t u r e . It is a covalent, t h r e e - d i m e n s i o n a l n e t w o r k solid.
26
It is n o t possible. This indicates t h a t solution is very dilute. Hence, Total h y d r o g e n ion concentration, [H + ]
E„ log /c2 — log fcj = 2.303 R
7.
9.
S o d i u m zincate 12. (a) : p H = 5, [H + ] = 10"5 m o l L"1 O n 100 times dilution, [H + ] = 10"7 m o l L"1 .-. p H = 7
1 0 " 7 + 10^ 7 .
= 1.46 x 1 0 " 3 3 m 6.
: 124 x 10" 4 S m 2 m o l " 1
logfc2-logfcj +
E„
1
1
l 1
1
2.303 R
= log(1.60xl0"5)209000J m o l L 2.303 x 8.314 J m o l L^K" 1 600 K
700 K
log k2 = 4.796 + 2.599 = - 2.197 k2 = 6.36 x 10"3 s _ 1 14. (a) : First ionisation energies of the e l e m e n t s of first transition series (Ti —» Cu) increase as the a t o m i c n u m b e r increases in n u c l e a r charge. 15. (a) : 3BaCl 2 + A1 2 (S0 4 ) 3 — > 3BaS0 4 + 2A1C13 Millimoles of BaCl 2 = 20 x 0.2 = 4 Millimoles of A1 2 (S0 4 ) 3 = 30 x 0.3 = 9 3 m o l e s BaCl 2 reacts w i t h 1 m o l A1 2 (S0 4 ) 3 1
4 m m o l BaCl, will react w i t h - x 4 2 3 4 = — mmol A12(S04)3 1 m o l A1 2 (S0 4 ) 3 f o r m s 3 m o l e s B a S 0 4 . 4 3 4 - m m o l A 1 2 ( S 0 4 ) 3 will form J x - m m o l B a S 0 4 = 4 mmol BaS04 Millimole of B a S 0 4 f o r m e d = 4
w=
NO,
millimole x molecular wt. 1000 6
1000
NO,
Cone. HNO3 + conc. H 2 S 0 4
22. (b):
16. (a) : Silica gel adsorbs water. 17. (d):
NH,
NC + CHCI3 + KOH
>[
| + 3KC1 + 3 H 2 0 (R)
(minor)
Benzene isonitrile O
NH2
NH— s —
SO 2 CI
o
(S)
(P)
N-Phenylbenzenesulphonamide (soluble in alkali) N 2 C1"
NH, NaNQ 2 /HCl 273-278 K
(P)
^
(T)
NO Z
(major) (Y) 23. (b) : Here, carbon and oxygen combines to form two compounds, carbon monoxide and carbon dioxide. The masses of oxygen (i.e., 16 g and 32 g) w h i c h c o m b i n e w i t h a fixed m a s s of carbon (12 g) bear a simple ratio, i.e., 16 : 32 or 1 : 2 which obeys the law of multiple proportions.
24. (a) 25. (b): Generally, w i t h i n a g r o u p , electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p-orbital (F) leads to greater repulsion than 3p-orbital (CI). Hence, CI has the most negative electron gain enthalpy, 26. (a): Here only galena is PbS (a sulphide ore). Cassiterite is Sn0 2 (oxide ore). Magnetite is Fe 3 0 4 (oxide ore) and malachite is C u ( 0 H ) 2 - C u C 0 3 (carbonate ore). The froth floatation process is used to concentrate sulphide ores, based on preferential wetting properties with frothing agent and water. 27. (a) 28. (c) : H 2 S 2 0 8 + H 2 0 Peroxodisulphuric acid
H,SO.
H 2 SO 5
Sulphuric Peroxomonosulphuric acid acid .
(U) Orange dye
CH 2 NO 2 (i) reduction
18. (b): H 2 NNH 2 -2HC1 does not contain carbon, hence does not give positive results for Lassaigne's test for nitrogen.
(ii) diazotisation
CH,
19. (d) : Insulin contains zinc, haemoglobin contains iron and vitamin B12 contains cobalt. 20. (a) : In ortho-nitrophenol, - O H is linked to - N O z by means of intramolecular H-bonding. So, it is highly volatile. 21. (c) : T h e p r o d u c t i o n of d i h y d r o g e n can b e increased by reacting CO of syngas m i x t u r e s with steam in the presence of iron chromate as catalyst.
H+ (Pinacol type of rearrangement)
30. (b) : H C H O — > Oxidised [Ag(NH 3 ) 2 f — > Reduced 31. (d): Xe in XeF 4 ,XeF 6 ,Xe0 3 and Xe0 4 possesses sp3rf2, sp3d3, sp3 and sp3 hybridisation respectively. Contd. on page no. 91 CHEMISTRY TODAY
MAY'13
27
FOCUS
Engineering Aspirants 2013 -
Are you Prepared for JEE Advanced? (Here are the few Tips to get your optimum) The JEE Advanced (formerly known as IIT-JEE) is an annual entrance examination to get Admission in IITs. It is also one of the toughest engineering entrance exams in the world. Only 1.5 lac students will be short listed from JEE Main 2013 to appear for the JEE Advanced 2013 on June 2nd, 2013. A serious aspirant ideally must have completed the syllabus by now. Schedule of JEE (Advanced), 2013 The examination will be held on Sunday, June 02, 2013 as per the schedule given below: Paper 1 9:00 to 12:00 hrs. (1ST) Paper 2 14:00 to 17:00 hrs. (1ST) EXAMINATION PATTERN: There will be two question papers, each of three hours duration. Both the question papers will consist of three separate sections on Chemistry, Physics and Mathematics. Questions will be of objective type, designed to test comprehension, reasoning and analytical ability of Students. All the questions will be Multiple Choice Type (MCQ) Negative marking scheme will be followed in the checking of examinations. A Student can opt for question paper in any of the language viz. English or Hindi. SYLLABUS COVERAGE: JEE Syllabus of Class XI & XII contributes about 45% and 55% of IIT-JEE question-papers respectively. While preparing all the chapters of Physics, Chemistry and Mathematics, based on our past experience stress may be given in particular on the following topics: Mathematics : Quadratic Equations & Expressions, Complex Numbers, Probability, Vectors & 3D Geometry, Matrices in Algebra; Circle, Parabola, Hyperbola in Coordinate Geometry; Functions, Limits, Continuity and Differentiability, Application of Derivatives, Definite Integral in Calculus. Physics: Mechanics, Fluids, Heat & Thermodynamics, Waves and Sound, Capacitors & Electrostatics, Magnetics, Electromagnetic Induction, Optics and Modern Physics. Chemistry: Qualitative Analysis, Coordination Chemistry & Chemical Bonding in Inorganic Chemistry, Electrochemistry, Thermodynamics, Chemical Equilibrium in Physical Chemistry and Organic
CHEMISTRY TODAY | MAY '13 28
Chemistry Complete as a topic. TIPS FOR JEE Advanced, 2013: CHEMISTRY (please see Tips on Mathematics in 'Mathematics Today' & on Physics in 'Physics For You') 1. Conceptual clarity, application skills and awareness to the prescribed syllabus should be the main focus. 2. Practice the most relevant numerical daily to develop speed. Pay special attention to the topics like Mole concept, Chemical Equilibrium and Electrochemistry. 3. Take a careful and patient approach for Organic chemistry giving particular attention to the topics like Stereochemistry, GOC (General Organic Chemistry) and Functional Group Analysis. 4. In Inorganic chemistry most of the questions which are asked are really conceptual, concerned with structures, processes and applications. Special attention is a must to the topics like Chemical Bonding and Coordination Chemistry. Cracking the JEE (ADVANCED) 2013 ""Stay focussed and maintain a positive attitude ^Develop speed. Refer to reputed mock-test series to build a winning exam temperament. Solve the past year's IIT-JEE papers. Focus on your weak areas and improve u p o n your concepts. ^Practise of JEE level questions is necessary as it improves your reasoning and analytical ability. *Remember it is quality of time spent and not the quantity alone. Hence give short breaks of 5 to 10 minutes every 1-2 hours of serious study. Completely relax when you take a break. Practice meditation to develop inner calm, poise, confidence and power of concentration. *Don't overstress yourself. Five to six hours of sleep every night is a must, especially three-four days before IIT-JEE to keep you physically and mentally fit. While short naps may help to regain freshness, avoid over-sleeping during the day. ^Finally, don't be nervous if you find the paper tough since it is the relative performance that counts. Put your best analytical mind to work, and believe in your preparation. Authored by Ramesh Batlish, FIITJEE Expert
mm
PRACTICE PAPER
You have different HC1 solutions having concentration 3 M and 1 M respectively. H o w m a n y mL of both solutions should be mixed to form 1 L of 1.2 M ? (a) 900 mL of 1 M and 100 mL of 3 M (b) 100 mL of 1 M and 900 mL of 3 M (c) 500 mL of each (d) 333.33 mL of 3 M and 666.66 mL of 1 M 2.
A m o n g the following, the least stable resonance structure is O
(a)
(b)
I O
o (c)
3.
o (d)
I O
For the chemical equilibrium CaCO 3(s) >Ca0 (s) + C0 2 f e ) AH°r can be determined f r o m which one of the following plots?
t o
(a)
o
u *
(b) 1/T
O
(b) N a 2 S 2 0 3
(c) ZnS
dil. H C 1 .
Roasting
(d) Na 2 S
>
dil. H C I .
A sample of liquid in a thermally insulated container is stirred for 1 hr by a mechanical attachment to a motor in the surroundings, which of the following thermodynamic quantity for the system is zero? (a) Work (W) (b) Change in internal energy (AE) (c) Change in enthalpy (AH) (d) N o n e of these. In b r o w n ring Fe a n d N O exist as Fe+ and NO + rather than Fe2+ and NO. These forms can be differentiated by (a) measuring the conductivity of an aqueous solution of the complex (b) by careful crystallization of the complex u n d e r special conditions and measuring the magnetic m o m e n t in solid state (c) estimating the concentration of iron (d) determining elevation in boiling point using the complex as solute.
o (d)
(a)
o
bB o 1/T4.
dil. H C 1 .
Photoelectric emission is observed from a surface for frequencies U, and t>2 of the incident radiations (\)j > t>2). If the m a x i m u m K.E. of the photoelectrons in two cases are in the ratio 1 : K, then the threshold frequency t>0 is given by
t>0 O
(C)
(a) N a 2 S 0 3
In which of the following reaction, S 0 2 gas is not released?
(c)
iC-1 KX>2-X> K-1
(b) (d)
K-Oj -1>2 K-1 K
Correct R, S configuration of the following c o m p o u n d is
CHEMISTRY TODAY | MAY '13
29
Me
(c)
Y is
a n d X is
CH—CH3 CH2—CH~ Q
>
6
(a) R at P a n d R at Q (c) S at P a n d R at Q 9.
(CH3)3CC1 AICI3
Anisole
'Br
-CH= CH-Me
(b) (d)
s
a n d Y is
(d) X is
-H
OH 13. The salts of w h i c h pair are expected to h a v e s a m e colour in their freshly p r e p a r e d a q u e o u s solutions? (a) VOCIj, CuCl 2 (b) C11CI2, FeCl 2 (c) FeCl^ VOCl 2 (d) MnCl 2 , FeCl 2
RatPandSatQ SatPandSatQ
CljFeCl,
HBr >x Heat
The p r o d u c t X in the above series of reactions is
14v W h e n 2 g of a non-volatile h y d r o c a r b o n containing 94.4% carbon is dissolved in 100 g of benzene, the v a p o u r p r e s s u r e of b e n z e n e is l o w e r e d f r o m 74.66 torr to 74.01 torr. W h a t will be the molecular f o r m u l a of the h y d r o c a r b o n ? (a) C 7 H 8 (b) C 8 H 1 6 (c) C 1 4 H 1 0 (d) C 1 0 H 6
OCH3 .Br
15. The reaction 2K + L > M + N goes to completion a n d follows the following rate law -±[K] = k[K]2[L] at For this reaction the values of x a n d y in the following table are
C(CH 3 ) 3
C(CH 3 ) 3
10. A white crystalline salt P reacts w i t h dilute HC1 to liberate a suffocating gas Q a n d also f o r m s a yellow precipitate. The gas Q t u r n s p o t a s s i u m dichromate acidified w i t h dilute H 2 S 0 4 to a green coloured solution R. P, Q a n d R are respectively (a) Na 2 S 2 0 3 , S0 2 / Cr 2 (S0 4 ) 3 (b) Na 2 S0 3 , SOj, Cr 2 (S0 4 ) 3 (c) Na 2 S0 4 , S0 2 , Cr 2 (S0 4 ) 3 (d) Na 2 S, SOj, Cr 2 (S0 4 ) 3
Na/CH 3 OH
(a) X is
H z O, AgOH
M o x 106 (mol dm" 3 )
Half-life(f 1 / 2 )
1 2 3 4
300 300 5 10
40 60 300 300
62.6 x 625
->X
18. Toluene reacts w i t h excess of Cl 2 in presence of sunlight to give a p r o d u c t w h i c h o n hydrolysis followed b y reaction w i t h N a O H gives
• Y
a n d Y is J a n d Y is [
y
(b) 62.6 a n d 312.5 (d) 31.3 a n d 312.5
17. If Cl 2 gas is p a s s e d into a q u e o u s solution of KI containing s o m e CC14 a n d the m i x t u r e is shaken, then (a) c o p p e r layer b e c o m e s violet (b) lower layer b e c o m e s violet (c) h o m o g e n e o u s violet layer is f o r m e d (d) N o n e of the above
O (b) X is [
(s)
16. Pollutant of automobile exhausts that affects n e r v o u s system a n d p r o d u c e s m e n t a l diseases is (a) Lead (b) M e r c u r y (c) Nitric oxide (d) S u l p h u r dioxide
-Br Br
[K]0 x 106 (mol dm" 3 )
(a) 62.6 a n d 625 (c) 31.3 a n d 625
11. The electronic configurations of f o u r elements are given below. 1. 2s 2 2p 5 2. 3s 2 3p5 4 4. 3s 3p* 3. 2s' 2p Arrange these elements in the correct o r d e r of the m a g n i t u d e (without sign) of their electron affinity. (a) 1 < 2 < 4 < 3 (b) 2 < 1 < 4 < 3 (b) 1 < 3 < 4 < 2 (d) 3 < 4 < 1 < 2 12.
Set
.COOH
j
(a)
CHEMISTRY TODAY | MAY '13 30
»
COONa (b)
25. Three acids are given:
.COOC1 (d) N o n e of the above.
(C)
COOH
H,C
Na
CH,
19. Which one of the following complexes is an outer orbital complex? (a) [Fe(CN) 6 ] 4_ (b) [Mn(CN)Jj~ (c)
[CO(NH 3 ) 6 ]
3+
(d)
[Ni(NH 3 ) 6 ]
2+
(i) NaNOo/HCl 1
20.
CH, and
(Atomic numbers; M n = 25, Fe = 26, Co = 27, Ni = 28) > (Y).
(ii) H 2 /Ni
(V)
(X)
H
3
C —
C
O
O
H
S CH3 (Z) The correct order of ease of acid-catalysed esterification is (a) X>Z>Y (b) Y>X>Z (c) Z > X > Y (d) Y > Z > X
26. Which one of the following statements is true? (a) Less is the critical t e m p e r a t u r e of a gas, greater is its adsorption. (b) Higher is the critical t e m p e r a t u r e of a gas, greater is its adsorption. (c) Greater is the van der Waals constant V of a gas, less is its adsorption. (d) A gas cannot be adsorbed above its critical temperature. 27. Heating an a q u e o u s chloride to dryness will (a) Al(OH)Cl 2 (c) A12C16
21. Isoelectric point of given a m i n o acid is, R—CH2— C H — C O O H (pK b = 11.7)
(a) 6.4 (c) 11
£
22. The composition of a s a m p l e of wustite is Fe 0 93 O. W h a t is the percentage of iron present as Fe3+ ? (a) 15% (b) 25% (c) 35% (d) 45% 23. W h a t is the p r o d u c t of the reaction of H 2 0 2 with Cl 2 ? (a) O + H O C l (b) HCl + 0 2 (c) H z O + HCl (d) HC1 + H 2
Me The m a i n p r o d u c t is
(a) - <
X™ X CH,
aluminium
28. W h a t will be the reduction potential of copper at p H = 14 if Ksp of Cu(OH) 2 is 1.0 * 10"19 and
N H 3 ( p K „ = 8.7) + (b) 5.5 (d) 3.2
24. M e 2 C = C H C H 2 - C H 2 - C = C H - C H 3
solution of give (b) A1 2 0 3 (d) A1C13
CU2+/CU = 0 . 3 4
(a) + 0.34 V (c) - 0 . 3 3 V
V ?
(b) - 0 . 2 2 V (d) - 0 . 3 4 V
29. The density of a mixture of N 2 and 0 2 at NTP is 1.3 g L 1 . Calculate the partial pressure of 0 2 . (a) 2.8 atm (b) 0.25 atm (c) 0.28 atm (d) 1.28 atm 30. Structure of C 8 H 8 C1 2 , which on aqueous alkali hydrolysis gives a p r o d u c t which does not give positive iodoform test, but gives silver mirror test is
— ^ Product.
' \ ri
(a)
(b)
(b)
? C6H5—C—CH3 CI
(d) C 6 H 5 CH 2 CHC1 2 CH,
(c)
31. A m o n g s t the following, the lowest degree of p a r a m a g n e t i s m per mole of the c o m p o u n d at 298 K will be s h o w n by CHEMISTRY TODAY | MAY '13
31
(a) M n S 0 4 • 4 H 2 0 (c) F e S 0 4 • 6 H z O
37. A q u e o u s 10% N a H C 0 3 solution is u s e d as reagent for identifying X. W h i c h o n e of the following c o m p o u n d s yields X o n hydrolysis? (a) C H 3 C 0 2 C 2 H 5 (b) C 2 H 5 - 0 - C 2 H 5 (c) C H 3 C H O (d) C H 3 C H 2 O H
(b) C u S 0 4 • 5 H z O (d) N i S 0 4 • 6 H 2 0
32. H o w m a n y types of F—S—F b o n d s are p r e s e n t in SF4? (a) 2 (b) 4 (c) 3 (d) 5
38. The n u m b e r of oc-particles emitted p e r second b y 1 g of a R a is 3.7 * 1010. The decay constant is (a) 1.39 x 1CT11 s"1 (b) 13.9 * 10"11 s 1 lu (c) 13.9 x 10" s" (d) 13.9 x 10" s~
33. W h e n a v a p o u r at a t m o s p h e r i c p r e s s u r e w a s gradually h e a t e d f r o m 298 K; its colour w a s f o u n d to d e e p e n at first a n d t h e n f a d e as the t e m p e r a t u r e w a s raised above 433 K. At 873 K, the v a p o u r becomes almost colourless, b u t its colour d e e p e n e d w h e n the p r e s s u r e w a s raised k e e p i n g the t e m p e r a t u r e constant. The v a p o u r w a s
39. S 0 2 is obtained quantitatively f r o m g y p s u m (a) b y treating g y p s u m w i t h conc. H N 0 3 at 700°C (b) b y heating anhydrous g y p s u m at 1200°C (c) b y h e a t i n g a n h y d r o u s g y p s u m w i t h conc. HC1 at 1200°C (d) b y h e a t i n g g y p s u m strongly w i t h conc. H 2 SO 4 .
(a) the b r o m i n e (b) A mixture of nitrogen dioxide a n d dinitrogen tetroxide (c) p u r e dinitrogen tetroxide (d) p u r e nitrogen dioxide.
40. O n e of the p r o d u c t s f o r m e d d u e to the reaction b e t w e e n K M n 0 4 a n d HC1 is (a) red liquid (b) M n 0 2 (c) greenish yellow gas (d) HC10 4 .
OH 34.
+ Br,
H,C
ASSERTION AND REASON Directions : In the following questions (41-60), a statement of assertion is followed b y a statement of reason. M a r k the correct choice as :
O.
\
Br
CH 3
+
~
5
~
(a) If b o t h assertion a n d r e a s o n are t r u e a n d reason is the correct explanation of assertion.
G
The rate of reaction is m o d e r a t e l y affected b y the n a t u r e of G. Choose the correct decreasing rate of reaction w i t h various substituents as G. (a) - N 0 2 > - B r > - H > - C H 3 > - O C H 3 (b) - O C H 3 > - C H 3 > - H > - B r > - N 0 2 (c) - O C H 3 > - B r > - C H 3 > - H > - N 0 2 (d) - N O z > - O C H 3 > - B r > - H > - C H 3 35. MC12 + XC12 (excess) xo
(c) If assertion is true b u t reason is false. (d) If b o t h assertion a n d reason are false. 41. Assertion : N 0 2 t u r n s acidic Kl-starch p a p e r blue. Reason : N 0 2 t u r n I" into I 2 .
MC1 4 + Xi
Heating above » 400°C
A n ore of X could be (a) cinnabar (c) siderite
(b) If b o t h assertion a n d reason are t r u e b u t r e a s o n is not the correct explanation of assertion.
42. Assertion : Cu + ion is colourless. R e a s o n : Four w a t e r molecules are coordinated to Cu + ion.
- o , + x
(b) malachite (d) h o r n silver.
43. Assertion : Pyrrole is m o r e basic t h a n pyridine. R e a s o n : In pyrrole, nitrogen is sp3 h y b r i d i z e d .
36. The correct order of basicities of the following c o m p o u n d s is NH CH,-C CH3-CH2-NH2 \ NH,
(CH3)2NH 3
(a) 2 > 1 > 3 > 4 (c) 3 > 1 > 2 > 4 32
CHEMISTRY TODAY | MAY ' 1 3
44. Assertion : A colloidal sol of As 2 S 3 is coagulated faster b y 0.1 M BaCl 2 t h a n by 0.1 M NaCl. Reason : BaCl 2 gives d o u b l e the n u m b e r of CI ions t h a n NaCl. 45. Assertion : Ethyl xanthate is u s e d as a collector in f r o t h floatation process. R e a s o n : Collectors d e p r e s s the floatation p r o p e r t y of o n e of the c o m p o n e n t s of the ore.
O II CH3-C-NH2 4 (b) 1 > 3 > 2 > 4 (d) 1 > 2 > 3 > 4
46. Assertion : The rate of a reaction n o r m a l l y increases b y a factor of 2 or 3 for every 10° rise in temperature.
J
Reason : Increase in t e m p e r a t u r e increases the n u m b e r of collisions. 47. Assertion : Li+ ion h a s the lowest mobility in a q u e o u s solution. Reason : Lithium h a s higher ionization enthalpy. 48. Assertion : The electron gain e n t h a l p y of N is +ve while that of P is - v e . R e a s o n : This is d u e to the smaller atomic size of N in w h i c h there is a considerable electron-electron repulsion a n d h e n c e the additional electron is n o t accepted easily. 49. Assertion : Aniline o n t r e a t m e n t w i t h CS 2 ; HgCl 2 d o not give m u s t a r d oil odour. R e a s o n : R—NCS, alkylisothiocyanate d o not have m u s t a r d oil odour. 50. Assertion : T h i o p h e n e p r e s e n t in commercial b e n z e n e as a n i m p u r i t y can b e r e m o v e d b y s h a k i n g the m i x t u r e w i t h cold concentrated H 2 S 0 4 . Reason : T h i o p h e n e is a heterocyclic aromatic compound.
58. Assertion : A spectral line will be observed for a 2px - 2p y transition. Reason : The energy is released in the f o r m of w a v e of light w h e n electron d r o p s f r o m 2p x to 2p y orbital. 59. Assertion: Compressibility factor (Z) for non-ideal gases is always greater t h a n 1. Reason : Non-ideal gases always exert higher p r e s s u r e t h a n expected. 60. Assertion : Saline h y d r i d e s are non-volatile, n o n - c o n d u c t i n g a n d crystalline solids. Reason : Saline h y d r i d e s are c o m p o u n d s of h y d r o g e n w i t h m o s t of the p-block elements. SOLUTIONS 1.
51. Assertion : D i m e t h y l a m i n e h a s repulsive interaction w i t h O H " ion in the solution. Reason : The ionisation constant of d i m e t h y l a m i n e gets decreased in the alkaline N a O H solution. 52. Assertion : C o n c e n t r a t e d a q u e o u s solution of CuCl 2 is g r e e n in colour. Reason : The solution contains t w o complex ions i.e., [CU(H 2 0) 4 ] 2+ a n d [CuCl 4 ] 2 " in equilibrium. 2+
(a) : Let x m L of 3 M a n d (1000 - x) m l of 1 M HC1 are mixed. M 1 y 1 + M 2 v 2 = M3v3 3.r + 1 ( 1 0 0 0 - * ) = 1.2* 1000 or 3x +1000 - * = 1200 or x = 100 m L Thus, 100 m L or 3 M a n d 900 m L of 1 M HC1 s h o u l d be mixed. (a): Structure (a) is least stable because the same charges (positive charges) are present at nearest positions.
3.
(a): K ^ - A e * ™ In Kcq = In A
AH°
RT AH° 2.303 log Knj •• 2.303 log ART AH° 1 log„, = logio -A 2.303R T
2+
53. Assertion : In a galvanic cell: Z n | Zn 1| Cu 1 C u after complete discharging, the equilibrium gets established. Reason : In galvanic cell, Z n | Zn 2+ 1| Cu 2+ 1 C u after complete discharging (Zn 2+ ) b e c o m e s equal to (Cu 2+ ).
A g r a p h b e t w e e n log 1 0 K^ a n d — will give straight AH° a n d intercept = log u y 4 2.303 This is e q u a t i o n of a straight line (y = mx + c). line slope = -
54. Assertion : KCOC1, (RC0)20 a n d RCOOR' all react w i t h G r i g n a r d reagent to f o r m 3° alcohols. Reason : RCOC1 reacts w i t h jR2Cd to f o r m ketones b u t ( R C 0 ) 2 0 a n d RCOOR' d o not react at all.
-AH° 2.303R
55. Assertion : B e n z a l d e h y d e u n d e r g o e s aldol condensation. Reason : A l d e h y d e s that d o not h a v e a - h y d r o g e n u n d e r g o aldol condensation. 56. Assertion : F~ ion is a w e a k ligand a n d f o r m s o u t e r orbital complex. Reason : F~ ion cannot force the electrons of dz2 a n d dx2_ y i orbitals of the inner shell to occupy dxyr dyz a n d d„-orbitals of the s a m e shell. 57. Assertion : F e S 0 . 7 H 0 always contains Fe,3+ 4
2
ions. Reason : This is d u e to aerial oxidation of Fe 2+
1 /T
•
4.
(d): Na 2 S + 2HC1
5.
(d): For a thermally insulated container, Aq = 0. But work (W), change in internal energy (AE) and change in enthalpy (AH) are non-zero.
NaCl + H2S
6.
(b): W h e n the combination is Fe 2+ a n d NO, there will be 5 u n p a i r e d electrons in the complex (4 in Fe 2+ a n d 1 in N O ) a n d w h e n the combination is Fe+ a n d NO + , there will be only 3 u n p a i r e d electrons 33 CHEMISTRY TODAY | MAY '13
which can be differentiated by measuring the magnetic moments. 7.
and groups 17 and 16 respectively. Thus, the least negative EA is of 3 due to electron-electron repulsion in the compact 2p-orbital. The more negative EA is of 4, then 1 (halogen) and most - v e is of 2 (halogen). Thus, the correct order is 3 < 4 < 1 < 2.
(b): KEl = hvx - hv>0, KE2 = hv2-hx>0 KE 1 /i(u,-u 1 1 n) = — or —=— KE2 K h(v2-v0) K or Kx>i - Kv0 = u 2 - u() or KU] - X)2 = Kx>0 - u 0 Xuj - u 2 = u 0 (K - 1) K'U1 - u 2 t>o = K - l
12. (d)
(4) M e
CH-CH, 8.
(3) r uHT (c): C H — ,=C
i>
+4
- r^i CH=CH-Me (1)
H
6
AP po = I AP = (74.01 - 74.66) torr P° = 74.66 torr If M is the molar mass of hydrocarbon, then n22 2/ M x2n x + n2 (100 / 78) + (2 / M)
14. (c): Raoult's law =
Me I
CH-CH3
CH 2 —CH-
CH=CH—Me
Hence,
0 OCH, 9.
(d):
+2
13. (a): In VOCl 2 , V4+ has d] configuration. In Cu Cl 2 , Cu 2+ has d9 configuration. Both have 1 unpaired electron, so these have same colour in aqueous solution.
(CH 3 ) 3 CC1_^
Cl 2 /FeCl 3
(100 / 78) + (2 / M)
Thus atomic ratio is Nc : NH :
—OCH 3 group is a stronger o, pC ( C H , ) , directing group J than C(CH 3 ) 3
OCH
2/ M
74.66
M = 177.6 g m o l 1 m c : m H :: 94.4 : 5.6
OCH,
Anhyd. A1C13
74.66 - 74.01
94.4 12
'
=> 7.87:5.6 => 1 . 4 : 1 =}• 7 : 5 Hence empirical formula is C 7 H 5 Molar empirical mass = 89 g mol - 1 molar mass No. of C 7 H 5 units = empirical mass
OH HBr, Heat
1
177.6 89
(cleavage of ethers)
Hence, molecular formula = C 14 H 10 C(CH 3 ) 3
C(CH 3 ) 3
15. (b):In more than one reactant system, the half-life time (f1/2) is the time for half of the reactant present in small amount. For set (1) and (2), L is in small amount and according to rate law, the reaction is of first order w.r.t. L. So in these sets the half-life is independent of initial concentration and so the f1/2 for set(2) will be same as that for set (1) i.e. x = 62.6 s. In set (3) and (4), K is present in small quantity and f1/2 will depend on [K], According to rate law expression the order w.r.t. K is 2.
10. (a): Na 2 S 2 0 3 + H Q —> NaCl + H 2 0 + S02 T Suffocating gas (Q) (P) + s Yellow
K 2 Cr 2 O y + S0 2 + H 2 S0 4 (Q)
K 2 S 0 4 + Cr 2 (S0 4 ) 3 + H 2 0 Green (R)
11. (d): Elements with electronic configurations 1 and 3 belong to second period and groups 17 and 16 respectively whereas the elements with electronic configurations 2 and 4 belong to third period 34
. hl2
-k7a
CHEMISTRY TODAY | MAY' 13 )
1
or
M
JAqU
^1/2)4
tAik
625
10
NO,
20. (a): F
or 2
5
(1/2)4
625 •'• ( f i/2)4 = — = 312.5 s i.e., y = 312.5
H,C
16. (a): Lead coming from automobile produces mental diseases.
>C6H5-COOH
NaOH
> C 6 H 5 COONa
-H,0 '
19. (d): (a) Fe2+;[Ar] [T|| T 1 T | t | t | •
[ 4V
4s
Fe2+ in; [Ar] | t l f f l f t l strong ligand 3d 6 CNT
] •
c
4p
4s
[Fe(CN) 6 ] ; [Ar] l U l U U M
E d2sp3 inner orbital complex
(b) Mn 2+ ;
[Ar] I T 1 T1 T 1 T i n 3d
[Mn(CN) 6 ] ;
F T T
•
5
4d x
[Ar] [til Til ? 1 * 1
H
» I » 1=
[Ar] ITU T 1 T I T p r i
•
3d6 [CO(NH3)6)3+;
3p
4s
[Ar]|N|N|TL
R
F
F
[AR] M I M M TTTI • 3d 8
R T T 4s
4p
In this case also NH 3 is a strong ligand but electrons remain unpaired since only one orbital is left vacant in 3d. Thus, [Ni(NH3)6f; [AriitirmnimijF 3d 8
4s
4p 3
4d 2
sp d outer orbital complex
(X)
\
N
NH,
HoC
21. (b): At isoelectric point pi = |[ P JC a +
= | [8.7 + (14 -11.7)]
1 11 = - [8.7 + 2.3] = — = 5.5 2 2 22. (a): Let, no. of O 2 " ions in Fe0 93 O = 100 Total no. of Fe2+ and Fe3+ ions = 93 Let no. of Fe2+ ions in the sample = x and no. of Fe3+ ions in the sample = y x + y = 93 ^ ...(i) Total - v e charge carried by 100 O 2 ions = 200 Total +ve charge carried by x Fe2+ ions = 2x Total charge carried by y Fe3+ ions = 3y Thus, 2x + 3y = 200 ^ ...(ii) Solving (i) and (ii), we get x = 79, and y = 14 .•. No. of Fe2+ ions, x = 79, and .-. No. of Fe3+ ions, y = 14 % of iron present as Fe3+ 14 -x 100 = 15.05% 79 + 14 Cl2 + H z O - > HC1 + HOC1 HCIO + H 2 Q 2 > HC1 + H 2 Q + P 2
23. (b):
2HC1 + O,
Overall reaction : Cl2 + H , 0 -
24. (d): Me r~*H+ > C = C H - C H 2 - -CH7-C= : C H - C H , Me-' 5 6| 3 4 Me 3° carbocation is expected to form. Internal attack is expected to produce 3° carbocation
d2sp3 inner orbital complex
(d) Ni 2+ ;
HoC
NO,
M
3p
1 •
orbital complex
(c) Co3+;
(i) N a N 0 2 + HC1 0 - 5°C ; (ii)H 2 /Ni
-3NaCl
3d 6
/ A
H3C
-3HC1'
-H,0
\
exhausts
17. (b): When chlorine gas is bubbled into an absolution of KI, some of the F ions are oxidized to I2. The I2 molecules combine with I ions to form brown IJ, the aq. solution is above a larger of CC14 in which I2 is quite soluble. 2KI + Cl2 -> 2C1 + I2; I2 + CCI4 Violet colour (lower layer because CC14 is heavier than water) 3C12(hv 3NaOH 18. (b): C 6 H 5 —CH > C6H5-CCI, [C6H5-C(OH)3]
(CH 3 ) 2 NH Nucleophilic aromatic substitution
Me Me^ 2
-CH2-CH2- CH2—C=CH—CHj 3
CH 3
4
6| Me
7 (C7 attacks on C 2 )
CH 3
.CH,
s
V-CH3 CH,
CH,
35 CHEMISTRY TODAY | MAY '13
25. (b): As the n u m b e r a n d size of the substituents a r o u n d the — C O O H g r o u p or the — O H g r o u p in alcohols increases, the rate of esterification slows d o w n . Thus, the correct o r d e r of esterification is Y>X>Z.
31. (b): Mn 2 + in M n S 0 4 - 4 H 2 0 h a s d 5 configuration
26. (b): H i g h e r is the critical t e m p e r a t u r e of a gas, m o r e easily it is liquefiable a n d greater is its adsorpition.
d s configuration (two u n p a i r e d electrons). T h u s
[OH~] = 1M K_ = [Cu 2+ ] [OH"] 2 K,sp lxlQ-19 [Cu2+] = 1 [OH"]2
0.0591,
n
1
log—-— [Cu 2 + ]
= 0.34 • • ^ l o g l O " = 0.34 -
u n p a i r e d electrons) a n d Ni 2+ in N i S 0 4 - 6 H 2 0 h a s C U S 0 4 - 5 H 2 0 h a s lowest d e g r e e of p a r a m a g e t i s m .
O n raising the t e m p e r a t u r e above 298 K, the colour d e e p e n e d because of increase in N 0 2 content. This conversion of N 2 0 4 into N 0 2 b e c o m e complete at a b o u t 413 K. W h e n the t e m p e r a t u r e is raised above 433 K the colour f a d e s a w a y d u e to dissociation of N 0 2 into a m i x t u r e of N O a n d 0 2 a n d this is complete at a b o u t 873 K (colourless). If at this t e m p e r a t u r e , the p r e s s u r e is increased, the equilibrium shifts b a c k w a r d leading to the f o r m a t i o n of N 0 2 according to Le-Chatelier's principle.
= 1 x 10 Cu
E,Cu'VCu ~ E °Cu 2 VCU
0.0591X19
34. (b): H 3 C
= - 0 . 2 2 V.
'G
_
29. (c): Let x N a n d xc>2 m o l e fraction of N 2 a n d 0 2 be present in the mixture respectively. Average molecular m a s s of mixture (M) f r o m general gas equation.
OH
< c r Greater the stability of H C
M = — x RT = — x 0.0821 x 273 = 29.137 P 1 X n 2 M n + X 0 2 M 0 2 = 29.137 x Na x 28 + Xo 2 x 32 = 29.137 Since Xq2 = 1 28x N2 + ( l - x N 2 ) 3 2 = 29.137 2&cN2 + 3 2 - 3 2 X N j = 29.13 7 32 - 4 x N 2 = 29.137 3 2 - 2 9 . 1 3 7 = 4XN2 Mole fraction of N j , (x N J = 0.71 Mole fraction of O2, (x0j = 1 - 0.71 = 0.28 Partial p r e s s u r e of 0 2 = 0.28 * 1 = 0.28 atm.
35. (a): 2HgCl 2 + SnCl 2
SnCl 2 + Hg 2 Cl 2 - 4 SnCl 4 + 2 H g l j
„
HT g O
Heating above 400°C
S
1 ^
> T TH g ( s ) + - 0 2 ( x )
Hence, ore is HgS, cinnabar. 36. ( b ) : NH,
NH
^>C6H5ch2CH^
i
—H,0
C6H5 —CH2-
OH
i o d o f o r m test.
CHEMISTRY TODAY | M AY ' 1 3
C H 3 - C - N H 2 + H+ I
^ H
CHO
Final p r o d u c t will give silver mirror test b u t n o t
36
Hg 2 Cl 2 + SnCl 4 Mercurous chloride
/.OH
NaOH( ,
, higher the
leaving t e n d e n c y a n d h i g hO e rHthe rate of reaction.
30. (d): C6H5CH2CHC12-
electron);
Fe 2+ in F e S 0 4 - 6 H 2 0 h a s d6 configuration (four
33. (b): N O , a n d N 2 0 4 are p r e s e n t in equilibrium as r e p r e s e n t e d below: Warm above xt ^ 298K 433K-873K , „ T„ N 2°4 , -2NQ2^=r 2NO + O j r Colourless Brown H.gn pressure C o l o u r l e s s -
2 8 . (b): P H = 14, / . P O H = 0
+ 2
h a s d9 c o n f i g r ration (one u n p a i r e d
32. (a): T w o axial a n d t w o equatorial b o n d s .
27. (b): A1C13 + 3 H 2 0 Al(OH) 3 + 3HC1 2Al(OH) 3 — ^ A1 2 0 3 + 3 H 2 0
Cu
(five u n p a i r e d electrons); Cu 2 + in C u S 0 4 - 5 H 2 0
<
NH,
2
» CH3-C^-NH2 III
> CH 3 — C — N H 2 II
<—>
C H3 — C_ = N H2 INJTL 3 -V rv
The conjugate acid (III) obtained by addition of a proton to (I) is stabilized by two equivalent resonance structures and hence compound (I) is the most basic. Further 2° amines are more basic than 1° amines while amides are least basic due to derealization of lone pair of electrons of N over the C = 0 group. Thus, the order : 1 > 3 > 2 > 4.
47. (b): Because of extensive hydration, the effective size of Li+ is the largest in aqueous solution.
37. (a): Acetic acid liberates C 0 2 gas on reaction with 10% N a H C 0 3 solution. Ethyl acetate on hydrolysis gives acetic acid.
50. (b): On shaking with concentrated H 2 S0 4 thiophene being more reactive undergoes sulphonation and the thiophene-2-sulphonic acid thus formed dissolves in concentrated H 2 S0 4 .
acid .
CH 3 COOC 2 H 5 +H 2 O S
, No. of atoms disintegrating per second 38. (a): A. = : Total number of atoms present 1 =
3.7 xlO 1 0 —
6.02 xlO 2 3 226
=
226 x 3.7 x 1010 -
6.02 xlO 2 3
Hofmann's
mustard
oil
CaS04
Addition of OH~ increases the common ion, OH" concentration, shifts reaction backward respective [(CH 3 ) 2 NH] gets increased, but ionisation constant remains constant. Second statement is correct as these two are parts of equilibrium mixture. 52. (a)
= 1.39 x 10" u s"1 39. (b):
gives
51. (c): ( C H 3 ) 2 N H + H 2 0 ^ ( C H 3 ) 2 N H 2 + 0 H "
CH 3 COOH + NaHCO;, ~ > CH 3 COONa + C0 2 T + H 2 0
,
49. (d): Aniline reaction.
CH 3 COOH+C 2 H S OH
(X)
Or
48. (a): N required extra energy in order to add an extra electron.
1200 c
°
> 2CaO + 2SO z + 0 2
Anhydrous
40. (c): A greenish yellow gas is formed. 2KMn0 4 + 3H 2 S0 4 + 10HC1 -> K 2 S0 4 + 2MnS0 4 + 8H z O + 5C12T Greenish yellow
41. (a): N 0 2 is an acid radical and oxidizing reagent, it oxidises I" into I2. 21" + 2 N 0 2 + 4H + > I2 + 2NOT + 2H Z O 42. (c): Cu + ion has 3 d w outer electronic configuration where no d-d transition is possible. 43. (d): Pyridine is more basic than pyrrole. In pyridine, the lone pair of electrons on N is available for protonation but in pyrrole they are contributed towards aromatic sextet formation. The N-atom in pyrrole is sp2 hybridised. 44. (b): As 2 S 3 being negatively charged, is coagulated faster by Ba2+ ions than by Na + ions. 45. (c): Collectors attach themselves by polar groups to grains of ores which then become water repellent and pass on into the froth. 46. (b): With increase in temperature by 10°C, the fraction of molecules possessing energy equal to or greater than threshold value becomes 2 or 3 times.
53. (c): Complete discharging of cell suggests equal potential at the cathode and anode. 54. (b): RCOC1, (RCO) z O and RCOOR' all add two molecules of Grignard reagent to give 3° alcohols. 55. (d): Aldehydes having a methyl or methylene group in the a-position or more correctly having atleast one hydrogen atom in the a-position undergo dimerisation in presence of a base at low temperature to form ^-hydroxy aldehydes called aldols. 56. (a) 57. (a): FeS0 4 -7H 2 0
^ o
> 2FeS0 4 Fe 2 0 3 + S0 2 + S0 3 (Fe3+)
58. (d): 2px and 2py orbitals are degenerate orbitals, i.e., they are of equal energy and hence no possibility of transition of electron. 59. (d): Z can be greater than 1 or less than 1. Nonideal gases exert ower pressure than expected due to backward pull by other molecules. 60. (c): Saline or ionic hydrides are compounds of hydrogen with most of the s-block metals. However, with p-block metals hydrogen forms molecular or covalent hydrides.
37 CHEMISTRY TODAY | MAY '13
PRACTICE PAPER
JEE Advanced PAPER- 1 SECTION-
3.
A vessel at 1000 K contains C 0 2 w i t h a p r e s s u r e of 0.5 atm. Some of the C 0 2 is converted into C O o n the a d d i t i o n of graphite. The value of K if the total p r e s s u r e at equilibrium is 0.8 a t m is (a) 1.8 a t m (b) 3 a t m (c) 0.3 a t m (d) 0.18 atm
4.
W h i c h of the following chemical unit is certainly to be f o u n d in a n e n z y m e ?
Straight Objective Type This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
1.
2.
The t e m p e r a t u r e of blast f u r n a c e to p r o d u c e iron f r o m its ore F e 2 0 3 varies f r o m 500°C at the t o p of the furnace to about 1900°C at the b o t t o m of the furnace. The reaction b e t w e e n the ore F e 2 0 3 a n d CO at the lowest t e m p e r a t u r e (~ 500°C) is (a) 3 F e 2 0 3 + C O -> 2 F e 3 0 4 + C O z (b) F e 2 0 3 + 3CO —> 2Fe + 3CO z (c) F e 2 0 3 + 3CO 2FeO + 3 C O z (d) F e 2 0 3 + C O + C a C 0 3 -> F e 2 0 3 + C O + C 0 2 + CaO In general, the first ionisation energy decreases in a regular w a y on d e s c e n d i n g the m a i n g r o u p s . A d e p a r t u r e f r o m this trend is observed in g r o u p 13 as f o l l o w s :
(a)
HO'
H ^ (c)
5.
HO
Q
(b)
O
I II ...N—C—...
(d)
Plots for 2s-orbital are given below.
-1
Element B Al Ga In
Gallium h a s higher ionisation energy in comparison to A1 because (a) gallium is h a r d e r t h a n a l u m i n i u m (b) gallium is preceded b y ten transition elements, w h e r e 3d shell is being filled w h i c h m a k e s Ga larger t h a n it w o u l d be (c) gallium is preceded b y ten transition elements, w h e r e 3d shell is being filled w h i c h m a k e s Ga smaller t h a n it w o u l d be (d) effective nuclear charge for Ga is m u c h higher than Al. 38
T
Ionisation energy (in k j m o l ) 1st 3rd 2nd 801 2427 3659 577 1816 2744 1979 2962 579 1820 558 2704
CHEMISTRY TODAY | MAY ' 1 3
X, Y a n d Z are respectively (a)
l
F, l F 2 and 4 n r T 2 1
2
2
(c) knr T , *F a n d Y 6.
7.
(b) V 2 , T a n d (d)
x 2
¥ , 4nr
2
4nr2lV2 a n d *F
D u r i n g the t r a n s f o r m a t i o n of "X to J y b y a a n d (3-decay, the n u m b e r of P-particles emitted are (a)
-7-
(c)
d+
(b) a-b
2
-+ c
(d) 2c -d+a-b
W h i c h of the following sets of species does n o t follow octet rule?
(a) CO, PC15, PCI3, AICI3
(a) x>z >y>w (c) y>x>w>z
(b) CO, B 2 H 6 , N H 3 , H 2 0
(c) AICI3, BF3, PC15, SF6 (d) H 2 0, NH3, COZ, AICI3 8.
9.
Bottles containing Q H 5 I a n d C 6 H 5 C H 2 I lost their original labels. They w e r e labelled A a n d B for testing. A a n d B were separately taken in test tubes a n d boiled w i t h N a O H solution. The e n d solution in each tube w a s m a d e acidic w i t h dilute H N O 3 a n d s o m e A g N 0 3 solution w a s a d d e d . Solution B gave a yellow precipitate. Which o n e of the following statements is t r u e for the experiment? (a) A d d i t i o n of H N O 3 w a s u n n e c e s s a r y (b) A w a s C 6 H 5 I (c) A w a s C 6 H 5 C H 2 I (d) B w a s C 6 H 5 I The basic characters of the given substituted anilines (I to IV)
(b) (d)
x>w>y>z w>z>y>x
12. A complex is p r e p a r e d by mixing CoCl 3 a n d NH 3 . 0.1 M solution of the complex w a s f o u n d to freeze at -0.372°C. The f o r m u l a of the complex is (a)
[CO(NH3)6]C13
(b)
[CO(NH3)5C1]C12
(c)
[CO(NH3)4C12]C1
(d)
[Co(NH3)3CL3]
13. ( M n S 0 4 + H 2 0 2 ) solution is u s e d to detect N H 3 gas, w h i c h p r o d u c e s a black ppt. of MnO(OH) 2 . The s a m e observation m a y be obtained w h e n a solution A is a d d e d to the above solution. Then A m a y be (a) K O H (b) N a O H (c) Ca(OH) 2 (d) only (a) and (b). 14. Which o r d e r is/are correct regarding point? (a) CH, C H V
CH;
H > (b) CH 3
CH,
/CI
CI
= < H CI
/ H / o =C\
x
0 H
boiling
>
H
CI
NH,
IV. CI
are such that (a) I > II > IV > III (c) II > I > III > IV
(b) I > II > III > IV (d) II > I > IV > III
10. A 1.5276 g s a m p l e of CdCl 2 w a s converted to metallic c a d m i u m a n d c a d m i u m - f r e e p r o d u c t s b y a n electrolytic process. The w e i g h t of the metallic c a d m i u m w a s 0.9367g. If the atomic w e i g h t of chlorine is taken as 35.453, w h a t m u s t b e the atomic w e i g h t of C d f r o m this experiment? (a) 112.41 (b) 113.1 (c) 112.20
(d) 112.61 S E C T I O N - II
Multiple Correct Answer Type This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE are correct.
11. Four c o m p o u n d s are n o t e d below. A r r a n g e t h e m in o r d e r of decreasing reactivity t o w a r d s nitration. CH2-C1
CH2 - CN
15. The correct characteristics of lyophilic and lyophobic colloid(s) is/are (a) Lyophilic sols are highly h y d r a t e d while lyophobic sols are not. (b) Lyophilic sols h a v e lower surface tension a n d viscosity t h a n that of the dispersion medium. (c) Lyophobic sols can be easily coagulated but lyophilic sols are not. (d) Gold sol is irreversible sol because gold cannot be obtained back. SECTION - III Integer Answer Type This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).
16. The difference b e t w e e n n u m b e r of lone pairs and n u m b e r of S—S b o n d s in S 8 molecule is CHEMISTRY TODAY | MAY '13
39
17. The rate of a reaction starting with initial concentrations 2 x 10"3 M and 1 x l(T 3 M are equal to 2.40 x 10"4 M s"1 and 0.60 x UP 4 M s _1 respectively. The order of reaction is, 18. The number of possible isomers of an octahedral complex, [Co(C 2 0 4 ) 2 (NH 3 ) 2 r is
0.1 g of another non-volatile substance Q in the same amount of benzene raises the boiling point of benzene by 0.4°C. The ratio of molecular masses of P and Q is x : 1 where x is 20.
^ s ?
.CI
+ H2O
H,0 CH,OH
In this reaction we get ... alcohols (stereoisomers not (a) one (b) (c) three (d)
19. A solution containing 0.1 g of a non-volatile organic substance P (molecular mass 100) in 100 g of benzene raises the boiling point of benzene by 0.2°C, while a solution containing
types of substituted considered) two four
PAPER-II SECTIONStraight Objective Type This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
21. The energy needed for Li^j —> Li3+(g) + 3e~ is 1.96 x 104 k j mol" 4 , if the first ionization enthalpy of Li is 520 k j mol -1 , second ionization enthalpy of Li is approximately (Ionization enthalpy of H is 2 18 x 10 -18 J/atom) (a) 5950 k j mol" 1 (b) 5950 J atom" 1 -1 (c) 7270 k j mol (d) 7 2 7 . 0 k j m o r 1 22. Given the potentials: M n ^ , + 2e~
following
standard
reduction
(a)
CH3-CH2-CH2-CH2-C-OH
(b)
O II CH3-CH-CH2-C-OH
CH,
o
I 2 + 2e~ -> 2 I H ) ; £° = + 0.53 V 0 2 f e ) + 4 H ^ } + 4e~ 2 H z O ; E° = + 1.23 V Electrolysis of an aqueous solution of Mnl 2 would give (a) H 2 , I 2
(b)
(c) Mn, 0 2
(d) Mn, I 2 following
conjugated
diketo
II Which of the following is the correct decreasing order of E° values? (a) 1 > II > III (b) III > II > I (c) I >111 >11 (d) II > I > III
CHEMISTRY TODAY | M A Y '13
(c)
CH3-CH2-CH-C-OH
CH 3 (d)
o
II (CH3^C-CH2-OH
H2,OZ
° = o = o = °
40
O
Mn ( s ) ; £° = -1.18 V
2H z O + 2e~ -> H 2(g) + 2 0 H W ) ; E° = - 0.83 V
23. Consider the compounds.
24. An organic acid A(C 5 H 10 O 2 ) reacts with Br2 in presence of phosphorus to produce a resolvable compound, which on dehydrobromination gives an unsaturated acid. The unsaturated acid does not show geometrical isomerism and on decarboxylation gives alkene. Alkene on ozonolysis gives two compounds, one gives positive Schiff's test while the other doesn't. The organic acid 'A' is
25. The composition of the equilibrium mixture for the equilibrium CL2 ^ 2C1 at 1400 K may be determined by the rate of diffusion of mixture through a pin hole. It is found that at 1400 K, the mixture diffuses 1.16 times as fast as krypton diffuses under the same conditions. Find the degree of dissociation of CL2 equilibrium. (a) 0.116 (b) 0.1374 (c) 1.374 (d) 0.980 26. A white coloured inorganic salt gives the following reactions: (i) It is soluble in water and the solution has sweet taste. The solution turns black in presence of H 2 S.
CH, I (b) H , C - C H - C - H V 3 II O
(ii) The salt w h e n h e a t e d gives acetone a n d a yellow coloured residue w h i c h is u s e d in paints. (iii) The solution of the salt gives a w h i t e precipitate w i t h dilute HC1 w h i c h is soluble in hot water. The inorganic salt is (a) C a ( C H 3 C O O ) 2 (b) Z n ( C H 3 C O O ) 2 (c) Pb(CH 3 COO) 2 (d) Ba(CH 3 COO) 2 27. N a C N is s o m e t i m e s a d d e d in the Froth floatation process as a d e p r e s s a n t w h e n ZnS a n d PbS minerals are extracted because (a) N a C N causes r e d u c t i o n b y precipitation (b) PbS f o r m s soluble complex Na 2 [Pb(CN) 4 ] while ZnS f o r m s f r o t h (c) Pb(CN) 2 is precipitated while n o effect o n ZnS (d) ZnS f o r m s soluble complex, Na 2 [Zn(CN) 4 ] while PbS f o r m s froth.
(c)
diastereomers enantiomers
and
H3C-C
O
H
O
O
CH 3
H-C-H II O
30. The c o m p o u n d S is
CH, (a)
O
H3C-CH-CH-C-H
H2C- CN
o ^ > c - cII - H
H3C \ (b)
H3C
H2C - CN
(c)
(ii) Br 2 /CCl 4 (ii) H 2 /Pd
(a) (c)
H - C - H II O
(d) H,C - CH - CH, - C - H a n d 3 2 I II
(i) H 2 /Pd/CaC0 3
CHJ—C = C—CH3
H3C-CH-CH2-C-H CH3
28. The p r o d u c t of the following I a n d II sequence are related as 1/
and
CH, CN I H3C-CH-CH-CH-OH I H2C - OH
H3C
(b) identical (d) geometrical isomers.
(d)
H3C
>
CN I C - C H - OH
H2C- OH
S E C T I O N - II Paragraph Type This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
P a r a g r a p h f o r Q u e s t i o n s 29 a n d 30 Two aliphatic a l d e h y d e s P a n d Q react in the presence of a q u e o u s K 2 C 0 3 to give c o m p o u n d R, w h i c h u p o n treatment w i t h H C N p r o v i d e s c o m p o u n d S. O n acidification a n d heating S gives the p r o d u c t s h o w n below: OH
P a r a g r a p h f o r Q u e s t i o n s 31 a n d 32 Fluorine, the first m e m b e r of g r o u p 17, differs f r o m the other m e m b e r s of the g r o u p in m a n y respects d u e to following reasons: (i) Its size is very small (ii) It h a s very h i g h electronegativity (iii) Its dissociation energy in the molecular form (X2) is the least. 31.
W h i c h h a s the m a x i m u m molar enthalpy of vaporisation? (a) H F (b) HC1 (c) HBr (d) HI
32.
Which h a l o g e n d o e s not f o r m a n y oxyacid? (a) Iodine (b) Bromine (c) Chlorine (d) Fluorine P a r a g r a p h f o r Q u e s t i o n s 33 a n d 34
The w o r k d o n e d u r i n g adiabatic reversible process is 29.
The c o m p o u n d P a n d Q respectively are CH, I
nR\Tt -Tf ] given b y W = •
Thus, if Tf > T„ then since
Y-l
(a) H 3 C - C H - C - H O
and
H3C-C-H
1 - y = - v e (•.• Y > 1), the w o r k is d o n e o n gas. Also if Tf < T„ W = - v e , t h u s w o r k is d o n e b y the gas. The
O 41 CHEMISTRY TODAY | M AY '13
adiabatic process are m o r e steeper t h a n isothermal process a n d slope of adiabatic process = y « slope of isothermal process i.e., slope of adiabatic process > slope of isothermal process (since y > 1). Also the adiabatic process obey PV = constant, w h e r e a s in isothermal process PV = constant.
O 38.1
w h e n t e m p e r a t u r e of m moles of H e is raised f r o m T to 2T is (a) 8RT (b) 4RT (c) 16RT (d) zero SECTION - III Multiple Correct Answer Type This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE are correct.
35. Which of the following m e t h o d s give rise to l-bromo-2-phenyl ethane? (a) Benzene + oxirane + H + followed b y reaction with HBr (b) Benzene + oxirane + H + followed by reaction with P/Br 2 (c) P r o p a n e + HBr followed b y reaction w i t h benzene in the presence of A1C13 (d) Styrene + H 3 0 + + followed b y reaction w i t h P/Br 2 . 36. Given that Ni 2+ / N i : - 0 . 2 5 V, Cu 2 + / C u == 0.34 V A g + / Ag = 0.80 V and Zn2+/Zn=-0.76 V Which of the following reactions u n d e r s t a n d a r d conditions will not take place in the specified direction? 2+ (a) N i 2 + w r Cu, • Ni (s) + Cu' (b) Cu(s) + 2Ag+(fllJ) —» Cu :2+ (c) Cu(S) + 2H (d) Zn ( s ) + 2 H
+
+
2Ag ( s )
(aq)
—> Cu («?) ( ,„) ^+ H2(g) ; Z n (aij) + H2(gy
37. Ferroelectricity is exhibited b y (a) b a r i u m titaniate (BaTi0 2 ) (b) s o d i u m p o t a s s i u m tartarate (Rochelle's salt) (c) p o t a s s i u m d i h y d r o g e n p h o s p h a t e ( K H 2 P 0 4 ) (d) lead zirconate ( P b Z r 0 3 ) .
42
CHEMISTRY TODAY | MAY ' 1 3
C H ; N
^ > Products. Thepossible O
(a)
CMe, (b)
o
(c)
(d)
"CMe,
39. W h i c h of the following statement(s) is/are correct? (a) S 0 3 is a stronger oxidising agent a n d m o r e acidic t h a n S 0 2 . (b) Selenium f o r m s only t w o oxo acids i.e., selenous acid (H 2 Se0 3 ) a n d selenic acid (H 2 Se0 4 ). (c) The acidic strength a n d oxidising p o w e r of oxo acids is greater in + 6 oxidation state t h a n in + 4 oxidation state. (d) The t h e r m a l stability of oxides of g r o u p 16 elements decreases in the order. S02 > Se02 > Te02 > Po02 40. W h i c h of the following statement/s is/are not correct? (a) Physical a d s o r p t i o n is directly related to the critical t e m p e r a t u r e of the gas (adsorbate) (b) O n e g r a m of charcoal at 25°C will always a d s o r b the s a m e a m o u n t of a particular gas at a particular pressure. (c) At particular t e m p e r a t u r e , absorption always increases w i t h increase of pressure. (d) In adsorption, the concentration of a d s o r b a t e is always greater at the surface of the adsorbent. SOLUTIONS PAPER-I 1.
(a): At the t o p of the blast f u r n a c e t e m p e r a t u r e ~ 500°C) the reaction is 3 F e 2 0 3 + C O —> 2 F e 3 0 4 + C O z
2.
(c): Ineffective shielding causes decrease in size.
3.
(a):
2+
(fl())
C M 6 3
p r o d u c t s are (major + minor)
33. In a n adiabatic expansion of air (assume it a mixture of N 2 a n d 0 2 ) , the v o l u m e increases b y 5%. The percentage c h a n g e in p r e s s u r e is (a) 7% (b) 6% (c) 4% (d) 3% 34. In a t h e r m o d y n a m i c process h e l i u m gas Obey's T the law — y j = constant. The heat given to a gas
jj
c 02(g) + c(s)
2CO,to
Initial p r e s s u r e 0.5 a t m At. e q m . (0.5 - x) 2x :. (0.5 - x) + 2x = 0.8 or x = 0.3 atm Pco2 = 0.5 - 0.3 = 0.2, pco = 2x = 2x .0,3 - 0.6 a t m K = P c o / P c o ? = (0-6)2 / 0.2 = 1.8 a t m
(lowest
4.
(c): Enzymes being proteins contain the group -NH-CO-.
5.
(b): Y will be definitely /' because value of \|/ cannot be negative, thus X will be \|/2 and Z will be 47ir2\|/- Z represent radial probability function; its value will be zero at origin.
6.
( c ) : ®X -4>jY
+x
2
He +
Here, x = oc-particles emitted and y = (3-particles emitted. a = b + 4x _a-b x
c = d + 2x - y y = d-c + 2x a-b y=d-c+—y=d+ 7.
13. (a,b,c) 14. (a,b,c,d) : As magnitude of dipole increases, b.p. increases i.e.,
H3C H
/
c=c
\
H
H
/
c=c X
CH 3 H
15. (a,c): (b) is wrong because lyophilic sols have higher viscosity than that of the dispersion medium. (d) is wrong because gold sol is irreversible as gold obtained back cannot be converted into sol by directly shaking with the dispersion medium. 16. (8): The difference between number of lone pairs and number of S—S bonds in S 8 molecule is 8. S 8 molecule is shown here. Every sulphur atom is in sp3 hybridized state. :S: :S:
a-b
NHK
'Xs
(c): BF3, A1C13 are electron deficient molecules while in PC15 and SF6 molecules P and S possess more than 8 electrons in their valence shells.
8.
(b): Since B gives yellow ppt. with A g N 0 3 / H N 0 3 , B must be C 6 H 5 CH 2 I and hence A is C 6 H 5 I.
9.
(d): —OCH3 is a better electron donating group than — CH 3 and N 0 2 group, N 0 2 is better electron withdrawing group than —CI.
U
(ro)i
fMoli1
('OH
UA>]2J
T
logK?o)l/( r o)2) logUAo],/!^],} _ log(2.40xlO" 4 /0.60xlO" 4 ) log(2 x 10"3 / 1 x 10~3) =
log4
= 2
log 2 The order of reaction = 2. 18. (3):
NH,
.11. (d): w>z>y>x - CN is more electron-withdrawing than - CI. 12. (c): AT f = i * KfX m = i * 1.86 x 0.1 or 0.372 = 0.186 * i or i — 2 This shows that the complex gives two ions in solution. Thus, the formula of the complex is [Co(NH3)4Cl2]Cl. [Co(NH 3 ) 4 Cl 2 ] CI ^ [Co(NH 3 ) 4 Cl 2 ] + + CI"]
T
17. (2): Let the initial rate be : r0 = k[A0]a
i (0.016667) =0.008333 mol
The atomic weight of Cd 0.9367 g . . = s. = 112.41 g/mol 0.008333
H
:S:
Thus, there are 8 S—S bonds and 16 lone pairs. Difference = 16 - 8 = 8.
10. (a): Weight of CdCl 2 = 1.5276 g Weight of Cd in CdCl 2 = 0.9367 g Weight of CI in CdCl 2 = 0.5909 g 0.5909 " ( C 1 ) = 35.453 g / m o l = 0.016667 mol From the formula CdCl 2 we see that the number of moles of Cd is exactly half the number of moles of CI. n(Cd) = - n ( C l )
H3C
CH 3
moment
NH, Mirror
cis-d-isomer
cis-/-isomer
NH, ox
Co
ox
NH3 trans-isomer 43 CHEMISTRY TODAY | MAY '13
o II Hence,
is ( C H 3 ) 2 C H - C H 2 - C - O H
25. (b): Equilibrium represented as : of dissociation of Cl 2 may be ClA2(g) 2l 2Cl te) t =0 a 0 2aa tec, - a) Total moles = a( 1 - a) + 2aa = a( 1 + a) aM,CI, M,CI,
Q 1 = (m B ) Q 2 100 mB = 50 (m B ) P :(m B ) Q = 100:50
=
-
a( 1 + a ) M,Kr RiKr
= 2:1
M
1.16 =
M,a,
(1.16) x 7 1 - 1 = a, a = 0.1374 84 26. (c): The solution of lead acetate is sweet in taste, w h e n HC1 is added to its solution, it forms PbCl 2 which is soluble in hot water. A Pb(CH 3 COO) 2 > CH 3 COCH 3 + PbO + C 0 2
PAPER-II 21. (c): JE3 of Li,•(g) Enthalpy of Li = 2.18 x 10"18 x 9 J atom""1 = 2.18 x 10"18 x 9 x 10"3 x 6.023 x 1023 k j mol" 1 = 11817 k j mol" 1 And also /, +1 2 +13 In all write IE1 + IE2 + IE = 19600 k j mol "1 and IEi = 520 k j mol""1 ^ = 19600 - (11817 + 520) = 7263 k j mol" 1
27. (d): NaCN is added as a depressant for zinc sulphide (ZnS). N a C N forms a layer of zinc complex, Na 2 [Zn(CN) 4 ] on the surface of ZnS and thereby prevents it from forming the froth while PbS forms froth.
22. (a): Either Mn 2+ or H 2 0 will get reduced. Higher the reduction potential, easier to reduce. Thus H 2 0 will be reduced to give H 2 at cathode. Either I or H 2 0 will get oxidised. Higher the oxidation potential, easier to oxidise. 21 I2 + 2£>"; E° = - 0.53 V 2H z O -> 0 2 + 4H + + 4e~; E° = -1.23V Thus, I will be oxidised to give I 2 at anode.
28. (a): CH,-C=C-CH,
(i) H 2 /Pd/CaCQ 3
Unsaturated acid (no geometrical isomer) I +ve Schiff's test aldehyde
44
II - v e Schiff's test ketone
H
t
.CH3
/
/ C = C
\
H3c.
vC = = C, / C H 3
/ Br
H
Br 2 /CCl 4
X
R
H 2 /Pd
CH, H - -Br Br-H
HBr-
CH 3
CH, -Br
-H CH,
-HBr *
29. (b): P + Q
Alkene trisubstituted alkene
— CO :L
(ii) Br2/CCI4
"3CX
23. (d): Higher the E° value, higher the reduction potential and greater the tendency to be reduced. II on reduction gives two aromatic rings which together have more resonance energy and stability than the present diketo compound. Reduction of I gives a simple aromatic ring having two - O H groups at ortho position-relatively more stable than corresponding para form. Resovablecompound » (bromo derivative)
mix
84(1 + a )
20, (3)
Hi) • 11 a \ iA. t o ; . A ( C 5 H 1 0 O 2 )
(1 + a )
K
2C°3(a?)
> R
HCN
OH
Ozonolysis
O ^ O
CHEMISTRY TODAY | M A Y 13 5
Br
R reacts with HCN so, it must contain ^ C = 0 group. Since the compound S gives the given product formed on treatment with H + followed by heating thus S must have a - CN group which on acidification gives - COOH group with one carbon more than that in R. Therefore, P and Q are (CH 3 ) 2 CHCHO and H C H O respectively. H I H 3 C—CH—C- - H + H - c = o
or
PV513 = constant
or
PVy = constant
v 7 = — for He 3
J
Thus, process is adiabatic .\ Q = 0 35. (b): p >
[ > H
+
CH,3 11O
2-methylpropanal
methanal
(P)
(Q)
OH K2C
°3W:
O
^
O
H
|p/Br2
.Br
H3C H3C
H
>c-c=
HC^
0
H3C H,C > r " H2C
H2C OH
(S)
(R)
H3C H3C > 1 H 2 C.
OH
iH+<4 OH I CH
OH I CH CN
36. (a, c) : These reactions (a) and (c) will have - v e emf and hence cannot take place in the specified direction. 37. (a,b,c) O 38. (a,b,d):
CMe, O
30. (d)
31.
(a)
M&.C migration
Ph—C—CH,—CMe,—
32. (d)
O 1
33. (a): PV' = constant 105V
Pi-
rn
= constant
105 0-x p 100 or
Ei = 105 p 100
or
105 B± = 100 P
=1
7/5
p1 = 0.93p 7% decrease in p T 34. (b): - — = constant p 2/5 or •••
>
T = constant x p',2/5 PV=RT PV = R x constant p.2/5 ^2/5
x V = constant
Ph—C—CH 2 N 2 I CMe,
Me 3 C—C—CH 2 —Ph
O / \ Ph—C—CH 2 I CMe 3
39. (a,b,c) : Thermal stability of the oxides of group 16 elements decreases in the order SO z > Te0 2 > Se0 2 > P o 0 2 and not in the order as mentioned. TeO z decomposes at > 733°C while SeO z has - 600°C as the decomposition temperature. Po0 2 decomposes at 500°C. 40. (b,c,d): (b) is wrong because adsorption depends u p o n the surface area. (c) is wrong because at about 0°C, adsorption remains constant after a particular pressure. (d) is wrong because there can be negative adsorption also.
QUICK REVISION OF INORGANIC CHEMISTRY ;-BLOCK ELEMENTS
Halides 2
x
p-Block e l e m e n t s are characterised b y the ns np valence shell electronic c o n f i g u r a t i o n , w h e r e x = 1-6, n = 2 - 6 But i n n e r core f o r d i f f e r e n t e l e m e n t s m a y b e different.
• •
Group 13 : The Boron Family This g r o u p is the m o s t metallic in n a t u r e of the p - b l o c k e l e m e n t s a n d o n l y b o r o n is n o n metallic.
•
Property
B
Al
Ga
In
T1
Atomic n u m b e r (Z)
5
13
31
49
81
Atomic r a d i u s (A)
0.82
1.18
1.26
1.44
1.48
R a d i u s of trivalent ion (A)
0.20
Electronegativity
2.0
1.5
1.6
1.7
1.8
Oxidation states
+3
+3
+1, +3
+1, +3
+1, +3
0.50
0.62
0.80
•
All the h a l i d e s of g r o u p 13 e l e m e n t s are k n o w n except Tl(III) i o d i d e . T h e trihalides of all the e l e m e n t s of this g r o u p act as Lewis acids. B > Al > Ga > In —> electron accepting t e n d e n c y BF 3 a n d BC13 —> gases at r o o m t e m p e r a t u r e . BBr 3 —> volatile liquid; BI 3 —> solid L e w i s acidity : BF 3 < BC13 < BBr 3 < BI 3 B e c a u s e e l e c t r o n - d e n s i t y a r o u n d b o r o n in BF 3 increases d u e to pn-pn b a c k - b o n d i n g .
Q-, 0
F
0
0.95
0
Ionisation energy : B > Al < Ga > In > T1
•
Hydrides • • B f o r m s covalent h y d r i d e s , B„H„ + 4 a n d B„H„ + 6 called b o r a n e s . • Al —> polymeric h y d r i d e —> (A1H3)„ - d e c o m p o s e s • o n heating. • InH 3 a n d T1H 3 —» extremely unstable. • B, Al a n d Ga h a v e the t e n d e n c y to f o r m c o m p l e x anionic h y d r i d e s like Li[AlH 4 ], Li[BH 4 ], Li[GaH 4 ] • etc. d u e to t h e p r e s e n c e of a n a s c e n t p-orbital in t h e i r o u t e r m o s t s h e l l s d u e t o w h i c h t h e s e readily accepts electron p a i r f r o m the h y d r i d e ion (H~) as • AH 3 + H > [AH 4 ]- [Here, A = Al, Ga a n d B] Oxides and hydroxides • Oxides ( M 2 0 3 type): B 2 0 3 —> acidic A1 2 0 3 a n d G a 2 0 3 —> a m p h o t e r i c l n 2 0 3 a n d T1 2 0 3 —» s t r o n g basic • Hydroxides (M(OH) 3 t y p e ) : T1 —> Tl(OH) 3 a n d T I O H
•
pit-pjt back - bonding
T h e t e n d e n c y of b a c k - d o n a t i o n decreases to BC13 a n d BBr 3 b e c a u s e the e n e r g y difference b e t w e e n fully-filled orbital of h a l o g e n (Cl-3p, Br-4p) a n d v a c a n t orbital of B-2p increases. T r i h a l i d e s of Al exist in d i m e r i c f o r m b o t h in v a p o u r state a n d in n o n - p o l a r solvent. [AlFe] 3 ", [GaCl 6 ] 3 " a n d [InCl 6 ] 3 " a r e f o r m e d b y Al, Ga a n d In d u e to the availability of v a c a n t d-orbital in central atoms. D u e to the small size a n d inability to a c c o m m o d a t e e n e r g y , BC1 3 d o e s n o t c h a n g e f r o m c o v a l e n t c o m p o u n d to ion f o r m a t i o n in a q u e o u s solution (as occurs in A1C13). Al, Ga, In a n d T1 ions exist as o c t a h e d r a l a q u a i o n s i.e. [M(OH 2 ) 6 ] 3 + in a q u e o u s s o l u t i o n a n d m a n y salts like halides, s u l p h a t e s , nitrates a n d p e r c h l o r a t e s exist as h y d r a t e s . A l u m i n i u m s u l p h a t e s f o r m d o u b l e salts called a l u m w i t h salts of o t h e r metals. A l u m s [ M 2 S 0 4 - M 2 ( S 0 4 ) 3 - 2 4 H 2 0 ] --> u s e d f o r s o f t e n i n g of h a r d w a t e r a n d as m o r d a n t in d y i n g a n d p r i n t i n g of textiles.
S o m e important chemical reactions A1 2 0 3 + 6HC1 > 2A1C13 + 3H z O A1 2 0 3 + 2 N a O H > 2NaA102 + H2Q B , 0 3 + 6NaOH • -> 2Na,BO, + 3 H 2 0
4BC13 + 3LiAlH 4
F I F-B •
> 3LiCl + 3A1C13 + 2B 2 H 6 heat
Na 2 B 4 0 7 -10H 2 0 •
» Na 2 B 4 0 7 + 1 0 H 2 0
2NaB02 + B203 2NaBH 4 +1 2 — - > B 2 H 6 + 2NaI + H 2 2A1N + 6 H 2 0 -— » 2Al(OH) 3 + 2 N H 3 C r 2 0 3 + 2A1 — - > A1 2 0 3 + 2Cr 1+
Al + 3NH 4 OH -
-> Al(OH)3(s)
NaOH
Properties H 3 BO 3 + N O 2
H,BO, > Na3B03 + H2 Mg 3 B 2
B2O3
B 2 0 3 + 3H 2 T
BN + H ,
Boranes The c o m p o u n d B 2 H 6 -2NH 3 is ionic and exist as [H 3 N BH 2 < - NH 3 ] + [BH 4 ]and forms borazine on heating. Boron n i t r i d e ( i n o r g a n i c g r a p h i t e ) s t r u c t u r e resembles graphite.
Borax
x
;B H^
/ H
2c-2e B - H terminal bond
\
3c-2e~ B - H - B bridging bond (Banana bond)
BF H
BF 3 , b e i n g electron deficient c o m p o u n d , can accept a lone pair of electron, t h u s behaves as Lewis acid and also used as catalyst in organic reactions. It can form complexes like
with H 2 S
Transparent bead
B 2 0 3 + CoO - - > CO(B0 2 ) 2 : blue bead B 2 0 3 + M n O • —> Mn(BO Z ) 2 : pink bead Boric acid or orthoboric acid, H 3 B 0 3 • A w e a k m o n o b a s i c acid. It d o e s n o t d o n a t e protons like most acids but rather it accepts pair of electron from OH". It is therefore, a Lewis acid, and is better written as B(OH) 3 . B(OH) 3 + 2 H 2 0 ^ H 3 0 + + [B(OH)4]", pK„ = 9.25 • The titration between B(OH) 3 and a strong alkali solution (e.g. N a O H ) is not possible, because sodium metaborate Na+[B(OH)4]" or [NaB0 2 -2H 2 0] formed by reaction between B(OH) 3 and N a O H g e t s h y d r o l y s e d to r e g e n e r a t e B(OH) 3 a n d NaOH. B(OH) 3 + N a O H Na + [B(OH) 4 ] • H o w e v e r , if s o m e q u a n t i t y of p o l y h y d r o x y c o m p o u n d , like catechol, glycerol, mannitol, sugar etc, is a d d e d to the titration solution, the complex c o m p o u n d gives H + ions which makes H 3 B 0 3 to behave as a strong acid. • Orthoboric acid contains triangular BO;)3" units. In the solid the B(OH) 3 units are hydrogen bonded together into two dimensional sheets with almost hexagonal symmetry. Aluminium oxide and hydroxide . Al(OH) 3 y - A1 2 0 3
a - A1 2 0 3
Corundum
•
Borazine or borazole or inorganic benzene B,N,H, 3BC13 + 4NH 4 C1 3NaBH 4 + 4NH 4 C1 Boranes are electron deficient compounds. H
S-H I H
Na 2 [B 4 0 5 (0H) 4 ] -8H 2 0 (Borax) Water softener and cleansing agent. Borax bead test Na 2 B 4 Q 7 A,740°C> 2 N a B Q 2 + b 2 o 3
BORON 2 isotopic forms -> 10B (19%) and n B(81%). • Preparation B 2 0 3 ( s ) + 3M g ( s )
I
F I F-B • I F
with N H ,
> NaA10 2 + 2H z O
2BCl 3fe) + 3 H 2 ( i ? ) — > r2B}<— Electrolytic reduction of fused borates or By thermal decomposition tetra-fluoroborates of boranes and BX 3 at (KBF4) temperature above than 900°C
H I N-H I H
•
Alums General formula : M(I) M'(III) ( S 0 4 ) 2 1 2 H 2 0 and [M(H 2 0) 6 ] + , [M'(H 2 0) 6 ] 3 + and S0 4 2 " in 1 : 1 : 2 ratio. e.g. M(I) = K+ or NH 4 + , M'(III) = Al3+ Used as m o r d a n t in dyeing. Cement : Ca 3 Al 2 0 6 - Tricalcium aluminate
Group 14 : The Carbon Family Property
c
Si
Ge
Sn
Pb
Atomic n u m b e r (Z)
6
14
32
50
82
Atomic r a d i u s (A)
0.77
1.17
1.22
1.41
1.44
® Q tmmm O
.2
^
:
S
—1 ^
-"FFIF
BIOMOLECULES
'
i
:-P"
-
r~~~
J:G|L • • . -Y-
—
|
—IT.
•J.IHJJ.UI I.I I MI
_ HHHi
I
f. Yield a large number of monosaccharide units, e.g. starch, cellulose, glycogen, etc. ~
'
T
H
H
— • Cellulose: In plants cell wall, straight chain, composed of P-D-glucose units joined • by giycosidic linkage between Cx of one glucose unit and C4 of the next glucose unit. • Glycogen (Animal starch) : In animals brain, liver and muscles. More branched than amylopectin. • —
Z
'r • flWWCTWPWl ""' ' *. I—"—~—^ —" " " UABIfl&iBS3 • Sucrose (Non-reducing) : D - (+) 1 '™ Dextrose, Grape sugar, C6H1206 glucose and D - (-) - fructose held Give 4 molecules on • prepared by hydrolysis of sucrose and starch. through their respective carbonyl groups hydrolysis. • contains one-CHOgp, one 1° - OH,four2°-OHgps by giycosidic linkage between C, of ctGen.formula: • Two isomeric forms : a-glucose and (3-glucose. They undergo glucose and C2 of P-fructose. On C24H42021 mutarotation. , , , . .. , , , . ... . ., T-, . ,. . , . . hydrolysis it changes from dextro (+) to e.g. stachyose. • They differ in configuration at chiral C, -» Anomers. , _ ' laevo(-)-> Invert sugar. ——— — 1 • Maltose (Reducing): Two a-D-glucose i units in which Cj of one glucose linked to BTBWBWffBS C4 of another glucose unit. Give 2 molecules on hydrolysis. Gen. formula : , . . , , . . ^ • Lactose (Reducing) (Milk sugar): P - D Give 3 molecules on hydrolysis. galactose and P - D - glucose with c H q 12 22 11 Gen. formula: C18H32016 linkage between C, of galactose and C4 2 units are joined by giycosidic linkage. ofglucose. eg raffinose .
^
1
..:: Yields 2-10 monosaccharide ^"""••••"•I •BMMMMB^H mmufftlmti u n i t s on h y d r o l y s i s , Contain aldehyde group (-CHO) j" Contain keto group (> C = 0) Laevulose, Fruit sugar, C6H1206 classification based on number e.g. aldotrioses (glyceraldehyde), e.g. Ketotrioses (dihydroxyacetone), —• • C2 is chiral. ! of monosaccharide units given Aldohexoses (glucose, galactose, Ketopentoses (Ribulose), • Two isomers : a-D-fructopyranose and on hydrolysis. | mannose) J Ketohexoses (fructose) P-D-fructopyranose.
^
—:— • •—— —i-t——— S— ——— £ Importance •'.' "IS " Optically active polyhvdroxy aldehydes or ketones or the compounds ,r , . , r j sr j j j j r « Starch-food reserve in plants. I which produce such units on hydrolysis. : . ; , , , „;iTig,.—_— ,,.. ... ....,. . ....:..:............: • • •.. • • ....,,•;.. • .^j,, : • Glycogen-food reserve in animals. ; jE r^;- rp—— • Cellulose - structural material of cell 1 Bi.auBisi/iiVMX . ;•-.—• •. 1 in walls of bacteria and plants. On the basis of their behaviour Monosaccharides : cannot ! According to type of towards hydrolysis. be hydrolysed further. ifunctional group present
maaiUMJ
X
CQ
.2
.
:
unbranched
chain polymer and amylopectin (80-85%)
• Essential amino acids (10) : Which cannot be synthesized in the
—— - . - . - - -
• pH at which no net migration of amino acid under electricfield—» Isoelectric point.
;
Structure
————i
I 15B3flfTSI»IffHl-
• Types | ,
-
•flgjiifflg^W
— '
— -
{"(Pentose su ar + Base * rPhosphTricaadT
; : :
*
——:
•—
HUCMB
""'
——
!
.
i "'!"
—
'
1 — — purines = pyrimidines
(C + T>
fl^^SgS • Double helical structure : By • Sugar is P-D-ribose. Watson, Crick and Wilkins. H • 3 bases are same + fourth one is Uracil (U) Pairing between C = G and instead of Thymine (T). A=T
Pyrimidines: Cytosine(C), Thymine (T).
Purines : Adenine (A), Guanine (G), • Chargaff Rule: Total amount of
• Sugar is P-D-2-deoxyribose. ^ • 4 bases are:
——
"'"":"" "
formed due to intramolecular H- bonds Formed due to intermolecular litilfiUiifiiUiilV between the C = O of one amino acid H-bonds between two Due to coagulation, native shape of the protein is residue and the N - H of the 4th amino neighbouring polypeptide destroyed and biological activity is lost (2° and 3° acid residue in the chain, e.g., a-keratin in chains in a zig-zag manner, structures destroyed, 1 ° remains intact). hair, myosin in muscles. e.g., silk protein fibroin.
• Primary: Sequence of a-amino acids. • Secondary: Shape due to H-bonding. J • Tertiary : Overall folding of the polypeptide chains (secondary structure). • Quaternary : Spatial arrangement of subunits w.r.t. each other. p—— —jjEBBIBBMI—
—
• Fibrous: Thread like e.g. keratin, collagen, myosin, etc. . Ninhydrin test: for detection of a-amino acids. • Globular : Spheroidal e.g. enzymes, hormones, Hb, _ , . , . . ., ,, „ • Sanger s or DNP method: for determining N-terminal amino acid residue. antibodies, albumin, etc.
^CH^^^BESHi
L_>
^
.
_*.
Classification
a-amino acids (R-C-COOH) I
NH 2
(water soluble)
(water insoluble): branched chain polymer.
^T^T}-^ body and must be obtained through diet. 1 • Non-essential amino acids (10): (Dispensable) can be synthesized by body.
acids connected to each other by
peptidebondorpeptidelinkage.
Based on molecular structure.
Peptides
Hydrolysis
p^HHHHBB
• •
•
•
• •
Electronegativity
2.5
1.8
1.8
1.8
1.8
•
Oxidation states
+4
+4
+2, +4
+2, +4
+2, +4
•
The variable oxidation states are seen d u e to inert pair effect. C and Si —> non-metals, Ge —s> semi-metal, Sn and Pb —> metals. Metallic character increases d o w n the group. The covalent nature in the compounds of Sn4+ and Pb 4+ ions is d u e to high polarisation produced by them. As we move d o w n the group stability of lower o x i d a t i o n s t a t e i n c r e a s e s d u e to i n e r t p a i r effect. Catenation : C » Si > Ge = Sn » Pb Except C, all other elements form complexes due to presence of vacant d-orbitals in them i.e., these behaves as Lewis acids. e.g. SiF4 forms) [SiF6]2-, octahedral structure
•
Stability of tetrahalides CX 4 > SiX4 > GeX 4 > SnX 4 > PbX 4 Elements in this group are relatively unreacted b u t reactivity increases d o w n the g r o u p . Pb often appears more noble than expected due to a surface coating of oxide and partly due to high over potential for the reduction of H + to H 2 at a lead surface. Reactivity of the elements of group 14 (IVA)
C, S, Ge, Pb unaffected by H z O.
H2O
•
•
Allotropic forms C —> crystalline and microcrystalline Si —> crystalline and amorphous Ge —> 2 crystalline forms Sn —» 3 solid forms —> grey tin, white tin and rhombic tin Artificial d i a m o n d s p r e p a r e d b y M o i s s o n ' s method. Artificial graphite —> Acheson process S e m i c o n d u c t o r g r a d e Si is p r e p a r e d b y t h e reduction of highly p u r e SiCLj/SiHCLi with H 2 or by the pyrolysis of SiH 4 mainly
In case of N(CH 3 ) 3 , geometry is pyramidal but in case of N(SiH 3 ) 3 it is planar because the lone pair of N-atom is transferred to the empty d-orbitals of silicon, (pn-dn overlapping). Oxides
Nature
CO, CO^ Si0 2
acidic
GeO, GeO z , SnO, Sn0 2 , Pb304
amphoteric
PbO, P b 0 2
basic
> S n 0 2 + 2H 2
C, Si, Ge u n a f f e c t e d b y d i l u t e acids. Dilute acids
Pb does not dissolve in dilute H 2 S 0 4 due to formation of PbS0 4 coating. D i a m o n d is u n a f f e c t e d b y c o n c e n t r a t e d acids, b u t g r a p h i t e changes to mellitic acid (Ci 2 H 6 O n ) also called benzene hexa-carboxylic acid (A) COOH HOOCSJX.COOH HOOC^Y^COOH
Concentrated acids
Multiple bonding • C has strong tendency to show pn-pn multiple bond either with itself (C = C, C = C) or with atoms like N, O (C = N, C = O etc.) Carbon does not have d-orbitals and never forms dn-pn or dn-dn bonds. Silicon on the other h a n d f o r m s dn-dn bonds. •
Sn + 2H z O (steam)
sp3d2 hybridisation •
Reactivity
Reagent
COOH (A) with hot concentrated H N 0 3 and to graphite oxide with hot concentrated HF/HNO 3 . Si is oxidised and changes to SiF4 by hot concentrated H N 0 3 / H F . Pb does not dissolve in concentrated HC1 d u e to f o r m a t i o n of P b C l 2 coating. Carbon is unaffected by alkalies.
Alkalies
Complex formation
Sn and Pb are slowly attacked by cold alkali, and rapidly by hot alkali, giving stannates Na 2 [Sn(OH) 6 | and plumbates Na 2 [Pb(OH) 6 ], Si, G e , Sn a n d P b c a n s h o w coordination n u m b e r more than 4. Example : Si - Ge (6), Sn - Pb (8)
D i a m o n d is unreactive, but graphite reacts forming (CF)„.
R I - > H O -Si - O H + 2HC1 I R
CI - Si - CI + H 2 O
Si a n d Ge f o r m volatile SiX 4 a n d GeX 4 . Halogens
Sn a n d Pb are less reactive. Sn reacts with Cl 2 and Br 2 in cold, a n d with F 2 and I2 on w a r m i n g . Lead reacts with F 2 in cold a n d with Cl 2 on heating forming PbX 2 .
'T
Si-O
I f 1 Si-O--
\R
R
)n
silicones
Important chemical reactions
xo2 + m a —> xa 4 + m o 2
2 P b 0 2 + 4HNO3
(x = Ge, T b )
> 2Pb(N0 3 ) 2 + 2 H z O + 0 2
4CO + Ni ——» Ni(CO) 4 CC14 + H z O
FeorCu
> COCl 2 + 2HC1
X + 2NaOH + HzO
Si(OH) 4 + 4HC1
SnO + 2 N a O H - — > N a 2 S n 0 2 + H z O RMgCl + SiCl 4 2RC1 + Si
P b 3 0 4 + 8HC1 -
•> 3PbCl 2 + 4 H 2 0 + Cl 2
PbCl 2 + 2HC1
(X = Si, Sn)
Sn02 + 2NaOH
- > 2Hg + SnCl 4
P b ( N 0 3 ) 2 + 2HC1
N a 2 X 0 3 + 2H 2
SiCl 4 + 4 H 2 0 —
Hg 2 Cl 2 + SnCl 2
> H 2 SiF 6 + 2H z O
Fe203 + CO
> 2Fe + 3 C 0 2
> Na2Sn03 + H 2 0
Cu powder
> H 2 PbCl 4
Si0 2 + 6HF
2C
2CO + 4N 2
+ 0 2 + 4N 2
red hot
> R - SiCl 3 + MgCl 2 (R is an alkyl group)
> PbCl 2 + 2 H N 0 3
producer gas
from air
H y d r i d e s of Ge are called germanes, [Ge„H2„ + 2] n = u p t o 5 are known.
> R,SiCl
Group 15 : The Nitrogen Family They are called pnictogens.
Atomic number Electronic configuration
Element
7
[He] 2s2 2p3
P h o s p h o r u s (P)
15
[Ne] 3s2 3p3
Arsenic (As)
33
[Ar] 3d
3
4s 4p
+3, -3, +5
A n t i m o n y (Sb)
51
[Kr] 4dw 5s2 5p 3
+3, - 3 , +5
Bismuth (Bi)
83
[Xe] 4 / 1 4 5 d w 6s2 6p 3
+3, +5
10
General characteristics of group 15 elements • The elements of g r o u p 15 are less metallic than t h e c o r r e s p o n d i n g e l e m e n t s of g r o u p 14. O n m o v i n g d o w n the group, the metallic character increases. N P As Sb J5i_ Non-metals
•
•
Oxidation No.
Nitrogen (N)
Metalloids
+3, - 3 , +4, +5
N2O4
p2o3 p2o4
AS 2 0 4
N2O5
P2O5
AS 2 0 5
N2O3
Metal
Allotropy : Except n i t r o g e n a n d b i s m u t h , all elements of this group show allotropy. Phosphorus exists in three allotropic forms, i.e., white, red and black p h o s p h o r u s . Arsenic a n d antimony exist in t w o allotropic forms, i.e., yellow a n d grey. Oxides : All these elements form oxides of the type X 2 0 3 , X 2 0 4 a n d X2Os.
2
+1, +2, +3, - 3 , +4, +5
•
AS 2 0 3
Sb 2 0 3
Bi 2 0 3
Acidic
Sb 2 0 4
Bi 2 0 4
nature increases
Sb 2 O s
Bi 2 O s >r
Acidic nature decreases'* As electronegativity increases, acidic character of the oxide increases. A m o n g the oxides of same element, higher the oxidation state, greater the acidic character. Oxoacids : The strength a n d stability of oxoacids h a v i n g t h e e l e m e n t s in t h e s a m e o x i d a t i o n s t a t e s d e c r e a s e s g r a d u a l l y w i t h d e c r e a s e in electronegativity of the central atom.
HNOZ HNO3 •
H3As03 H3Sb03 H3AS04 H3Sb04 > Stability decreases H y d r i d e s : M H 3 type h y d r i d e s NH3
Ammonia
H3PO3 H3P04
PH3
ASH 3
SbH 3
BiH 3
Phosphine
Arsine
Stibine
Bismuthine
> Basic character decreases. > Thermal stability decreases. As the size increases, the electron density gets diffused over a larger region (due to bigger size) and hence the ability to d o n a t e the electron pair (basic nature) decreases. Nitrogen (N 2 ) • Preparation : O 2NH 3 + 3CuO — > N 2 + 3Cu + 3 H 2 0
O Purified air + Hot Cu Properties : o,
NO
3000°C Li, Mg, A1
N,
• Nitrogen
> U N, M g 3 N 2 , A l N
H,
: NH, CaC, (1000°C)
> C a C N2,
Uses : Because of unreactive nature, it is used to provide an inert atmosphere in certain metallurgical operations and during welding. Used as blanketing gas.
S o m e i m p o r t a n t c o m p o u n d s of g r o u p 15 e l e m e n t s
Group 16 : The Oxygen Family •
The first four elements of this g r o u p are non-metallic in character (O, S, Se, Te) a n d are collectively called as chalcogens, i.e. ore forming elements.
Element
Atomic number
Electronic configuration
Oxidation states
Oxygen (O)
8
[He] 2s2 2p i
-2
Sulphur (S)
16
[Ne] 3s2 3p*
-2, +2, +4, +6
Selenium (Se)
34
[Ar] 3d
4s 4p*
-2, +2, +4, +6
Tellurium (Te)
52
[Kr] 4 d w 5s2 5p4
-2, +3, +4, +6
Polonium (Po)
84
[Xe] 4 / 1 4 5d106s2 6p4
+2, +4
General characteristics of group 16 elements • O x y g e n h a s a g r e a t t e n d e n c y to f o r m pn-pn double bond but other elements do not, due to their larger size. However S and other elements have vacant d-orbitals and form pn-dn multiple bonds. But their tendency to form pn-dn multiple bonds decreases from S to Te. • Allotropy : All the e l e m e n t s of the g r o u p 16 exhibit allotropy. O Oxygen exists as 0 2 and 0 3 . O Sulphur exists in a number of allotropic forms, such as rhombic, monoclinic, plastic sulphur. O Selenium : Red and grey (both non-metallic). Element
Fluorine
10
•
•
2
O Tellurium : Crystalline and amorphous. O Polonium : a and P (both metals). Hydrides : Elements of group 16 form hydrides of general formula H 2 M such as H z O, H 2 S, H 2 Se, H 2 Te and H 2 Po. Water is colourless, odourless liquid while hydrides of the other elements of this group are colourless, bad smelling, poisonous gases. Acidic nature of hydrides increases while thermal stability decreases down the group. All hydrides except H 2 0 act as reducing agent and reducing character increases d o w n the group. Halides: The elements of group 16 form a variety of halides. Formulae of k n o w n halides of this g r o u p are given in the table. Bromine
Chlorine
Iodine
Oxygen
F2O
ci2o, cio2, ci2o7
Br 2 0
Sulphur
SF6, S2F10, SF4, S2F2
SC14, SCI* S2C12
S2Br2
Selenium
SeF6, SeF4, Se 2 Fe 2
SeCl4, SeCl2, Se 2 Cl 2
SeBr4, SeBr2*, Se2Br2
Tellurium
TeF 6 , Te2F10, TeF 4
TeCl2*, TeCl 4
TeBr 4 , TeBr 2
Tel 4
Polonium
POF6
PoCl 4 , PoCl,
PoBr 4 , PoBr 2
PoI 4
i2o5 -
* Unstable in o x y - h y d r o g e n a n d oxy-acetylene w e l d i n g torches.
Dioxygen/Oxygen (0 2 ) • Oxygen is the first element of group 16. Joseph Priestly p r e p a r e d it by heating mercury oxide w i t h the help of the sun rays and gave it the name oxygen. • Preparation : 725 K
O
2HgO
O
2KC10 3(5) - ^ 2 K C l
o
electrolysis n
2H z O v.
Na
> 2Hg + 0 2 ( s )
+
2 H•2(g) + ° 2
->H 2 O
P 4 (hot) CS2 (hot)
-*P2O5
-> co2 + so2
In oxides
-2
In peroxides
-1 -1/2
In dioxygen
0
In oxygen difluoride
+2
In oxygen monofluoride
+1
O z o n e (0 3 ) • Oxygen is never completely converted into ozone and we always obtain a mixture of oxygen and ozone. This mixture is called ozonised oxygen. Electric discharge
CH„
C 0 2 + H z O + 890 k j Uses : It is used in steel making, as oxidiser in rocket propellants and in fuel cells. It is used
Oxidation state of O
In superoxides
30 2 ( g )
strong heating^ H 2 , 1073 K
o,
Compounds
•
302 ^ 20 3 , AH = +284.5 k j Properties: Pure ozone is a blue gas, with a strong irritating smell. It is about 1.5 times heavier than air. It is more soluble than oxygen in water. 53 CHEMISTRY TODAY | MAY '13
NO
-> H 2 SO 4
PbS
> KO,
CH,=CH, H z O/Zn
-> H C H O
(HC1 + SnCl 2 )
•
Key Points Fluorides are the most stable, while iodides are the least stable among halides of group 16 elements. S u l p h u r exists as S 8 and has puckered ring structure. Volatility order is H z O < H 2 S > H 2 Se > H 2 Te. Se and Te are called photosensitive elements.
-> PbSO,
KOH
o,
at the crowded places, for preservation of meat in cold storages.
O 2 + NO 2
H,S
> SnCL
Uses : It is used for sterilizing drinking water, for bleaching delicate fabrics, for purifying air
Oxoacids of s u l p h u r Oxidation state, basicity and salt type
Oxoacid
Properties
Structure -
S = +4, d i b a s i c , a n d forms two series of salts, sulphites (S03 ) and bisulphites ( H S 0 3 )
Sulphurous acid H 2 SO 3
HO O
Sulphuric acid H 2 S 0 4 S = +6, d i b a s i c a n d forms two series of salts, (oil of vitriol) sulphates (S04 ) and bisulphates ( H S 0 4 )
-
stable, acts as oxidising agent and dehydrating agent
-
stable crystalline a n d hygroscopic solid powerful oxidizing agent
H e / N o OH
O
P e r o x o m o n o s u l p h u r i c S = +6, monobasic and acid H 2 S 0 5 or C a r o ' s forms single type of salt, acid peroxy disulphates
II
.s O
Peroxodisulphuric acid S = + 6 a n d f o r m s H 2 S 2 O S or M a r s h a l l ' s s i n g l e t y p e of s a l t , peroxydisulphates acid (SAb
acts as reducing agent as well as oxidising agent free acid does not exist
-
' OH
O-OH
ft 8 c / O/H\ > - o
-
° 8
-
stable crystalline a n d hygroscopic solid strong oxidizing agent
OH
Group 17 : The Halogens Family The general valence shell electronic configuration of group 17 is ns2 np5 (where n - 2 to 6). Element
Atomic number
Fluorine (F)
9
Chlorine (CI)
17
[He] 2s2 2p5 2
5
10
2
[Ne] 3s 3p
-1 - 1 , +1, +3, +5, +7
35
[Ar] 3d
Iodine (I)
53
[Kr] 4dw 5s2 5p5
85
14
[Xe] 4 /
4s 4p
5
Bromine (Br) Astatine (At)
Oxidation states
Electronic configuration
5d
w
-1, +1, +3, +5, +7 -1, +1, +3, +5, +7 2
6s 6p
5 -
•
Interhalogen compounds : The halogens, on account of difference in their electronegativities, combine with each other and form a binary covalent c o m p o u n d s of AB„ type which are called interhalogen compounds. A is always bigger atom and B is smaller atom and n may have values 1, 3, 5, 7 corresponding to oxidation states of halogens.
54
CHEMISTRY TODAY | M A Y '13
•
All the i n t e r h a l o g e n c o m p o u n d s except IF 7 can be p r e p a r e d b y direct c o m b i n a t i o n of h a l o g e n s u n d e r a p p r o p r i a t e conditions. Example : Cl 2 + F 2
> 2C1F
equal volume
• 250 C
Cl 2 + 3F 2
° > 2C1F3
excess
° > BrF 5
Group 18 : Inert Gases
Interhalogen compounds Total no. of electron pairs around X
Bond pairs
4
1
XY
XY3
5
XY5
6
XY7
7
Key Points Electron affinity : F < CI > Br > I O x i d i s i n g p o w e r : F 2 > Cl 2 > Br 2 > I 2 R e d u c i n g p o w e r : I" > Br - > CI" > F~ I 2 is soluble in KI solution d u e to f o r m a t i o n of KI 3 (or I3-). Interhalogen c o m p o u n d s are m o r e reactive than the constituent h a l o g e n s b e c a u s e X - Y b o n d is w e a k e r t h a n X - X or Y - Y.
200 C
BrF 3 + F 2
Molecule
• • • •
Lone pairs
Shape
3
Linear
CIF, BrF, BrCl, ICl, IBr
2
T-shaped
ClF3,BrF3, ICI3
5
1
Square pyramidal
BrF5, IF5
Pentagonal
IF7
0
Atomic number
H e l i u m (He)
2
Is 2
N e o n (Ne)
10
Is 2 2s 2 2p b
A r g o n (Ar)
18
Is 2 2s 2 2p 6 3s2 3p6
K r y p t o n (Kr)
36
Is 2 2s 2 2p 6 3s 2 3p 6 3dw 4s 2 4 pb
Xenon (Xe)
54
Is 2 2s2 2p6 3s2 3p6 3 d w 4s 2 4p 6 4 d w 5s2 5p 6
Radon(Rn)
86
Is 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p26 4d6w 4 f 4 5s 2 5p6 5dw 6s 6p
Example
3
7
Element
Electronic configuration
C o m p o u n d s of x e n o n Compound XeF 2 , Xenon difluoride
Preparation
Properties
Ni tube X e +
F
(2 : 1) Xe + F2 (2-1)
H2
2 - 1 5 5 ^
X e F
Electric > discharge
2
> Xe + 2HF
H,0
1
XeF 2 — - — > Xe + 2HF + - 0 2
X e F
^
2
— — > 2IF + Xe
hx> 2
Structure
(Hg vapour)
2
(+2), XeF2
2HC1
Linear, sp3d
> Xe + 2HF + Cl 2 XeF 4 , Xenon tetra fluoride
„ X E + P
(1:5) x
n -7) Y ' '
2
Nitube O 6ORI 600 R C
„ >
XEF
Electric^ discharge, -78°C
-
4
-
Xe03, Xenon trioxide
Acts as f l u o r i n a t i n g agent. C o l o u r l e s s , c r y s t a l l i n e solid w i t h m e l t i n g p o i n t = 117.1°C XeF 4 + 2H 2 Xe + 4 H F U n d e r g o e s d i s p r o p o r t i o n a t i o n in w a t e r 2XeF 4 + 3 H 2 0 — > Xe + X e 0 3 + 6HF + F 2 Gives a d d u c t XeF 4 + SbF 5 > [XeF 3 ] + [SbF 6 ] _ Acts as oxidising a n d f l u o r i n a t i n g a g e n t Dissolves in a n h y d r o u s H F
C o m p l e t e h y d r o l y s i s of - Colourless, h i g h l y explosive a n d p o w e r f u l XeF 4 a n d XeF 6 oxidising agent 2XeF 4 + 3 H 2 0 — > - W i t h w a t e r f o r m s xenic acid X e 0 3 + Xe + F 2 + 6HF Xe03 + HzO — > H2Xe04 XeF 6 + 3 H 2 0 - Undergoes disproportionation w h e n dissolved X e 0 3 + 6HF in alkali ulf' t ' r 2 X e 0 3 + 4 0 H " — > Xe + 0 2 + XeOt + 2 H 2 0
F 1 \ 1 \
1 1 / \ / /
Xe 1 / V\\ 1 / / \ \ 1 / | ] \ F
F (+4), XeF4
Square planar, sp3 fi
0 O
4
X
0 (+6),Xe0 3 Trigonal pyramidal, sp 3
55 CHEMISTRY TODAY | MAY '13
XeOF* xenon oxydifluoride
P a r t i a l h y d r o l y s i s of XeF 4 XeF 4 + H z O
-
Unstable
-80°C
Xe =
» XeOF, + 2 H F
0
F (+4), XeOF 2 T-shaped, sp3d
Xe0 2 F 2 x e n o n dioxydifluoride
Partial hydrolysis of XeOF 4 or XeF 6 XeOF 4 + H 2 0 — > Xe02F2 + 2HF XeF 6 + 2 H z O — » Xe02F2 + 4HF A c t i o n of S i 0 2 o n XeOF 4 2XeOF 4 + S i 0 2 — > 2 X e 0 2 F 2 + SiF 4
Colourless solid w i t h melting p o i n t 29.8°C U n d e r g o e s h y d r o l y s i s readily Xe02F2 + H 2 0 — > X e 0 3 + 2HF
d- and f- Block Elements •
The rf-block e l e m e n t s are also called as transition e l e m e n t s as t h e y s h o w t r a n s i t i o n a l p r o p e r t i e s b e t w e e n s- a n d p-block elements. T h e s e e l e m e n t s are called d-block e l e m e n t s since in t h e m 3d, 4 d , 5d a n d 6d subshells are i n c o m p l e t e a n d the last electron enters the (n -1 )d orbitals, i.e., penultimate shell. Their g e n e r a l electronic c o n f i g u r a t i o n is (h - 1 )dl ~ 10 ns° ' 2 w h e r e n is t h e o u t e r m o s t shell.
O
O X e
'N> F (+6), Xe0 2 F 2 Distorted trigonal bipyramidal, sp3d
Series
Elements
First transition series Sc (At. n o . 21) t o Z n ( A t . no. 30) or 3d series S e c o n d t r a n s i t i o n Y ( A t . n o . 39) t o C d ( A t . n o . 48) series or 4d series Third transition series or 5d series
La (At. n o . 57), Hf (At. n o . 72) to H g (At. no. 80)
Fourth transition series or 6d series
A c (At. n o . 89), Rf (At. n o . 104) to U u b (At. no. 112)
S o m e i m p o r t a n t c o m p o u ri d s of d - b l o c k e l e m e n t s Preparation
Compounds
Physical properties
Uses
K2Cr207
F r o m s o d i u m d i c h r o m a t e O r a n g e red, crystalline solid, i n d y e i n g , p h o t o g r a p h y , ( o b t a i n e d f r o m c h r o m i t e oxidising agent, m e l t i n g p o i n t leather i n d u s t r y =398°C ore) N a 2 C r 2 0 7 + 2KC1 — > K2Cr207 + 2NaCl
KMn04
F r o m p o t a s s i u m m a n g a n a t e D a r k violet, crystalline solid, oxidising agent, m e l t i n g p o i n t 2 K 2 M n 0 4 + Cl 2 — > 2 K M n 0 4 + 2KC1 = 523 K
as d i s i n f e c t a n t , g e r m i c i d e , Baeyer's reagent (alkaline KMn04)
CuS04-5H20 (blue vitriol)
Blue ( h y d r a t e d salt), at F r o m c u p r i c oxide 423 K b e c o m e s a n h y d r o u s , C u O + H 2 SO 4 — > C u S 0 4 + H 2 O readily dissolves in w a t e r
as electrolyte, fungicide
AgN03 (Lunar caustic)
White crystalline, d e c o m p o s e s for m a k i n g inks, hair dyes, F r o m silver m e t a l t o g i v e s i l v e r a t h i g h p h o t o g r a p h y , s i l v e r i n g of 3Ag + 4 H N 0 3 — > glass AgNOa + NOT + H 2 0 temperature
Hgl2
It exists in t w o f o r m s , i.e., red to p r e p a r e Nessler's reagent, F r o m m e r c u r i c chloride a n d yellow, dissolves in excess f o r m a k i n g o i n t m e n t s f o r HgCl 2 + 2KI — > H g l 2 + 2KC1 of KI to f o r m a complex, called t r e a t i n g skin infections as N e s s l e r ' s r e a g e n t
56
CHEMISTRY TODAY | MAY '13
mordant,
Some important reactions SO s
Oxidising nature
^
Fe3+<-
Fe
Change in oxidation state
K2Cr207 orange
2+
Cr 3 + (green) » K 2 C r 0 4 (yellow)
KOH
H+/H2O2
CrOs
passed over NaOH
CrO,C'l z z
NaCl (Chromyl
orange red va
chloride test)
4 * yellow solution
P°urs
(deep violet)
o2< "2°2
conc. H,SO, 2 4
Fe 2+
Fe3+<-
(acid in excess)
Oxidising nature
Change in oxidation state
KMnO,
i0ir
MnQ2<
green
- > M n 2, Q 7
if KMnO, IX.V] 1 iw^is1 in excess
H,0
Baeyer's reagent S <~
H,0 ^
M n
(Alkaline)
° 2
I Decolourise ( C = C ) compounds KCN
AgN03 - ^ » [ A j C N white ppt.
0 2 + Fe304
^ <—
A
-
>
K[Ag(CN) 2 ]
soluble (potassium argentocyanide)
Fe203
H
5 Pale violet
- Block Elements T h e general electronic c o n f i g u r a t i o n of t h e / - b l o c k e l e m e n t s is (n - 2 ) / 1 " 1 4 ( n - 1 )rf 0 _ 1 ns For l a n t h a n o i d , n is 6 w h i l e its v a l u e is 7 for actinoid series. Lanthanoids
Ce
Pr
Nd
Actinoids
Th
Pa
U
Pm L
N
P
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
(Lanthanoid contraction): In l a n t h a n o i d s t h e r e is r e g u l a r d e c r e a s e in a t o m i c a n d i o n i c r a d i i w i t h increase in a t o m i c n u m b e r f r o m La to Lu. T h i s r e g u l a r d e c r e a s e is k n o w n as l a n t h a n o i d contraction. O Cause of lanthanoid contraction : A s w e m o v e along the l a n t h a n o i d series, the nuclear charge increases b y o n e unit at each successive element a n d a n e w electron enters the 4/-subshell. D u e to the peculiar s h a p e of /-orbitals, there is imperfection shielding of 4/-electrons b y other electrons f r o m the nuclear attraction. A s a result of this, the size of l a n t h a n o i d a t o m s k e e p o n decreasing f r o m La to Lu. LnCi A Ln2(C03); LnN LnX 3 <-
> LnH2 + LnH,
vvj, Ln(OH)3 + H 2
Ln 2 S 3 Ln 3 + + H ,
UBr,
U 0 2 + H2<—52-°
U(OH)4
+
H.
Passive d u e to oxide layer UC1 4 + UC1 6
Key Points A c t i n o i d s f o r m oxo-ions such as tUt °O2f ,, ruv^ PuO 2: , UO + , U 0 2 , etc. T h e s e i o n s are stable in acid and aqueous solutions while lanthanoids do n o t f o r m oxo-ions. Elements with atomic n u m b e r greater than 92 ( a t o m i c n u m b e r of u r a n i u m ) a r e c a l l e d t r a n s u r a n i u m e l e m e n t s . All these e l e m e n t s are man-made.
57 CHEMISTRY TODAY | M AY '13
COORDINATION COMPOUNDS C o o r d i n a t i o n c o m p o u n d s are t h o s e c o m p o u n d s w h i c h retain their identities even w h e n dissolved in w a t e r or a n y other solvent a n d their p r o p e r t i e s a r e c o m p l e t e l y d i f f e r e n t f r o m t h o s e of t h e constituents. __ coordination sphere
I
~ ' _ [•Cp.;(NH^i:Clj i
O
> counter ions
" Vligands ^coordination number
central metal ion
N o m e n c l a t u r e : The m a i n rules for n a m i n g the complexes are : O If the coordination c o m p o u n d is ionic, t h e n the positive part of coordination c o m p o u n d is n a m e d first just like simple salts. O The ligands of the coordination s p h e r e are to be n a m e d first followed b y the metal ion. The ligands can be neutral, anionic or cationic. O If any ligand is p r e s e n t m o r e t h a n once, t h e n their repetition is indicated b y prefixes like di,
O O
tri, tetra, etc. W h e n b i d e n t a t e or p o l y d e n t a t e ligands are coordinated to the metal ion, their repetition is indicated b y prefixes like bis, tris, tetrakis, etc. W h e n m o r e t h a n one ligand is p r e s e n t in the complex, t h e n the ligands are n a m e d in the alphabetical order. After n a m i n g the ligands, the central metal ion is to be n a m e d followed b y its oxidation state in R o m a n n u m b e r s in brackets. If the complex is n e u t r a l or p r o v i d e s a cationic complex ion, t h e n the central metal ion is to be n a m e d as it is. If the complex p r o v i d e s anionic complex ion then the n a m e of central metal ion e n d s as 'ate'. After the n a m i n g of central m e t a l ion, anion w h i c h is in the outer s p h e r e is to be n a m e d . If the complex contains t w o or m o r e metal atoms, it is t e r m e d polynuclear. The b r i d g i n g ligands w h i c h link the t w o metal a t o m s together are indicated b y the prefix
Type of I s o m e r i s m
STRUCTURAL ISOMERISM W h e n c o m p o u n d s h a v e different ligands
STEREOISOMERISM W h e n different positions a n d a r r a n g e m e n t s of
w i t h i n their coordination spheres.
ligands in space.
Ionisation Isomerism
Geometrical Isomerism
C o m p o u n d s that give different ions in
Arises in heteroleptic complexes d u e to different
solution are ionisation isomers.
possible geometric a r r a n g e m e n t s of the ligands.
e.g. [CoBr(NH 3 ) 5 ]S0 4 ,
In square planar complex
[Co(S0 4 )(NH 3 ) 5 ]Br Hydrate Isomerism C o m p o u n d s that differ in n u m b e r of water molecules attached to metal a t o m are h y d r a t e isomers. e.g. [Cr(H 2 0) 6 ]Cl 3 , [CrCl(H20)5Cl]Cl2.H20 Linkage Isomerism S h o w n b y c o m p o u n d s containing a m b i d e n t a t e ligands. [CO(N0 2 )(NH 3 ) 5 ]C1 2 [Co(ONO)(NH 3 ) 5 ]C1 2 Coordination Isomerism Occurs w h e n b o t h positive a n d negative ions are complex ions a n d t w o isomers differ in ligand's distribution in t w o complex ions, e.g. [Co(NH 3 ) 6 ][Cr(CN) 6 ] [Cr(NH 3 ) 6 ][Co(CN) 6 ] 52 CHEMISTRY TODAY | M A Y '13
[Ma2b2\
cis : 2 ligands a a r r a n g e d adjacent to each other trans : 2 ligands a a r r a n g e d opposite to each other.
CI \
Pt
C1X
/NH3 X
NH,
Cl\
/NH3 Pt
H3N^
X
C1
In o c t a h e d r a l c o m p l e x e s [Ma3b3] • Facial ( f a c ) : 3 d o n o r a t o m s of the same ligands occupy adjacent positions at t h e corners of a n octahedral face. • M e r i d i o n a l ( m e r ) : W h e n the positions are a r o u n d the m e r i d i a n of the o c t a h e d r o n . Optical Isomerism Arises d u e to lack of element of symmetry. ^ E n a n t i o m e r s are n o n - s u p e r i m p o s a b l e mirror imagt of each other. The t w o f o r m s are chiral. • dextro (d): rotate p l a n e of polarised light to right. * laevo (/): rotate p l a n e of polarised light to left.
d
*-y
d.xy
dz2/ dx2-y2/ dXy, dxz, dyZ\
d, "" ,
d-xy d x z dyZ Free metal
Complex
[Cr(NH3)6]
3+
Octahedral field
No. of 3d Type electof rons in hybrid metal orbital ion 2
3
d sp
6
[Fe(CN)6] ~ [FeF6]3~
[Fe(CN)6]*" 3
3+
[CO(NH3)6]
3
dXz Square planar field
Geometry
No. of unpaired electrons in the complex ion
Octahedral
3
d2sp3
Octahedral
0
5
d2sp3
Octahedral
1
5
sp3d2
Octahedral
5
Octahedral
0
6
2
d sp
3
[CoF,]3"
6
Octahedral
4
[Ni(CN)4]2~
8
dsp
2
Square planar
0
[CU(NH3)4]2+
9
dsp2
Square planar
1
[CU(CN)4]3_
10
sp3
Tetrahedral
0
[MnfCN^] 3 "
4
d2sp3
Octahedral
2
3+
4
sp3d2
Octahedral
4
10
sp3
Tetrahedral
0
3
Tetrahedral
2
[Mn(H20)3] Ni(CO) 4 2
[Ni2Cl4] ~
8
sp3d2
sp
Spectrochemical series : I" < Br" < SCN~ < CI" < F" < OH" < C 2 0 4 ~ < O 2 - < H 2 0 < NCS- < N H 3 < en < N 0 2 < CN" < CO Weak field ligand
Increasing order of A0
>
Strong field ligand
o o
CHEMISTRY TODAY | MAY 13
59
1.
2.
3.
Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar? (a) Ca < Ba < S < Se < Ar (b) Ca < S < Ba < Se < Ar (c) S < Se < Ca < Ba < Ar (d) Ba < Ca < Se < S < Ar A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is (a) 6 (b) 2 (c) 5 (d) 4 Energy of an electron is given by E = -2.178xlO""
18
J
Wavelength of light
required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be {h = 6.62 x 10"M J s and c = 3.0 x 108 m s"1) 7 (a) 8.500 x H r m (b) 1.214 x l 0 " m 7 7 (c) 2.816 x 10" m (c) 6.500 x 10" m 4.
An unknown alcohol is treated with the "Lucas reagent" to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism? (a) tertiary alcohol by S N 2 (b) secondary alcohol by S N 1 (c) tertiary alcohol by S N 1 (d) secondary alcohol by S N 2
5.
A gaseous hydrocarbon gives u p o n combustion 0.72 g of water and 3.08 g of C 0 2 . The empirical formula of the hydrocarbon is (a) C 7 H 8 (b) C 2 H 4 (c) C 3 H 4 (d) C 6 H 5
6.
A solution of (-) - 1 - chloro - 1 - phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of
60
CHEMISTRY TODAY | MA Y ' 1 3
(a) free radical (c) carbene
(b) carbanion (d) carbocation.
7.
Which one of the following molecules is expected to exhibit diamagnetic behaviour? (a) S 2 (b) C 2 (c) N 2 (d) 0 2
8.
Which of the following exists as covalent crystals in the solid state? (a) Phosphorus (b) Iodine (c) Silicon (d) Sulphur
9.
Given E°Cr3-VCrE°cr2opc^
- 0.74 V; EMnOj/Mn-,2+ = 1.51 V = 1.33 V; E°a/C1- = 1-36 V
Based on the data given above, strongest oxidising agent will be (a) M n 0 4 (b) CI" (c) Cr 3+ (d) Mn 2+ 10. Arrange the following compounds in order of decreasing acidity. OH OH OH OH
CI
CH3
NO 2
(I)
(II)
(III)
(a) IV > III > I > II (c) I > II > III > IV
OCH3 (IV)
(b) II > IV > I > III (d) III > I > II > IV
11. The molarity of a solution obtained by mixing 750 mL of 0.5 M HC1 with 250 mL of 2 M HC1 will be (a) 0.975 M (b) 0.875 M (c) l.OOM (d) 1.75 M 12. For gaseous state, if most probable speed is denoted by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speed are
(a) C*
c : C = 1 :1.225 :1.128
c* c* c*
C : C = 1.225:1.128:1
(b) (c) (d)
C : C = 1.128:1:1.225 C : C = 1:1.128:1.225
13. Which of the following is the wrong statement? (a) Ozone is diamagnetic gas. (b) ONC1 and O N O are not isoelectronic. (c) 0 3 molecule is bent. (d) Ozone is violet-black in solid state. 14. An organic compound A u p o n reacting with N H 3 gives B. O n heating, B gives C. C in presence of KOH reacts with Br 2 to give CH 3 CH 2 NH 2 . A is (a) CH 3 CH 2 COOH (b) CH 3 COOH (c) CH 3 CH 2 CH 2 COOH (d) CH 3 —CH—COOH CH 3 15. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest EM3+/M2+ value? (a) Co (Z = 27) (b) C r ( Z = 24) (c) Mn (Z = 25) (d) F e ( Z = 26) 16. In which of the following pairs of molecules/ions, both the species are not likely to exist? (a) H 2 ,He 2 2 +
(b) H 2 , He22~
(c) H 2 , H e f -
(d) H 2 2+ ,He 2
17. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was (a) Phosgene (b) Methylisocyanate (c) Methylamine (d) Ammonia 18. Consider the following reaction. x M n 0 4 + yC 2 0 4 ~ + zH + -> xMn2++2yC02 + | H 2 0
21. The rate of a reaction doubles w h e n its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R = 8.314 J KT1 mol" 1 and log 2 = 0.301) (a) 60.5 k j mol" 1 (b) 53.6 k j mol" 1 1 (c) 48.6 k j moF (d) 58.5 k j mol - 1 22. The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na + will be (a) + 2.55 eV (b) - 2 . 5 5 e V (c) - 5.1 eV (d) - 1 0 . 2 eV 23. The coagulating power of electrolytes having ions Na + , Al 3+ and Ba2+ for arsenic sulphide sol increases in the order : (a) Al 3+ < Na + < Ba2+ (b) Al 3 + < Ba 2+ < Na + + 2+ 3+ (c) N a < B a < A l (d) Ba 2+ < N a + < Al3+ 24. The order of stability of the following carbocations is
CH2=CH-CH2; CH3-CH2-CH2; II n (a) III > I > II (c) II > III > I
25. Stability of the species Li* Li2 and Li2+ increases in the order of (a) L i 2 < L i 2 < L i 2 (b) Li2 < Li2 < Li2 (c) LiJ < Li 2 < Li 2 (d) L i 2 < L i 2 < L i 2 26. C o m p o u n d (A), C 8 H 9 Br, gives a white precipitate w h e n warmed with alcoholic A g N 0 3 . Oxidation of (A) gives an acid (B), C 8 H 6 0 4 . (B) easily forms anhydride on heating. Identify the compound (,4). ,CH 2 Br ,CH2Br (a) CH,
The values of x, y and z in the reaction are, respectively (a) 5, 2 and 8 (b) 5, 2 and 16 (c) 2, 5 and 8 (d) 2, 5 and 16 19. Which of the following complex species is not expected to exhibit optical isomerism? (a) [Co(en)(NH 3 ) 2 Cl 2 ] + (b) [Co(en) 3 ] 3+ (c) [Co(en) 2 Cl 2 ] + (d) [Co(NH 3 ) 3 Cl 3 ] 20. H o w many litres of water must be added to 1 litre of an aqueous solution of HC1 with a p H of 1 to create an aqueous solution with p H of 2? (a) 9.0 L (b) 0.1 L (c) 0.9 L (d) 2.0 L
(b) III > II > I (d) I >11 >111
.C 2 H 5 (c) Br CH, 27. Which of the following arrangements does not represent the correct order of the property stated against it? (a) Sc < Ti < Cr < M n : number of oxidation states (b) V2+ < Cr 2+ < Mn 2+ < Fe2+ : paramagnetic behaviour 61 CHEMISTRY TODAY | M AY '13
(c) Ni 2+ < Co 2+ < Fe 2+ < Mn 2 + : ionic size (d) Co 3+ < Fe 3+ < Cr 3+ < Sc3+ : stability in aqueous solution.
(a): Moles of water produced = Moles of C 0 2 produced =
28. Experimentally it was f o u n d that a metal oxide has formula M0.9sO. Metal M, is present as M2+ and M 3+ in its oxide. Fraction of the metal which exists as M 3 + would be (a) 5.08% (b) 7.01% (c) 4.08% (d) 6.05% 29. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be (R = 8.314 J/mol K) (In 7.5 = 2.01) (a) q = + 208 J, w = + 208 J (b) q = + 208 J, zv = - 208 J (c) q = - 208 J, w = - 208 J (d) q = - 208 J, w = + 208 J 30. Synthesis of each molecule photosynthesis involves (a) 6 molecules of ATP (b) 18 molecules of ATP (c) 10 molecules of ATP (d) 8 molecules of ATP
of
glucose
in
SOLUTIONS 1.
(d): Ionization enthalpy decreases from top to bottom in a group while it increases from left to right in a period. 390-180 42 /
3.
(b): E = - 2 - 1 7 8 x l O "
18
Z
E = - 2 . 1 7 8 xlO" 1 8
2
1
=5
2- J - l
U 1
nlj 1
x = 0.07
and
— = 0.04 2 y = 0.08
18
= 0.04
3.08 = 0.07 44
Equation for combustion hydrocarbon, Q H y is
=>
0.72
of
an
x _ 0.07 _ 7 y ~ 0.08 " 8 .•. The empirical formula of the hydrocarbon is C7H8. 6.
(d): A carbocation intermediate is formed during racemisation.
7.
(d): 0 2 is expected to be diamagnetic in nature but actually it is paramagnetic.
8.
(c)
9.
(a): Greater the reduction potential of a substance, stronger is the oxidising agent. MnC>4 is the strongest oxidising agent.
10. (d): Electron donating groups (—CH 3 and — OCH 3 ) decrease while electron w i t h d r a w i n g groups ( — N 0 2 and —CI) increase the acidity. Since — O C H 3 is a stronger electron donating group than — CH 3 and — N 0 2 is stronger electron w i t h d r a w i n g g r o u p than —CI, therefore order of decreasing acidity is III > I > II > IV. 11. (b): M m i x V m i x = MlVl M
M„
i
y
i
+ M
2
V
+ M2V2 2
V„
2
(2 ) " ( I ) 2
0.5 x 750 + 2 x 250 E = +2.178 x 10" 18 x - = 1.6335 x 10" 18 J
__
_ ~ E ~
or 4.
62
A*mix =
1000
M m i x = 0.875 M
E= — X 6 62 x 10
J s x 3 x 10 8 m
12. (d): C* :C :C =
2 RT
8 RT
3RT
M
nM
M
1.6335xlO" 1 8 J
X = 12.14 x 10"8 m X = 1.214 x 10" 7 m
( c ) ; In Lucas test, turbidity appears immediately with tertiary alcohol by S N 1 mechanism.
CHEMISTRY TODAY | MAY ' 1 3
unknown
= V2
'3.14
C* :C :C = 1 :1.128 :1.225 13. (None): All the statements are correct.
14. (a): O
22. (c): Electron gain enthalpy = - Ionisation potential = - 5 . 1 eV
O NH,
CH3-CH2-C-OH —-> CH3-CH2-C-ONH4 (A)
(B)
i
o
Br2/KOH
CH3-CH2-NH2<
CH3-CH2-C-NH2 (Q
15. (a) 16. (d): Species with zero bond order does not exist. H 2 2+ : o(ls)° Bond order = 0 H e 2 : a(ls) 2 cr*(ls)2 2 - 2
Bond order =
=0
23. (c): For a negatively charged sol, like As2S3, greater the positive charge on cations, greater is the coagulating power. 24. (a): Greater the number of resonating structures a carbocation possess, greater is its stability. 25. (c): Species Li2 Li2
Li2 0.5 Higher the bond order, greater is the stability. CH 2 —Br
,COOH [O],
26. (a):
17. (b)
Bond order 1 0.5
CH 3 +
18. (d):2Mn0 4 "+ 5 C 2 O f + 16H -> 2Mn 2+ + 10C0 2 + 8 H 2 0 x = 2, y = 5, z = 16
COOH
(B)
(A)
\o
19. (d): [Co(NH 3 ) 3 Cl 3 ] will not exhibit optical isomerism due to presence of plane of symmetry. 20. (a): Initial concentration of aq. HC1 solution with p H 1 = 10"1 M Final concentration of this solution after dilution = 10~2 M r MV = M1(V 1 +V 2 ) 1 0 "
1
0.1
x
1
=
1 0 "
2
( 1
+
V )
II
o (C)
27. (b): Number of unpaired electrons in Fe2+ is less than Mn 2+ , so Fe2+ is less paramagnetic than Mn 2+ . 28. (c): Let the fraction of metal which exists as M 3+ be x. Then the fraction of metal as M2+ = (0.98 - x) 3x + 2(0.98 - x) = 2 x + 1.96 = 2 x = 0.04
-=1+V
0.01
10 = 1 + V V=9L 21. (b): As r = k[A]n
% of M 3+ = Since
W rev = -2.303 wRTlog 10 t
2.303 R log2 =
ioo = 4.08%
29. (b): As it absorbs heat, q = + 208 J
(Given) _2 = _2 K , k2
0.98
2
v^iy
- t ,
W rev = -2.303 x (0.04) x 8.314x310 log 10
T{T2
E„ 310-300 2.303x8.314x10 - 3 310 x 300
J
0.3010 x 2.303 x 8.314 x 10~ x 93 x 10 E„ = 10 Ea = 53.6 k j mol" 1
( 375 —
Wrev = - 207.76 = - 208 J 3
30. fb): 6 C 0 2 + 18ATP + 12NADPH + 6RuBP 6RuBP + Glucose + 18ADP + 18P + 12NADP + One molecule of glucose is formed from 6 C 0 2 by utilising 18ATP and 12NADPH.
63 CHEMISTRY TODAY | MAY '13
Useful for
TARGET
UP-CPMT, J & K CET,
PMTs
Karnataka CET, CMC Vellore, AMU, AIIMS, WB JEE,
PRACTICE QUESTIONS 1.
2.
C o n s i d e r t h e m o d e s of t r a n s f o r m a t i o n s of a g a s from state 'A to state 'B' as -B shown in the following vP-V diagram. Which one of the following is true ? 6. (a) AH = q along A C (b) AS is same along both A —> B and A —> C —> B (c) w is same along both A —> B and A —> C —> B (d) w > 0 along both A B and A C. Consider the following half cell reactions. Cu 2+ + 2e~ ^ Cu; E° = 0.34 V Ag + + e- ^ Ag; E° = 0.80 V Which one of the following is true? (a) C o p p e r can displace silver f r o m A g N 0 3 solution. (b) Silver can displace copper from C u ( N 0 3 ) 2 solution. (c) Silver is more reactive than copper. 2+ + (d) Cu /Cu and Ag /Ag with positive E° values cannot be coupled to form a Galvanic cell.
3.
4.
5.
64
If a 0.00001 molar solution of HC1 is diluted thousand folds the p H of the resulting solution will be (a) 5 (b) 7 (c) 8 (d) 6.98
CHEMISTRY TODAY | M AY ' 1 3
alkene, Q, which reacts with HC1 to give R. R o n treatment with aqueous KOH gives compound S, C 3 H 8 0 . P and S are (a) functional isomers (b) position isomers (c) chain isomers (d) stereoisomers. The salt (a) (c)
aqueous solution of which of the following will have the lowest pH? NaC10 3 (b) NaCIO NaC10 4 (d) NaC10 2
7.
In the reaction: P 2 O s + 2 H N 0 3 -> 2 H P 0 3 + x, the term x is (a) N 2 0 4 (b) H 2 0 (d) P N 0 3 (c) N 2 O S
8.
In the reaction : I 2 + I~ -> I f , the Lewis base is (a) (b) I . (d) none of these. (c) I3"
r
The correct order of the increasing ionic character is (a) BeCl 2 < MgCl 2 < BaCl 2 < CaCl 2 (b) BeCl 2 < MgCl 2 < CaCl 2 < BaCl 2 (c) BeCl 2 < BaCl 2 < MgCl 2 < CaCl 2 (d) BaCl 2 < CaCl 2 < MgCl 2 < BeCl 2 10. Identify the product in the reaction. C6H6
HN0
3/H2S0 4
i n t e r i r i e ( j j a t e compound Sn/HC1
NO 2 (c)
> Product.
(b) C 6 H 4 (NO 2 ) 2
(a) C 6 H 5 N 0 2
If 100 g iron powder (at. wt. of Fe = 56) is heated with 32 g of sulphur (at. wt. of S = 32). To prepare one mole of iron sulphide how much iron powder will remain unreacted? (a) 4 g (b) 32 g (c) 44 g (d) Nil A primary alcohol, C 3 H 8 0 (P) on heating with sulphuric acid undergoes dehydration to give an
MGMCET
NH 2
Q H ^
(d) C 6 H 4 C"
^NH2 CH2-CH2COOH
^ 11. C o m p o u n d Y is
—
-> X
ale. KOH
*Y
(a)
0 2 < 0 2 < Of
(b) Of (c)
( O )
(a) KL3 (c) K2L2
CH2-CH2-COCl ( O )
12. W h e n c a l c i u m a c e t a t e a n d c a l c i u m f o r m a t e t o g e t h e r a r e s u b j e c t e d to d r y d i s t i l l a t i o n , t h e p r o d u c t is (a) a c e t a l d e h y d e (c) f o r m a l d e h y d e
(b) acetone (d) n o n e of these.
13. A t 27°C a g a s w a s c o m p r e s s e d to half of its v o l u m e . To w h a t t e m p e r a t u r e m u s t it n o w b e h e a t e d so t h a t it occupies its original v o l u m e ? (a) 54°C (b) 327°C (c) 600°C
(d) 327 K
14. T h e o x i d e t h a t g i v e s h y d r o g e n p e r o x i d e o n t r e a t m e n t w i t h a dilute acid is (a) P b 0 2 (c) M n 0 2
<02<02<
(b) N a 2 0 2 (d) TiO z
15. Relative acidity of the f o l l o w i n g is in the o r d e r (a) R C O O H > H 2 C 0 3 > C 6 H 5 O H > H 2 0 > R O H (b) R C O O H > R O H > H 2 C 0 3 > C 6 H 5 O H > H z O (c) R O H > R C O O H > H 2 C 0 3 > C 6 H 5 O H > H 2 0 (d) R C O O H > C 6 H 5 O H > R O H > H 2 C 0 3 > H z O 16. W h i c h o n e of the f o l l o w i n g is n o t coloured? (a) Cu 2 + (b) Cu + 2+ (c) Ni (d) Fe 3+ 17. W h i c h one of the f o l l o w i n g is incorrect? (a) Boron halides are all m o n o m e r i c while those of Al are dimeric. (b) Boron halides a n d a l u m i n i u m halides exist as m o n o m e r i c halides. (c) B o r o n h a l i d e s a n d a l u m i n i u m h a l i d e s are Lewis acids. \ (d) B 2 0 3 alone is acidic while A1 2 0 3 is amphoteric. 18. Bond o r d e r of t h e species 0 2 , 0 2 , 0 | + a n d increases in the o r d e r
Of
19. A c o m p o u n d f o r m e d b y e l e m e n t s K a n d L crystallises in the cubic a r r a n g e m e n t in which K a t o m s are at the corners of a cube a n d L atoms are at t h e face centres. W h a t is the f o r m u l a of compound?
CH(OH)-CH2 - C O O H
(d)
Of
0 2 < O f < 0 2 < Of
(d) Of
(c)
<02<02<
< Of
02~
(b) L3K (d) KL2
20. Which of the following will exhibit highest boiling point? (a) C H 3 C H 2 O C H 2 C H 2 C H 3 (b) C H 3 C H 2 C H 2 C H 2 C H 2 O H (c) C H 3 C H 2 C H 2 C H ( C H 3 ) O H (d) C H 3 C H 2 C ( C H 3 ) 2 O H 21. The a m i n e w h i c h can react w i t h C 6 H 5 S0 2 C1 to f o r m a p r o d u c t insoluble in alkali shall be (a) a p r i m a r y a m i n e (b) a s e c o n d a r y a m i n e (c) a tertiary a m i n e (d) b o t h p r i m a r y a n d secondary amines. 22. A d d i t i o n of conc. HC1 to saturated BaCl 2 solution precipitates BaCl 2 because (a) it follows Le Chatelier's principle (b) of c o m m o n ion effect (c) ionic p r o d u c t [Ba 2+ ][C1 - ] remains constant in a s a t u r a t e d solution (d) a t c o n s t a n t t e m p e r a t u r e , t h e p r o d u c t [Ba2+][C1~]2 r e m a i n s constant in a saturated solution 23. K M n 0 4 acts as an oxidising agent in the neutral m e d i u m and gets reduced to MnO z . The equivalent w e i g h t of K M n 0 4 in n e u t r a l m e d i u m will be (a) mol. wt./2 (b) mol. wt./3 (c) mol. wt./4 (d) mol. wt./7 24. W h i c h one h a s the highest p a r a m a g n e t i s m ? (a) Ni(CO) 4 (b) [Co(NH 3 ) 6 ]Cl 3 (c) [Ni(NH 3 ) 6 ]Cl 2 (d) [CU(NH 3 ) 4 ]C1 2 25. T h e w a v e l e n g t h of r a d i a t i o n emitted w h e n a n electron in a h y d r o g e n a t o m m a k e s a transition f r o m an e n e r g y level w i t h n = 3 to a level with n = 2 is -1312 [Given that E„ = — 2 — M mol - 1 ] (a) 6.56 x 10"7 m (c) 65.6 x 10- 7 m
(b) 65.6 n m (d) N o n e of these. 65 CHEMISTRY TODAY | MAY '13
26. Which highest (a) 0.1 (c) 0.1
of the following solutions boils at the temperature? M Glucose (b) 0.1 M NaCl M BaCl 2 (d) 0.1 M Urea
27. For the preparation of f-butyl ethyl ether, which method/reaction is to be preferred? CH 3 I (a) CH 3 CH 2 —Br + Na O—C—CH 3 CH 3
CH 3 I (b) CH 3 CH 2 0~ Na + + CH 3 —C—Br CH 3 CH 3 H2SO4, A
(c) CH 3 —C—OH + H O — C 2 H 5 — - — - — > CH 3
CH 3 28. Compound P(C 5 H 10 O) forms phenyl hydrazone and gives a negative Tollen's and iodoform test. Compound P on reduction gives n-pentane. The compound P will be
(
O II H-C-CH2-CH2-CH2-CH3
33. H o w m a n y e l e c t r o n s a r e d e l i v e r e d at t h e cathode during electrolysis by a current of 1 A in 60 seconds? (a) 3.74 x 1020 (b) 6.0 * 1023 (c) 7.48 x 1021 ( d ) 6 o x 10 20
35. Which of the following has the highest dipole moment?
CH 3 H H^
c=o
(b)
I
I
I
I
c= c CH 3 H
CH 3
CI
30. Among the following acids, the one which can form an anhydride most easily is ^
-
C O O H
(b)
I
c
I (d) C = C I I
CH, H
CH 3 CI
(a)
COOH
CH2-CH2COOH Cd)
CHEMISTRY TODAY | MAY '13
•
I CH 3 —(CH 2 ) 1 5 —N + —CH 3 Br~
CH 3 (b) C H 3 - ( C H 2 ) 1 4 - C H 2 - N H 2 (c) CH 3 —(CH 2 ) 1 6 —CH 2 0S0 2 Na + (d) OHC — (CH 2 ) 1 4 —CH 2 —COO" Na + 37. The a b s o l u t e c o n f i g u r a t i o n of the f o l l o w i n g compound is
CH 3 COOH
COOH
CH 3
I
CH 3
29. Which of the following is an amphoteric oxide? (a) C r 0 3 (b) C r 2 0 3 (c) V 2 0 3 (d) TiO
66
i r j
32. Vapour pressure of benzene at 30°C is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene, its vapour pressure is decreased to 120.2 mm. The molecular weight of the solute is (a) 35.67 g (b) 357.8 g (c) 432.8 g (d) 502.7 g
(d) None of the above.
a
(d)
36. Which one of the following is not a surfactant?
o
«
0 » O^H
+
^
H I (c) C :
H3C-CH2-C-CH2-CH3
COOH | COOH
0
H
o
(a)
0
(a)
(b) H 3 C—C—CH 2 —CH 2 —CH 3
(c)
(a)
34. The heat of neutralization of a strong base and a strong acid is 57 kj. The heat released w h e n 0.5 mole of H N 0 3 solution is a d d e d to 0.20 mole of N a O H solution, is (a) 11.4 kj (b) 34.7 kj (c) 23.5 kj (d) 58.8 kj
CH 3 I (d) CH 3 — C—Br + HO—CH 2 CH 3
(a)
31. Which of the following is aromatic?
CH2-CH2COOH
H-
2
C1
H
ClN C 2, H 5
(a) 2S, 3R (c) 2R, 3S
46. If the concentration of CrOf" ions in a saturated
(b) 2S, 3S (d) 2R, 3R
solution of silver chromate is 2 x 10"4 M, what is the solubility p r o d u c t of silver chromate?
38. The treatment of benzene with zso-butene in the presence of sulphuric acid gives (a) iso-butylbenzene (b) fert-butylbenzene (c) n-butylbenzene (d) no reaction. 39. Which of the following can exhibit geometrical isomersim? (a) [MnBr 4 ] 2 ~ (b) [Pt(NH 3 ) 3 Cir (c) [PtCl 2 .P(C 2 H 5 ) 3 ] 2 (d) N o n e of these
(b) 8 x 10-12
(a) 4 x 10-8 (c) 16 x 10"
12
(d) 32 x lO-i 2
47. Which is correct statement among the following? (a) Starch is a polymer of a-glucose. (b) Amylose is a component of cellulose. (c) Proteins are c o m p o u n d s of only one type of amino acid. (d) In cyclic structure of fructose, there are four
40. The O N O angle is m a x i m u m in (a) N 0 3 (b) NO z " (c) N O z (d) N0 2 +
carbon a n d one oxygen atom.
41. The rate c o n s t a n t k, f o r t h e g i v e n r e a c t i o n is 2.3 x 10~2 s"1. N 2 O s ( g ) -> 2 N 0 2 f e ) + 1/2 0 2 t e )
48. For the square planar complex, [M(A)(B)(C)(D)] (where, M = central metal; A, B, C a n d D are m o n o d e n t a t e ligands) the n u m b e r of possible geometrical isomers are (a) 1 (c) 3
(b) 2 (d) 4
Which equation given below describes the c h a n g e of [ N 2 0 5 ] w i t h t i m e ? [ N 2 O 5 ] 0 a n d [ N 2 0 5 ] f c o r r e s p o n d to c o n c e n t r a t i o n of N 2 0 5 initially a n d at time f. (a) [N 2 0 5 ] f = [N 2 O 5 ] 0 + kt (b) [N 2 O 5 ] 0 = [N 2 0 5 ] ( ekt (c) log[N 2 0 5 ] f = log[N 2 O 5 ] 0 + kt
49. D u r i n g electrolysis of water the v o l u m e of 0 2 liberated is 2.24 dm 3 . The volume of hydrogen liberated, u n d e r same conditions will be
(d)
50. In an isothermal process,
i n ^ s k ^ k t . [N2o5]t
(a) 2.24 d m 3 (c) 4.48 d m 3 (a) q = 0 a n d AE = 0 (c) <7 = 0 a n d AE * 0
42. The two-third life (f 2 / 3 ) of a first order reaction in which k = 5.48 x 10"14 s"1 is / , (a)
2.303 5.48 xlO" 2.303
, „ t J log 3 14
x
(b)
2.303
, „ — log 2
1
2.303
(b) q * 0 and AE = 0 (d) <7 * 0 and AE * 0
SOLUTIONS 1.
5.48 x K T 1 4 ,
(b) 1.12 d m 3 (d) 0.56 d m 3
( b ) : E n t r o p y is a state f u n c t i o n i.e. the change in e n t r o p y d e p e n d s u p o n the initial a n d final
,
2
5 ^ 0 0 ^ 3 (d) 3 43. At w h a t temperature the r.m.s velocity of chlorine gas will be equal to that of S 0 2 at STP? (a) 57.59°C (b) 0°C (c) 273°C (d) 302.9°C
states of the system a n d not on h o w that change
(C)
44. W h i c h of t h e f o l l o w i n g ions h a s t h e g r e a t e s t coagulating power w h e n a d d e d to negative AS2S3 sol? (a) Na + (b) Mg 2 + (c) Al 3+ (d) P O f " 45. Insulin contains about 0.08% by weight of sulphur. W h a t w o u l d be the approximate molecular m a s s of insulin if one s u l p h u r a t o m is p r e s e n t in a molecule of insulin? (a) 40,000 (b) 2560 (c) 20,000 (d) 25,600
is b r o u g h t about. 2-
(a) : The silver electrode h a s h i g h e r reduction potential. The resulting galvanic cell is Cu(S) I Cu2+(aq) 11 Ag+(„9) I Ag
3.
( d ) : For 0.00001 M HC1, [H + ] = 10"5 M. When HC1 is diluted thousand fold, [H+] = 10"8 M Dissociation of water, H 2 0 v
H+ + O H '
Conc. if no dissociation
10~8
C
(10-8 + x)
Conc. after dissociation C - x 14
8
=> x + 10~ x - 10-
x
8
Kw = 10~ = (10" + x) x 2
0
14
x
=0
=> x = 9.5 x 10~8 [neglecting the - v e root]
67 CHEMISTRY TODAY | MAY '13
x = [OH - ] = 9.5 x 10"8 mol dm" 3 K„
10,-14
[OH~]
9.5x10-°
[H + ] = -
CI CH2—CH2COOH cyp ^
11. ( b ) :
[H+] = 1.005 x 10-7 mol dm~ 3
( b ) : CH 3 CH 2 CH 2 OH P
C H 3 C H = C H 2 + H2O Q
C H 3 C H : = CH 2 + HC1Q
-> C H 3 C H - C H 3 I CI R
Cinnamic acid
12. (a) : W h e n calcium f o r m a t e is distilled w i t h calcium acetate, acetaldehyde is formed. HCO; /OOCjCH, I .Ca + Ca. HCO; q / OOC ;CH 3 Calcium formate
C H 3 C H - C H 3 + KC1 I OH
CI R
S
7.
(c) : P 2 0 5 + 2 H N 0 3
8.
(a) : ion can act as L e w i s b a s e i.e. it can donate electron pair to some species e.g. I 2 . The electron d o n o r species are t e r m e d as L e w i s base.
• 2HPO s + N 2 0 5
Acetaldehyde
13. (b): Given that Tx = 273 + 27 = 300 K V1 = V/2 Vj/TJ = V2/T2
=> V/2/300 = V/T2
:. T2 = 600 K
or 327°C
14. (b): Sodium peroxide gives hydrogen peroxide on treatment with dil acid. N a 2 0 2 + dil.H 2 S0 4
> H 2 0 2 + Na2S04
15. (a) : The o r d e r of acidity of RCOOH, H 2 C 0 3 , C 6 H 5 OH, H 2 0 , and ROH is <
RCOOH > H 2 C 0 3 > C 6 H 5 OH > H z O > ROH I n c r e a s i n g Ka v a l u e
r +12 —> 13
I n c r e a s i n g acidic s t r e n g t h
(b): In the alkaline earth metal group, d o w n the group as the size increases the ionic character of compound increases.
NO 2
•NO:
Sn/HCl
16. (b): Cu + has 3d 10 outer electronic configuration (completely filled li-orbital) with zero unpaired electrons, as a result d-d transitions (responsible for absorption of light radiation and hence imparting colour) are not possible. Thus compounds of Cu + are colourless. 17. (b): Trihalides of boron are monomeric, whereas halides of A1 like A1C13 exist as dimers, A12C16.
m-dinitrobenzene
NH
-NH2 3-aminoaniline
CHEMISTRY TODAY | MA Y ' 1 3 68
" a t i o n > 2CH 3 CHO + 2 C a C 0 3
Let original volume of gas = V = V2
P and S are position isomers, (c) : NaC10 4 / on hydrolysis, gives strongest acid HC10 4 as compared to other salts. This is strongly ionized and gives highest concentration of H + . Therefore aqueous solution of NaC10 4 will have lowest p H value.
HNQ3/H2S04:; 100°C
Calcium acetate
Disti
C H 3 C H - C H 3 + KOH (ai?) .
10. (d):
'
CH = CHCOOH
32
32 g of S and 56 g of Fe is required to prepare one mole of FeS. 100 - 56 = 44 g of iron p o w d e r will remain unreacted.
9.
- HC1
(c) : Fe + S —»FeS 56
6.
ale. KOH .
/ HVZ \ \ reaction J
p H = - l o g [ H + ] = 6.99 8.
CH2CH—COOH
CI
/
A
V
cr
/
Dimer
A 1
\
/ B CI
X
x
X, X = Br, CI, F Monomer
18. ( b ) : 0 2 = e l s 2 a*ls2 a2s2 a*2s2 a2p2 n2p2 = n2py n*2p\ = 7t*2pi
e.exlQ-^xSxlO8 182.2 xlO 3
x 6 x l 0 ,23
B.O. = ~ (10 - 6) = 2 = 0.652 x 10-6 m Similarly B.O. for :
52 x iq-7
m
26. (c) : van't Hoff factor for BaCl 2 (= 3) is highest
0 2 = 2.5, O f = 3, O f = 1
among given solutes.
Hence, the increasing order of bond order is Of < o2 < o2 < o f
27. (a) 28. (c) : P ( C 5 H 1 0 O )
19. (a) : No. of K atoms = i x 8 = 1 (At corners) No. of L atoms = ^ x 6 = 3
= 6
> f o r m a t i o n of p h e n y l
hydrazone i.e. P must be a carbonyl compound.
(Face centres)
P
> negative Tollen's test,
i.e. It must be a ketone.
.-. Formula of the compound is KL3. 20. ( b ) : Boiling point of alcohols is higher than ethers d u e to H-bonding. In alcohols, the boiling point decreases w i t h b r a n c h i n g d u e to decrease in surface area. Hence, n-pentanol will have highest boiling point. 21. ( b ) : Secondary amine when reacts with QH5SO2CI (Hinsberg's reagent) forms dialkyl sulphonamide which does not dissolve in alkali. 22. (b)
P
• no iodoform test
i.e. P does not contain CH 3 CO - group. P must be H 3 C C H 2 - C O - C H 2 C H 3 . 29. ( b ) : C r 2 0 3 is an amphoteric oxide. C r 0 3 is an acidic oxide. V 2 0 3 is basic oxide. TiO is basic oxide. 30. (c) : The effect of heat on dicarboxylic acids can be summarised in the form of Blanc's rule which
23. ( b ) : M n 0 4 + 2H z O + 3e Equivalent mass =
• Mn02 + 40H"
Molecular mass
states that dicarboxylic acids in which the two carboxyl groups are attached to the same carbon atom on heating undergo decarboxylation to form
158
monocarboxylic acids; 1,4- and 1,5- dicarboxylic
= 52.66
acids on heating alone or distillation with acetic 24. (c) :
anhydride form cyclic anhydrides while 1,6- and
3d
1,7- dicarboxylic acids on heating alone or on
2+
[Ni(NH 3 ) 6 ] -
distillation w i t h acetic a n h y d r i d e give cyclic
• 4s
Id
4p
ketones with loss of one carbon atom as carbon dioxide. COOH
s f d
2
Options (a) and (b) have zero unpaired electron. Option (d) has one unpaired electron. h c
AT = 25. (1 a ) : A£
v - Ev = =E 3 2
~
1 3 1 2
—
+
743 K
* H C O O H + CO,
COOH
AE = 182.2 k j m o P =>
, hex 6 x 10 23 1= r 182.2 x 10
=
182.2 x ! 0 J 6 xlO 2 3
co2
.COOH
1 3 1 2
^ ^ C O O H -5248 + 11808 36
CO + H 2 o +
J atom
1
—
CH2-CH2COOH i CH2-CH2COOH
CO 573 K
CH2CH2\ CH 2 CH 2 /
c=o
31. ( b ) : Aromaticity can be predicted by the use of Huckel's rule which says
69 CHEMISTRY TODAY | MAY ' 1 3
(i)
H3C
(4n + 2) rc-electrons are required in delocalised system to give it aromaticity.
H
H3C
H c
/
H3C^
(ii) There must be conjugation in the species. (iii) The molecule must be planar.
H
=
c
C
H
CH 3
(Some polarity)
(Polarity is compensated)
CI
CH 3 c = c
H3C^ Here, total n u m b e r of electrons available for delocalisation = 6
36. (b) : To act as surfactant, there should be both polar and non-polar ends in the molecules.
It is expected to be aromatic. 32. (b) by
CH 3
Relative lowering of vapour pressure is given
C2H5
= Vapour pressure of pure solvent = Vapour pressure of solution = Mass of solute = Molecular mass of solute = Mass of solvent = Molecular mass of solvent. solution
121.8-120.2 121.8
15 hn 250/78
Nomenclature is according to Cahn-Ingold-Prelog method i.e., priority rule. 38. ( b ) :
CH,3
:
20
electrons
Heat of neutralization (AH) = 57 k j
Volume of H N 0 3 = 0.5 mol Volume of N a O H = 0.2 mol When HNO3 solution is added to N a O H solution, with 0.2 mole of O H ions of N a O H solution. Heat released = AH x 0.2 = 57 x 0.2 = 11.4 k j —
+\CH,-C-CH,
> I
-H +
CH,3 I C - C H ,J 1 HCH3
C - C H ,3 I CH,
Cr
f-butylbenzene
40. (d)
39. (c)
a (initial) 1 41. ( d ) : k = y In t a-x (after time t) k x t = In
[N2Q5]Q [N 2 O s l f
42. (a) : For a first order reaction, , 2.303 , k=log 2/3
a a-x
2.303
CHEMISTRY TODAY | MAY '13
^
CH,
0
The electronegative oxygen a t o m is a s t r o n g electron withdrawing group. So there is polarity in the carbonyl compounds. This polarity is even greater than in the asymmetric alkene. 70
®
[3° carbocation]
CH,
then 0.2 mole of HNO3 solution will combine
C
-> C HJ, - C - CH,
iso-butene
. m = 357.8
6.023x10 2 3 x 60 electrons 96500 = 3.74 x 10
35. (a) : J J /
CH,
I C H 3 - C = C H 2 +H 2 SO 4 -
33. (a) : Charge = current x time Charge, Q = 1 A * 60 s = 60 C Now, 96500 C charge is carried by 6.023 * 1023 electrons. 60 C charge
- CI - H
Cl-
P° - P _ w/m p° w / m+W / M
34. (a)
2
37. ( b ) : H-
Where, P° P w m W M For dilute
^Cl
(Polarity compensated)
(4n + 2)n electrons means 2, 6, 10 ... n electrons.
h/3
-
5.48 xlO" 1 4
h/3
-
5.48 x 10 - 1 4
2.303
1 °S 1 - 2 / 3 log 3
3 RT
43. (a) : r.m.s velocity =
M 3RT
3R x 298
M,cu
M,•so,
(at STP, T = 298 K)
ntest
298 M,•ci,
M,•so,
1.
=>
T = — x 71 = 330.59 K 64
=>
T = 330.59 - 2 7 3
If you have taken any of the exams given b e l o w a n d possess plenty of grey cells, p h o t o g r a p h i c m e m o r y then you are the right candidate for this contest. A l l y o u have t o d o is w r i t e d o w n as many questions ( w i t h all choices) y o u can remember, neatly on a paper w i t h name of the exam, your name, address, age, your p h o t o g r a p h a n d mail t h e m to us.
i
T = 57.59°C (•." 273 K = 0°C) 44. (c) : A c c o r d i n g to H a r d y - S c h u l z e rule Ions c a r r y i n g c h a r g e o p p o s i t e to that of the sol are effective in causing coagulation of the sol. (ii) Coagulating p o w e r of a n electrolyte is directly p r o p o r t i o n a l to the valency of the active ions (i.e., ions c a u s i n g coagulation).
(i)
S i n c e A s 2 S 3 is a n e g a t i v e s o l , t h u s P 0 4 ~ is i n e f f e c t i v e in c a u s i n g c o a g u l a t i o n , a n d o u t of the cations, Al 3+ h a s the m a x i m u m c o a g u l a t i n g power.
2.
Engineering : UPSEE, M P PET, J & K CET, A M U , W B JEE, Bihar CECE, Odisha JEE, Maharashtra CET... 3.
2Ag + + Cr0 4 2 " 4 x 10-4 2 x lO"4
Ksp = | Ag + ] 2 [Cr0 4 2 "] = [4 X 10-4]2 [2 X 10-4] = 32 x 10-12
4.
48. (c) : y,B
and lastly the pitfalls D o n ' t send i n c o m p l e t e question. Our panel of experts w i l l cross-check your questions. You have t o send it w i t h i n a m o n t h of giving t h e particularexam.
47. (a) : Starch is a p o l y m e r of a - D - g l u c o s e .
A*
the benefits Plenty! Each c o m p l e t e q u e s t i o n w i t h answer w i l l make y o u richer by Rs. 1 5 * / - . M o r e t h e questions, t h e merrier it w i l l be. We w i l l make you f a m o u s by p u b l i s h i n g your name ( p h o t o if possible). Also y o u can derive psychological satisfaction f r o m t h e fact t h a t your questions w i l l b e n e f i t t h o u s a n d s o f readers.
x 32 = 40,000
0.08 Hence, m o l e c u l a r m a s s of insulin is 40,000 g.
46. ( d ) : A g 2 C r 0 4
the exams PMT : AIIMS, JIPMER, A M U .... etc.
45. (a) : 0.08 g of S is p r e s e n t in 100 g insulin. 32 g of S in rp r e s e n t in
whocanparticipate
Af
Mailto:
TheEditor, MTG Learning M e d i a Pvt. Ltd. Plot 99, Sector44
49. (c) : 2 H 2 0
Institutional A r e a , G u r g a o n - 1 2 2 0 0 3 (HR).
2H 2 + 0 2
V o l u m e of H 2 a n d 0 2 is liberated in the ratio of 2 : 1. H e n c e if the v o l u m e of 0 2 liberated = 2.24 d m 3 , the v o l u m e of H 2 liberated = 4.48 d m 3 . 50. ( b ) : For a n i s o t h e r m a l process, AT = 0 Hence, AE = 0, a * 0. OO
Tel.: 0 1 2 4 - 4 9 5 1 2 0 0 * Conditions apply o Payment will be made after the MCQs are published. o Kindly note that each question should be complete, o Payment will be made only for complete questions, o Preference will be given to the reader sending the maximum complete and correct questions. Other conditions apply. The decision of the Editor, MTG shall be final and binding.
71 CHEMISTRY TODAY | MAY ' 1 3
CONCEPT BOOSTER "First of all, 'ALL THE BEST' to my students who have appeared in JEE MAIN EXAM for their results. I also convey my best wishes to the students who are appearing in the NEET EXAM, 2013. "CONCEPT BOOSTER - NEET SPECIAL" will help all the NEET 'aspirant-; to check their grip in tackling tricky questions which is an integral part of any competitive exam. So, don't get tensed and whatever time you are getting carry on your practice to solve tricky questions. BEST OF LUCK ONCE AGAIN.
If Fe2+, Fe 3+ and Fe block are kept together then 3+ (a) Fe increases 3+ (b) Fe decreases (c) Fe / Fe remains unchanged (d) Fe decreases.
GENERAL, PHYSICAL AND INORGANIC CHEMISTRY
The correct decreasing order of first ionisation enthalpies of five elements in the second period is (a) B e > B > C > N > F (b) N > F > C > B > B e (c) F > N > C > B e > B (d) N > F > B > C > B e The bonds present in ion are (a) six equivalent Cr—O—Cr bond (b) six equivalent Cr—Cr bond (c) six equivalent O-Obond (d) none of these. 3.
bonds
and
one
Cr—O
bonds
and
one
Cr—O
bonds
and
one
X + K 2 C 0 3 + air h e a t > Y; Y + Cl 2 which of the following is correct?
- > Z (Pink)
(a) X = Black, MnO z ; Y = Blue, K 2 Cr0 4 / Z = KMn04 (b) X = Green, Cr 2 0 3 ; Y = Yellow, K 2 Cr0 4 , Z = K 2 Cr 2 0 7 (c) X = Black, M n O z ; Y = Green, K 2 M n 0 4 , Z = KMn04 (d) X = Black, Bi 2 0 3 ; Y = Colourless, KBi0 2 , Z = KBi0 3 4.
6.
Which of the following will give a pair of enantiomorphs? (a) [Cr(NH 3 ) 6 ] [Co(CN) 6 ] (b) [Co(en) 2 Cl 2 ] CI (c) [Pt(NH 3 ) 4 ] [PtCl 6 ] (d) [Co(NH 3 ) 4 Cl 2 ] N 0 2
7.
The correct order of C—O bond length among CO, CO§", C 0 2 is (a) C 0 < C 0 | " < C 0 2 (b) C 0 | - < C 0 2 < C 0 (c) C O < C 0 2 < C O | - (d) C 0 2 < C 0 | " < C 0
8.
In c o m p o u n d s of type £C13, where £ = B, P, As or Bi, the angles CI — £ — CI for different £ are in the order (a) B > P = As = Bi (b) B > P > A s > B i (c) B < P = As = Bi (d) B < P < As < Bi
9.
The correct order of equivalent conductance at infinite dilution of LiCl, NaCl a n d KC1 is (a) LiCl > NaCl > KC1 (b) KC1 > NaCl > LiCl (c) NaCl > KC1 > LiCl (d) LiCl > KC1 > NaCl
the structure of dichromate Cr-O
The correct order of increasing bond length of F 2 , N 2 , Cl 2 and 0 2 is (a) 0 2 < N 2 < C 1 2 < F 2 (b) N 2 < 0 2 < F 2 < Cl 2 (c) C 1 2 < 0 2 < N 2 < F 2 (d) F 2 < Cl 2 < 0 2 < N 2 Standard electrode potentials are Fe 2+ /Fe ;E° = - 0.44 V Fe3+/Fe2+; E° = + 0.77 V
10. W h e n 1 mole of monoatomic ideal gas at T K undergoes adiabatic change u n d e r a constant external pressure of 1 atm, volume changes from 1 L to 2 L. The final temperature in kelvin would be T__ „ , 2 (b) mT + (a) 3x0.0821 22 / 3 2 (d) T (c) T 3x0.0821
* Institute of Chemistry (IOC)- Asansol, Durgapur, Dhanbad, Kolkata 72
CHEMISTRY TODAY | MAY' 13
'ARUNAVASARKAR
ORGANIC CHEMISTRY
"• Ot ° ON /
(c) P h — C — C H — P h (d) P h — C— CH— Ph II I II I O H O D
Me OH
"> X ; Identify'X'.
«a
OH (a) NH,
OH
^ ^
a
15. (CH 3 ) 2 C
(a) S N 2 (C)
12. Which of the following options are correct for the given reaction?
a
(CH 3 ) 2 C = CHCH 2 SPh + NaCl Which mechanism is followed in the above reaction?
(d) None of these.
so3H + Bro
16. C H 2 — CH— CD,CI + CHXT -
-> P. P, can be
SO,H
CHi
°'i
>x.
(a)
C H 2 — C H — CD,CI I 2 I OCH3 OCH3
(b)
CH2—CH-CD,C1 I I OH OCH3
-Br (b)
Br Br A / S O 3 H
w
(b) S N 1 (d) S N 1 '
SN2'
'X' will be
FeBr,
(a)
CH — CH 2 + Na S P h - ^ i i *
NHCOMe
COOCHJ
NH2
CI
(c) C H 2 — CH— C D , O C H 3
u
(d) N o substitution occurs. H ,'.
13.
Ag + /AgNQ 3
Y ^ B r °
CH3OCH2—CH—CD2 O
H
Me
(d)
>'X'.
H
'X' should be H
H
(a)
TT
/ Me
(C2H5)2O'
OH
Me
'CH, (a)
O
~COOC 2 H 5 CH(CH 3 ) 2
— CH2CHO
14. Ph 2 C — C H , " I I
H
CH, C2H5 COOC 2 H 5
Product ' P .
OH
(C)
'P' will be
O
H
(b)
P
u
V_^
C
(
C H
3)
3
(d) Reaction is not possible due to the presence of bulky group at the attacking site.
H C
.OC,H K CH,
(b) C H
Me (d) None of these.
D (a) Ph 2 C
C(CH 3 ) 2
'Z' can be
H C \
(b) Phn
H
OH
NaOCjH;
Br
H
(C)
Br 17. CH
c
-
C
O
-
H
18. The major product (Q) formed in the reaction sequence is CHEMISTRY TODAY | MAY '13
73
c
CH, C 6 H 5 COCl dil. NaOH
I. CH,MgI
->P
(b)
N" H,C
(b) C 6 H 5 — CO(CH 3 ) 2
(a) C 6 H 5 COCH 3
(c)
C
H
6
(d) CH 3 — CH - N H 2 and (CH 3 CH 2 ) 2 NH 22. The compound having the highest dipole moment is O (a) ^
C-Cr'O'Cr N H
II : strongest base, IV : strongest base, III: strongest base, II : strongest base,
23.
IV
I : strongest III: strongest IV : strongest III: strongest
o
COO"
-SO 2 NH 2
/
III
II
o
(c)
19. From the following compounds which is the strongest base and which is the strongest acid?
(a) (b) (c) (d)
11 N^CH,
OH
(d) C6H5-
I
and C H 3 C H 2 N H 2
(c) CH 3 CH = N H and CH 3 CH 2 NH 2
O II -C-
5
Js.
N
H-
NaNO z
- N H 3 dil. HC1 CH,
(a) (+)CH 3 CHOHCOOH (b) (±)CH 3 CHOHCOOH (c) (-)CH 3 CHOHCOOH COO^Na +
acid acid acid acid.
-NH,
(d) H -
20. Consider the following diazonium ions.
CH, Me 2 N
// %
O2N
N,
1/
No
(II)
(I)
CH30
%
I T \
CH-
N,
$
—N 24. (IV)
(III)
The order of reactivity towards diazo-coupling with phenol in the presence of dil. NaOH is (a) I < I V < I I < I I I (b) I < I I I < I V < I I (c) I I I < I < I I < I V (d) I I I < I < I V < I I
(I)
(a) (b) (c) (d)
21. The products (X) and (Y) are formed in two separate reactions of acetaldehyde. CH3CHO CH3CHO
NH,
HO—CEt2
-»x
NH, H2 NI A
> Y
25.
X and Y are respectively CH,
A
(a) H3C
HN
and (CH 3 CH,) 2 NH NH CH,
H 74
CHEMISTRY TODAY | MAY' 13
pKa of I > pKa of II pK„ of I < pKa of II pK„ of I = of II Can't be compared due to slow rate of dissociation of II.
/ Me
C H
~
C 1
7
(R-isomer)
(a) (b) (c) (d)
Product will be again (R)-isomer Product will be (S)-isomer Both (R)- and (S)-isomer will be produced Optically inactive cyclic product will be obtained.
SOLUTION
1.
2.
C02=> : 0 =
(c): A s w e m o v e along the period, the nuclear charge increases a n d size of a t o m decreases. So, the o r d e r of ionization e n e r g y is F > N > C. But ionization e n t h a l p y of b o r o n is low as c o m p a r e d to beryllium as first e~ of b o r o n is to b e r e m o v e d f r o m p-orbital, while in beryllium, it has to be r e m o v e d f r o m s-orbital. O O (a):
TO= C—O: M o r e is the single b o n d character in resonance hybrid, m o r e is the b o n d length. Hence, C O < C O z < COf". a
2K2Mn04 + 2C02T
Cl
(b): Bond length decreases as the multiplicity of b o n d increases. Therefore, N = N < 0 = 0 < F - F Further, chlorine is bigger in size t h a n fluorine, therefore b o n d length of F 2 < Cl 2 .
6.
Q
As
2 K M n 0 4 + 2KC1 (Pink) Z
5.
P 2 ' T r i g o n a l geometry; Bond a n g l e - W
(pcy
(Green) Y
2K2Mn04 + ci2-
S
Cl sp 3 ; P y r a m i d a l
CI ^
(c): 2 M n 0 2 + 2 K 2 C 0 3 + 0 2 (air)
C l
n — n ^ (BQj)
,>Cr. -cr O '/ . o
(Black) X
:Q— C = Q :
C=0:
^
Cl
S
3
P ' Pyramidal
Bi Q / / I ^Q
sp3, P y r a m i d a l
O r d e r of b o n d angles is
(b): E°Fe2+/Fe — 0.44 V, E ° F e W = + 0.77 V If a cell is constructed combining these t w o electrodes, oxidation occurs at Fe 2+ /Fe electrode. Fe > Fe 2+ + 2e~ (anode) 2[Fe 3+ + e~ > Fe 2+ ] (cathode) Cell reaction is Fe + 2Fe 3+ >• 3Fe 2+ 2+ 3+ If Fe , Fe a n d Fe blocks are k e p t together, then Fe 3+ reacts w i t h Fe to yield Fe 2+ . Fe 3+ decreases, (b): Either a pair of crystals, molecules or c o m p o u n d s that are m i r r o r images of each other b u t are not identical a n d that rotate the p l a n e of polarized light equally, b u t in opposite direction are called as e n a n t i o m o r p h s .
BC13 > PC13 > ASC13 > BiCl 3 . 9.
(b)
10. (d): For adiabatic chemge, AE = Wad AE = nCv (T2 - Tj)
•••
nCv(T2-T1)
=
-P(V2-V1)
= -1(2-1) R —r(T2-T!) = - l y-1 C„ -C,,1 = jR, r P
R = c—.-.y-l ^V
now, n = 1 Y =5/3 (for m o n o a t o m i c gas) R = 0.0821 L-atm mol" 1 K"1 7.
(c): C O
:C =
:C=0:
0:
cof-
C o'
76
c \
<—> N
o
CHEMISTRY TODAY | MAY' 13
o
/
S
0.0821, -(r2-T!)=-i 1-1
O
O"
O
o
<—>
c
o
Bond angle <109°28' a n d decreases f r o m PCI3 to BiCl 3 .
/ / N
o -
(3x0.0821)
R R =— c :.Cv = v - 1 V I
0 \
c
/ M e
OH
O"
11. (b):
>
N
+/ NH
NH,
V r / (-h 2 0) °
NH\
C
(-OH)
/Me
NHCOMe
II O •Br — Br — F e B r ,
12. (b): : B r — : B r + F e B r ,
- BrFeBr 4 O
O ll/O3
O"
so;
SO3H _H+
Br FeBrJ
BrO I S03+
a
O ll/O-
Br BrO
B?b
EE
a complex
Concept : Above reaction is an example of 'ipso-substitution'. In some cases of electrophilic aromatic substitution (displacement) of one ring by another occurs. This process is 'ipso-substitution' (latin 'ipso' means 'on itself). H
H Ag" / —^.'Phn
13. (c): Ph in""' /
f
Me
£
C:\kX ;>T>Br :OH ; V O H ^ Me/
T ^Br: OH uH
J
Ag (-AgBr) (PhCHMe Shift)
H Ph
C J # Me
C H 2, C H O
<
: Ph \
J r C H ., — — C — H / Me /
Concept: Bulkier P h C H M e shift takes place without hydride shift. The configuration of the migrating chiral alkyl g r o u p is retained because it doesn't become free during rearrangement. 14. (b): Its a type of pinacol-pinacolone rearrangement. A close look to the reaction mechanism will give us the actual product. Ph
H
/
Ph/H\
H I
H +
-H
r> P h — C — C H Ph
O—H
+
> Ph—C — C —H I II Ph O
This —OH will not be attacked as it will form the less stable carbocation.
Concept: N o atom of —'D' i.e., deuterium will be attacked as the reaction is purely intramolecular. 15. (c): W h e n steric factors prevent normal S N 2 attack at an allylic carbon, the nucleophile attacks at the '/-carbon i.e., an allylic rearrangement takes place u n d e r S N 2 conditions. This mechanism is termed as S N 2' mechanism (bimolecular nucleophilic substitution with allylic rearrangement).
SN2
16. (d): CH3O- + C H 2 — C H — CD 2 C1 — >
r *
C H 3 O C H 2 — C H — CD 2 — CI
SN2
> C H 3 O C H 2 — C H — CD 2 CHEMISTRY TODAY | MAY '13 7 7
C o n c e p t : The epoxide ring opens u p as a result of nucleophilic attack (S N 2) on the less substituted carbon. The resulting alkoxy oxygen displaces chlorine (an internal S N 2) to form again an epoxide ring. O (?Br
17. (a): H 3 C
>x
>CH,
/CH
-Br"
— H
C(CH 3 ) 2 I Br
3
C (CH3)2
/
I Br %)c
'CH
CH,
3
OC,H-
C H T ? -
2
H
5
C(CH3)2
f
COOC2H
COOC2H
5
-C(CH 3 ) 2
5
C(CH3)2 CH,
C,H=OH
COOC2H5 HC(CH 3 ) 2 C o n c e p t : The cyclopropane ring opens u p with simultaneous expulsion of bromine (Step IV). The resulting (3, y-unsaturated ester undergoes base-catalyzed isomerization to form a thermodynamically more stable a , P-unsaturated ester.
Y
18. (a):
OH
£-ICI
CFT-C-CL
o
> C
6
H
5
O
- C - N .
^ - » C
C
o
6
H
5
- C - C H
Most basic part (because lone pairs are quite available)
N' H
3
<•
- most acidic hydrogen O O" I
O S
S =
— N H
<—>
II O
O
N H
0 II = NH 1 O" S
<—>
78
CHEMISTRY TODAY | MAY' 13
(p)
CH 3 MgI
H,0* C
6
H
5
O <-
- C - N .
I 20. (b): Diazonium ion acts as an electrophile in coupling reaction; greater the electron w i t h d r a w i n g power, higher the electrophilicity. (c):
CH
3
CHO
N H ?
I3 — N H — ( H
p - i
- C - N
CH,
21.
O S
5
OMgl
(Q)
19. (c)
H
sr
o
<
6
fl ~o
> CH
3
—
C H — N H
CH
3
NH< ^>CH
CH=O •
3
the inductive effect of the chlorine atoms on the position of electrons in the —COOH group and therefore on acidity should be same since the same bonds intervene, but the field effect disfavours the dissociation of I but favours the dissociation of II because the — Cl atoms are closer in space to the —COOH group in I than they are in II and because of this I is less acidic than II. Concept : The conjugate base I is destabilized by repulsive interaction because the negative
— C H = N H -
CH3CH2NH2 (V) 22.
(b):
Aromatic system as Huckel's rule [(4n + 2)n] is executed. So, charge separation becomes easiest among the given options.
O"
23.
5+
(b):
cooH-
HNO,
CH,
—COO" group on the other hand the conjugate base of II is stabilized by electrostatic force of attraction because the positive ends of the two
coo-
-NH.3 NaNO / > z dil. HCI
H-
-N 2 CI-
8+
CH3 (Diazo reaction)
(±) C H 3 C H O H C O O H <•
5-
C — Cl dipoles are closer to the —COO" group, 25.
— COOH
24. (a): The inductive effects may be directly transmitted through space or solvent molecules rather than along a chain. Such effects are often called field effects. Field effects depends on the geometry of the molecule but the inductive effect depends only on the nature of the bonds. For example, in the isomeric dichloroacids I and II,
(a): HOCEt,, > C - c i - f ^ ( H20) Me** > " / \ _ O—C I^J
CEt, Me
/
2012 Mftntt
Volumes of the following
1
Cl •
Me <-
^ ^
Neighbouring Group Participation (NGP)
H
HO/C\t2 Me
HP
C— OH
H
/
OH
(R-isomer)
OO
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CHEMISTRY TODAY | MAY '13
79
K$ M i l NISTCMMIII NIG PROBLEMS PHYSICAL CHEMISTRY S O M E BASIC C O N C E P T S OF C H E M I S T R Y
1.
2.
A 500 mL sample of the effluent f r o m a water softener required 6 drops of standard soap solution to produce a permanent lather. The soap solution h a d been calibrated against an artificial hard water solution containing 0.136 g of CaCl 2 per litre. On the average, it required 28 drops of standard soap solution to lather 500 mL of the artificial solution. Then, the hardness of the effluent sample (in terms of p p m ) of CaCO s is (a) 123 p p m (b) 26 p p m (c) 0.123 p p m (d) 38 p p m Two oxides of a certain metal were separately heated in a current of hydrogen until constant weights were obtained. The water produced in each case was carefully collected and weighed. 2 g of each oxide gave, respectively 0.2517 g and 0.4526 g of water. This observation illustrates (a) law of conservation of mass (b) law of constant proportions (c) law of multiple proportions (d) law of reciprocal proportions
3.
H3PO4 is a tribasic acid and one of its salts is N a H 2 P 0 4 . What volume of 1 M N a O H should be added to 12 g N a H 2 P 0 4 (mol. wt. 120) to exactly convert it into N a 3 P 0 4 ? (a) 100 mL (b) 300 mL (c) 200 mL (d) 80 mL
4.
An element ' M ' combines with oxygen, hydrogen, chlorine and fluorine to form respective compounds containing 56.36, 91.17, 22.54 and 35.22 percent of ' M ' , respectively. The vapour densities of these c o m p o u n d s (H = 1) are 110, 17, 68,75 and 44 respectively. Atomic weight of the element 'M' is (a) 39 (b) 31 (c) 92 (d) 23
5.
An element 'L' forms various compounds, their weight percent of 'L' has been indicated below: C o m p o u n d s of 'L'
80
CHEMISTRY TODAY | MAY' 13
I
II
III
IV
Vapour density of c o m p o u n d s
14
38
14
22
Weight percent
42.8
15.8
85.7
81.4
Atomic weight of the element ' L ' is (a) 9 (b) 12 (c) 15 6.
(d) 24
In Haber's process, 30 L of dihydrogen a n d 30 L of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of the gaseous mixture in the end? (a) 20 L N H ^ 25 L N 2 and 20 L H 2 (b) 10 L NH 3 , 25 L N 2 and 15 L H 2 (c) 20 L NHg, 10 L N 2 and 30 L H 2 (d) 20 L NH 3 , 25 L N 2 and 15 L H 2 S T R U C T U R E OF A T O M
7.
In Bohr series of lines of hydrogen spectrum, the third line f r o m the red end corresponds to which one of the following inter-orbit j u m p s of the electron for Bohr orbit in an atom of hydrogen? (a) 5 —> 2 (b) 4 —»1 (c) 2 - > 5 (d) 1 3 - > 2
8.
The value of Planck's constant is 6.63 x 10~34 J s. The velocity of light is 3.0 x 10 s m s"1. Which value is closest to the wavelength (in nm) of a q u a n t u m of light with frequency of 8 x 10 15 s"1? (a) 2 x 10~25 (b) 5 x 10~18 (c) 4 x 1 0 (d) 3 x 107
9.
The uncertainty in the velocity of a cricket ball of mass 100 g, w h e n uncertainty in its position is of the order of lA, w o u l d be (a) > 5.27 x 10"24 m s"1 (b) > 2.35 x 10"23 m s"1 (c) > 3.16 x 10"22 m s"1 (d) > 8 . 5 1 x 1 0 ' 24 m s " 1
10. W h e n a certain metal was irradiated with a light of frequency 3.2 x 1016 Hz, the photoelectrons emitted h a d twice the kinetic energy as did photoelectrons emitted w h e n the same metal was irradiated with light frequency 2.0 x 1016 Hz. Hence, threshold frequency is (a) 8 x 1017 H z (b) 0.8 x 1015 H z 15 (c) 8 x 10 H z (d) 8 x 1016 H z
11. Light of wavelength A, shines on a metal surface with intensity x and the metal emits y electrons per second of average energy, z. What will h a p p e n to y and z if x is doubled? (a) y will be doubled and z will become half. (b) y will remain same and z will be doubled. (c) both y and z will be doubled. (d) y will be doubled but z will remain same. 12. The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass n u m b e r A is given by 1.25 x 1(T13 x AV3 cm. Radius of atom is 1 A. If the mass n u m b e r is 64, then the fraction of the atomic volume that is occupied by the nucleus is (a) l . O x K T 3 (b) 5.0 xlO" 5 2 (c) 2.5xlO" (d) 1.25xlO" 1 3 CHEMICAL BONDING AND MOLECULAR STRUCTURE
13. When water is stirred with glucose, strong hydrogen b o n d s are initially formed between glucose molecules and water molecules, but as more water is added, these hydrogen b o n d s are broken. Which graph represents the observed temperature changes?
(a)
Volume of water added
Volume of water added
(d)
Volume of water added
Volume of water added
14. Trimethylamine Me 3 N, reacts with boron trifluoride, BF3, to form a c o m p o u n d of formula Me 3 N-BF 3 . H o w m a y this reaction be written in terms of the shapes of the reactants and products? r Me Me
F
(a) Me .. ^T ..• Me j. + rF - nB V F ~N
Me (b) M e , .
fT 'N'
A "
.-• F (c) Me ... ^ ... Me + F - B . 'N' >F
F
Me ... ^ ... Me 'N"
Me B (d) Me-.. T' ... Me + F "' A '"F •NF
B F "'* £ " F
F 15. Which set of properties could apply to a non-ionic c o m p o u n d which has a giant lattice? Physical state at room temp.
Electrical conductivity of the molten compound
(a) Liquid (b) Liquid
does not conduct - 1 1 4 does not conduct melts over a temperature range conduct well 808 does not conduct 1610
Melting point (°C)
16. What conclusions can be drawn from the following reactions? H 2 + 4.5 eV -> H + H H + 13.6 eV —> + e~ (a) It is more difficult to break u p an H 2 molecule than to break u p a hydrogen atom. (b) It is easier to break u p an H 2 molecule than to break u p a hydrogen atom. (c) The average energy of formation of H and p+ are the same. (d) N o n e of these. 17. The radius and charge of each of six ions are s h o w n in the table Ion Radius/nm
+ r L 0.14 0.18
M2+
x-
0.15
0.14
Z2"
Y~
0.18
0.15
..- Me >Me
18. What fraction of charge is present on I in
B F '** ^ " F F
B F ' " 4""F F Me
The ionic solids JX, LY and MZ are of the same lattice type. What is the correct order of their lattice energies placing the one with the highest numerical value first? (a) JX>LY>MZ (b) JX>MZ>LY (c) LY>MZ>JX (d) MZ>JX>LY
Me-N
B. ... Me + F '"
..„ ... "• ... Me Me... ^ N ,-F F - B >F
Me ,. ^ ... Me 'N '
Me
(c) Solid (d) Solid
(b)
(C)
Me
8+
8-
covalently b o n d e d H — I , if the dipole moment of HI is 0.38 C and the b o n d length is 1.61 A? (a) 5% (b) 16% (c) 33% (d) 79% CHEMISTRY TODAY | MAY '13
81
STATES OF MATTER
19. A vessel has nitrogen gas and water vapour at a total pressure of 1 atm. The partial pressure of water vapour is 0.3 atm. The contents of this vessel are transferred to another vessel having one third the capacity of the original vessel, completely at the same temperature. The total pressure of the system in the n e w vessel is (a) 2.7 atm (b) 1 atm (c) 3.33 atm (d) none of these. 20. For two gases, A and B with molecular weights MA and M B , it is observed that at a certain temperature T, the mean velocity of A is equal to the ulms of B. Thus, the m e a n velocity of A can be m a d e equal to the m e a n velocity of B, if (a) A is at temperature T and B at T'; T>T' (b) both A and B are raised to a higher temperature (c) both A and B are lowered in temperature (d) none of these. 21. A mixture of 50.0 mL of N H 3 and 60.0 mL of O z gas reacts as 4 N H % ) + 5 0 2 f e ) ^ 4 N 0 f e ) + 6H 2 0 f e ) If all the gases are at the same temperature and pressure, and the reaction continues until one of the gases is completely consumed, what volume of water vapour is produced? (a) 48 mL (b) 60.0 mL (c) 72 mL (d) 75.0 mL 22. A flask contains 12 g of a gas of molecular mass 120 at a pressure of 100 atm was evacuated by means of a p u m p until the pressure was 0.01 atm. Which of the following is the best estimate of the n u m b e r of molecules left in the flask (N = 6 x 10 23 mol" 1 )? (a) 6 x 1019 (b) 6 x 1018 (c) 6 x 1017 (d) 6 x 10 13 23. Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction : C H C H 6 6fe) + 3H2C?) 6 12(g) A mixture of C 6 H 6 and excess H 2 has a pressure of 60 m m of H g in an u n k n o w n volume. After the gas had been passed over a nickel catalyst and all the benzene converted to cyclohexane, the pressure of the gas was 30 m m of H g in the same volume at the same temperature. The fraction of C 6 H 6 (by volume) present in the original volume is (a) 1/3 (b) 1/4 (c) 1/5 (d) 1/6 24. Two glass bulbs A and B are connected by a very small tube having a stop-cock. Bulb A has a volume of 100 cm 3 and contained the gas while bulb B was empty. On opening the stop-clock, the
82
CHEMISTRY TODAY | MAY' 13
pressure fell d o w n to 40%. The volume of the bulb B must be (a) 75 cm 3 (b) 125 cm 3 3 (c) 150 cm (d) 250 cm 3 THERMODYNAMICS
25. For a pure substance, a plot of Gibb's free energy (G) against temperature at constant pressure is G G
26. Consider the following reaction : When 10.0 mL of 1.00 M A g N 0 3 solution is a d d e d to 10 mL of 1.00 N NaCl solution at 25°C in a calorimeter, a white ppt. of AgCl is formed a n d the temperature of aqueous mixture increases to 32.6°C. Assuming that the specific heat of aqueous mixture is 4.18 J/(g°C), that the density of the mixture is 1.00 g/mL and that calorimeter itself absorbs a negligible a m o u n t of heat, the value of AH in k j is (a) - 64 k j (b) - 6 . 4 k J ( c ) - 0 . 6 4 k J ( d ) 0.64 k j 27. Boron can u n d e r g o the following reactions with the given enthalpy changes : 2B(S) + 3/20 2 ( g ) > B 2 0 3 ( s ) ; AH = -1260 k j 2B(S) + 3H 2 f e ) » B 2 H % ) ; AH = - 30 k j If in a container (at constant pressure) which is isolated f r o m the surrounding, mixture of H 2 (gas) and O z (gas) are passed over excess of B(s), then what molar ratio of oxygen and hydrogen will be used so that the temperature of the container do not change? (a) 1 : 42 (b) 1 : 84 (c) 42 : 1 (d) 15 : 3 28. An ideal gas is allowed to expand both reversibly a n d irreversibly in an isolated system. If T, is the initial temperature and T* is the final temperature, which of the following statements is correct? (a) (7»irrev > 0 ) ) rev (b) Tf > Tj for reversible process but Tf = T, for irreversible process. (c) (Tf)rev irrev (d) Tf= Tj for both reversible and irreversible processes. 29. One gram sample of N H 4 N 0 3 is decomposed in a b o m b calorimeter. The temperature of the calorimeter increases by 6.12 K. The heat capacity
of the system is 1.23 kj/g/deg. What is the molar heat of decomposition for N H 4 N 0 3 ? (a) - 7.53 kj/mol (b) - 3 9 8 . 1 kj/mol (c) - 1 6 . 1 kj/mol (d) - 6 0 2 kj/mol 30. In an oven, using 10 kg coal (assume the coal is 80% carbon in weight), insufficient oxygen is supplied such that 60% of carbon is converted to C 0 2 a n d 40% carbon is converted to CO. The heat generated, w h e n coal is b u r n t in this fashion would b e : Given, C (s) + 0J2(g)~ •CO.2 fe)' AH = - 394 k j :
i
C
(s)
+
2 °2fe) ' • CO,
(a) 183200 k j (c) 185200 k j
AH = - 111 k j
(b) 187200 k j (d) 181200 k j EQUILIBRIUM
31. A solution contains 0.05 M each of NaCl and N a 2 C r 0 4 . In both, solid A g N 0 3 is gradually added. Mark out the correct statement. (Given Ksp(AgCl) = 1.7 * 10"10 M 2 and K Sf ,(Ag 2 CrO 4 ) = 1.9*10" 1 2 M 3 ) (a) CrOf ions are precipitated first. (b) Cl" ions are precipitated first. (c) Both Cl - and CrO|~ ions are precipitated together. (d) The second ion starts precipitating w h e n half of the first ion is precipitated. 32. For a general reaction + S^y the specific rate constant is k. 2.0 x 10 -3 m o r ' L s"1 at a certain ^(forward)
temperature. Reaction starts with equimolar amounts of P and Q. Reaching at equilibrium, it is observed that P is twice that of R. The specific rate constant for the backward reaction is (a) 1.5 x 102 mol-1 L s"1 (b) 5.0 x 10^ mol"1 L s"1 (c) 8.0xlO"3mol"1 L s 1 (d) none of the above. 33. For a hypothetical reaction + B
(g) ; a graph between log K and T 1 is straight line as follows, where 6 = tan" 1 0.5 and OA = 10.
\)
°
T' 1
Assuming AH° is independent of temperature, the equilibrium constant at 298 K and 798 K are respectively, (a) Kx = 9.96 x 109, K2 = 2.76 x 1010 (b) Kx = 3.79 x 108, K2 = 9.98 x 109 (c) Kj = 7.96 x 109, K2 = 7.96 x 109 (d) Ki = 9.98 x 109, K2 = 9.96 x 109 34. Before equilibrium is set-up for the chemical reaction, N 2 0 4 2 N 0 2 , vapour density of the gaseous mixture was measured. If D is the
theoretical value of vapour density, variation of x with D/d is shown by the graph. What is value of D/d at point Y? (a) 0 (b) 1.5 (c) 1 _D
(d) 0.5
d 35. Solution of aniline hydrochloride is X due to hydrolysis of Y. X and Y are (a) basic, C 6 H 5 NH 3 + (b) acidic, C 6 H 5 NH 3 + (c) basic, Cl" (d) acidic, Cl . 36. If Ksp of Mg(OH) 2 is 1.2 x 10" n . Then the highest p H of the 0.1 M solution of Mg 2+ ion from which Mg(OH) 2 is not precipitated is (a) 4.96 (b) 6.96 (c) 7.54 (d) 9.04 SOLID STATE
37. In a cubic lattice each edge of the unit cell is 400 pm. Atomic weight of the element is 60 and its density is 6.25 g/c.c. Avogadro n u m b e r = 6 x 1023. The crystal lattice is (a) face-centred (b) primitive (c) body-centred (d) end-centred. 38. In a cubic packed structure of mixed oxides, the lattice is m a d e u p of oxide ions, one fifth of tetrahedral voids are occupied by divalent (X2+) ions. While one half of the octahedral voids are occupied by trivalent ions (Y3+), then the formula of the oxide is (a) XY 2 0 4 (b) X 2 Y0 4 (c) X 4 Y 5 O 10 (d) X 5 Y 4 O 10 39. ZnS exists in two crystalline structures: (i) Zinc blende and (ii) Wurtzite Mark the correct statement about their structure. (a) In zinc blende S 2 " ions are arranged in ccp while Zn 2 + occupy half of the tetrahedral sites. (b) In zinc blende S 2_ ions are arranged in hep and Zn 2 + ions occupy all octahedral voids. (c) In Wurtzite structure S 2 ions are arranged in ccp and Zn 2 + ions occupy half of tetrahedral voids. (d) In Wurtzite structures S 2_ ions are in hep arrangement and Zn 2+ ions are occupying all octahedral voids. 40. Select the incorrect statements. (a) In closest packing of x-atoms, there are 3x tetrahedral holes, a n d x-octahedral holes. (b) The co-ordination no. of U : O in U 0 2 is 8 :4. (c) In hexagonal packing of oxide ion sapphire (the aluminium oxide) Al 3+ ions crystallize in CHEMISTRY TODAY | MAY '13 8 3
2/3 rd of the octahedral hole, then the formula of sapphire is A1 2 0 3 . (d) If anion constitutes ccp a n d all the octahedral voids are occupied by cation (A), the crystal has the same molecular formula AB, if the cation (/I) occupies the alternate tetrahedral voids. 41. Ferrous oxide has a cubic structure. The length of edge of the unit cell is 5 A . The density of the oxide is 4.0 g cm" 3 . Then the n u m b e r of Fe 2+ a n d O 2 " ions present in each unit cell will be (a) Four Fe 2+ and four Oz~ (b) Two Fe 2+ and two O 2 " (c) Four Fe 2+ and two O 2 " (d) Two F e 2 + a n d four O 2 " 42. A solid AB has NaCl type structure. If the radius of cation A + is 170 pm, calculate the m a x i m u m possible radius of the anion B~. (a) 210.3 p m (b) 397.4 p m (c) 410.6 p m (d) 347.9 p m SOLUTIONS
43. Phenol associates in benzene to a certain extent to form dimer. A solution containing 2.0 * 10"2 kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (Kj for benzene = 5.12 K kg mol" 1 ) (a) 73.4% (b) 50.1% (c) 42.3% (d) 25.1% 44. When 0.6 g of an organic acid dissolved in 200 mL of water is extracted with 50 mL of ether, it is found on analysis, that 0.4 g of the organic acid goes into the ether layer. Which of the following represents the correct ratio of the solubility of the acid in ether and water? SethetL = a l 2 5 (b) ^c3 § L . = 0.5 (a) J ^water
water
(c)
=2 -'water
(d)
48. A current of dry air was bubbled through a bulb containing 26.66 g of an organic substance in 200 g of water, then through a bulb at the same temperature containing pure water and finally through a tube containing fused CaCl 2 . The loss of weight of water bulb = 0.0870 g and the gain of the weight of CaCl 2 tube = 2.036 g. The molecular weight of the organic substance is (a) 56 (b) 41 (c) 78 (d) 67 ELECTROCHEMISTRY
49. Concentration polarization arises because of the (a) different concentrations of solutions in the two half-cells (b) changes in the concentration of electrolyte around the electrode from bulk concentration (c) reversible nature of the cell (d) variation in temperature during measurements. 50. During electrolysis in the cell shown, 1.05 g copper is deposited on electrode X. Which one of the following statements is correct? (Relative atomic masses Cu = 63, Ag = 108). H|-
^ether _ g ^water
46. Solubility curve Na 2 SO 4 T0H 2 O of Ml* in water with tl temperature is given 34°C as shown in the figure. cH Temp.From the figure, we can say that (a) solution process is exothermic
CHEMISTRY TODAY | MAY' 13
47. 0.100 g of non-volatile solute is a d d e d to 1 L of each acetic acid, benzene, naphthalene and water. Which of the following is correct decreasing order of freezing point of the corresponding solutions? (a) Naphthalene > benzene > acetic acid > water (b) Water > acetic acid > benzene > naphthalene (c) Naphthalene > acetic acid > benzene > water (d) Water > benzene > acetic acid > naphthalene.
c
45. The degree of dissociation (a) of a weak electrolyte AXB is related to van't Hoff factor (i) by the expression i-1 1-1 (b) oc = (a) a = x+y +l (x + y-1) x +y +l x+y-1 (d) oe = (c) a = i-1 i-1
84
(b) solution process is exothermic till 34°C and endothermic after 34°C (c) solution process is endothermic till 34°C'and exothermic thereafter (d) solution process is endothermic.
W coppe: CuSO.
^
^' f » silver
U l ^ AgN0 3 (aqueous)
(aqueous) (a) The mass of silver electrode W increases by 3.60 g. (b) The mass of silver deposited on electrode Y during the electrolysis is 1.80 g. (c) Z is a salt-bridge. (d) The concentration of copper ions in solution decreases during electrolysis. 51. Equivalent conductance of 1 M propanoic acid is 10 ohm 1 cm 2 & "' and that at infinite dilution 200 ohm" 1 cm 2 eq" 1 . p H of the acid solution is (d) 6.8 (a) 7 (b) 3.3 (c) 1.3
52. The molar ionic conductivities of NH4 and OH" at infinite dilution are 72 and 198 ohm" 1 cm 2 respectively. The molar conductivity of a centinormal N H 4 O H solution at the same temperature is found to be 9 ohm" 1 cm 2 . The percentage dissociation of N H 4 O H at this concentration will be (a) 3.33% (b) 7.14% (c) 12.5% (d) 4.54% 53. Two weak acid solutions HA, and H 4 2 each with the same concentration and having pK a values 3 and 5 are placed in contact with hydrogen electrode (1 atm, 25°C) and are interconnected through a salt bridge. The emf of the cell is (a) 0.21V (b) 0.059 V (c) 0.018 V
R=^
2 2x
(b)
(c) R = 2X(K 1 + K2)
R
=
C6H5N
2CH3COCH3 + C2H6 COCH,
III. S,O s " + 21"
2SOA
+1,
For a pseudo first order reaction, CH 3 COOC 2 H 5 + H z O the net rate is given by
CH 3 COOH + C 2 H 5 OH
— = 4 x 10" 4 L mol- 1 s 1 t(CH 3 COOC 2 H 5 )] dt - 3 x 104 L mol" 1 s 1 [CH 3 COOH][CH 2 H 5 OH] Then, the equilibrium constant of the reaction of the reaction would be (a) 1.3 x 10"7 mol L"" 1 1 (b) 1.3x10" ' mol L " s" (c) 1.33 x 10"9 mol L"1 (d) 1.33 xlO" 8 mol L"1 86
59. G r a p h b e t w e e n log k a n d
tan 9 be
V2.303 J
Hence, Ea
1
xv
T
= 5, JibCO will 0
L/T
5 cal 2.303 (d) none of these. (b)
(a) 2.303 x 2 cal
60. The reaction A —» B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hr. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B? (a) 1 h r (b) 0.5 hr (c) 0.25 hr (d) 2 h r SURFACE CHEMISTRY
H
IV. JRCOOR' + N a O H — > RCOONa + R'OH (a) Only I and IV (b) Only I and II (c) Only I and III (d) All of these. 56.
(b)
C1C6H4N COCH,
2
nC0 n-1 (n-1 (d) fc \n + l J
nC0 2 n£o_ n +1
(c) - 2 cal
55. Which of the following given reaction(s) is/are of the first order?
II.
(c)
2x
CHEMICAL KINETICS
Cl
The concentration of B at the point of intersection of the two curves as shown in the figure for, A —> nB, is
2(k1 + K2)
(d) R =
(CH3)3COOC(CH3)3
58.
(d) 0.021V
KJ + K 2
I.
Consider following reactant sample I. 1 mol of A and 1 mol of B in a 1 L vessel II. 2 mol of A a n d 2 mol of B in a 2 L vessel III. 0.2 mol of A and 0.2 mol of B in a 0.1 L vessel Which of the reactant sample reacts at the highest rate? (a) I (b) II (c) III (d) All are at equal rate.
(a)
54. Kj and k 2 are specific conductance of the solutions A and B in the same conductivity cell. If equal volumes of solutions A and B are mixed, what will be the resistance of the mixture using the same conductivity cell, whose cell constant is x ? (Assume there is no change in the degree of dissociation on mixing) (a)
57.
CHEMISTRY TODAY | MAY' 13
x/m
61.
|
Temperature constant
x/m
Temperature constant
t
(/ (nf Which is the correct statement? (a) (I) represents Freundlich's adsorption isotherm while (II) represents Langmuir's adsorption isotherm. (b) (I) represents Langmuir's adsorption isotherm while (II) represents Freundlich's adsorption isotherm. (c) Both Freundlich and Langmuir adsorption isotherms have been found to be applicable in the adsorption of gases on solids only. (d) Change of pressure affects a lot and further adsorption can take place in both the isotherms.
62. Potassium stearate is obtained by the saponification of an oil or a fat. It has formula CH 3 - (CH 2 ) 16 - COO"K + . The molecule has a lyophobic end (CH 3 - ) and a lyophilic end COO~K+. Potassium stearate is an example for (a) lyophobic colloid (b) lyophilic colloid (c) multimolecular colloid (d) associated colloid or micelle.
(a) 7.5 x 1015 d p m (c) 5 x 1010 d p m
SOLUTION 1.
m
b
(c)
Hardness of effluent sample =
66. The mass of helium atom of mass number 4 is 4.0026 amu, while that of the neutron and proton are 1.0087 and 1.0078 respectively on the same scale. Hence the nuclear binding energy per nucleon in the helium atom is nearly (a) 5 MeV (b) 7 M e V (c) 10 MeV (d) 14 MeV 67. In the radioactive change, z+iQ"' -iS" Z-2-jR" the radiations emitted in sequence are (a) a, (3, y (b) (3, a, y (c) y, a, (3 (d) (3, y, a 68. 1 g atom of an cc-emitting nucleide £ X (f 1/2 = 10 hr) was placed in a sealed container. 4.52 * 1023 He atoms will accumulate in the container in (a) 4.52 hr (b) 9.40 hr (c) 10.0 hr (d) 20.0 hr 69. The radio nucleide 90* Th undergoes two successive P-decays followed by one oc-decay. The atomic number and the mass number respectively of the resulting radionucleide are (a) 94 and 230 (b) 90 and 230 (c) 92 and 230 (d) 92 and 234 70. The 11/2 of radioactive K-40 is (A. = 2.5 x 10~7 min -1 ). The decay 2.0 g of the sample is about
5.27 years activity of
3.
(c)
4. (b)
x 123 = 26 ppm
2.
(c)
7.
(a): The lines at the red end belong to Balmer series. These are obtained for jumps to n = 2 from outer orbits (1st line from n = 3, 2 line from n = 4, 3 r d line from n = 5).
8.
(c)
NUCLEAR CHEMISTRY
65. The counting rate observed from radioactive source at t = 0 second was 1600 counts/s and at t = 8 s it was 100 counts/s. The counting rate observed as count per s at t = 6 s will be (a) 400 (b) 300 (c) 200 (d) 150
x 0.136
= 0.123 parts of C a C 0 3 0.136 g of CaCl 2 in 1000 parts of H z O = 0.123 in 106 parts = 123 Hardness of CaCl 2 = 123 p p m
(b) - = aP m
(d) m aP m a 64. The gold numbers of P, Q, R and S are 0.04, 0.002, 10 and 25 respectively. The protective powers of P, Q, R and S are in the order (a) P > Q> R> S (b) Q> P > R> S (c) S > R > Q > P (d) R> P > Q> S
(b): The hardness of the effluent sample is — t h of the hardness of the CaCl 2 solution. Hardness of CaCl 2 solution = ? I l l parts of CaCl 2 = 100 parts C a C 0 3 0.136 parts (mol. wt.) of CaCl 2 = ^
63. The equation for Langmuir adsorption isotherm under high pressure is (a)
(b) 5 x 105 dpm (d) 7.5 x 102" d p m
9.
5. (b)
6. (b)
(a)
KE 10. (c): ho = hx>0 + KE, u - u0 = h Given, (KE)2 = 2(KE)1 u 2 - u 0 = 2(Uj - u 0 ) or 3.2 x 1016 —Uq = 2(2 x 10 lb - u 0 ) or u n = 0.8 x 1016 Hz = 8 x 1015 Hz 11. (d):When intensity is doubled, number of electrons emitted per second is also doubled but average energy of photoelectrons emitted remains same. 12. (d): Radius of nucleus = 1.25 x 10"13 x /11/3 cm = 1.25 x 10"13 x 64 1/3 = 1.25 x 10~13 x 4 cm = 5 x 10~13 cm Radius of atom = 1 A = 10 cm (4/3)71(5 xlO" 1 3 ) 3 Volume of nucleus Volume of atom (4/3)jt(10,-8\3 )' -39 125x10 = 125 x l O " 1 ^ 1.25x101-13 10"24
13. (c)
14. (c)
18. (a): 5,charge
15. (d)
16. (b)
17. (d)
Dipole moment
0.38xl0" 1 8 €
= 2.4 x 10" n esu 1.61x10" 2.4xl0~"esu % of charge = x 100 = 5% 4.8xlO~ 1 0 esu 19. (d): Total pressure = 1 atm partial pressure = 1 - 0.3 = 0.7 atm CHEMISTRY TODAY | MAY '13
87
When capacity is changed, vapour pressure does not change, only the partial pressure of 0.7 N 2 changes, so VN2
=
=
0.233 atm
30. (b): As coal has 80% carbon in weight weight of carbon in 10 kg coal
So, total pressure - 0.233 + 0.3 = 0.533 atm 3 RT 8 RT 20. (d) : M„ 71Ma 8 71MA
3_ MB
M a MB
8 RT
for
=
37iT_ _
T_
T_
8 M;B
M0 ' T' ~
80
= 10 x — = 8 kg = 8000 6g 5 100 As 60% of C is converted to C 0 2 thus wt. of C converted into C 0 2
±
3K
3 RT' y KMB
kMa
T MA
.
T "
_8_
MB
;
r >T
3k'
21. (c): 4NH 3 + 5 0 2 —> 4NO + 6H z O 4 mol 5 mol 50 mL 60 mL Equal mol contains equal volume u n d e r given conditions. 6x50 H 2 O f e ) obtained from 50 mL of N H 3 = 75 mL H^20( (g„v) obtained from 60 mL of 0 <2
6x60 5
:
72 mL
Hence, 0 2 is limiting reactant. Hence, 72 mL water vapour is produced. 22. (b): PjV = n 1 RT, P2V = n2RT 12 0.01 l n _ 5 n22 = x = 10 120 100 «i p i 1 mol contains 6.023 x 1023 molecules 10"5 mol contains 6.023 x 1018 molecules. 23. (d)
=
:.
m=-S
1 x 10 - 2 AH = - 64 kj.
0.05 Ag + ions are required to precipitate CrOf
P + Q R + S a Initial moles a 0 0 At equilibrium a x a a x=— 3 x2 a2 (3x)2 = Kc = 2 (a-x)(a-x)' 9x{2x) 9x4x2 1 kf X,c = — = 0.25, but iC = — c 4 kb
= 64 kj/mol AgCl
K
or
28. (a)
29. (d): Molecular mass of N H 4 N 0 3 = 80 Molar heat of decomposition (for , . CxATxM calorimeter) = m
CHEMISTRY TODAY | MAY' 13
= 1
12 From eq. (ii) 12 g (1 mole) of C on combusion liberates = 111 k j of heat 3200 g of C on combustion liberates = 111 x3200 12 Total heat liberated would be = 157600 + 29600 = 187200 k j 31. (b): Ag + ions required to precipitate CL Ksp (AgCl)
32. (c):
Moles of Ag5 + = 10.0 x - J — = 10 - 2 mol 1000 Moles of AgCl = 10"2 mol. Heat evolved per mole of AgCl 6.4 x 10
394 x4800
'Ksp(Ag2CrQ4)^1/2 0.05
26. (a): Mass = Volume x density = 20 g Change in temperature, AT = 7.6°C Heat evolved = 4.18 x 20 x 7.6 = 6.4 x 102 J
88
= 8000 x — = 4800 g6 100 and 40% of C to CO thus wt. of C converted into 40 = 3200 g CO = 8000 x 100 N o w from eq. (i) 12 g (1 mole) of C on combustion liberates = 394 k j of heat 4800 g of C on combustion liberates
24. (c)
25. (a): G = H - T S Y = c + mx
27. (b)
AH=1-23x6-12x8°=602kJ 1 As heat of decomposition is always negative. - 602 k j •••
bomb
f
• 2 X 1 0 " 3 =8x10-3 0.25
33. (d): Slope = tan 6 = AH°
0.5 = 2.303 x 8.314
AH0 2.303jR
AH° = 9.574 J mol - 1 Intercept = log 10 A = 10, A = 1010 log6 K = 10
— 2.303 x 8.314 x 298
9.96 x 109 As AH° is independent of temperature, equilibrium constant remains almost constant. It means even at 798 K, equilibrium constant 9 K = 9.98 x 10 34. (c) :N 2 0 4 ( ? ) :2NO. K =
, 4 x 6.023 x 10 23 x 125 x 10 2 4 Z= — =4.16 = 4 72 Four molecules of ferrous oxide, FeO per unit cell- it means 4Fe2+ and 40 2 ". 42. (c):ForNaCl type structure, r+ — = 0.414 - 0.737 For maximum possible radius of A
2fe)
= 0.414; ra- ••
Degree of dissociation x = ——— or x = - - l d d At Y, x = 0 ,.
35. (b): C 6 H 5 NH 2 - HC1 + H 2 O
KfW-1000
43. (a):M 2 (obs):
— - 1 = 0or — = 1 d d C 6 H 5 N H ; O H + HC1
4 weak base
i strong arid
Therefore the solution formed will be acidic, due to the hydrolysis of Q H g N H j ion. 36. (d): As KSP of Mg(OH) 2 = [Mg2+] [OH"] 2
170 = 410.6 0.414
W-AT,/
5.12 x 2.0 x l 0 ~ 2 x 1000 = 148.4 1.0x0.69 Calculated molecular mass of phenol = 94 M 2 (cal) _ 94 = 0.633 M 2 (obs) 148.4 2C 6 H 5 OH —
(C 6 H 5 OH) 2
l-a
a/2 (X
(X
Total species = ( l - a ) + — = 1 - —
r_T r_l2
1.2x10"" ln [OH R = — — = 1.2 X LO"10 .-.
[OH~] = 1.1 x 10" [H+] = 10"14/1.1 x 10"5 = 0.91 x 10"9 p H = - l o g [H+] = log 0.91 x 10"9 = 9.04 37. (a): Let the number of atoms in a unit cell = x 60 x x Mass of x atoms i.e., one unit cell = — 6x10 Volume of the unit cell = (edge length) 3 = (400 x 10"12 x 100)3 = (400 x 10"10 cm) 3 = (4 x 10~8 cm) 3 = 64 x 10"24 cm 3 ~ , Mass of unit cell Density = 6.25 Volume of unit cell 60 x x 6.25 = 23 6 x 10 x 64 x 10 - 2 4 6.25x6x64x10" •=4 60 Since the unit cell contains 4 atoms, so it is facecentred cubic unit cell. 38. (c) 39. (a): In zinc blende structure, S2~ ions are arranged in ccp while Zri2+ ions occupy half of the tetrahedral sites. 40. (a) ZxM d x NA x V 41. (a): d = — — or Z = NaXV
M
a = 5 A = 5x10"® cm Molecular mass, M = 56 +16 = 72 g
a . or — = l - i 2 a = 2 ( l - f ) = 2 ( 1 - 0.633) = 0.734 = 73.4% i-
5
or
l-a/2
44. (d): Solubility in water = 0.2/200 mL Solubility in ether = 0.4/50 mL or 1.6/200 mL Sether _ 1 - 6 ^water
45. (a)
0.2
46. (c)
47. (c)
48. (a): Loss in weight or water bulb in which water vapour from solution flowed in and vapour from pure water escaped p°-p. Gain in weight of CaCl 2 tube being the weight of water vapour from pure water °c p° Loss in wt. of water bulb _ p° - p __ 0.087 _ n2 Gain in wt. of CaCl 2 tube p° 2.036 tij 0.08 _ 2 6 . 6 6 / M 2 => M 2 = 56.16 ^ 2.036 ~ 2 0 0 / 1 8 49. (b) 50. (a):When the same amount of electricity is passed, then the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalent. W l _ E l _ zi 108 _ x W 2 - £ 2 - z 2 ^ 31.5 ~ 1.05 => x = 3.6 g silver deposited 51. (c): a = ^ J i = = o.05 A° eq 200 CHEMISTRY TODAY | MAY '13
89
[ H + ] = c a = l x 0.05 = 0.05 p H = - log [H + ] = - log (5 x 10 2) = 2 log 10 - log5 = 2 - 0.6990 = 1.3. 52. (a): A°m (NH 4 OH) = ^ Ac
9
+
+
In II nd case, fraction of A reacted =
= 72 + 198 = 270 £2_1 cm 2
1
53. (b): Pt I H 2 (l atm) IHA 2 1 I HA, I H 2 ( l atm) I Pt At anode E H / H + = E° H / H + + 0.059 (pH) 2 . (v p H = -log[H + ])
We know, [H + ] = C a =
64. (b): As w e k n o w lesser is gold number, more is the protective power of lyophilic colloid.
p H 1 = ^ P K ( , 1 - i l o g C , pH2 = i p X f l 2 - i l o g C po
_ po cell ~ c H/H
+
, po H+/H
: 0.059
- \?Kai
0.059
( 5 - 3 ) = 0.059 V
54. (d): When equal volumes of A and B are mixed both the solutions get doubly diluted, hence individual contribution of A = — and that of B=
2
K2
4
= 0.75 For a first order reaction, time taken for the same fraction of reaction is independent of initial concentration. 61. (b) 62. (d) 63. (a): a and b are two Langmuir parameters. At very high pressure, the above isotherm acquires the limiting form. — = — (at very high pressure) m b
At cathode E H + / H = E °H+/H - 0.059 ( p H ) r
c
0.675 0.9
Total specific conductance of the mixture
65. (c): NQ N „
r0 = 1600 = 1N(I _1600 when t = 8 ~ 100 ' 2.303 1600 log X 100 or X = 0.3466; n o w r 0 = 1600; r 6 = ? ; t = 6 2.303, 1600 = 200 6 =: -logX ~ r6 66. (b): He atom has 2p + 2n. Hence Am = (2 x 1.0078 + 2 x 1.0087) - 4.0026 = 0.0304 a m u Energy released = 0.0304 x 931.5 = 28.3 MeV
Thus, R =
2x
Binding energy per nucleon =
KJ + K 2
56. (d)
55. (b) 57. (c): Vessel
III has
highest
concentration
of
reactants, c = — = — = 2 mol/L (for each reactant) v 0.1 Higher the concentration of reactants higher is the rate of reaction. 58. (c): [A] left = [B] formed = n x [A] decayed C0e~kt = nxC0(
e-kt=-
l-e~kt)
[B]formed = n x C 0 x ( l - -e~
n +1
kt
)
r, (, n \ nc0 = n x C0n x 1 = I n+l) n +1 En 59. (c): log A: = log A 2.303RT Slope =
-E„ 2.303J?
1 2.303
(given)
= 7 MeV (approximately) 67. (b): Increase of Z by 1 with no change in m means emission of one P-particle. Decrease in Z by 2 and m by 4 means emission of one a-particle. N o change in Z and m means emission of y-rays. 68. (d): N 0 = 1 gram atom = 6.023 x 10 23 atoms Atoms disintegrated = 4.52 x 10 23 Atoms left = 6.023 x 10 23 - 4.52 x 1023 = 1.5 x 10 23 , 0.693 0.693 , X= = hr hn io 2.303x10 6.023 x 10 23 t= log 23 0.693 1.5x10 = 33.23 log 4 = 20.1 hr 69. (b): 2 $ T h
-2P , 234 a
>
92/l
-«
s 230y. > 90 i n
70. (a): X = 2.5 x 10"7 min" 1 Mass of K-40 = 2.0 g
Ea = -2.303 R x slope = R = - 2 cal 60. (a): In I st case, fraction of A reacted = ^ = 1=0.75 0.8 4 90
CHEMISTRY TODAY | MAY' 13
N = — x 6.02 x l O 2 3 =3.01 xlO 2 2 40 Rate of decay = XN = 2.5 x 10"7 x 3.01 x 1022 = 7.5 x 1015 d p m
O O
Contd. from page no. 27
CuCl
*C6H-CI
CuBr
C 6 H 5 N 2 + CF
-+C 6 H 5 Br
CU2(CN)2
33. (c) : - O H + AC 2 0 34. ( a ) : ^ = 4 =
M„
Sandmeyer's reaction
*C6H5CN
^
MD
—>
(2 x)z =(a-x)(100-a-x)
2.1x10
(2x0.9) =(a - 0.9)(100 -a-
1/16
U
n N t t n N n : =i
0.9)
On usual calculations, a = 79 % of N 2 in air = 79% % of 0 2 in air = 100 - 79 = 21% 43. (d) 44. (d)
+2
:
[NO] [N2][02]
2.1 x 10 = 16
35. (a) : Oxidation state of Pt in [PtCl 4 ] 5d Pt 2+ 5d 8 i.e.
[Ptci4;
Kc =
- O A c + CH 3 COOH.
M.
M A_ _
N.2(g) 2NO,fe) a 100 - a 2x (a - x) (100 a x) 2x 1.8 Given : x = 0.9 100 100 Substituting the values in the relation Initially At equb.
32. (a) :
:
45. (a) : [Ni(NH 3 ) 6 ] :
dsf dsp2 - h y b r i d i s a t i o n lead to s q u a r e p l a n a r geometry.
[Zn(NH 3 ) 6 ] [Cr(NH 3 ) 6 ] 3+
36. ( b ) : Tl+ ions are more stable than Tl3+ due to inert pair effect which results in reluctance of s-electrons to unpair to show higher oxidation state.
[Co(NH 3 ) 6 ] 3+
sp3d2 (outer), octahedral, paramagnetic sp3d2 (outer), octahedral, diamagnetic d2sp3 (inner), octahedral, paramagnetic d2sp3 (inner), octahedral, diamagnetic
37. (d) 38. (b) : Nylon 6,6 is an example of fibres. W 39. ( b ) : r| = — ;r| = 1 that m e a n s W = q2 and q2 is q2 given as 3000 J.
46. ( b ) : 3KIO s + 3S0 2 3KI + 3SO, During this reaction, oxidation number of iodine changes from +5 in K I 0 3 to - 1 in KI. 47. ( d ) : C H 2 - C H C H 3 + PC15 O
40 ( b ) : Substance A is N H 4 N 0 3 which on boiling with N a O H gives N H 3 (gas B) NH4N03 + NaOH - > NaNOs + NH3 + H 2 0 (A) (B) N H 3 gives b r o w n p p t . w i t h Nessler's reagent (K 2 HgI 4 ). 2K 2 HgI 4 + N H 3 + 3KOH - > H 2 N H g O H g I + 7KI + 2H z O Substance A on heating gives N z O (gas C) which rekindles glowing splinter but is not converted into N 0 2 by air oxidation. NH4N03
(A) vie-dihalide
CH 3 CH 2 CHO + PC15 -
CH3-CH2CH (B) ^C1 ^em-dihalide
48. (a) : All the elements of group 2 except beryllium (Be) c o m b i n e w i t h h y d r o g e n to f o r m t h e i r hydrides of, M H , type. 49. (a) : The mechanism involves the following steps: 1
r-C1 X
N z O + 2H z O
NaNH 2 /liq.NH 3 _
H-NH2. Benzyne
(C) 41. (b) 42. (c) : Let the total n u m b e r of moles of N 2 and 0 2 initially = 100 and n u m b e r of moles of N 2 initially = a Further, let the n u m b e r of moles of N O present at equilibrium = x
• CH2-CH-CH3 I I CI CI
NH 7
N H ,z Thus for this reaction, presence of a - H is necessary which is not present in other three options. 50. (a)
OO CHEMISTRY TODAY | MAY '13 9 1
