CHEMICAL BONDING
CONTENTS 3.1
Cause and Modes of chemical combination
3.2
Electrovalent bond
3.3
Covalent bond
3.4
Co-ordinate covalent or dative bond
3.5
Polarity of covalent bond
3.6
Dipole moment
3.7
Change of ionic character to covalent character and Fajan’s rule
3.8
Quantum theory (Modern theory) of covalent bond and Overlapping
3.9
Hybridization
3.10
Resonance
3.11
Bond characteristics
3.12
VSEPR (Valence shell electron pair repulsion) theory
3.13
Molecular orbital theory
3.14
Hydrogen bonding
3.15
Types of bonding and Forces in solids Assignment (Basic and Advance Level) Answer Sheet of Assignment
M olecules are the smal smal lest lest par ti cles of matter which are capable of independent existence and whi ch show the char acteri acteri sti c propert propert ies of the th e substance. M olecul ol ecules es are ar e stable combinati ons of a group of atoms. M olecules are ar e stabl stabl e because because they have low er energy as compared to that of t he consti consti tuent atoms. The basic basic qu esti esti ons whi ch we search search ar e, what i s the nature of the force which keeps various atoms together together i n any mol ecul e? What i s the shape of the molecule ? What ar e the bond lengths between between atoms ? H ow much energy i s released released dur in g the f orm ation of a molecule ? What are the parameters which govern the properti es of an y molecule ?
CHEMICAL BONDING
Atoms of different elements excepting excepting noble gases donot have complete octet so they combine with other atoms to form chemical bond. The force which holds the atoms or ions together within the molecule is called a chemical bond and and the process of their combination is called Chemical Bonding. Bonding. Chemical bonding depends on the valency of atoms. Valency was termed as the number of chemical bonds formed by an atom in a molecule or number of electrons present in outermost shell i.e., i.e., valence electrons. Valence electrons actually involved in bond formation are called bonding electrons. The remaining valence electrons still available for bond formation are referred to as non-bonding electrons.
p+n
p+ n
p +n
Cl (2, 8, 7)
Na (2, 8, 1)
+
–
p+ n
Bond formation takes place
3.1 Cause and Modes of chemical combination. combination .
Chemical combination takes place due to following reasons. (1) Chemical bonding takes place to acquire a state of minimum energy and maximum stability s tability.. (2) By (2) By formation of chemical bond, atoms convert into molecule to acquire stable s table configuration of the nearest noble gas. Modes : Chemical bonding can occur in the following manner. Transfer of electrons from one Ionic bond atom to another Mutual sharing of electrons between the atoms
Covalent bond
Mutual sharing of electrons provided entirely by one of the
Co-ordination bond
3.2 Electrovalent bond. bond.
When a bond is formed by complete transfer of electrons from one atom to another so as to complete their outermost orbits by acquiring 8 electrons (i.e., (i.e., octet) or 2 electrons (i.e., ( i.e., duplet) in case of hydrogen, helium etc. and hence acquire the stable nearest noble gas configuration, the bond formed is called ionic bond, electrovalent bond or polar bond. Compounds containing ionic bond are called ionic, electrovalent or polar compounds. Example :
Na
Cl
Na
Cl
or
Na Cl
Some other examples are: MgCl are: MgCl 2, CaCl 2, MgO, MgO, Na2 S , CaH 2, AlF 3, NaH , KH , K 2 O , KI , RbCl , NaBr, CaH 2 etc. (1) Conditions for formation of electrovalent bond (i) Number of valency electrons : The atom which changes into cation (+ ive ion) should possess 1, 2 or 3 valency electrons. The other atom which changes into anion ( – ive ion) should possess 5, 6 or 7 electrons in the valency shell. (ii) Electronegativity (ii) Electronegativity difference : A high difference of electronegativity electronegativity (about 2) of the two atoms is necessary for the formation of an electrovalent bond. Electrovalent Electrovalent bond is not possible between similar atoms. atoms.
CHEMICAL BONDING
(iii) Small decrease in energy : There must be overall decrease in energy i.e., energy must be released. For this an atom should have low value of Ionisation of Ionisation potential and and the other atom should have high value of electron affinity. affinity. (iv) Latt (iv) Lattice ice energy energy : Higher Higher the lattice lattice energy, energy, greater greater will be the ease of forming forming an ionic compoun compound. d. The amount of energy released when free ions combine together to form one mole of a crystal is called lattice energy (U). Magnitude of lattice energy
Charge of ion size of ion
AB( s) U A ( g ) B ( g )
Determination of lattice lattice energy (Born (Born Haber cycle) When a chemical bond is formed between two atoms (or ions), the potential energy of the system constituting the two atoms or ions decreases. If there is no fall in potential energy of the system, no bonding is possible, the energy changes involved in the formation fo rmation of ionic compounds from their constituent elements can be studied with the help help of a thermochemical cycle called Born Haber cycle. Example : Example : The formation of 1 mole of NaCl from sodium and chlorine involves following steps : Step I : Conversion of metallic sodium into gaseous sodium atoms: Na( s) S Na( g ) , where S = 1 mole
sublimation energy i.e., i.e., the energy required for the conversion of one mole of metallic sodium into gaseous sodium atoms. Step II : Dissociation of chlorine molecules into chlorine atoms : Cl 2 ( g ) D 2Cl (g ) , where D = Dissociation energy of Cl 2 so the energy required for the formation of one mole of gaseous chlorine atoms D / 2 . Step III: Conversion of gaseous sodium atoms into sodium ions : Na( g ) IE Na ( g ) e , where 1 mole
IE = Ionisation energy of sodium. Step IV: Conversion of gaseous chlorine atoms into chloride ions : Cl ( g ) e Cl ( g ) EA , where EA =
Na( s) + S
1 2
Cl 2 (g )
H f
1/2 D
Cl (g)
Na( g )
1 mole
Electron affinity of chlorine. Step V : Combination of gaseous sodium and chloride ions to form solid sodium so dium chloride crystal. Na ( g ) Cl ( g ) NaCl NaCl ( s) U , where U lattice energy
+
– e
+ IE
Na ( g )
– EA
–
+
+e
–
NaCl – U
Cl ( g )
(Born Haber
1 mole
of NaCl 1
The overall change may be represented as : Na( s) Cl 2 ( g ) NaCl NaCl ( s), H f , where H f is the heat of 2
formation for 1 mole of NaCl NaCl ( s) . According to Hess's law of constant heat summation, heat of formation of one mole of NaCl should be same whether it takes place directly in one step or through a number of steps. Thus, 1
H f S D IE EA U 2
(2) Types of ions The following types of ions are encountered :
CHEMICAL BONDING
(i) Ions with inert gas configuration configuration : The atoms of the representative elements elements of group I, II and III by complete loss of their valency electrons and the elements of group V, VI, and VII by gaining 3,2 and 1 electrons respectively form ions either with ns 2 configuration or ns 2 p 6 configuration. (a) Ions with 1 s 2 (He) configuration : H , Li , Be 2 etc. The formation of Li and Be 2 is difficult due to their small size and high ionisation potential. ionisation potential. (b) Ions with ns 2 p 6 configuration : More than three electrons are hardly lost or gained in the ion formation Cations : Na , Ca 2 , Al 3 etc. Anions : Cl , O 2 , N 3 , etc. (ii) Ions with pseudo inert gas configuration : The Zn 2 ; ion is formed when zinc atom loses its outer 4s electrons. The outer shell configuration of Zn 2 ions is 3 s 2 3 p 6 3d 10 . The ns 2 np 6 nd 10 outer shell configuration is often called pseudo noble gas configuration which is considered as stable one. Examples: Zn 2 , Cd 2 , Hg 2 , Cu Ag , Au , Ga 3 etc (iii) Exceptional configurations : Many d- d- and f block elements elements produce ions ions with with configurations configurations different than the above two. Ions like Fe 3 , Mn 2 , etc., attain a stable configuration half filled d orbitals Fe 3
3 s 2 3 p 6 3d 5 ;
Mn Mn 2
3 s 2 3 p 6 3d 5
Examples of other configurations are many. Ti 2 Cr 2
(3 s 2 3 p 6 3d 2 ) (3 s 2 3 p 6 3d 4 )
; V 2 ; Fe 2
(3 s 2 3 p 6 3d 3 ) (3 s 2 3 p 6 3d 6 )
However, such ions are comparatively less stable (iv) Ions with ns 2 configuration : Heavier members of groups III, IV and V lose plose p-electrons electrons only to form ions with ns 2 configuration. Tl , Sn 2 , Pb 2 , Bi 3 are the examples of this type. These are stable ions. (v) Polyatomic ions : The ions which are composed of more than one atom are called polyatomic ions. These ions move as such in chemical reactions. Some common polyatomic ions are NO 3 (Nitrate) NH 4 (Ammonium); PO 43 (phosphate);
SO 42 (Sulphate)
CO 32 (Carbonate) ;
SO 32 (Sulphite), etc.
(vi) Polyhalide Polyhalide ions ions : Halogens or interhalogens combine with halide ions to form polyhalide ions. I 3 , ICl 4 , ICl 2 etc. Fluorine due to highest electronegativity and absence of d-oribitals does not form polyhalide ions. The atoms within the polyatomic ions are held to each other by covalent bonds. The electro valencies of an ion (any type) is equal to the number of charges present on it. (3) Method of writing formula of an ionic compound In order to write the formula of an ionic compound which is made up of two ions (simple or polyatomic) having electrovalencies electrovalencies x and y respectively, the following points are followed :
CHEMICAL BONDING
(i)
Write the symbols of the ions side by side in such a way that positive ion is at the left and negative ion at the right as AB. (ii) Write their electrovalencies in figures on the top of each symbol as A x B y (iii) Divide their valencies by H.C.F (iv) Now apply criss cross rule as
x A
y , i.e., formula A y B x B
Examples : Name of compound
Calcium chloride Potassium phosphate
Exchange of valencies 2
1
Ca
Cl
1
2
CaCl 2 K 3 PO4
3
K
Calcium oxide
Formula
PO4 2
CaO or
1
1
Ca
O
CaO
Name of compound
Aluminium oxide Magnesium nitride Ammonium sulphate
Exchange of valencies 3
2
Al
O
2
3
Mg
N
Formula
Al 2 O 3
Mg 3 N 2
1
2
NH 4
SO 4
( NH 4 ) 2 SO 4
(4) Difference between atoms and ions The following are the points of difference between atoms and ions. Atoms
Ions
1. Atoms are perfectly neutral in nature, i.e., number of protons equal to number of electrons. Na (protons 11, electrons 11), Cl (Protons – 17, electrons –17)
Ions are charged particles, cations are positively charged, i.e., number of protons more than the number of electrons. Anions are negatively charged, i.e., number of protons less than the number of electrons. Na+ (protons 11, electrons 10), Cl – (protons 17, electrons 18)
2.
Except noble gases, atoms have less than 8 electrons in the Ions have generally 8 electrons in the outermost orbit, i.e., outermost orbit ns2np6 configuration. Na 2,8,1; Ca 2,8,8,2 Na+ 2,8; Cl – 2,8,8 Cl 2,8,7; S 2,8,6 Ca2+ 2,8,8
3.
Chemical activity is due to loss or gain or sharing of electrons as to acquire noble gas configuration
The chemical activity is due to the charge on the ion. Oppositely charged ions are held together by electrostatic forces
(5) Characteristics of ionic compounds (i) Physical state : Electrovalent compounds are generally crystalline is nature. The constituent ions are arranged in a regular way in their lattice. These are hard due to strong forces of attraction between oppositely charged ions which keep them in their fixed positions. (ii) Melting and boiling points : Ionic compounds possess high melting and boiling points. This is because ions are tightly held together by strong electrostatic forces of attraction and hence a huge amount of energy is required to break the crystal lattice. For example order of melting and boiling points in halides of sodium and oxides of IInd group elements is as, NaF NaCl NaBr NaI , MgO CaO BaO
(iii) Hard and brittle : Electrovalent compounds are har in nature. The hardness is due to strong forces of attraction between oppositely charged ion which keep them in their alloted positions. The brittleness of the crystals is due to movement of a layer of a crystal on the other layer by
CHEMICAL BONDING
application of external force when like ions come infront of each other. The forces of repulsion come into play. The breaking of crystal occurs on account of these forces or repulsion. (iv) Electrical conductivity : Electrovalent solids donot conduct electricity. This is because the ions remain intact occupying fixed positions in the crystal lattice. When ionic compounds are melted or dissolved in a polar solvent, the ions become free to move. They are attracted towards the respective electrode and act as current carriers. Thus, electrovalent compounds in the molten state or in solution conduct electricity. (v) Solubility : Electrovalent compounds are fairly soluble in polar solvents and insoluble in nonpolar solvents. The polar solvents have high values of dielectric constants. Water is one of the best polar solvents as it has a high value of dielectric constant. The dielectric constant of a solvent is defined as its capacity to weaken the force of attraction between the electrical charges immersed in that solvent. In solvent like water, the electrostate force of attraction between the ions decreases. As a result there ions get separated and finally solvated . The values of dielectric constants of some of the compounds are given as : Compound Dielectric constant
Water Methyl AIc Ethyl AIc. Acetone Ether 81 35 27 21 4.1 Capacity to dissolve electrovalent compounds decreases
Lattice energy and solvation energy also explains the solubility of electrovalent compounds. These compounds dissolve in such a solvent of which the value of solvation energy is higher than the lattice energy of the compound . The value of solvation energy depends on the relative size of the ions. Smaller the ion more of solvation, hence higher the solvation energy. Note : Some ionic compounds e.g., BaSO 4 , PbSO 4 , AgCl , AgBr , AgI , Ag 2 CrO 4 etc. are sparingly soluble in water because in all such cases higher values of lattice energy predominates over solvation energy. (vi) Space isomerism :The electrovalent bonds are non-rigid and non-directional. Thus these compound do not show space isomerism e.g. geometrical or optical isomerism. (vii) Ionic reactions : Electrovalent compounds furnish ions in solution. The chemical reaction of these compounds are ionic reactions, which are fast. Ionic bonds are more common in inorganic compounds.
Ag Cl K NO 3 K Cl Ag NO 3 (Precipitat e )
(viii) Isomorphism : Electrovalent compounds show isomorphism. Compound having same electronic structures are isomorphous to each other.
(ix) Cooling curve : Cooling curve of an ionic compound is not smooth, it has two break points corresponding to time of solidification. A liquid
M.PtTemp
B
C D Time
CHEMICAL BONDING
(x) Electrovalency and Variable electrovalency : The capacity of an element to form electro-valent or ionic bond is called its electro-valency or the number of electrons lost or gained by the atom to form ionic compound is known as its electro-valency. Certain metallic element lose different number of electrons under different conditions, thereby showing variable electrovalency. The following are the reasons: (a) Unstability of core : The residue configuration left after the loss of valency electrons is called kernel or core. In the case of the atoms of transition elements, ions formed after the loss of valency electrons do not possess a stable core as the configuration of outermost shell is not ns 2 np 6 but ns 2 np 6 d 1 to 10 . The outer shell lose one or more electrons giving rise to metal ions of higher valencies. Example : Fe 2 3 s 2 3 p 6 3d 6 ,4 s 0 (not stable) Fe 3 3 s 2 3 p 6 3d 5 ,4 s 0
(stable)
(b) Inert pair effect : Some of heavier representative elements of third, fourth and fifth groups having configuration of outermost shell ns 2 np 1 , ns 2 np 2 and ns 2 np 3 show valencies with a difference of 2, i.e., (1 : 3) (2 : 4) (3 : 5) respectively. In the case of lower valencies, only the electrons present in p–subshell are lost and ns2 electrons remain intact. The reluctance of s-electron pair to take part in bond formation is known as the inert pair effect. 3.3 Covalent bond.
Covalent bond was first proposed by Lewis in 1916. The bond formed between the two atoms by mutual sharing of electrons so as to complete their octets or duplets (in case of elements having only one shell) is called covalent bond or covalent linkage. A covalent bond between two similar atoms is non-polar covalent bond while it is polar between two different atom having different electronegativities. Covalent bond may be single, double or a triple bond. Example : Formation of chlorine molecule : chlorine atom has seven electrons in the valency shell. In the formation of chlorine molecule, each chlorine atom contributes one electron and the pair of electrons is shared between two atoms. both the atoms acquire stable configuration of argon.
**
* Cl *
Cl
** ( 2, 8 , 7 )
( 2, 8 , 7 )
Cl
Cl or Cl Cl
*
(2 , 8 , 8 )
( 2, 8 , 8 )
Formation of HCl molecule : Both hydrogen and chlorine contribute one electron each and then the pair of electrons is equally shared. Hydrogen acquires the configuration of helium and chlorine acquires the configuration of argon.
* *
H (1)
Cl
* * ( 2, 8 , 7 )
* *
H (2)
*
Cl
( 2, 8 , 8 )
or H Cl
CHEMICAL BONDING
Formation of water molecule : Oxygen atom has 6 valency electrons. It can achieve configuration of neon by sharing two electrons, one with each hydrogen atom. H (1)
**
O H
** (2, 8 , 7 )
H
O
*
H
or H O H
(1)
( 2)
( 2, 8 , 8 )
(2)
Formation of O2 molecule : Each oxygen atom contributes two electrons and two pairs of electrons are the shared equally. Both the atoms acquire configuration of neon. ** * * ** ( 2, 6 )
O O
O
( 2, 6 )
( 2, 8 )
O
or O O
( 2, 8 )
Formation of N 2 molecule : Nitrogen atom has five valency electrons. Both nitrogen atoms achieve configuration of neon by sharing 3 pairs of electrons, i.e., each atom contributes 3 electrons. * *
N *** N
(2, 5 )
(2, 5 )
* *
N **
( 2, 8 )
N or N N (2, 8 )
Some other examples are : H 2 S , NH 3, HCN , PCl 3, PH 3, C 2 H 2 , H 2, C 2 H 4, SnCl 4 , FeCl 3 , BH 3 , graphite, BeCl 2 etc. (1) Conditions for formation of covalent bonds (i) Number of valency electrons : The combining atoms should be short by 1, 2 or 3 electrons in the valency shell in comparison to stable noble gas configuration. (ii) Electronegativity difference : Electronegativity difference between the two atoms should be zero or very small. (iii) Small decrease in energy : The approach of the atoms towards one another should be accompanied by decrease of energy. (2) Characteristics of covalent compounds (i) Physical state : These exist as gases or liquids under the normal conditions of temperature and pressure. This is because very weak forces of attraction exist between discrete molecules. Some covalent compounds exist as soft solids. (ii) Melting and boiling points : Diamond, Carborandum ( SiC ), Silica ( SiO2), AlN etc. have giant three dimensional network structures; therefore have exceptionally high melting points otherwise these compounds have relatively low melting and boiling points. This is due to weak forces of attraction between the molecules. (iii) Electrical conductivity : In general covalent substances are bad conductor of electricity. Polar covalent compounds like HCl in solution conduct electricity. Graphite can conduct electricity in solid state since electrons can pass from one layer to the other. (iv) Solubility : These compounds are generally insoluble in polar solvent like water but soluble in non-polar solvents like benzene etc. some covalent compounds like alcohol, dissolve in water due to hydrogen bonding. (v) Isomerism : The covalent bond is rigid and directional. These compounds, thus show isomerism (structural and space). (vi) Molecular reactions : Covalent substances show molecular reactions. The reaction rates are usually low because it involves two steps (i) breaking of covalent bonds of the reactants and (ii) establishing of new bonds while the ionic reactions involved only regrouping of ions. (vii) Covalency and Variable covalency : The number of electrons contributed by an atom of the element for sharing with other atoms is called covalency of the element. The variable covalency
CHEMICAL BONDING
of an element is equal to the total number of unpaired electrons in s , p and d-orbitals of its valency shell . Covalency = 8 – [Number of the group to which element belongs] Examples : Nitrogen 7 N = Covalency of N = 3 2 p
2s
The element such as P , S , Cl , Br, I have vacant d -orbitals in their valency shell. These elements show variable covalency by increasing the number of unpaired electrons under excited conditions. The electrons from paired orbitals get excited to vacant d -orbitals of the same shell. Promotion energy : The energy required for excitation of electrons. Promotion rule : Excitation of electrons in the same shell Phosphorus : Ground state Covalency 3 3state p 3s : Excited
Phosphorus
Covalency-5
3 p PCl 3 is more stable due to inert3spair effect.
Sulphur
:
Ground state
Sulphur : 1st excited state
Excited state
3s
3 p
3s
3 p
3d
Covalency-2 (as in SF 2)
3d
Covalency-4 (as in SF 4) 3d
2nd excited state
Covalency-6 (as in SF 6)
3 p
3s
3d So variable valency of S is 2, 4 and 6. Iodine can have maximum 7 unpaired electrons in its orbitals. It's variable valencies are 1, 3, 5 and 7. Four elements, H, N, O and F do not possess d-orbitals in their valency shell. Thus, such an excitation is not possible and variable valency is not shown by these elements. This is reason that NCl 3 exists while NCl 5 does not. (3) The Lewis theory : The Lewis theory gave the first explanation of a covalent bond in terms of electrons that was generally accepted. The tendency of atoms to achieve eight electrons in their outermost shell is known as lewis octet rule. Octet rule is the basis of electronic theory of valency. It is suggested that valency electrons themselves are responsible for chemical combination. The valency electrons in atoms are shown in terms of Lewis dot formulae. To write Lewis formulae for an element, we write down its symbol surrounded by a number of dots or crosses equal to the number of valency electrons. Lewis dot formulae are also used to represent atoms covalently bonded in a molecule. Paired and unpaired valency electrons are also indicated. Lewis symbols for the representative elements are given in the following table :
Group
1 IA
2 IIA
13 IIIA
X
X
CO is not an exception to octet rule
: C O :
Lewis symbol X
14
15
16
17
IVA
VA
VIA
X
X
X
: or : C O ..
VIIA X
CHEMICAL BONDING
(4) Failure of octet rule : There are several stable molecules known in which the octet rule is violated i.e., atoms in these molecules have number of electrons in the valency shell either short of oct et or more than octet. In BF 3 molecules, boron atom forms three single covalent bonds with three fluorine atoms, i.e., it attains six electrons in the outer shell.
F
*
*
B *
+
3
F
|
F
F
B
*
*
F B F
|
F
PCl 5 molecule : Phosphorus atom have five electrons in valency shell. It forms five single covalent bonds with five chlorine atoms utilising all the valency electrons and thereby attains 10 electrons in the outer shell.
Cl
Cl
* *
*
P * *
+
5 Cl
Cl
P
Cl
Cl
Cl
Cl
Cl
P
Cl
Cl
Cl
(i) Sugden’s concept of singlet linkage explains the stability of such molecules. In PCl 5 , three chlorine atoms are linked by normal covalent bonds and two chlorine atoms are linked by singlet linkages, thus, phosphorus achieves 8 electrons in the outermost shell.
Cl
Cl
Cl
Cl
P
Cl
Cl
Cl
P
Cl
Cl
Cl
This structure indicates that the nature of two chlorine atoms is different than the other three as singlet linkage is weaker than normal covalent bond. The above observation is confirmed by the fact that on heating, PCl 5 dissociates into PCl 3 and Cl 2 . PCl 5
⇌
PCl 3 Cl 2
Similarly, in SF 6 four singlet linkages are present while in IF 7 , six singlet linkages are present.
F
F
F
F
S
F
F
S
F
F
F
F
F
F
(ii) Sidgwick’s concept of maximum covalency This rule states that the covalency of an element may exceed four and octet can be exceeded. The maximum covalency of an element actually depends on the period of periodic table to which it belongs. This rule explains the formation of PCl 5 and SF 6.
CHEMICAL BONDING
This also explains why nitrogen does not form NF 5 or NCl 5 because nitrogen belongs to second period and the maximum covalency of nitrogen is three. (iii) Odd electron bond : In 1916 Luder postulated that there are number of stable molecules in which double bonds are formed by sharing of an odd number of electrons, i.e., one, three, five, etc., between the two bonded atoms. the bonds of this type are called odd electron bonds.
**
The normal valence bond structure of oxygen molecule, O O ** , fails to account for the paramagnetic nature of oxygen. Thus, structure involving three electrons bond has been suggested by Pauling. The following structure : O O . Explains the paramagnetic nature and high dissociation energy of oxygen molecule. N
Some other examples are : N
O
Nitric oxide
(NO)
O
O
Nitrogen dioxide ( NO 2 )
O
O
–
Superoxide ion (O 2 )
The number of singlet bonds = Total number of bonds – Number of electrons required to complete the octet. Properties of Odd Electron bond (i) The odd electron bonds are generally established either between two like atoms or between different atoms which have not more than 0.5 difference in their electronegativities. (ii) Odd electron bonds are approximately half as strong as a normal covalent bond. (iii) Molecules containing odd electrons are extremely reactive and have the tendency to dimerise. (iv) Bond length of one electron bond is greater than that of a normal covalent bond. Whereas the bond length of a three electron bond is intermediate between those of a double and a triple bond. (v) One electron bond is a resonance hybrid of the two structures i.e., A B A B Similarly, a three electron bond is a resonance hybrid of the two structures i.e., A B A B (5) Construction of structures for molecules and poly atomic ions : The following method is applicable to species in which the octet rule is not violated. (i) Determine the total number of valence electrons in all the atoms present, including the net charge on the species (n1). (ii) Determine n2 = [2 × (number of H atoms) + 8 × (number of other atoms)]. (iii)Determine the number of bonding electrons, n3, which equals n2 – n1. No. of bonds equals n3/2. (iv) Determine the number of non-bonding electrons, n4, which equals n1 – n3. No. of lone pairs equals n4/2. (v) Knowing the central atom (you’ll need to know some chemistry here, math will not help!), arrange and distribute other atoms and n3/2 bonds. Then complete octets using n4/2 lone pairs. (vi) Determine the ‘formal charge’ on each atom. (vii) Formal Charge = [valence electrons in atom) – (no. of bonds) – (no. of unshared electrons)] (viii) Other aspects like resonance etc. can now be incorporated. Illustrative examples :
CHEMICAL BONDING
(i) CO32 ; n1 4 (6 3) 2 24 [2 added for net charge] n 2 = (2 × 0) + (8 × 4) = 32 (no. H atom, 4 other atoms (1’C ’ and 3 ‘O’) n 3 = 32 – 24 = 8, hence 8/2 = 4 bonds n 4 = 24 – 8 = 16, hence 8 lone pairs.
Since carbon is the central atom, 3 oxygen atoms are to be arranged around it, thus, O |
O C O , but total bonds are equal to 4. . .
O
.O :
|
.
|
. .
C O : Hence, we get O C O . Now, arrange lone pairs to complete octet : O . .
(ii) CO 2 ; n1 = 4 + (6 × 2 ) = 16 n2 = (2 × 0) + (8 × 3) = 24 n3 = 24 – 16 = 8, hence 4 bonds n4 = 16 – 8 = 8, hence 4 lone-pairs Since C is the central atom, the two oxygen atoms are around to be arranged it thus the structure would be; O – C – O, but total no. of bonds = 4 ..
..
Thus, O = C = O. After arrangement of lone pairs to complete octets, we get, : O C O : and thus ..
..
final structure is : O C O : 3.4 Co-ordinate covalent or Dative bond.
This is a special type of covalent bond where the shared pair of electrons are contributed by one species only but shared by both. The atom which contributes the electrons is called the donor while the other which only shares the electron pair is known as acceptor. This bond is usually represented by an arrow ( ) pointing from donor to the acceptor atom. For example, in BF 3 molecule, boron is short of two electrons. So to complete its octet, it shares the lone pair of nitrogen in ammonia forming a dative bond as shown in figure
H
F
H * N ** B
H
F
H F
F
H * N B F
H
H F
F
|
|
|
|
H N B F H F
Formation of a co - ordinate bond between NH 3 and BF 3
Examples : CO, N 2O, H 2O2, N 2O3, N 2O4, N 2O5, HNO3, NO 3 , SO2, SO3, H 2 SO4, SO 42 , SO 22 , H 3 PO4 , H 4 P 2 O7 , H 3 PO 3 , Al 2 Cl 6 (Anhydrous) , O 3 , SO 2 Cl 2 , SOCl 2 , HIO3 , HClO 4 , HClO 3 , CH 3 NC , N 2 H 5 , CH 3 NO 2 , NH 4 , [Cu(NH 3 )4 ]2 etc.
(1) Characteristics of co-ordinate covalent bond
CHEMICAL BONDING
(i) Melting and boiling points : Their melting and boiling points are higher than purely covalent compounds and lower than purely ionic compounds. (ii) Solubility : These are sparingly soluble in polar solvent like water but readily soluble in non-polar solvents. (iii) Conductivity : Like covalent compounds, these are also bad conductors of electricity. Their solutions or fused masses do not allow the passage to electricity. (iv) Directional character of bond : The bond is rigid and directional. Thus, coordinate compounds show isomerism. Table: Electron dot formulae and Bond formulae or Dash formulae of some compounds No.
Molecular formula
Electron dot formula
Dash formula or Bond formula
1.
Sodium chloride
NaCl
2.
Magnesium chloride
MgCl 2
– Mg++ Cl – Cl
Cl Mg Cl
3.
Calcium chloride
CaCl 2
– ++ Cl – Cl Ca
Cl Ca Cl
4.
Magnesium oxide
MgO
5.
Sodium sulphide
Na2 S
– Na+ Cl
–– Mg++ O
Calcium hydride
CaH 2
Na S Na
H –Ca++ H –
H Ca H
– 3+ –
F Al 3 F F
H Cl
F Al F – F
7.
Aluminium flouride
AlF 3
8.
Hydrogen chloride
HCl
H Cl H O H
9.
Water
H 2 O
10.
Hydrogen sulphide
H 2 S
11.
12.
13.
14.
15.
Ammonia Hydrogen cyanide Methane
Ethane
Ethene
NH 3 HCN CH 4
C 2 H 6
C 2 H 4
Mg O
–– Na+ S Na+
6.
Na Cl
H S H
H H N H H C N
H O H H S H
H |
H N |
H H C N
H
H H C H H
H C H
H H H C C H H H
H C C H
H H C C H H
| |
H H H |
|
|
|
H H H H |
|
|
|
C C H H
CHEMICAL BONDING
16.
Ethyne
C 2 H 2
17.
Phosphene
PH 3
18.
19.
Phosphorous trichloride Sodium hydroxide
H C C H
H C C H
H P H H
PCl 3 NaOH
H P H |
H
Cl P Cl Cl
Na+ * O H
Cl P Cl |
Cl
–
Na O H
–
K C N
20.
Potassium cyanide
K + *C N
KCN
21.
Calcium carbonate
CaCO 3
22.
Carbon mono oxide
CO
23.
Nitrous oxide
N 2 O
*
Ca++ O C O O
H 2 O 2
25.
Dinitrogen trioxide
N 2 O 3
26.
Hydronium ion
27.
Nitrogen dioxide
N 2 O 4
28.
Nitrogen pentaoxide
N 2 O5
HNO 3
O N O N O O O
30.
Nitrate ion
NO 3
31.
Nitrous acid
HNO 2
32.
Nitrite ion
NO 2
33.
Sulphur dioxide
SO 2
Sulphur trioxide
SO 3
– * N
O O O
– *
H O N O
O N O N O
O
O
H O N O O
H * O N O O
O N N O O H O N N O O O
|
H
|
O N N O O O
H O O H or H O O
H O H
+ H O H H
34.
N N O
Nitric acid
H O O H or H O O H N O O N O
2
Ca O C O || O
C O
N N O
Hydrogen peroxide
29.
2–
C O
24.
H 3 O
*
O N O
O N O O
H O N O
O N O
O S O
O S O
O S O O
O S O O
CHEMICAL BONDING
35.
Peroxysulphuric acid (Marshal acid)
H 2 S 2 O 8
O O O O S O * H H * O S O O
36.
Hypochlorous acid
HOCl
Cl H * O
37.
Chlorous acid
HClO 2
Cl H * O O
38.
39.
40.
Perchloric anhydride
Cl 2 O7
H 2 SO 4
Sulphuric acid
Sulphate ion
SO 42
Sulphurus acid
H 2 SO 3
42.
43.
Sulphite ion
SO
Phosphoric acid
2 3
H 3 PO4
H O Cl
O O O Cl O Cl O O O
* *
O H O S O H O – – * *
O O S O O
41.
O O H O S O O S O H O O
H * O S O H * O
– –
O S O O
* *
O H * O P O * H O * H
H O Cl O O O O
Cl O Cl
O H O S O H O O
O S O O
H O S O H
O
O S O O
O
H O P O H |
O |
H O
44.
45.
46.
Pyrophosphoric acid
Persulphuric (caro’s acid)
Thiosulphurous acid
H 4 P 2 O7
acid H SO 2 5
H 2 S 2 O 3
O O H * O P O P O * H O O * * H H *
O H O S O O * H O
* *
S H O S O H O
O O O
O
H O P O P O H |
|
O
O
H
H
|
|
O H O S O O H O S
H O S O H O
CHEMICAL BONDING
47.
48.
Phosphorous acid
Aluminium chloride
H 3 PO3
Al 2 Cl 6 (Anhydrous)
H H * O P O * H O
Cl Cl Cl Al Al Cl Cl Cl
H |
H O P O H O
Cl Al Cl
Cl Cl
Al
Cl Cl
O
49.
Ozone
O
O3
O
Sulphuryl chloride
SO 2 Cl 2
52.
53.
54.
55.
Sulphonyl chloride
Iodic acid
**
Cl
Perchloric acid
Hyposulphurous acid
Pyrosulphurus acid
HClO 4
Chloric acid
H O I O O
*
O H O Cl O O
* *
H 2 S 2 O 4
S H O S O O H O
H 2 S 2 O 5
* *
HClO 3
|
O S Cl
S H O O S O O H O
56.
|
O S Cl Cl
O H * O I O
HIO3
Cl O S O
* * * * * *
SOCl 2
O
* * *
51.
O
**
Cl ** O S O * Cl * * * *
50.
O
H * O Cl O O
|
Cl
O
H O Cl O O S
H O S O O H O S
H O O S O O H O H O Cl O O
3.5 Polarity of covalent bond.
A covalent bond in which electrons are shared equally between bonded atoms, is called non polar covalent bond while a covalent bond, in which electrons are shared unequally and the bonded atoms acquire a partial positive and negative charge, is called a polar covalent bond. The atom having higher electronegativity draws the bonded electron pair more towards itself resulting in partial charge separation. This is the reason that HCl molecule in vapour state contains polar covalent bond
Polar covalent bond is indicated by notation : H – Cl
CHEMICAL BONDING
(1) Bond polarity in terms of ionic character : The polar covalent bond, has partial ionic character.
Which usually increases with increasing difference in the electronegativity ( EN ) between bonded atom H – F H – Cl H – Br H – I EN 2.1 4.0 2.1 3.0 2.1 2.8 2.1 2.5 Difference in EN 1.9 0.9 0.7 0.4 Ionic character decreases as the difference in electronegativity decreases
(2) Percentage ionic character : Hennay and Smith gave the following equation for calculating the percentage of ionic character in A– B bond on the basis of the values of electronegativity of the atoms A and B. Percentage of ionic character = [16 ( A B ) 3.5 ( A B )2 ] .
Whereas ( x A x B ) is the electronegativity difference. This equation gives approximate calculation of percentage of ionic character, e.g., 50% ionic character corresponds to ( x A x B) equals to 1.7. 3.6 Dipole moment. “The product of magnitude of negative or positive charge (q) and the distance (d) between the centres of positive and negative charges is called dipole moment ”.
It is usually denoted by . Mathematically, it can be expressed as follows : = Electric charge bond length As q is in the order of 10 –10 esu and d is in the order of 10 –8 cm, is in the order of 10 –18 esu cm. Dipole moment is measured in “ Debye” ( D) unit. 1 D 10 18 esu cm = 3.33 10 30 coulomb metre. Generally, as electronegativity difference increases in diatomic molecules, the value of dipole moment increases. Greater the value of dipole moment of a molecule, greater the polarity of the bond between the atoms. Dipole moment is indicated by an arrow having a symbol ( ) pointing towards the negative end. Dipole moment has both magnitude and direction and therefore it is a vector quantity. Symmetrical polyatomic molecules are not polar so they do not have any value of dipole moment . H
F
O
C
O
C
B F
H
F
H H
= 0 due to symmetry , PbCl 4 , SF 6 , SO 3 , C 6 H 6 , naphthalene
Some other examples are – CCl 4,CS 2 and all homonuclear molecules (H 2, O2, Cl 2 etc) Note : Amongst isomeric dihalobenzenes, the dipole moment decreases in the order : o > m > p. A molecule of the type MX 4, MX 3 has zero dipole moment, because the -bonding orbitals used by M ( Z < 21) must be sp3, sp2 hybridization respectively (e.g. CH 4, CCl 4, SiF 4 , SnCl 4, BF 3 , AlCl 3 etc.) sp 2
3
Unsymmetrical polyatomic molecules always have spnet value of dipole moment, thus such molecules are polar in nature. H 2O, CH 3Cl, NH 3, etc are polar molecules as they have some positive values of dipole moments. Cl O H
H Water = 1.84 D
S O
O Sulphur dioxide
N
C H
H H Methyl chloride
H
H H Ammonia = 1.46 D
CHEMICAL BONDING
0 due to unsymmetry
Some other examples are – CH 3Cl , CH 2Cl 2, CHCl 3, SnCl 2, ICl , C 6 H 5CH 3 , H 2O2, O3, Freon etc. Applications of dipole moment
(i) In determining the symmetry (or shape) of the molecules : Dipole moment is an important factor in determining the shape of molecules containing three or more atoms. For instance, if any molecule possesses two or more polar bonds, it will not be symmetrical if it possesses some molecular dipole moment as in case of water ( 1.84 D) and ammonia ( 1.49 D). But if a molecule contains a number of similar atoms linked to the central atom and the overall dipole moment of the molecule is found to be zero, this will imply that the molecule is symmetrical, e.g., in case of BF 3 , CH 4 , CCl 4 etc., (ii) In determining percentage ionic character : Every ionic compound having some percentage of covalent character according to Fajan’s rule. The percentage of ionic character in a compound
having some covalent character can be calculated by the following equation. The percent ionic character
Observed dipole moment Calculated dipole moment assuming 100% ionic bond
100
(iii) In determining the polarity of bonds as bond moment : As q d , obviously, greater is the magnitude of dipole moment, higher will be the polarity of the bond . The contribution of individual bond in the dipole moment of a polyatomic molecule is termed as bond moment. The measured dipole moment of water molecule is 1.84 D. This dipole moment is the vectorial sum of the individual bond moments of two O- H bonds having bond angle 104.5o. Thus, obs 2 O H cos 52 .25 or 1.84 = 2 O – H × 0.6129 ; O – H = 1. 50 D Note
: EN dipole moment, so HF > HCl > HBr > HI , Where, EN = Electronegativity EN bond polarity, so HF > H 2O > NH 3 > H 2 S. If the electronegativity of surrounding atom decreases, then dipole moment increases.
NCl 3 NBr 3 NI 3
(iv) To distinguish cis and trans forms : The trans isomer usually possesses either zero dipole moment or very low value in comparision to cis–form H C Cl
H C Cl
||
H C Cl
Cis – 1, 2 – dichloro ethene
||
Cl C H 1. 9 D
Trans – 1, 2 –dichloro ethene,
0
Calculate the % of ionic character of a bond having length = 0.92 Å and 1.91 D as its observed dipole moment. (a) 43.25 (b) 86.5 (c) 8.65 (d) 43.5 Solution: (a) Calculated considering 100% ionic bond [When we consider a compound ionic, then each ionic sphere should have one electron charge on it of 4.80 10 10 esu (in CGS unit) or 1.60 10 19 C (in SI unit)] = 4.8 × 10–10 × 0.92 × 10–8 esu cm = 4.416 D Example:
CHEMICAL BONDING
% Ionic character
1.91 4.416
100 43 . 25 .
Important Tips
The dipole moment of CO molecule is greater than expected. This is due to the presence of a dative (co-ordinate) bond. Critical temperature of water is higher than that of O2 because H 2O molecule has dipole moment. Liquid is not deflected by a non uniform electrostatic field in hexane because of = 0
3.7 Change of ionic character to covalent character and Fajan’s rule.
Although in an ionic compound like M + X – the bond is considered to be 100% ionic, but it has some covalent character. The partial covalent character of an ionic bond has been explained on the basis of polarization. Polarization : When a cation of small size approaches an anion, the electron cloud of the bigger anion is attracted towards the cation and hence gets distorted. This distortion effect is called polarization of the anion. The power of the cation to polarize nearby anion is called its polarizing power. The tendency of an anion to get distorted or polarized by the cation is called polarizability. Due to polarization, some sort of sharing electrons occurs between two ions and the bond shows some covalent character. Fajan’s rule : The magnitude of polarization or increased covalent character depends upon a number of factors. These factors were suggested by Fajan and are known as Fajan’s rules.
Factors favouring the covalent character (i) Small size of cation : Smaller size of cation greater is its polarizing power i.e. greater will be the covalent nature of the bond . On moving down a group, the polarizing power of the cation goes on decreasing. While it increases on moving left to right in a period. Thus polarizing power of cation follows the order Li Na K Rb Cs . This explains why LiCl is more covalent than KCl . (ii) Large size of anion : Larger the size of anion greater is its polarizing power i.e. greater will be the covalent nature of the bond . The polarizability of the anions by a given cation decreases in moving left to right in a period. While it increases on moving down a group. Thus polarzibility of the anion follows the order I Br Cl F . This explains why iodides are most covalent in nature. (iii) Large charge on either of the two ions : As the charge on the ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases with the result its ability for forming the covalent bond increases. FeCl 3 has a greater covalent character than FeCl 2. This is because Fe 3+ ion has greater charge and smaller size than Fe 2+ ion. As a result Fe 3+ ion has greater polarizing power. Covalent character of lithium halides is in the order LiF < LiCl < LiBr < LiI. (iv) Electronic configuration of the cation : For the two ions of the same size and charge, one with a pseudo noble gas configuration (i.e. 18 electrons in the outermost shell) will be more polarizing than a cation with noble gas configuration (i.e., 8 electron in outer most shell). CuCl is more covalent than NaCl, because Cu+ contains 18 electrons in outermost shell which brings greater polarization of the anion. Important tips
CHEMICAL BONDING
en (Electronegativity difference)
Ionic character Covalent character
Increase of polarization brings more of covalent character in an ionic compound. The increased covalent character is indicated by the decrease in melting point of the ionic compound Decreasing trends of melting points w ith increased covalent character NaF < NaCl < NaBr < NaI (size of anion increases) o M.Pt ( C) 988 801 755 651 BaCl 2 < SrCl 2 < CaCl 2 < MgCl 2 < BeCl 2 (size of cation decreases) M.Pt ( oC) 960 872 772 712 405 Lithium salts are soluble in organic solvents because of their covalent chara cter Sulphides are less soluble in water than oxides of the same metal due to the covalent nature of sulphur .
3.8 Quantum theory (Modern theory) of covalent bond and overlapping.
(1) A modern Approach for covalent bond (Valence bond theory or VBT) (i) Heitler and London concept . (a) To form a covalent bond, two atoms must come close to each other so that orbitals of one overlaps with the other. (b) Orbitals having unpaired electrons of anti spin overlaps with each other. (c) After overlapping a new localized bond orbital is formed which has maximum probability of finding electrons. (d) Covalent bond is formed due to electrostatic attraction between radii and the accumulated electrons cloud and by attraction between spins of anti spin electrons. (e) Greater is the overlapping, lesser will be the bond length, more will be attraction and more will be bond energy and the stability of bond will also be high. (ii) Pauling and slater extension (a) The extent of overlapping depends upon: Nature of orbitals involved in overlapping, and nature of overlapping. (b) More closer the valence shells are to the nucleus, more will be the overlapping and the bond energy will also be high. (c) Between two sub shells of same energy level, the sub shell more directionally concentrated shows more overlapping. Bond energy : 2 s 2 s < 2 s 2 p < 2 p 2 p (d) s -orbitals are spherically symmetrical and thus show only head on overlapping. On the other hand, p -orbitals are directionally concentrated and thus show either head on overlapping or lateral overlapping. (iii) Energy concept (a) Atoms combine with each other to minimize their energy. (b) Let us take the example of hydrogen molecule in which the bond between two hydrogen atoms is quite strong. (c) During the formation of hydrogen molecule, when two hydrogen atoms approach each other, two types of interaction become operative as shown in figure. Attraction The force of attraction between the molecules of one atom + + Repulsion and electrons of the other atom. The force of repulsion Attraction between the nuclei of reacting atoms and electrons of the reacting atoms
CHEMICAL BONDING
)
(d) As the two hydrogen atoms approach each other from the infinite distance, they start interacting with each other when the magnitude of attractive forces is more than that of repulsive forces a bond is developed +ve between two atoms. O (e) The decrease in potential energy taking place during P.E. decreases as E formation of hydrogen molecule may be shown graphically ( ) –ve d 0 (f) The inter nuclear distance at the point O have minimum energy or maximum stability is called bond length. (g) The amount of energy released (i.e., decrease in potential energy) is known as enthalpy of formation. (h) From the curve it is apparent that greater the decrease in potential energy, stronger will be the bond formed and vice versa. (i) It is to be noted that for dissociation of hydrogen molecule into atoms, equivalent amount of energy is to be supplied. (j) Obviously in general, a stronger bond will require greater amount of energy for the separation of atoms. The energy required to cleave one mole of bonds of the same kind is known as the bond energy or bond dissociation energy. This is also called as orbital overlap concept of covalent bond.
(2) Overlapping (i) According to this concept a covalent bond is formed by the partial overlapping of two half filled atomic orbitals containing one electron each with opposite spins then they merge to form a new orbital known as molecular orbital. (ii) These two electrons have greater probability of their presence in the region of overlap and thus get stabilised i.e., during overlapping energy is released . Examples of overlapping are given below : Formation of hydrogen molecule : Two hydrogen atoms having electrons with opposite spins come close to each other, their s-orbitals overlap with each other resulting in the union of two atoms to form a molecule. + H-atom
– +
–
+
+
H -atom
+
H 2-molecule
Formation of fluorine molecule : In the formation of F 2 molecule p-orbitals of each flourine atom having electrons with opposite spins come close to each other, overlapping take place resulting is the union of two atoms. 2 p x
2 p x 2 py
2 py
2 p z +
F -atom
2 p z
F -atom Formation of fluorine molecule
F 2-molecule
CHEMICAL BONDING
Types of overlapping and nature of covalent bonds ( - and - bonds) : overlapping of different type gives sigma ( ) and pi ( ) bond. Various modes of overlapping given below : s – s overlapping : In this type two half filled s-orbitals overlap along the internuclear axis to form -bond .
s-p overlapping : It involves the overlapping of half filled s-orbital of one atom with the half filled porbital of other atom This overlapping again gives -bond e.g., formation of H – F molecule involves the overlapping of 1s orbital of H with the half filled 2 pz – orbital of Fluorine. + s-orbital
p-orbital
s– p overlap -bond
s – p overlapping
p- p overlapping : p- p overlapping can take place in the following two ways. (i) When there is the coaxial overlapping between p-orbitals of one atom with the p-orbitals of the other then - bond formation take place e.g., formation of F 2 molecule in which 2 pz orbital of one F atom overlap coaxially with the 2 pz orbitals of second atom. -bond formation take place as shown below : + pz-orbital
pz -orbital
p z – p z overlap -bond
p – p
(ii) When p–orbitals involved in overlapping are parallel and perpendicular to the internuclear axis. This types of overlapping results in formation of pi bond. It is always accompanied by a bond and consists of two charge clouds i.e., above and below the plane of sigma bond. Since overlapping takes place on both sides of the internuclear axis, free rotation of atoms around a pi bond is not possible. +
+ +
– p-orbital
Internuclear
–
-bond
p-orbital p-p overlapping
Table : Difference in
and
bonds
Sigma ( ) bond
Pi ( ) bond
It results from the end to end overlapping of two s-orbitals or two p-orbitals or one s and one p-orbital.
It result from the sidewise (lateral) overlapping of two porbitals.
Stronger
Less strong
Bond energy 80 kcals
Bond energy 65 kcals
More stable
Less stable
Less reactive
More reactive
CHEMICAL BONDING
Can exist independently
Always exist along with a -bond
The electron cloud is symmetrical about the internuclear axis.
The electron cloud is above and below the plane of internuclear axis.
Important Tips
To count the and bonds in molecule having single, double and triple bond first we write its expanded structure. H |
H \
2
/
ex. H H , O O , N N ,
H
C N
H
|
|
C C C N , N C C C N , | (3 , 6 )
C
H C
C H
|
||
H C
C H C
C N
|
(8 , 8 )
H (3 , 12 )
All the single bonds are -bond. In a double bond, one will be and the other type while in a triple bond, one will be and other two . OH |
The enolic form of acetone has 9 , 1 and two lone pairs CH 2 C CH 3 (enol form of acetone) H
It is the bond that actually takes part in reaction,
\ C C / / \
H
H
H |
H |
H 2 H C C H , but the shape of molecule is
H
|
|
H
H
decided by -bond. The number of sp2-s sigma bonds in benzene are 6. When two atoms of the element having same spin of electron approach for bonding, the orbital overlapping and bonding both does not occur. thanSidewise lateral, or sidewise overlapping. Head on overlapping is more Headstronger on overlapping p-p > s-p > s-s > overlapping p-p
3.9 Hybridization.
The concept of hybridization was introduced by Pauling and Slater. It is defined as the intermixing of dissimilar orbitals of the same atom but having slightly different energies to form same number of new orbitals of equal energies and identical shapes. The new orbitals so formed are known as hybrid orbitals. Characteristics of hybridization
(1) Only orbitals of almost similar energies and belonging to the same atom or ion undergoes hybridization. (2) Hybridization takes place only in orbitals, electrons are not involved in it . (3) The number of hybrid orbitals produced is equal to the number of pure orbitals, mixed during hybridization. (4) In the excited state, the number of unpaired electrons must correspond to the oxidation state of the central atom in the molecule. (5) Both half filled orbitals or fully filled orbitals of equivalent energy can involve in hybridization. (6) Hybrid orbitals form only sigma bonds. (7) Orbitals involved in bond formation do not participate in hybridization.
CHEMICAL BONDING
(8) Hybridization never takes place in an isolated atom but it occurs only at the time of bond formation. (9) The hybrid orbitals are distributed in space as apart as possible resulting in a definite geometry of molecule. (10) Hybridized orbitals provide efficient overlapping than overlapping by pure s, p and d-orbitals. (11) Hybridized orbitals possess lower energy. Depending upon the type and number of orbitals involved in intermixing, the hybridization can be of various types namely sp, sp2, sp3, sp3d , dsp2, sp3d 2, sp3d 3. The nature and number of orbitals involved in the above mentioned types of hybridization and their acquired shapes are discussed in following table Type of hy brisation
Character
Geometry of molecules as per VSPER theory
No. of bonde d atoms
No. of lone pairs
Actual shape of molecules
180o
s-character=50%, p -character= 50%
sp
A Linear
2
0
Linear
Example
CO2 , HgCl 2 , BeF 2 , ZnCl 2 , MgCl 2 , C 2 H 2 , HCN , BeH 2 C 2 H 2 , CS 2 , N 2 O, Hg 2 Cl 2 , [ Ag { NH 3 ) 2 ]
BF 3 , AlCl 3 , SO 3 , C 2 H 4 ,
s-character= 33.33%, p-character=66.67%
sp 2
120o
A
3
0
Trigonal Planar
C 2 Cl 4 , C 2 H 2Cl 2, [ HgI 3 ],
Trigonal Planar, 120o
<120o
NO3 , CO32 , HCHO C 6 H 6 , CH 3 graphite, [Cu( PMe 3 )3 ]
2
1
V-shape (bent)
NO 2 , SO 2 , SnCl 2 CH 4 , SiH 4 , SO 42 , SnCl 4 , ClO4 , BF 4 , NH 4 , CCl 4 ,
SiF 4 , H 2 NH 2 ,
109o28 A
s-character = 25%, p-character = 75%
sp 3
4
0
Tetrahedral
Tetrahedral , 109.5 o
[ BeF 4 ] , XeO 4 , [ AlCl 4 ] , SnCl 4 , PH 4 ,
Diamond, silica, Ni(CO)4 , Si(CH 3 )4 , SiC , SF 2 , [ NiCl 4 ]2 , [ MnO4 ][VO4 ]3
dsp
2
s-character = 25% p – character = 50% d – character = 25%
< 109.5o
3
1
Trigonal pyramidal
104.5o
2
2
V -shape (bent)
4
0
Square planar
0
Trigonal bipyramida l
90o
sp 3 d
ClO 3 , POCl 3 H 3 O , XeO 3 H 2O, H 2 S , PbCl 2 , OF 2 , NH 2 ClO 2 [Cu( NH 3 )4 ] 2 , [ Ni(CN )4 ] 2 [ Pt (NH 3 )4 ]2
Square planar 90o
s-character = 20%, p-character = 60%, d -character = 20%
NH 3 , PCl 3 , PH 3 , AsH 3
120o
A
Trigonal bipyramidal
5
PCl 5 , SbCl 5 , XeO 3 F 2 , PF 5 AsF 5 , PCl 4 , PCl 6 , [Cu(Cl )5 ]3 [ Ni(CN )5 ] 3 , [ Fe(CO )5 ]
CHEMICAL BONDING
1
3
2
Irregular tetrahedral T-shaped
2
3
Linear
TeCl 4 , SF 4 ClF 3 , IF 3 I 3 , XeF 2 SF 6 , PF 6 , SnCl 6 , MoF 6 ,
s-character = 25% p – character = 50% d – character = 25%
sp 3 d 2
4
6
0
Octahedral
( BaCl 6 ) , ( PF 6 ) , [ Fe(CN )6 ]4 , [ Fe( H 2 O)6 ] 3
Octohedral
Square pyramidal Square planar
5
1
4
2
7
0
Pentagonal bipyramida l
6
1
Distorted octahedral
ICl 5 , BrF 5 , IF 5 XeOF 4 XeF 4 , ICl 4
90o
3
s-character= 14.28%, p-character= 42.86%, d -character= 42.86%
3
sp d
72o
A
IF 7 , [ ZrF 7 ]3 [UF 7 ]3 [UO 2 F 5 ]3
Pentagonal bipyramidal
XeF 6
Short trick to find out hybridization : The structure of any molecule can be predicted on the basis of hybridization which in turn can be known by the following general formulation. H
1 2
(V M C A)
Where H = Number of orbitals involved in hybridization viz. 2, 3, 4, 5, 6 and 7, hence nature of hybridization will be sp, sp2, sp3, sp3d , sp3d 2, sp3d 3 respectively. V = Number electrons in valence shell of the central atom, M = Number of monovalent atom C = Charge on cation, A = Charge on anion Few examples are given below to illustrate this: Type : (A) When the central atom is surrounded by monovalent atoms only, e.g. BeF 2, BCl 3, CCl 4, NCl 3, PCl 5, NH 3, H 2O, OF 2, TeCl 4, SCl 2, IF 7, ClF 3, SF 4, SF 6, XeF 2, XeF 4, etc. Let us take the case of PCl 5. H
1 2
(5 5 0 0) 5 . Thus, the type of
hybridization is sp3d .
Type : (B) When the central atom is surrounded by divalent atoms only; e.g. CO2, CS 2, SO2, SO3, XeO3 1
etc. Let us take the case of SO3. H (6 0 0 0) 3 . Thus, the type of hybridization in SO3 is sp2. 2
Type : (C) When the central atom is surrounded by monovalent as well as divalent atoms, e.g. COCl 2, POCl 3, XeO2 F 2 etc. Let us take the case of POCl 3. H
1 2
(5 3 0 0) 4 .
Thus, the nature of hybridization in POCl 3 is sp3.
Type : (D) When the species is a cation, e.g. NH 4+, CH 3+, H 3O+ etc. Let us take the case of CH 3+. H
1 2
(4 3 1 0 ) 3 . Thus, the
hybridization in CH 3+ is sp2.
CHEMICAL BONDING
Type : (E) When the species is an anion, e.g. SO42–, CO32–, PO43–, NO2–, NO3–, etc. Let us take the case of SO42–. H 1 (6 0 0 2) 4 . Thus, hybridization in SO 42 is sp3 . 2
Type : (F) When the species is a complex ion of the type ICl 4–, I 3–, ClF 2–, etc. Let us take the case of ClF 2–. H
1 2
(7 2 0 1) 5 . Thus, in ClF 2–, Cl is sp3d hybridized.
Type : (G) When the species is a complex ion of the type [ PtF 6]2 –, [Co( NH 3)6]2+, [ Ni ( NH 3)4Cl 2] etc. In such cases nature of hybridization is given by counting the co-ordination number. Important Tips The sequence of relative energy and size of s – p type hybrid orbitals is sp < sp 2 < sp 3. The relative value of the overlapping power of sp, sp2 and sp 3 hybrid orbitals are 1.93, 1.99 and 2.00 respectively. An increase in s-character of hybrid orbitals, increases the bond angle. Increasing order of s-characters and bond angle issp 3 < sp 2 < sp. Normally hybrid orbitals (sp, sp2, sp 3, dsp2, dsp 3 etc.) form -bonds but in benzyne lateral overlap of sp2-orbitals forms a bond. Some iso-structural pairs are [ NF 3 , H 3O ], [ NO3 , BF 3 ], [SO 42 , BF 4 ] . There structures are similar due to same hybridization. In BF 3 all atoms are co-planar. In PCl 5 the state of hybridization of P atom is sp 3d. In its trigonal bipyramidal shape all the P-Cl bonds are not equal.
The bond formed between S and O atoms in SO 2 molecule is due to overlap between their p-orbitals ( P P bonding) or between p orbital of O-atom with d-orbital of S-atom (called p – d bonding) 16
S 1 s 2 2 s 2 2 p 6 3 s 2 3 p x2 3 p y1 3 p z1
(Ground state configuration)
1 s 2 2 s 2 2 p 6 3 s 2 3 p x1 3 p y1 3 p 1z 3d 1 (Excited state configuration) 8 O
1 s 2 2 s 2 2 p x2 2 p y1 2 p z1
p - d bonding
3 p
p - p bonding
d
2 p
2 p
S S
S-atom undergoes sp2 hybridization leaving one d half-filled 3p z orbital and one d-orbital unhybridized. Out of two half filled orbitals of O-atom, one is involved in formation of -bond with S-atom and the other in forming -bond.
3.10 Resonance.
The phenomenon of resonance was put forward by Heisenberg to explain the properties of certain molecules. In case of certain molecules, a single Lewis structure cannot explain all the properties of t he molecule. The molecule is then supposed to have many structures, each of which can explain most of the properties of the molecule but none can explain all the properties of the molecule. The actual structure is in between of all these contributing structures and is called resonance hybrid and the different individual structures are called resonating structures or canonical forms. This phenomenon is called resonance. To illustrate this, consider a molecule of ozone O3 . Its structure can be written as O O
O O
(a)
O
O O
(b )
O
O (c )
CHEMICAL BONDING
It may be noted that the resonating structures have no real existence. Such structures are only theoretical. In fact, the actual molecule has no pictorial representation. The resonating structures are only a convenient way of picturing molecule to account for its properties. As a resonance hybrid of above two structures (a) and (b. For simplicity, ozone may be represented by structure (c), which shows the resonance hybrid having equal bonds between single and double. Resonance is shown by benzene, toluene, O 3, allenes (>C = C = C<), CO, CO 2, CO 3 , SO 3, NO, NO2 while it is not shown by H 2O2, H 2O, NH 3, CH 4, SiO2. (1) Conditions for writing resonance structures
The resonance contributing structures : (i) Should have same atomic positions. (ii) Should have same number of bond pairs and lone pairs. (iii) Should have nearly same energy. (iv) Should be so written such that negative charge is present on an electronegative atom and positive charge is present on an electropositive atom. (v) The like charges should not reside on adjacent atoms. But unlike charges should not greatly separated. (2) Characteristics of resonance
(i) The contributing structures (canonical forms) do not have any real existence. They are only imaginary. Only the resonance hybrid has the real existence. E 1 (ii) As a result of resonance, the bond lengths of single and E 2 double bond in a molecule become equal e.g. O–O bond E 3 lengths in ozone or C –O bond lengths in CO32 – ion. y Resonance energy g (iii) The resonance hybrid has lower energy and hence greater r e n stability than any of the contributing structures. E E 0 (iv) Greater is the resonance energy, greater is the stability of Resonating structures the molecule. Concept of resonance energy. Energies E 1, (v) Greater is the number of canonical forms especially with E 2 and E 3 for three structures and E 0 is the experimentally determined bond energy nearly same energy, greater is the stability of the molecule. (3) Resonance energy
It is the difference between the energy of resonance hybrid and that of the most stable of the resonating structures (having least energy). Thus, Resonance energy = Energy of resonance hybrid – Energy of the most stable of resonating structure. (4) Bond order calculation
In the case of molecules or ions having resonance, the bond order changes and is calculated as follows: Bond order
Total no. of bonds between two atoms in all the structures
e.g., (i) In benzene Bond order
Total no. of resonating structures double bond single bond 2
2 1 2
1. 5
CHEMICAL BONDING
(ii) In carbonate ion
O
O
O
|
||
|
/
// \
O
C
C O
O
C
\
O
/
Bond order
2 1 1 3
\\
O
O
1 .33
Important Tips
Resonance structure (canonical form) must be pla nar. In case of formic acid H C
O H C
O
H C
, the two O H
O
O bond lengths are different but in formate ion C
O the two C O bond lengths are equal due to resonance. O
In resonance hybrid, the bond lengths are different from those in the contributing contr ibuting structure. Resonance arises due to delocalisation of electrons. electrons. Resonating structures have different electronic arrangements. The resonating structures do not have real existence. Resonance energy = [Experimental heat of formation] – [Calculated heat of formation of most stable canonical form].
Three contributing structure of CO 2 molecule are – O C O O C O O C O C
Resonance structure of carbon monoxide is
Resonance energy of benzene is 152 15 2.0 kJ / / mol .
O
C
O
3.11 Bond characteristics. characteristics. (1) Bond length “The average distance between the centre of the nuclei of the two bonded atoms is called bond length ”.
It is expressed in terms of Angstrom (1 Å = 10–10 m) or picometer (1pm = 10 –12 m). It is determined experimentally by X-ray diffraction methods or spectroscopic methods. In an ionic compound, the bond bo nd length is the sum of their ionic radii (d (d = = r++ r–) and in a covalent compound, it is the sum of their covalent radii ( e.g., for HCl , d = = r H + + rCl ). ). Factors affecting bond length (i) Size (i) Size of the atoms : The bond length increases with increase in the size of the atoms. atoms . For example, X are in the order , HI HBr bond length of H HBr HCl HCl HF
(ii) Electronegativity (ii) Electronegativity difference difference :
Bond length
1
EN
.
(iii) Multiplicity of bond : The bond length decreases with the multiplicity of the bond. Thus, bond length of carbon–carbon bonds are in the order , C C C C C – C (iv) Type of hybridisation : As an s-orbital is smaller in size, greater the s-character shorter is the hybrid orbital and hence shorter is the bond length. For example, example , sp 3 C – H sp 2 C – H sp C – H
(v) Resonance (v) Resonance : Bond length is also affected by resonance as in benzene and CO 2 . In benzene bond C bond length 1.54 Å and C = C bond length 1 .34 Å length is 1.39 Å which is in between C
CHEMICAL BONDING
In CO 2 the C-O bond C-O bond length is 1.15 Å . (In between C O and C O )
O C O O C O O C O
(vi) Polarity of bond : Polar bond length is usually smaller than the theoretical non-polar bond length. (2) Bond energy “The amount of energy required to break one mole of bonds of a particular type so as to separate them ”. Greater is the bond into gaseous atoms is called bond dissociation energy or simply bond energy ”.
energy, stronger is the bond . Bond energy is usually expressed in kJ mol – 1 . Factors affecting bond energy
(i) Size (i) Size of the atoms : Bond energy
1 atomic size
Greater the size of the atom, greater is the bond length
and less is the bond dissociation energy i.e. less is the bond strength. strength . (ii) Multiplicity (ii) Multiplicity of bonds : bonds : For the bond between the two similar atoms, greater is the multiplicity of the bond, greater is the bond dissociation energy. energy. This is firstly because atoms come closer and secondly, the number of bonds to be broken is more, C C C C C C , C C C N N N : Greater the number of lone pairs of electrons present (iii) Number (iii) Number of lone lone pairs of electrons electrons present : on the bonded atoms, greater is the repulsion between the atoms and hence less is the bond dissociation energy. energy. For example for a few single bonds, we have Bond Lone pair of electrons Bond energy (kJ mol kJ mol –1)
C – C 0 348
N – N 1 163
O–O 2 146
F – F 3 139
(iv) Percentage s –character : character : The bond energy increases as the hybrid orbitals have greater amount of s orbital contribution. Thus, bond energy decreases in the following order, order , sp sp 2 sp 3 (v) Bond (v) Bond polarity : Greater the electronegativity difference, greater is the bond polarity and hence greater will be the bond strength i.e., bond energy, energy , H F H Cl H Br H I , (vi) Among halogens Cl – Cl > F – F > Br – Br > I – I, (Decreasing order of bond energy) (vii) Resonance (vii) Resonance : Resonance increases bond energy. (3) Bond angle
In case of molecules made up of three or more atoms, the average angle between the bonded orbitals (i.e., between the two covalent bonds) is known as bond angle . Factors affecting bond angle
(i) Repulsion (i) Repulsion between between atoms or groups attached to the central atom : The positive charge, developed due to high electronegativity of oxygen, on the two hydrogen atoms in water causes repulsion among themselves which increases the bond angle, H angle, H –O– H from from 90º to 105º.
CHEMICAL BONDING
(ii) Hybridisation of bonding orbitals : In hybridisation as the s character of the s hybrid bond increases, the bond angle increases. increases. Bond type sp3 sp2 sp Bond angle 120o 180o 109º28 (iii) Repulsion due to non-bonded electrons : Bond angle
1 No. of lone pair of electrons
. By
increasing lone pair of electron, bond angle decreases approximately by 2.5%. CH 4 NH 3 H 2O Bond angle 109º 107o 105o (iv) Elec (iv) Electrone tronegativity gativity of the atoms atoms : If the electroneg electronegativity ativity of the central central atom decrease decreases, s, bond angle decreases. H 2 O Bond angle
104 .5 o
H 2 S
H 2 Se
H 2 Te
92 .2 o
91 . 2 o
89 . 5 o
In case the central atom remains the same, bond angle increases with the decrease in electronegativity electronegativity of the surrounding atom, e.g. PCl 3 Bond angle Example:
100 o
PBr 3 101 .5 o
PI 3 ,
AsCl AsCl 3
102 o
o
98 . 4
AsBr 3
AsI 3
100 . 5 o
101 o
Energy required to dissociate 4 grams of gaseous hydrogen into free gaseous atoms is 208 kcal at at 25oC the the bond energy of H of H – H bond bond will be (a) 104 Kcal (b) 10 .4 Kcal (c) 20 .8 Kcal (d) 41 .6 Kcal
Answer: (a) 4 gram gaseous hydrogen has bond energy 208 kcal
So, 2 gram gaseous hydrogen has bond energy
208 2
kcal = 104 kcal .
Important Tips
More directional the bond, greater is the bond strength and vice versa. For example : sp 3 sp 3 sp 2 sp 2 sp sp p p s p s s
The hybrid orbitals with more p-character are more directional in character and hence of above order. The terms ‘bond energy’ and ‘bond dissociation’ energy are same only for di -atomic molecules. The order of O–O bond length in O 2, H 2O2 and O 3 is H 2O2 > O2 > O 3 Because of higher electron density and small size of F atom repulsion between electron of it two F atoms, weakens the F-F bond. The bond length increases as the bond order decreases. Carbon monoxide (CO) has the highest bond energy(1070 kJmol 1 ) of all the diatomic molecules. Bond energy
of
N 2 (946 94 6kJmol 1 ) and that of H 2(436 kJmol -1 ) are other diatomic molecules with very h igh bond energies.
3.12 VSEPR (Valence shell electron pair repulsion) theory .
The basic concept of the theory was suggested by Sidgwick and Powell (1940). It provides useful idea for predicting shapes and geometries of molecules. The concept tells that, the arrangement of bonds around the central atom depends upon the repulsion’s operating between electron pairs(bonded or non bonded) around the central atom. Gillespie and Nyholm developed this concept as VSEPR theory.
CHEMICAL BONDING
The main postulates of VSEPR theory are (i) For polyatomic molecules containing 3 or more atoms, one of the atoms is called the central atom to which other atoms are linked . (ii) The geometry of a molecule depends upon the total number of valence shell electron pairs (bonded or not bonded) present around the central atom and their repulsion due to relative sizes and shapes. (iii) If the central atom is surrounded by bond pairs only. It gives the symmetrical shape to the molecule. (iv) If the central atom is surrounded by lone pairs (lp) as well as bond pairs (bp) of e then the molecule has a distorted geometry. (v) The relative order of repulsion between electron pairs is as follows : Lone pair-lone pair>lone pair-bond pair>bond pair>bond pair-bond pair-bond pair A lone pair is concentrated around the central atom while a bond pair is pulled out between two bonded atoms. atoms. As such repulsion becomes greater when a lone pair is involved. Steps to be followed followed to find the shape shape of molecules : (i) Identify the central atom and count the number of valence electrons. (ii) Add to this, number of other atoms. (iii) If it is an ion, add negative charges and subtract positive charges. Call the total N total N . (iv) Divide N by 2 and compare the result r esult with the following table and obtain the shape. Total N/2
Shape of molecule or ion
Example
2
Linear
HgCl 2 / BeCl 2
3
Triangular planar
BF 3
3
Angular
SnCl 2 , NO 2
4
Tetrahedral
CH 4 , BF 4
4
Trigonal Pyramidal
NH 3 , PCl 3
4
Angular
H 2 O
5
Trigonal bipyramidal
PCl 5 , PF 5
5
Irregular tetrahedral
SF 4 , IF 4
5
T-shaped
CIF 3 , BrF Br F 3
5
Linear
XeF 2 , I 3
6
Octahedral
SF 6 , PF 6
6
Square Pyramidal
IF 5
6
Square planar
XeF 4 , ICI 4
Geometry of Molecules/Ions Molecules/Ions having bond pair as well as lone pair of electrons Type of molecule
No. of bond pairs of electron
No. of lone pairs of electrons
AX 3
2
1
Hybridization
sp
2
Bond angle
<
120o
Expected geometry
Actual geometry
Examples
Trigonal planar
V-shape, Bent, Angular
SO2, SnCl 2, NO2–
CHEMICAL BONDING
AX 4
2
2
sp 3
< 109o 28
Tetrahedral
V-shape, Angular
H 2O, H 2 S , SCl 2, OF 2, NH 2–, ClO2– NH 3, NF 3 , PCl 3, PH 3, AsH 3, ClO3– , H 3O+
AX 4
3
1
sp 3
< 109o 28
Tetrahedral
Pyramidal
AX 5
4
1
sp 3 d
< 109o 28
Irregular tetrahedron
SF 4, SCl 4, TeCl 4
AX 5
3
2
sp 3 d
90o
T-shaped
ICl 3, IF 3, ClF 3
AX 5
2
3
sp 3 d
180o
Trigonal bipyramidal Trigonal bipyramidal Trigonal bipyramidal
Linear
XeF 2, I 3–, ICl 2–
AX 6
5
1
sp 3 d 2
< 90o
Octahedral
AX 6
4
2
sp 3 d 2
–
Octahedral
AX 7
6
1
sp 3 d 3
–
Pentagonal pyramidal
Square pyramidal Square planar Distorted octahedral
ICl 5, BrF 5, IF 5 XeF 4, ICl 4– XeF 6
3.13 Molecular orbital theory .
Molecular orbital theory was given by Hund and Mulliken in 1932. When two or more constituent atomic orbital merge together, they form a bigger orbital called molecular orbital (MO). In atomic orbital, the electron is influenced by only one nucleus whereas in case of molecular orbital, the electron is influenced by two or more constituent nuclei. Thus, atomic orbital is monocentric and molecular orbital is polycentric. Molecular orbitals follow Pauli's exclusion principle, Hund’s rule, Aufbau's principle strictly.
Atomic orbital (AO)
According to VBT
Molecular orbital (MO)
According to MOT
The main ideas of this theory are : (i) When two atomic orbitals combine or overlap, they lose their identity and form new orbitals. The new orbitals thus formed are called molecular orbitals. (ii) Molecular orbitals are the energy states of a molecule in which the electrons of the molecule are filled just as atomic orbitals are the energy states of an atom in which the electrons of the atom are filled. (iii) In terms of probability distribution, a molecular orbital gives the electron probability distribution around a group of nuclei just as an atomic orbital gives the electron probability distribution around the single nucleus. (iv) Only those atomic orbitals can combine to form molecular orbitals which have comparable energies and proper orientation. (v) The number of molecular orbitals formed is equal to the number of combining atomic orbitals. (vi) When two atomic orbitals combine, they form two new orbitals called bonding molecular orbital and antibonding molecular orbital. (vii) The bonding molecular orbital has lower energy and hence greater stability than the corresponding antibonding molecular orbital . (viii) The bonding molecular orbitals are represented by , etc, whereas the corresponding antibonding molecular orbitals are represented by *, * etc. (ix) The shapes of the molecular orbitals formed depend upon the type of combining atomic orbitals.
CHEMICAL BONDING
Formation of bonding and antibonding molecular orbitals (Linear combination of atomic orbitals – LCAO) When two atomic orbitals overlap they can be in phase (added) or out of phase (subtracted). If they overlap in phase, constructive interaction occurs in the region between two nuclei and a bonding Antibonding Subtractio orbital is produced. The energy of the bonding Repulsive orbitals is always lower (more stable) then the y energies of the combining atomic orbitals. When g r e n Atomic Atomic they overlap out of phase, destructive E orbital orbital interference reduces the probability of finding Addition an electron in the region between the nuclei and Bonding M.O. antibonding orbital is produced. The energy of Attractive Formation of bonding and antibonding molecular an antibonding orbital is higher (less stable) than the energies of the combining atomic orbitals. Thus, the number of molecular orbitals formed from atomic orbitals is equal to the number of atomic orbitals responsible for their formation. According to LCAO method, the following combination is not allowed. [consider the z-axis as the molecular axis). np x np y , np x np z or np y np z . Table : Difference in Bonding molecular orbitals and Antibonding molecular orbitals. Bonding molecular orbitals
Antibonding molecular orbitals
It is formed by linear combination of two atomic orbitals when their wave functions are added. i.e., b A B
It is formed by linear combination of two atomic orbitals when their wave functions are subtracted. i.e., a A B
Its energy is less than the combining atomic orbitals. It increases the electron density between the nuclei. It therefore stabilises the molecule. It has no nodal plane (plane where electron density is zero). It is symmetrical about internuclear axis.
Its energy is more than the combining atomic orbitals. It decreases the electron density between the nuclei. It therefore destabilises the molecule. It has nodal plane. It is symmetrical about internuclear axis and about a line perpendicular to it.
and – Molecular orbitals A sigma ( ) M.O. is one that has cylinderical symmetry around the internuclear axis. It does not show any change of sign or rotation through 180 o about the axis. Sigma M.O. has no nodal plane (in which electron density is zero) along the internuclear axis. The bonding M.O. is designated by and antibonding by * . + +
+
++
+
1s
1s
+
–
+
–
+
(1s) Bonding M.O.
+ –
CHEMICAL BONDING
Formation of (1s) and * (1s) M.O. from 1s atomic orbital
Sigma ( ) M.O. is also formed when two p-atomic orbitals overlap in head on (along their axes) position. –
–
+
+
+ +
2pz
–
+
–
–
2pz –
–
–
–
+ –
–
(2pz)
+
+
2pz +
+ – *(2pz)
+
Formation of (2 p z and * (2 p z ) molecular orbitals. -molecular
orbital is formed by the sideways overlapping of the p-atomic orbitals. It consists of two electrons clouds, one lying above and the other lying below a plane passing through the nuclei. It has nodal planes. Since the energy of a M.O. is directly related to the nuclei of nodal planes. The -MO is more energetic than the -MO. This explains why a -bond is a weaker bond than a -bond. +
+
+ +
+
– –
–
2px (or 2p y )
2px (or 2p y )
– –
(2px) or (2p y )
+
–
+ –
–
+
– +
2px
Nodal Plane
+
–
–
+
2p
x * (2p x) (or p y y) ) and * (2 p x ) orNodal (2 p y ) molecular (22p *Plane Formation of orbitals. (22p p y x) ) or (or or * (2p y )
Relative energies of Molecular orbitals : Initial energy of the atomic orbitals and the extent of their overlap is the criteria which determines the energy of the M.O. (1) It is obvious that molecular orbitals formed from lower energy atomic orbitals have lower energy than the molecular orbitals formed from higher energy atomic orbitals. (2) As the overlap is much more effective than -overlap, p-molecular orbital is of lowest energy, even though originally all the three p-orbitals are of equal energy. (3) The relative energies of the M.O. are obtained experimentally from spectroscopic data. (4) The sequence in the order of increasing energy for O 2 , F 2 and Ne 2 1 s * 1 s 2 s * 2 s 2 p z 2 p x 2 p y * 2 p x * 2 p y * 2 p z
CHEMICAL BONDING
(5) It may be noted that 2 p y an 2 p x bonding molecular orbitals are degenerate (i.e. have same energy). Similarly * 2 p y and * 2 p x antibonding molecular orbitals are also degenerate (have the same energy). (6) The main difference between the two types of sequences in energy level is that for molecules O 2 , F 2 and Ne 2 (Hypothetical) the 2 p z M.O. is lower in energy than 2 p x and 2 p y . (7) It has been found experimentally that in some of the diatomic molecules such as Li 2 , B 2 , C 2 and N 2 2 p z M.O. is higher in energy than 2 p y and 2 p x M.O.’s. Therefore, the order of increasing energy of these M.O.s changes to 1 s * 1 s 2 s * 2 s 2 p x 2 p y 2 p z * 2 p x * 2 p y * 2 p z . Cause of exceptional behaviour of MO’s in B 2 , C 2 and N 2
In atoms with Z upto 7, energy of 2s and 2 p atomic orbitals lie fairly close. As a result the interaction between 2s and 2 p orbitals is quite large. Thus 2 s and * 2 s MO’s become more stable with less energy at the cost of 2 p x and * 2 p x which gets unstabilised (higher energy). Stability of the molecules : Stabitity of the molecules can be explain in following ways. (1) Stability of molecules in terms of bonding and antibonding electrons Since electrons in bonding orbitals ( N b ) increase the stability of the molecule, on the whole, depends on their relative numbers. Thus (i) If N b N a , the molecule is stable. (ii) If N b N a , the molecule is unstable. (iii) Even if N b N a , the molecule is unstable. This is due to the fact the anti bonding effect is some what stronger than the bonding effect . (2) Stability of molecules in terms of bond order (i) The relative stability of a molecule is further evaluated by a parameter known as bond order. (ii) It can be defined as number of covalent bonds present between two atoms in a molecule. (iii) It is given by one half of the differences between the number of electrons in bonding molecular orbitals and those in antibonding molecular orbitals. (iv) Bond order
1 2
[No. of electrons in bonding molecular orbitals – No. of electrons in antibonding
molecular orbitals]. (v) The bond order of 1, 2 and 3 corresponds to single, double and triple bonds respectively. It may be mentioned that according to M.O theory, even a fractional bond order is possible, but cannot be negative. (vi) bond order Stability of molecule Dissociation energy
1 Bond length
(vii) If all the electrons in a molecule are paired then the substance is a diamagnetic on the other hand if there are unpaired electrons in the molecule, then the substance is paramagnetic. More the number of unpaired electron in the molecule greater is the paramagnetism of the substance. MO energy level diagrams of some molecules : Energy level diagrams of some molecules are given below. H 2 Molecule : H 2 molecule is formed from 1 s1 atomic orbitals of two H- atoms. The atomic orbitals (1 s 1 ) will combine to form two molecular orbitals (1s) and * (1s) . Two electrons are accommodated in (1s) and * (1s) remains vacant. Thus bond order for H 2 y g r e n e g n i s a e r c n
1 2
(2 0) 1 . It is stable and
* 1s
1s H
1s H
diamagnetic in nature.
CHEMICAL BONDING
He2 molecule : If two atoms of He (1 s 2 ) combine to form He 2 , the probability of the formation of molecular orbitals is as shown in the figure. * (1s2) g Filling of electrons is as follows n i s a e r c n I
He 2 (1s) 2 , * (1s) 2
Thus bond order
1 2
(2 2) 0
1s2 He
1s2 He (1s)2
Hence there is no possibility for the existence of He 2 molecule. N 2 molecule : Total number of electrons in N 2 are 14, of which 4 are in K shell and the 10 electrons are arranged as, KK ( 2 s)2 ( * 2 s)2 ( 2 p x )2 ( 2 p y )2 ( 2 p z )2 Bond order
1 2
* (2 p z )
(10 4 ) 3
2 p
2 p *(2 p x ) *(2 py)
(2 pz)
g n i s a e r c n I
(2 p x )= (2 py)
* (2s)
2s
2s
N (AO)
N (AO)
(2s) M.O. energy level diagram for N 2 molecule.
O2 molecule : Total number of electrons in O2 16
N2 (MO)
Electronic arrangement in M.O.’s , [ KK ( 2 s)2 ( * 2 s)2 ( 2 p z )2 ( 2 p x )2 ( z 2 p y )2 ( * 2 p x )1 ( * 2 p y )1 ]
Bond order
1 2
(8 4 ) 2
* (2 p z )
* 2 p x * 2 py
g n i s a e r c n I
2 p x = 2 py
2 p z
* 2s
O (AO)
O (AO) 2s
O2 (MO)
CHEMICAL BONDING
M.O. Energy level diagram for O2 molecule Table : Bond order and magnetic nature of some molecules & ions Molecule
Valence electrons
Molecular Orbital Configuration
N b
Na
Magnetic Nature
B.O.
H 2
1s
2
2
0
Diamagnetic
1
H 2
1s
1
1
0
Paramagnetic
0.5
H 2
1s
3
2
1
Paramagnetic
0.5
He 2
2 1s σ 1s
4
2
2
Molecule not exist
He 2
1s
3
2
1
Paramagnetic
0.5
He 2
KK '
2s
1
1
0
Paramagnetic
0.5
Be 2
KK '
2s
4
2
2
Diamagnetic
0
B 2
KK ' ( 2 s)2 ( * 2 s)2 ( 2 p x )1 ( 2 p y )1
6
4
2
Paramagnetic
1
C 2
KK ' 2 s 2 * 2 s 2 π 2 p x 2 π 2 p y 2
8
6
2
Diamagnetic
2
N 2
KK ' 2 s 2 *2 s2 π 2 p x 2 π 2 p y 2 2 p z 2
10
8
2
Diamagnetic
3
N 2
KK ' 2 s 2 *2 s2 π 2 p x 2 π 2 p y 2 2 p z 1
9
7
2
Paramagnetic
2.5
N 2
KK ' 2 s 2 *2 s2 π 2 p x 2 π 2 p y 2 2 p z 2 * 2 p x 1
11
8
3
Paramagnetic
2.5
O2
KK ' 2 s 2 *2 s2 2 p z 2 π 2 p x 2 π 2 p y 2 * 2 p x 1 * 2 p y 1
12
8
4
Paramagnetic
2
O 2
KK 2 s 2 *2 s2 2 p z 2 π 2 p x 2 π 2 p y 2 π* 2 p x 1 π* 2 p y 0
11
8
3
Paramagnetic
2.5
O 22
KK ' ( 2 s)2 ( * 2 s)2 ( 2 p z )2 ( 2 p x )2 ( 2 p y )2
10
8
2
Diamagnetic
3
O 2
KK σ 2 s2 σ* 2 s2 2 p z 2 π 2 p x 2 π 2 p y 2 * 2 p x 2 * 2 p y 1
13
8
5
Paramagnetic
1.5
O 22
KK 2 s2 *2 s2 2 p z 2 π 2 p x 2 π 2 p y 2 * 2 p x 2 * 2 p y 2
14
8
6
Diamagnetic
1
F 2
KK 2 s 2 *2 s 2 2 p z 2 π 2 p x 2 π 2 p y 2 π* 2 p x 2 π* 2 p y 2
14
8
6
Diamagnetic
1
2 s2 * 2 s2 2 p z 2π 2 p x 2 π 2 p y 2 π* 2 p x 2 π* 2 p y 2 π* 2 p z 2
16
8
8
Molecule not exist
0
10
8
2
Diamagnetic
3
Ne 2
σ
σ
σ
2
1
2
σ
σ
KK '
* 1s
1
σ
2
σ
* 1s
1
σ
σ
2
1
2
* 2s
σ
2
does
does
0
CO
KK ' 2 s 2 *2 s2 2 p z 2 π 2 p x 2 π 2 p y 2
NO
KK ' 2 s 2 *2 s 2 2 p z 2 π 2 p x 2 π 2 p y 2 π 2 p x
11
8
3
Paramagnetic
2.5
NO
KK ' 2 s 2 * 2 s2 2 p z 2 π 2 p x 2 π 2 p y 2
10
8
2
Diamagnetic
3
1
CHEMICAL BONDING
3.14 Hydrogen bonding.
In 1920, Latimer and Rodebush introduced the idea of “hydrogen bond” to explain the nature of association of substances in liquid state like water, hydrogen fluoride, ammonia, formic acid, etc. (1) Conditions for the formation of hydrogen bonding (i) High electronegativity of atom bonded to hydrogen : The molecule should contain an atom of high electronegativity such as F , O or N bonded to hydrogen atom. The common examples are H 2 O , NH 3 and HF .
H — X H — X H — X (ii) Small size of the electronegative atom : The size of the electronegative atom should be quite
small. This is due to the fact that the smaller the size of electronegative atom, the greater will be its attraction for the bonded electron pair. This will cause greater polarity in the bond between H and electronegative atom, which results in stronger hydrogen bond.
(2) Types of hydrogen bonding (i) Intermolecular hydrogen bond : Intermolecular hydrogen bond is formed between two different molecules of the same or different substances. For example: (a) Hydrogen bond between the molecules of hydrogen fluoride. (b) Hydrogen bond in alcohol and water molecules (ii) Intramolecular hydrogen bond (Chelation) : Intramolecular hydrogen bond is formed between the hydrogen atom and the highly electronegative atom ( F, O or N ) present in the same molecule. Intramolecular hydrogen bond results in the cyclisation of the molecules and prevents their association. Consequently, the effect of intramolecular hydrogen bond on the physical properties is negligible. For example : Intramolecular hydrogen bonds are present in molecules such as onitrophenol, o-nitrobenzoic acid, etc. O N || O
H O
O || C N || O
H | C
O O
H O
O H
Salicyldehyde
(o-hydroxy benzaldehyde)
nitrophenol nitrobenzoic acidhydrogen bonding depends on temperature. The extentOrtho of both intramolecular andOrtho intermolecular (3) Effects of hydrogen bonding Hydrogen bond helps in explaining the abnormal physical properties in several cases. Some of the properties affected by H-bond are given below: (i) Dissociation : In aqueous solution, hydrogen fluoride dissociates and gives the difluoride ion ( HF 2 ) instead of fluoride ion ( F ) . This is due to H -bonding in HF . This explains the existence of KHF 2 . On the other hand, the molecules of HCl , HBr , HI do not have H-bonding (because Cl , Br , I are not so highly electronegative). This explains the non-existence of compounds like KHCl 2 , KHBr 2 or KHI 2 . H- bond formed is usually longer than the covalent bond present in the molecule (e.g. in H 2O, O– H bond = 0.99 Å but H- bond = 1.77 Å). (ii) Association : The molecules of carboxylic acids exist as dimers because of the hydrogen bonding. The molecular masses of such compounds are found to be double than those calculated from their simple formulae. For example, molecular mass of acetic acid is found to be 120. (iii) High melting and boiling point : The compounds having hydrogen bonding show abnormally high melting and boiling points.
CHEMICAL BONDING
The high melting points and boiling points of the compounds ( H 2 O, HF and NH 3 ) containing hydrogen bonds is due to the fact that some extra energy is needed to break these bonds. A few examples are given below: (a) H 2 O is a liquid whereas H 2 S , H 2 Se and H 2 Te are all gases at ordinary temperature. The reason for this is that in case of water, hydrogen bonding causes association of the H 2 O molecules with the result that the boiling point of water is more than that of the other compounds. On the other hand, there is no such hydrogen bonding in H 2 S , H 2 Se and H 2Te. (b) NH 3 has higher boiling point than PH 3. This is again because there is hydrogen bonding in NH 3 but not in PH 3. (c) Ethanol has higher boiling point than diethyl ether because there is hydrogen bonding in the former but there is no hydrogen bonding in the later. Intramolecular hydrogen bonding is not possible in case of m - and p -isomers because of the size of the ring which would be formed. Thus, here the intermolecular hydrogen bonding takes place which causes some degree of association with the result the m -and p -isomers melt and boil at higher temperatures. (iv) Solubility : The compound which can form hydrogen bonds with the covalent molecules are soluble in such solvents. For example, lower alcohols are soluble in water because of the hydrogen bonding which can take place between water and alcohol molecules as shown below :
H O ................ H O ............... H O C 2 H 5
H
C 2 H 5
Similarly, ammonia ( NH 3 ) is soluble in water because of hydrogen bonding as represented below:
H H H
H N ....... H — O ..... .... ...... H H H
N ....... H — O
H
The intermolecular hydrogen bonding increases solubility of the compound in water while, the intramolecular hydrogen bonding decreases. This is due to the fact that the formation of internal hydrogen bond prevents hydrogen bonding between the compound and water which thus reduced solubility of the compound in water. O N || O
H O H O H H O
O N O p Nitropheno l
o- Nitrophenol Due to chelation, – OH group is not available to form hydrogen bond with water hence it is sparingly soluble in water.
– OH group available to form hydrogen bond with water, hence it is completely soluble in water.
(v) Volatility : As the compounds involving hydrogen bonding between different molecules (intermolecular hydrogen bonding) have higher boiling points, so they are less volatile. (vi) Viscosity and surface tension : The substances which contain hydrogen bonding exist as associated molecules. So their flow becomes comparatively difficult. In other words, they have higher viscosity and high surface tension.
CHEMICAL BONDING
(vii) Explanation of lower density of ice than water and maximum density of water at 277K : In case of solid ice, the hydrogen bonding gives rise to a cage like structure of water molecules as shown in following figure. As a matter of fact, each water molecule is linked tetrahedrally to four other water molecules. Due to this structure ice has lower density than water at 273 K . That is why ice floats on water. On heating, the hydrogen bonds start collapsing, obviously the molecules are not so closely packed as they are in the liquid state and thus the molecules start coming together resulting in the decrease of volume and hence increase of density. This goes on upto 277 K . After 277 K, the increase in volume due to expansion of the liquid water becomes much more than the decrease in volume due to breaking of H-bonds. Thus, after 277 K , there is net increase of volume on heating which means decrease in density. Hence density of water is maximum 277 K . H
0.90 (99 pm)
1.77 (177 pm)
O H
H
H O
H
H
Vacant Spaces
O H
H
O
O H
H
H O
H
H O
H
H
H
O
O
Cage like 2O in the ice H structure of H H Important Tips Hydrogen bonding is strongest when the bonded structure is stabilised by resonance. The bond length of hydrogen bond is the order of 250 to 275 pm. The bond that determines the secondary structure of protein is hydrogen bond. Pairs of DNA are held together by hydrogen bonds. Chlorine has the same electronegativity as nitrogen but does not form strong hydrogen bonds. This is because of the larger size than that of nitrogen.
3.15 Types of bonding and forces in solids. (1) Ionic bonding : Solid containing ionic bonds consists of any array or a net work of positive and
negative ions arranged systematically in a characteristic pattern. The binding forces are strong electrostatic bonds between positive and negative ions. e.g., Compounds of elements of group 1 and 2 with elements of group 16 and 17 e.g., NaCl , CaS etc. (2) Covalent bonding : The solid containing covalent bonding consists of an array of atoms that share electrons with their neighbouring atoms. The atoms are linked together by strong covalent bonds extending into three dimensional structure. e.g., Diamond, Silicon carbide, Silicon dioxide etc.
CHEMICAL BONDING
(3) Molecular bonding : The solid containing molecular bonding consists of symmetrical aggregates of
discrete molecules. However, these molecules are further bound to other molecules by relatively weak force such as dipole-dipole forces (Vander Waal forces), dispersion forces or H -bonds depending upon the nature of molecules. The existence of weak attractive forces among the non polar molecules was first proposed by S.D. Vander waal . Vander waal's forces are non-directional, non valence force of attraction. Vander Waal force molecular mass Boiling point Size of atom or molecule. The forces present in the crystals of naphthalene, Iodine and dry ices solid CH 4 , solid hydrogen are Vander Waals forces. SiO 2 Possesses giant covalent molecular structure due to tetravalency and catenation nature of Si . (4) Metallic Bond
The constituent particles in metallic solids are metal atoms which are held together by metallic bond. Lorentz proposed a simple theory of metallic bond. This theory is known as electron gas model or electron sea model. A metal atom consists of two parts, valence electrons and the remaining part (the nucleus and the inner shell electrons) called kernel. The kernels of metal atoms occupy the lattice sites while the space in-between is occupied by valence electrons. These electrons are not localized but are mobile. The attraction between the kernels and the mobile electrons, which hold the kernel together, is known as metallic bond. Low ionisation energy and sufficient number of vacant orbital in the valency shell are essential conditions for metallic bonding. Metallic bond is electrostatic in nature. The strength of the metallic bond depends on the number of valency electron and the charge on the nucleus. As the number of valency electron and the charge increase, the metallic bond becomes strong. Due to this fact alkali metals are soft and have low melting and boiling points while transition metals and hard and have high melting and boiling points. Strong metallic bonding is also favoured by smaller size of kernel. Ge, Cu, Zn has metallic bonding while brass etc does not have metallic bonding. Metals have properties like metallic lustre, thermal and electrical conductivity due to delocalized mobile electrons. Thermal conductivity of metal decreases with increase in temperature because the kernels start vibrating. Since the metallic bond is non-directional; metals can be twisted, drawn into wires or beaten into sheets. This is because the kernels can slip over each other when a deforming force is applied. The relative strength of various bonds is Ionic >Covalent>Metallic>H-bond>vander waal forces.
CHEMICAL BONDING
XII.
Chemical Bonding
Basic Level Chemical bond implies (a) Attraction (b) Repulsion (c) Neither attraction nor repulsion Atoms combine to form molecule because they tend to 2. (a) Decrease number of electrons in the outermost shell configuration (c) Increase number of electrons in the outermost shell outermost shell 3. When two atoms combine to form a molecule (a) Energy is released (b) (c) Energy is neither released nor-absorbed released or absorbed 1.
4.
[KCET 2002]
(d) Both (a) and (b) (b) Attain
an
inert
gas
(d) Attain 18 electrons in the [MP PET 1995]
Energy is absorbed (d) Energy may either be
Electronic theory of valency was ‘presented by’
(a) Pauling (b) Werner (c) Kossel and Lewis 5. Valency means (a) Combining capacity of an element (c) Oxidation number of an element 6. The valency of phosphorous in H 3 PO4 is (a) 2 (b) 3 (c) 4 (d) 5
(d) Heitler and London (b) Atomicity of an element (d) None of these [Delhi PMT 1991]
CHEMICAL BONDING
Formula of chloride of a metal M is MCl 3. Formula of its phosphate is (a) M ( PO 4 )2 (b) M 3 PO 4 (c) MPO 4 8. The electron dot representation of covalent molecules is also called (a) Lewisite structure (b) Bohr’s structure 7.
[CPMT 1979]
(d) M 2 PO 4 (c) Lewis structure
(d)
Millikan’s structure 9.
Carbon suboxide (C 3O2) has recently been shown as a component of the atmosphere of Venus. Which of the following formulation represent the correct ground state Lewis structure for carbon suboxide
(a) : O : C :: C : C : O : (b) : O :: C :: C : C :: O : (d) : O : C : C : C : O : 10. If the atomic no. of element X is 7. The best electron dot symbol for the element is .
(a) : X :
..
..
(c) : O :: C :: C :: C :: O : [BVP Pune 2003; NCERT 1973; CPMT 2003]
..
(b) X : (c) X (d) X : XI. Electrovalent Bonding
Basic Level 11. Which contributes significantly to ionic bonding
12.
13.
14. 15.
(a) Transfer of protons (b) Overlap of atomic orbitals (c) Sharing of electrons (d) Tendency to form closed shell configuration Electrovalent bond is formed by (a) Sharing of electrons (b) Donation of electrons (c) Transfer of electrons (d) None of these Ions are formed from neutral atoms by (a) Loss of electrons (b) Gain of electrons (c) Sharing of electrons (d) Loss or gain of electrons Most predominantly ionic compounds are obtained by the combination of the groups (a) 1 and 17 (b) 2 and 16 (c) 14 and 18 (d) 13 and 15 Element X is strongly electronegative and Y is strongly electropositive. Both are univalent. The compound formed would be [IIT 1980]
(a) X +Y – (b) X Y (c) X Y 16. Born-Haber cycle is used to determine (a) Electronegativity (b) Lattice energy these 17. The magnitude of the lattice energy of a solid increases if (a) The ions are large (b) (c) The ions are of equal size ions are small The crystal lattice of electrovalent compounds is composed of 18. (a) Atoms (b) Molecules (c) Oppositely charged ions 19. Most favourable conditions for ionic bonding are (a) Low charge on ions, large cation, large anion cation, small anion (c) High charge on ions, large cation, small anion cation, large anion 20. Which one of the following factors does not favour the formation of ionic bond (a) Low ionization energy of one atom one atom (c) High lattice energy of both atoms both atoms 21. Ionic bonds are formed between elements which have (a) No affinity for electrons
(d) X Y (c) Both
(d)
None of
The ions are small (d) Charges on the
(d) Both molecules and ions [DPMT 1994]
(b) Low charge on ions, large (d) High charge on ions, small (b) High electron affinity of (d) High electronegativity of
(b) Non-metallic property
CHEMICAL BONDING
(c) No electronegativity difference (d) Large electronegativity difference 22. An atom of one element A has 3 electrons in its outermost shell, and that B has 6 electrons in the outermost shell. The formula of the compound formed by these two is (a) A3 B6 (b) A2 B (c) A2 B3 (d) A3 B2 23. Two element X and Y have following electronic configuration X = 1s2, 2s22 p6, 3s23 p6, 4s2 and Y = 1s2, 2s22 p6, 3s23 p5 The compound formed by combination of X and Y is [BHU 1990, 99] (a) XY 2 (b) X 5Y 2 (c) X 2Y 5 (d) XY 5 [RPMT 1999] 24. Which is likely to have the highest melting point (a) He (b) CsCl (c) NH 3 (d) CHCl 3 25. Property of ionic compounds is [RPMT 2002] (a) High melting point and high boiling point (b) High melting point and low boiling point (c) Low melting point and high boiling point (d) Low melting point and low boiling point 26. Which will not conduct electricity (a) Aqueous KOH solution (b) Fused NaCl (c) Graphite (d) KCl in solid state [CPMT 1978] 27. In which of the compound, there is an electrovalent linkage (a) O 2 (b) CCl 4 (c) CHCl 3 (d) NaBr 28. An electrovalent compound does not exhibit space isomerism because of (a) Presence of oppositely charged ions (b) High melting points (c) Non-directional nature of the bond (d) Crystalline nature [MP PET 2001] 29. Which of the following species contain non-directional bond + (a) NH 3 (b) CsCl (c) NH 4 (d) BeF 2 [EAMCET 1986] 30. Ionic reactions are (a) Fast (b) Slow (c) Very slow (d) None of these [MP PET 2000] 31. Which of the following is the least ionic (a) C 2 H 5 Cl
(b) KCl
(d) C 6 H 5 N H 3 Cl
(c) BaCl 2
32. Which of the following hydride is ionic
(a) H 2 O
(b) NH 3
[REE 1999]
(c) CaH 2
(d) H 2 S
33. Which of the following bonds is the most ionic
[CET 1990]
(a) Cs Cl (b) Al Cl (c) C Cl 34. Which of the following is an ionic compound (a) SO 3 (b) ICl (c) KI (d) CHCl 3
(d) H Cl
35. The compound which does not contain ionic bond is
(a) NaOH (b) HCl
(c) K 2 S
(d) LiH
Ad Ad vance Level 36. An ionic compound A B is most likely to be formed when
(a) The ionization energy of A is high and electron affinity of B is low (b) The ionization energy of A is low and electron affinity of B is high (c) Both, the ionization energy of A and electron affinity of B are high (d) Both, the ionization energy of A and electron affinity of B are low 37. Which of the following statements concerning elements with atomic number 10 is true (a) It forms a covalent network of solids (b) It is monoatomic in nature (c) It has a very high value of electron affinity (d) It has extremely low value of ionisation energy 38. In which of the following solvents, KI has highest solubility? The dielectric constant ( ) of each liquid is given in parentheses
CHEMICAL BONDING
(b) (CH 3 )2 CO( 2) (c) 39. Which one is the highest melting halide (a) NaCl (b) NaF (c) 40. Ionic reactions occur in (a) Aqueous solution and organic solvents of high polarity (b) Non-polar or solvents of low polarity (c) Gaseous state (d) Solid state 41. The stability of ionic crystal depends principally on (a) High electron affinity of anion forming species (b) (c) Low I.E. of cation forming species (d) 42. Which of the following statements about LiCl and NaCl is wrong (a) LiCl has lower melting point than NaCl (b) LiCl dissolves more in organic solvents whereas NaCl does not (c) LiCl would ionise in water more than NaCl (d) Fused LiCl would be less conducting electric than fused NaCl (a)
C 6 H 6 ( 0)
CH 3 OH ( 32)
(d) CCl 4 ( 0)
NaBr
(d) NaI
The lattice energy of crystal Low heat of sublimation of cation forming solid
43. In K F , ionic radius of F is more than F while ionic radius of K is
(a) Less than K (b) More than F (c) Equal to F (d) None of these [CPMT 1997] 44. The energy that opposes dissolution of a solvent is (a) Hydration energy (b) Lattice energy (c) Internal energy (d) Bond energy [CBSE 1993] 45. Strongest bond is in (a) NaCl (b) CsCl (c) Both (a) and (b) (d) None of these 46. Amongst LiCl , RbCl , BeCl 2 and MgCl 2, the compounds with the greatest and the least ionic character respectively are [MNR 1993; UPSEAT 2002]
47.
(a) LiCl and RbCl (b) RbCl and BeCl 2 The bonding is electrovalent in (a) NaCl (b) Br 2
(c) RbCl and MgCl 2
(d) MgCl 2 and BeCl 2 [CPMT 2000]
(c) PF 5
(d) XeF 4
48. On the basis of concept of ionic potential ( ), the tendency to form covalent bond in a group
[BHU 1994]
(a) Increases (b) Decreases (c) Remains unchanged (d) Shows erratic change 49. The values of electronegativity of atoms A and B are 1.20 and 4.0 respectively. The percentage of ionic character in A– B bond is [DCE 2000]
(a) 50% (b) 72.24% 50. Which liquid is not deflected by a non-uniform electrostatic field (a) Water (b) Chloroform 51. When ionic compounds get dissolved in water (a) They involve heat changes (b) Inter ionic attraction is reduced (c) Ions show dipole-ion attraction with water molecules (d) All of these 52. Which forms a crystal of NaCl (a) NaCl molecules (b) Na+ and Cl – ions 53. The electronic configurations of four elements L : 1 s 2 2 s 2 2 p 4 ;
(c) 55.3%
(d) 43%
(c) Nitrobenzene
(d) Hexane
[CPMT 1972; NCERT 1976; DPMT 1996]
(c) Na and Cl atoms
(d) None of these
Q : 1 s 2 2 s 2 2 p 6 ,3 s 2 3 p 5
P : 1 s 2 2 s 2 2 p 6 ,3 s1 R : 1 s 2 2 s 2 2 p 6 ,3 s 2 The formulae of ionic compounds that could be formed between them are (a) L 2 P , RL, PQ, R 2 Q (b) LP , RL, PQ, RQ (c) P 2 L, RL, PQ, RQ 2 54. Strongest bond is (a) C – F (b) C – H (c) C – N
[NCERT 1983]
(d) LP , R 2 L, P 2 Q, RQ [AFMC 1991]
(d) C – O X. Covalent Bond
Basic Level
CHEMICAL BONDING
55. Which of the following statement is correct for covalent bond
[BHU 1997]
(a) Electrons are shared between two atoms (b) Direction is non-polar (c) It may be polar or non-polar (d) Valency electrons are attracted towards one atom [CPMT 1986] 56. The bond between two identical non-metal atoms has a pair of electrons (a) Unequally shared between the two atoms (b) Transferred fully from one atom to another atom (c) With identical spin (d) Equally shared between them [Pb. PET 2001] 57. Boron forms covalent compound due to (a) Small size (b) Higher ionization energy (c) Lower ionization energy (d) Both ‘a’ and ‘ b’ [IIT Screening 2001] 58. The number of S - S bonds in sulphur tri-oxide trimer ( S 3O9) is (a) Three (b) Two (c) One (d) Zero [Roorkee 1989] 59. In covalency (a) The transfer of electrons take place (b) Sharing of electrons take place (c) The electrons are shared by only one atom (d) None of these 60. The maximum number of covalent bonds by which the two atoms can be bonded to each other is (a) Four (b) Two (c) Three (d) No fixed number [NCERT 1975; CPMT 1981, 1983] 61. In a double bond connecting two atoms there is a sharing of (a) 2 electrons (b) 4 electrons (c) 1 electron (d) 6 electrons [Delhi PMT 2001] 62. The ICl molecule is (a) Purely covalent (b) Purely electrovalent (c) Polar with negative end on chlorine (d) Polar with negative end on iodine [CPMT 1988] 63. Covalent compounds are soluble in (a) Polar solvents (b) Non-polar solvents (c) Concentrated acids (d) All solvents 64. Covalent compounds in water are generally (a) Soluble (b) Insoluble (c) Dissociated (d) Hydrolysed [Pb CET 1992] 65. Which of the following properties would suggest that a compound under investigation is covalent (a) It conducts electricity on melting (b) It is a non-electrolyte (c) It has a high melting point (d) It is a compound of a metal and a non-metal [KCET 2002] 66. Covalent compounds have low melting point because (a) Covalent bond is less exothermic (b) Covalent molecules have definite shape (c) Covalent bond is weaker than ionic bond (d) Covalent molecules are held by weak Vander Waal’s force of attraction [JIPMER 1999] 67. Which of the following substance when dissolved in water wi ll give a solution that does not conduct electricity (a) HCl (b) KOH (c) CH 3COONa (d) NH 2CONH 2 68. At ordinary temperature and pressure among halogens, chlorine is a gas, bromine is a liquid and iodine is a solid. This is because (a) The specific heat is in the order I 2 > Br2 > Cl 2 (b) Intermolecular forces among molecules of chlorine are weakest and those of iodine the strongest (c) The order of density is I 2 > Br2 > Cl 2 (d) The order of stability is I 2 > Br2 > Cl 2 69. Which factor is most responsible for the increase in boiling points of nobl e gases from He to Xe (a) Decrease in I.E. (b) Monoatomic nature
CHEMICAL BONDING
(c) Decrease in polarisability polarisability 70. The chemical inertness of N 2 is attributed to (a) The presence of large number of bonding electrons in comparison to antibonding electrons (b) Its high heat of dissociation (c) Presence of a triple bond between nitrogen atoms which makes the molecule quite stable (d) All the statements are correct 71. The type of bonds presents in green vitriol ( FeSO4.7 H 2O) (a) Electrovalent and covalent coordinate (c) Electrovalent, covalent and coordinate 72. Which one possesses strong covalent bond (a) H Cl (b) Cl Cl (c) C Cl 73. Which has a covalent bond (a) Na 2 S (b) SnCl 4 (c) NaH
(d)
Increase
(b) Electrovalent
and
(d) Covalent and coordinate [EAMCET 1988]
(d) B Cl [AFMC 1988]
(d) MgCl 2
74. Among the alkaline earth metals the element forming predominantly covalent compound is
(a) Be (b) Mg (c) Sr (d) Calcium 75. Maximum covalency of an atom of an element is equal to (a) Number of unpaired electrons in the s and p orbitals of valency shell (b) Number of unpaired electrons in the p-orbitals of valency shell (c) Total number of electrons in the s- and p-orbitals of valency shell (d) Total number of electrons in the p-orbitals of valency shell 76. Which one is electron deficient compound (a) NH 3 (b) ICl (c) BCl 3 (d) PCl 3 77. Electron deficient species are known as (a) Lewis acids (b) Hydrophobic Lewis base 78. How many bonded electron pairs are present in IF 7 molecule (a) 6 (b) 7 (c) 5 (d) 8
in
[MNR 1986]
[Roorkee 1991]
(c) Nucleophile
(d)
79. Which of the following compounds has a 2 electron 3 centre bond 80.
81. 82. 83. 84. 85.
(a) BF 3 (b) NH 3 (c) B2 H 6 (d) In the electronic structure of acetic acid there are (a) 16 shared and 8 unshared electrons electrons (c) 12 shared and 12 unshared electrons electrons Which contains a triple bond (a) SO3 (b) HCN (c) NH 3 (d) Which of the following compounds does not follow octet rule (a) CO2 (b) PCl 3 (c) ICl (d) ClF 3 The octet rule is not valid for the molecule (a) CO2 (b) H 2O (c) O2 (d) CO Which of the following is an example of super octet molecule (a) ClF 3 (b) PCl 5 (c) IF 7 (d)
CO2 [AMU 1983]
(b) 8 shared and 16 unshared (d) 18 shared and 6 unshared
C 2 H 4
[IIT 1979]
All of these
The octet rule is not followed in (a) F 2 (b) NaF (c) CaF 2 (d) BF 3
Ad Ad vance Level 86. Bond energy of covalent O – H bond in water is
(a) Greater than bond energy of hydrogen bond (c) Less than bond energy of hydrogen bond
[EAMCET 1982]
(b) Equal to bond energy of hydrogen bond (d) None of these
CHEMICAL BONDING
87. In which one of the following cases, breaking of covalent bond takes place
(a) Boiling of H 2O (b) Melting of KCN 88. Which of the following has least covalent P – H bond
(c) Boiling of CF 4
(d) Melting of SiO2 [CPMT 1996]
(a) PH 3 (b) P 2 H 6 (c) P 2 H 5 (d) PH 6 89. Which of the following statements concerning a covalent bond is false (a) The electrons are shared between atoms (b) The bond is non-directional (c) The strength of the bond depends upon the extent of overlapping (d) The bond formed may be polar or non-polar 90. Which of the following is not a characteristic of covalent compounds (a) It has low melting point and boiling point (b) It is formed between two atoms having no or very small electronegativity difference (c) They have no definite geometry (d) They are generally insoluble in water 91. Which of the following statement is incorrect (a) Sodium hydride is ionic (b) Beryllium chloride is covalent (c) CCl 4 gives a white ppt. with AgNO3 solution (d) Bonds in NaCl are non-directional 92. Which has higher bond energy and stronger bond (a) F 2 (b) Cl 2 (c) Br2 (d) I 2
[Roorkee 1997]
IX. Co-ordinate bond or Dative bond
Basic Level 93. Co-ordinate compounds are formed by
(a) Transfer of electrons (b) Sharing of electrons (d) None of these 94. Which of the following contains a co-ordinate bond (a) N 2 H 5+ (b) BaCl 2 (c) HCl (d) 95. Dative Bond is present in (a) SO3 (b) NH 3 (c) BaCl 2 (d) 96. A lone pair of electrons in an atom implies (a) A pair of valence electrons not involved in bonding involved in bonding (c) A pair of electrons (d) 97. Which one of the following has a co-ordinate bond (a) NH 4Cl (b) AlCl 3 (c) NaCl 98. In the co-ordinate valency (a) Electrons are equally shared by the atoms shared by two atoms (c) Hydrogen bond is formed 99. For co-ordinate bond formation, the acceptor atom must have an orbital (a) With a unshared pair of electron (b) With a unpaired single electron (c) With no electron (d) With an electron of opposite spin to that of donor atom 100. Which are of the following molecule has a co-ordinate b ond (a) CH 3 NO 2 (b) AlCl 3 (c) NaCl
(c) Donation of electron pair
H 2O BF 3 [KCET (M&E) 2002]
(b) A
pair
2
[CPMT 1983]
(d) Cl 2 [CPMT 1989]
(b) Electrons of one atom are (d)
None of these
[CPMT 1994]
(d) CH 2 Cl 2 [RPMT 2002]
(a) BH 4 (b) CO 3 (c) H 3 O
(d) NH 4
102. A simple example of a co-ordinate covalent bond is exhibited by
(a)
C 2 H 2
electrons
A pair of valence electrons
101. Co-ordinate bond is absent in
of
(b) H 2 SO 4
103. Which of the following does not have a co-ordinate bond
[NCERT 1984]
(c) NH 3
(d) HCl
CHEMICAL BONDING
(a)
SO 2
(b) HNO 3 (c) H 2 SO 3 104. The compound containing co-ordinate bond is (a) H 2 SO 4 (b) O 3 (c) SO 3 105. The number of dative bonds in sulphuric acid molecule is (a) 0 (b) 1 (c) 2 106. Which combination is best explained by the co-ordinate covalent bond (a) H H 2 O
(b) Cl Cl
(c) Mg
1 2
O2
(d) HNO 2 [AFMC 1999]
(d) All of these [MP PET 2002]
(d) 4 [JIPMER 2001; CBSE 1990]
(d) H 2 I 2
107. Which of the following compound has co-ordinate (dative) bond
(a)
CH 3 NC
(b) CH 3 OH
[RPET 2003]
(c) CH 3 Cl
(d) NH 3 VIII.
Dipole moment
Basic Level 108. The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order K 2CO3
(I), MgCO3 (II), CaCO3 (III), BeCO3 (IV) [IIT 1996] (a) I < II < III < IV (b) IV < II < III < I (c) IV < II < I < III (d) II < IV < III < I 109. Which compound is non–polar (a) CHCl 3 (b) SiCl 4 (c) SnCl 2 (d) NH 3 110. BeF 2 has no dipole moment because of (a) Covalent bond (b) Linear and symmetrical structure (c) Non–linear structure (d) No charge displacement [Raj PMT 2000, Orissa JEE– 2002] 111. Which molecule has zero dipole–moment (a) NH 3 (b) H 2O (c) BCl 3 (d) CHCl 3 [MP PMT 2002, IIT screening – 2002] 112. Which of the following has zero dipole moment (a) ClF (b) PCl 3 (c) SiF 4 (d) CFCl 3 [AFMC 1994] 113. Molecules having dipole moment is (a) 2, 2– dimethyl propane (b) Trans–2 pentene (c) Neopentane (d) 2, 2, 3, 3–tetramethyl butane 114. Which of the following is not correct (a) Lone pair of electrons present on central atom can give rise to dipole moment (b) Dipole moment is vector quantity (c) CO2 molecule has dipole moment (d) Difference in electronegativities of combinating atoms can lead to dipole moment [IIT 1992] 115. The molecule having dipole moment is (a) 2, 2–dimethylpropane (b) Trans–3 hexene (c) Trans–2 pentene (d) 2, 2, 3, 3–tetramethylbutane [AFMC 1998] 116. Which one is most polar (a) CCl 4 (b) CH Cl 3 (c) CH 3 Cl (d) CH 3 OH [NTSE 1992] 117. The compound with the maximum dipole moment among the following is (a) p–dichlorobenzene (b) m–dichlorobenzene (c) o–dichlorobenzene (d) Carbon tetrachloride [IIT 1988] 118.Polarisation of electrons in acrolein may be written as (a)
CH 2 CH CH O
(b) CH 2 CH CH O
119. Which of the following has the highest dipole moment
(c) CH 2 CH CH O
(d) CH 2 CH CH O
[AIIMS 2002]
CHEMICAL BONDING
H
(a) H
\ /
C O
(b)
H
CH 3
|
|
C
C
|
|
CH 3
H 3 C H
H
(c)
|
|
C
C
|
|
CH 3
H |
(d)
(b) SbH 3
C
| C
|
CH 3
H
120. Which of the following has the highest dipole moment
(a) AsH 3
CH 3
|
CH 3
[CBSE PMT 1997, Haryana CEET 2000]
(c) PH 3
(d) NH 3
121. Which one of the following statements regarding intermolecular attraction is correct in the case of neon gas
(a) Dipole-dipole interaction (c) Instantaneous dipole-induced dipole interaction 122. The geometry of H 2 S and its dipole moment are
(b) (d) Ion-ion interaction
Ion-dipole interaction
(a) Angular and non-zero (b) Angular and zero 123. Which of the following molecules has non-zero dipole m oment (a) PCl 5 (b) BF 3
(c) Linear and non-zero
(d) Linear and zero
(c) SO 2
(d) CO 2
[IIT 1999]
124. Which of the following has a net dipole moment
(a) CCl 4
(b) BF 3
[Pb. CET 1997]
(c) NH 3
(d) CO 2
(c) BF 3
(d) NH 3
125. Dipole moment of NF 3 is smaller than
(a) CO
(b) CO 2
126. Which of the following bond has the most polar character
(a) C O (b) C Br 127. Which of the following has least polarity in bond (a) H F (b) H Cl 128. Which of the following bonds will be non-po lar (a) N H (b) C H 129. The type of bonding in HCl molecule is (a) Pure covalent (b) Polar covalent 130. Which of the following have both polar and non-polar bonds (a) C 2 H 6 (b) NH 4 Cl
[DCPMT 1982]
(c) C F
(d) C S [RPMT 1985]
(c) H O
(d) H S [Pb. CET 1985]
(d) O H
(c) F F
[NCERT 1983, 84; AIIMS 1992]
(c) Highly polar
(d) Hydrogen bonding [AIIMS 1997]
(c) HCl
(d) AlCl 3
131. Which one of the following molecule does not possess a perma nent electric dipole moment
(a) SO 2
(b) C 6 H 6
(c) PCl 3
[Pb. CET 1994]
(d) H 2 S
132. The cyanide ion, CN and N 2 are isoelectronic. But in contrast to CN , N 2 is chemically inert, because of
[IIT 1992]
(a) Absence of bond polarity (b) Absence of bond polarity (c) Unsymmetrical electron distribution (d) Presence of more number of electrons in bonding orbitals 133. Experiment shows that H 2 O has dipole moment whereas CO 2 has not. Point out the structures which best illustrate these facts O [DPMT 1984; NCERT 1983; CPMT 1984] (a) O C O ; H (c) O
H
O ; H O H
C 134. Which molecule does not show zero dipole moment (a) BF 3 (b) NH 3
(b) O C O ; H O H (d) C
O ; H O
O
O
(c) CCl 4
[RPET 1999]
(d) CH 4
135. Dipole moment is shown by
(a) 1, 4-dichlorobenzene (c) trans 1, 2-dichloroethene 136. Dipole moment is highest in (a) CHCl 3 (b) CH 4
[IIT 1986]
(b) cis 1, 2-dichloroethene (d)
trans 2, 3-dichloro-2-butene [CPMT 1993]
(c) CHF 3
(d) CCl 4
CHEMICAL BONDING
137.The molecule which has the largest dipole moment amongst the following is
(a) CH 4
(b) CHCl 3
(c) CCl 4
(d) CH 2 Cl 2
138. Which contains both polar and non-polar bonds
(a) NH 4 Cl
[IIT 1997]
(b) HCN
(c) H 2 O 2
(d) CH 4
139. The electronegativity values of C , H , O, N and S are 2.5, 2.1, 3.5, 3.0 and 2.5 respectively. Which of the following bonds is most
polar (a) C H
[EAMCET 1986]
(b) N H
(c) S H
(d) O H
140. The molecular size of ICl and Br 2 is approximately same, but boiling point of ICl is about 40 o higher than that of Br 2 , it is
because (a) ICl bond is stronger than Br Br bond (c) ICl is polar while Br 2 is nonpolar
(b) IE of iodine < IE of Br (d) I has larger size than Br
141. The critical temperature of water is higher than that of O 2 because H 2 O molecule has
(a) Fewer electrons than O 2
(b)
[IIT 1997]
Two covalent bonds
(c) V shape (d) Dipole moment 142. If molecule MX 3 has zero dipole moment, the sigma bonding orbitals used by M (atomic number < 21) are (a) Pure p (b) sp hybrid 143. The correct order of decreasing polarity is (a) HF SO 2 H 2 O NH 3
(c) sp2 hybrid
[IIT 1981]
(d) sp 3 hybrid
(b) HF H 2 O SO 2 NH 3
(c) HF NH 3 SO 2 H 2 O (d) H 2 O NH 3 SO 2 HF 144. In terms of polar character, which of the following order is correct (a) NH 3 H 2 O HF H 2 S (b) H 2 S NH 3 H 2 O HF (c) H 2 O NH 3 H 2 S HF 145. The order of dipole moments of the following molecules is (a) CHCl 3 CH 2 Cl 2 CH 3 Cl CCl 4
(d) HF H 2 O NH 3 H 2 S (b) CH 2 Cl 2 CH 3 Cl CHCl 3 CCl 4
(c) CH 3 Cl CH 2 Cl 2 CHCl 3 CCl 4 (d) CH 2 Cl 2 CHCl 3 CH 3 Cl CCl 4 146. Arrange the following compounds in order of increasing dipole moment Toluene (I), m -dichlorobenzene (II), o dichlorobenzene (III), p -dichlorobenzene (IV) (a) I, IV, II III (b) IV, I, II, III (c) IV, I, III, II (d) IV, II, I III [RPET 1994] 147. The correct order of dipole moment is (a) CH 4 NF 3 NH 3 H 2 O (b) NF 3 CH 4 NH 3 H 2 O (c) NH 3 NF 3 CH 4 H 2 O 148. Which have zero dipole moment (a) 1, 1-dichloroethene (b) Cis 1, 2-dichloroethene 149. An example of non-polar molecule is (a) SiF 4 (b) ClF 3
(d) H 2 O NH 3 NF 3 CH 4 [IIT 1987]
(c) Trans 1, 2-dichloroethene (d) None (c) PCl 3
(d) SO 2
Ad Ad vance Level 150. Which of the following about H 2 O molecule is not true
(a) The molecule has 0 (b) The molecule can act as a base (c) The substance shows abnormally high boiling point in comparison to the hydrides of other elements of oxygen group (d) The molecule has a bent shape 151. BF 3 and NF 3 both are covalent compounds but NF 3 is polar whereas BF 3 is non-polar. This is because (a) Nitrogen atom is smaller than boron atom
CHEMICAL BONDING
(b) N F bond is more polar than B F bond (c) NF 3 is pyramidal whereas BF 3 is planar triangular (d) BF 3 is electron deficiant whereas NF 3 is not 152. The dipole moment of chlorobenzene is 1.73 D. the dipole moment of
(a) 3.46 D
(b) 0.00 D
p -dichlorobenzene is expected to be
(c) 1. 73 D
[CPMT 1991]
(d) 1.00 D
153. The dipole moment of CHCl 3 is 1.05 debye while that of CCl 4 is zero, because CCl 4 is
(a) Linear (b) Symmetrical (c) Planar (d) Regular tetrahedral 154. Which bond angle, would result in the maximum dipole moment for the triatomic molecule XY 2 (a) 120 o
(c) 180 o
(b) 90 o
(d) 150 o
155. The dipole moment of HBr is 1 .6 10 30 coloumb metre and interatomic spacing is 1 Å . The % ionic character of HBr is [IIT 1997]
(a) 7
(b) 10
(c) 15
(d) 27 VII.
Polarisation and Fajan’s rule
Basic Level 156. Which of the following statement is correct
(a) Polarisation of an anion is maximum by high charged cation (b) Small sized cation minimises the polarisation (c) A small anion brings about a large degree of polarisation (d) A small anion undergoes a high degree of polarisation 157.The correct order of decreasing polarisability of the ions is (a)
Cl , Br , I , F
(b) F , I , Br , Cl
[CBSE 1990, DCE 1999]
(c) F , Cl , Br , I
(d) I , Br , Cl , F
158. According to Fajan’s rule covalent bond is favoured by
[CBSE 1990, DCE 1999]
(a) Large cation and small anion (b) Large cation and large anion (c) Small cation and small anion (d) Small cation and large anion [CBSE 1990, DCE 1999] 159. Which ion has a higher polarising power 2+ 3+ 2+ (a) Mg (b) Al (c) Ca (d) Na+ [CBSE 1990] 160. Among LiCl, BeCl 2, BCl 3 and CCl 4, the covalent bond character follows the order (a) LiCl BeCl 2 BCl 3 CCl 4 (b) LiCl BeCl 2 BCl 3 CCl 4 (c) LiCl BeCl 2 CCl 4 BCl 3
(d) LiCl BeCl 2 BCl 3 CCl 4
161. Which compound among the following has more covalent character
(a) AlCl 3 (b) AlI 3 162. Highest covalent character is found in which of the following (a) CaF 2 (b) CaCl 2 163. Which of the following has the least ionic character (a) FeCl 2 (b) ZnCl 2
[Roorkee 1990]
(c) MgI 2
(d) NaI [EAMCET 1990]
(c) CaI 2
(d) CaBr2
(c) CdCl 2
(d) MgCl 2
Ad Ad vance Level 164. Polarisation is the distortion of the shape of an anion by an adjacently placed cation. Which of the following statements is correct
(a) Maximum polarisation is brought about by a cation of high charge (b) Minimum polarisation is brought about by a cation of low radius (c) A large cation is likely to bring about a large degree of polarisation (d) Polarising power of cation is less than that of anion 165. Polarisation power of a cation increases when (a) Charge on the cation increases (b) Size of the cation increases
[AMU 1997]
CHEMICAL BONDING
(c) Charge on the cation decreases
(d) Has no relation to its size or charge VI. Overlapping,
and
bonds
Basic Level 166. The example of the p – p orbital overlapping is the formation of
(a) H 2 molecule (b) Cl 2 molecule (c) Hydrogen chloride molecule (d) Hydrogen bromide 167. Which statement is not correct (a) Double bond is shorter than a single bond (b) Sigma bond is weaker than -bond (c) Double bond is stronger than a sigma bond (d) Covalent bond is stronger than hydrogen bond 168. Which of the following bonds is most stable (a) 1s – 1s (b) 2 p – 2 p (c) 2s – 2 p (d) 1s – 2 p 169. - bonds are formed (a) After - bond (b) Before - bond (c) With - bond (d) None of these 170. Formation of - bond (a) Increases bond length (b) Decreases bond length (c) Distorts the geometry of molecule (d) Make homoatomic molecules more reactive 171. A covalent bond is formed between the atoms by the overlapping of orbitals containing (a) Single electron (b) Paired electron (c) Single electron with parallel spin (d) Single electron with opposite spin 172.s-orbital always forms (a) - bond (b) -bond (c) Both and -bonds (d) None of these 173. Which can not be explained by VBT – (a) Overlapping (b) Bond formation (c) Paramagnetic nature of oxygen (d) Shapes of molecules 174. Axial overlapping is also known as (a) Linear overlapping (b) End to end overlapping (c) Head on overlapping (d) All of these 175. Which atomic orbitals is always involved in sigma bonding only (a) s (b) p (c) d (d) f [DPMT 1990] 176. Strongest bond is (a) C C (b) C C (c) C C (d) All are equally strong [AMU 1983] 177. Acetate ion contains (a) One C, O single bond and one C, O double bond (b) Two C, O single bonds (c) Two C, O double bonds (d) None of these 178. A covalent bond may be formed by (a) s-s-overlap (b) s- p-overlap (c) p- p-overlap (d) All of these 179. The strength of bonds formed by 2s-2s, 2 p-2 p and 2 p-2s overlap has the following order (a) s-s > p- p > p-s (b) s-s > p-s > p- p (c) p - p > p - s > s - s (d) p - p > s - s > p - s [DCE 2001] 180. Number of bonds in SO2 (a) Two and two (b) Two and one and one lone pair (c) Two , two and one lone pair (d) None of these 181.How many and bonds are there in the molecule of X C O
[ X = Cl , Br or I ]
X
(a) One and one bond
(b) Two and two bond
(c) Three and three bond (d) Three and one bond
CHEMICAL BONDING
182. No. of bonds in P 4 O10 are
[AIEEE 2002]
(a) 6
183.
184. 185.
186.
187.
(b) 7 (c) 17 (d) 16 Assertion : (sigma) is a strong bond while (pi) is a weak bond [ AIIMS 2002] Reason : Atoms rotate freely about pi bound (a) If both Assertion & Reason are true and reason is explanations of assertion (b) If both Assertion and Reason are true but reason is not correct explanation of the assertion (c) If Assertion is true and Reason is false (d) If Assertion is false but Reason is true (e) Both Assertion and Reason are false An acetylene molecule has [MP PMT 1990; NCERT 1979; EAMCET 1978; CPMT 1988; MADT Bihar 1982; AMU 1985; MHCET 2000] (a) 3 and 2 bonds (b) 4 and 1 bonds (c) 2 and 3 bonds (d) All bonds A double bond between two carbon atoms consists of [MP PET 1993] (a) Two sigma bonds (b) Two pi bond (c) One sigma and one pi bond (d) A pair of electrons The triple bond between N atoms of nitrogen molecule ( N N ) consists of [MP PET 1996; RPMT 2000] (a) Three -bonds (b) Two -bonds and one bond (c) One -bond and two -bonds (d) Three -bonds Which of the following molecules has no bond (a) C 2 H 2 (b) C 2 H 4 (c) Cyclohexane (d) Benzene
188. The number of bonds in Xylene is
[MP PET 1996]
(a) 6 (b) 9 (c) 189. Which bond has the highest bond energy (a) Co-ordinate bond (b) Sigma bond (c) 190. Tetrahedral nature of bonding in C -atom was first shown by (a) Kekule (b) Vant Hoff and Le Bel (c) 191. The number and type of bonds between two carbon atoms in CaC 2 are (a) One sigma () and one pi ( ) bond (b) (c) One sigma () and one and a half pi ( ) bond (d)
12
(d) 18
Multiple bond
(d) Polar covalent bond [CPMT 1974; DPMT 1982]
Lewis
(d) Pasteur [IIT 1996]
One sigma () and two pi () bonds One sigma () bond
Ad Ad vance Level 192. In which of the following there exists a p – d bonding
(a) Diamond (b) Graphite 193. Which of the following has p – d bonding
(a) NO 3
(b)
CO 32
[AFMC 2001]
(c) Dimethyl amine
(d) Trisilylamine [CBSE 2002]
(c)
BO 33
2
(d) SO 3
194. Among the following ions the p – d overlap could be present in
(a) NO 3
(b) PO 43
[CBSE 2000]
(c) CO 32
(d) NO 21 V. Hybridization
Basic Level 195. Hybridization involves
(a) Addition of an electron pair (c) Separation of orbitals Which p-orbitals overlapping would give the strongest bond 196.
(b) Mixing up of atomic orbitals (d) Removal of electron pair [AIIMS 1987]
CHEMICAL BONDING
(a)
(b)
(c)
(d)
197. A hybrid orbital formed from s and p-orbital can contribute to
(a) A bond only (b) bond only (c) Either or bond 198. Type of hybridization of central carbon in propandiene is (a) sp3 (b) sp2 (c) sp 199. Which carbon is more electronegative (a) sp3 hybridised carbon (b) sp hybridised carbon (c) sp2 hybridised carbon (d) The electron attracting power of C is always same irrespective of its hybrid state 200. Which of the following will provide the most efficient overlap? (a) s - s (b) s - p (c) sp2 – sp2 201. On hybridization of one s and one p orbitals we get (a) Two mutually perpendicular orbitals (b) Two orbitals at 180o (c) Four orbitals directed tetrahedrally (d) Three orbitals in a plane 202. Beryllium atom in beryllium fluoride is (a) sp3 hybridized (b) sp2 hybridized (c) sp hybridized 203. As the s-character of hybridization orbital increases, the bond angle (a) Increases (b) Decreases (c) Does not change 204. In which of the following the angle between the two covalent bonds is greatest (a) H 2O (b) NH 3 (c) CH 4
(d) Cannot be predicted (d) None of these [Haryana 1999]
(d) sp - sp
[AIIMS 1985]
(d) unhybridized (d) Becomes zero [JIPMER 2001]
(d) CO2
205. The hybridization of Ag in the linear complex [ Ag ( NH 3 ) 2 ] is
[AIIMS 1983]
(a) dsp2 (b) sp (c) 206. Equilateral shape has (a) sp hybridization (b) sp2 hybridization (c) Shape of BF molecule is 207. 3 (a) Linear (b) Planar (c) 208. In the following molecule, the hybrid state of 1 and 3 carbon atoms is : (a) sp3 (b) sp2 209. The carbon atoms in graphite are (a) sp3 hybridized (b) sp hybridized 210. The hybridization in benzene is (a) sp (b) sp2 211. Which of the following has sp2-hybridization? (a) C 2 H 6 (b) C 2 H 4
sp2
(d) sp3 [CPMT 1988]
sp3
hybridization
(d)
sp3d hybridization [AIIMS 2000]
Tetrahedral CH 2 C CH 2 (c) sp
(d) Square pyramidal (d) sp3d
(c) sp2 hybridized
(d) None of these [UPSEAT 2003]
(c)
sp3
(d)
dsp2 [CPMT 1996]
(c) BeCl 2
212. The hybridization of S atom in SO2 is
(d) C 2 H 2 [Haryana CEET 1998; Pb CET 1998]
(a) sp (b) sp2 (c) sp3 213. In which one of the following molecules can the central atom said to adopt sp2 hybridization (a) BeF 2 (b) BF 3 (c) C 2 H 2
(d) sp3d [CPMT 1989]
(d) NH 3
214. In HCHO carbon atom has hybridization
sp2
(a) sp (b) 215. Which molecule is planar (a) NH 3 (b) CH 4 216. Which of the following has maximum bond angle (a) NH 3 (b) NH 4 217. Which compound does not possess linear geometry (a) CH 2 CH 2 (b) CH CH
[AIIMS 1987]
(c)
sp3
(d) None of these [PUCET 1998]
(c) C 2 H 4
(d) SiCl 4 [CBSE 2001; CET MP 1994]
(c) PCl 3
(d) BCl 3 [RPET 1999]
(c) BeCl 2
(d) CO 2
CHEMICAL BONDING
218. Which one of the following is a correct set with respect to molecule, hybridization and shape 2
(a) BeCl 2 , sp , linear
(b) BeCl 2 , sp , triangular planar
(c) BCl 3 , sp 2 , triangular planar 219. Pyramidal shape would be of (a) NO 3
[EAMCET Medical 2003]
2
(d) BCl 3 , sp 3 , tetrahedra l [RPET 1999]
(c) H 3 O
(b) H 2 O
(d) NH 4
220. The shape of CH 3 species is
[RPET 1999; CPMT 1997]
(a) Tetrahedral (b) Square planar 221. Which one of the following is a planar molecules
(c) Trigonal planar
(a) NH 3 (b) H 3 O 222. A sp3 hybrid orbital contains (a) 1/4 s character (b) 1/2s character 223. The shape of sulphate ion is (a) Square planar (b) Tetrahedral 224. The nature of hybridization in the NH 3 molecules is (a) sp (b) sp2 225. The structure and hybridization of Si (CH 3 )4 are, respectively
(c) BCl 3
(d) Linear [EAMCET Medical 2003]
(d) PCl 3 [DPMT 1984; BHU 1985; CPMT 1976]
(c) 2/3 s character
(d) 3/4s character [CPMT 1982]
(c) Trigonal bipyramidal
(d) Hexagonal
(c) sp3
(d) sp2d [CBSE 1996]
(a) Bent and sp (b) Trigonal and sp2 (c) Octahedral and sp3d 226. Select the correct hybridization in N-atom in NH 2 NH 2 molecule
(d) Tetrahedral and sp3
(a) sp (b) sp2 227. The central atom assumes sp3 hybridization in
(d) dsp2
(a) AsCl 3
(c) sp3
[MNR Allahabad 1994]
(b) SO 3
(c) BF 3
(d) NO 3
228. A molecule with four bonded electron pairs on the central atom a nd no lone pair is likely to be
(a) Linear (b) Tetrahedral 229. In piperidine NH , the hybrid state assumed by N is (a) sp (b) sp2 230. The central atom assumes sp3-hybridization in (a) PCl 3
(b) SO 3
(c) Octahedral
(d) Triangular planar [Haryana CEET 1998]
(c) sp3
(d) dsp2 [MNR 1994]
(c) BF 3
(d) NO 3
231. In which of the following, the central atom does not use sp3 hybrid orbitals in its bonding
(a) BeF 3
(b) OH 3
(c) NH 2
[MNR 1992]
(d) NF 3
232. Which of the following pair has the same structure
(a) PH 3 and BCl 3
(b) SO 2 and NH 3
[BHU 2001]
(c) PCl 5 and SF 6
(d) NH 4 and SO 42
233. The bond angle in water molecule is nearly [NCERT 1980; EAMCET 1981; MNR 1983, 85; AIIMS 1982; CPMT 1989; MP PET 1994, 96; MP PET/PMT 1998]
120o
(a) (b) 180o (c) 109o28’ 234. Which of the following hybridization results in non-plana r orbitals (a) sp 3
(b) dsp 2
(c) sp 2
(d) 104o30’ (d) sp
235. The type of hybrid orbitals used by chlorine atom in ClO 2 is
(a) sp 3
(b) sp 2
[IIT 1992]
(c) sp
(d) None of these
(c) 110 o , sp 3
(d) None of these
..
236. The bond angle and hybridization in ether CH 3 O CH 3 is . .
(a) 106 o 51' , sp 3
(b) 104 o 31' , sp 3
237. Pyramidal shape would be of
(a) NO 3
[RPET 1999]
(b) H 2 O
(c) H 3 O
(d) NH 4
CHEMICAL BONDING
238. Which of the following has tetrahedral sturcture
(a)
CO
[CPMT 2000]
(b) NH 4
(c) K 4 [ Fe(CN )6 ]
(d) None of these
239. Central atom of the following compound has one lone pair of electrons and three bond pairs of electrons
(a) H 2 S
(b) AlCl 3
(c) NH 3
(d) BF 3
*
240. The compound in which C uses sp3 hybrid orbitals for bond formation is *
(a) H C OOH
*
*
(b) ( NH 2 )2 C O
(c) (CH 3 )3 C OH
*
(d) CH 2 C O
241. The type of hybridization of boron in diborane is
[BHU 1999]
(a) sp-hybridization (b) sp2-hybridization (c) sp3-hybridization 242. Which one of the following compounds has smallest bond a ngle in its molecule (a) NH 3 (b) SO 2 (c) OH 2
(d) sp3d 2 hybridization [AIEEE 2003]
(d) SH 2
Ad Ad vance Level [RPMT 2000] sp 3 d 2 hybridized central atom is (a) Square planar (b) Trigonal bipyramidal (c) Octahedral (d) Square pyramidal 244. The shape of CO2 molecule is similar to (a) H 2 S (b) SO 2 (c) CS 2 (d) All 245. Among the following compounds the one that is polar and has the central atom with sp2 hybridization is (a) H 2 CO 3 (b) SiF 4 (c) BF 3 (d) HClO 2 246. Homolytic fission of C –C bond in ethane (CH 3 CH 3 ) given an intermediate in which carbon atom is
243. The geometry of the molecule with
(a) sp 3 hybridized (b) sp 2 hybridized (c) sp hybridized 247. In H 2 O 2 molecule the angle between two O H planes is (Crystalline phase) (a) 90 o (b) 101 o 248. Which of the following does not have a tetrahedral structure (a) BH 4
(c) 103 o
[CBSE 2002]
(d) 105 o [CBSE 1993]
(c) NH 4
(b) ( AlCl 3 )2
(d) sp 2 d hybridized
(d) H 3 O
249. Which one of the following compounds has sp 2 hybridization
(a) CO 2
(b) SO 2
[IIT 1997]
(c) N 2 O
(d) CO
250. Structurally, similar ions are
(a) CH 4 , PCl 4
[CBSE 2001]
(b) H 2 O, I 3
(c) SF 6 , ICl 4
(d) BeCl 2 , SnCl 2
251. The hybridization in PF 3 is
[DCE 2000]
(a) sp 3 (b) sp 2 252. Which one of the following has not triangular pyramidal shape (a) NH 3 (b) NCl 3 253. For the molecule
lP
(c) dsp 3
(d) d 2 sp 3 [Roorkee 1992]
(c) PF 3
(d) BCl 3
(iii) If B = R RAR = 105o
(iv) If B = S SAS = 107o
lP A
(i) If B = P PAP = 92o
B
(ii) If B = Q QAQ = 100o
B
The ‘P’ character of hybrid orbitals of A would be maximum in
CHEMICAL BONDING
(a) (i) (b) (ii) (c) 254. Atomic orbitals of carbon in diamond are (a) sp hybridized (b) sp2 hybridized (c) 255. OF 2 is (a) Linear molecule and sp-hybridized (b) (c) Bent molecule and sp3-hybridized (d) 256. Which of the following molecule does not show tetrahedral shape (a) CCl 4 (b) SiCl 4 (c)
(iii)
(d) (iv) [MP PET 2002]
sp3
hybridized
(d) None hybridized [Roorkee 1996]
Tetrahedral molecule and None of these
sp3-hybridized [RPET 1999]
SF 4
(d) CF 4
257. Which of the following is not tetrahedral
(a) SCl 4
(b) SO 42
[MP PMT 2001]
(c) Ni(CO) 4
(d) NiCl 42
258. The shape of ClO 4 ion is
[PUCET 1986]
(a) Square planar (b) Square pyramidal (c) Tetrahedral 259. A square planar complex is formed by hybridization of which a tomic orbitals (a) s, p x , p y , d yz (b) s, p x , p y , d x 2 y 2 (c) s, p x , p y , d z 2
(d) Trigonal bipyramidal [AIEEE 2002]
(d) s, p y , p z , d xy
260. The d -orbital involved in sp 3 d hybridization is
(a) d x 3 y 2
(b) d xy
(c) d z 2
(d) d zx 2
(c) d x 2 y 2
(d) d xz
261. The d -orbitals involved in dsp 2 -hybridization is
(a) d xy
(b) d z 2
262. In [Cu( NH 3 )4 ]SO 4 , Cu has the following hybridization
[AIIMS 1988]
(a) dsp 2 (b) sp 3 (c) sp 2 263. A molecule is square planar with no lone pa ir. What type of hybridization is associated with it
(d) sp 3 d 2 [Pb CET 1991]
(a) sp 3 d (b) sp 3 d 2 (c) dsp 3 (d) dsp 2 264. The AsF 5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are (a) d x 2 y 2 , d z 2 , s, p x , p y
(b) d xy , s, p x , p y , p z
(c) s, p x , p y , p z , d z 2
(d) d x 2 y 2 , s, p x , p y
265. In the complex [SbF 5 ]2 , sp 3 d hybridization is present. Geometry of the complex is
(a) Square (b) Square pyramidal 266. Hybridization state of chlorine in ClF 3 is
(c) Square bipyramidal
(a) sp 3 (b) sp 3 d 267. The structure of PF 5 molecule is
(c) sp 3 d 2
(a) Tetrahedral (c) Trigonal bipyramidal 268. The shape of IF 7 molecule is
(b) Square planar (d) Pentagonal bipyramidal
(a) Octahedral
[CPMT 1997]
[Pb PMT 2000]
(d) Tetrahedral [RPET 1999]
(d) sp 3 d 3 [AFMC 1995]
[AFMC 2002]
(b) Pentagonal bipyramidal
(c) Trigonal bipyramidal
(d) Tetrahedral
269. The hybridization of atomic orbitals of nitrogen in NO 2 , NO 3 and NH 4 are
(a) sp 2 , sp 3 and sp 2 respectively
[IIT Screening test 2000]
(b) sp, sp 2 and sp 3 respectively
(c) sp 2 , sp and sp 3 respectively (d) sp 2 , sp 3 and sp respectively 270. The correct order of the O O bond length in O 2 , H 2 O 2 and O 3 is
[BHU Varanasi 2000]
(a) O 2 O 3 H 2 O 2 (b) O 3 H 2 O 2 O 2 (c) O 2 H 2 O 2 O 3 (d) H 2 O 2 O 3 O 2 [BHU Varanasi 2000] 271. Which of the following is the correct reducing order of bond angle (a) NH 3 CH 4 C 2 H 2 H 2 O (b) C 2 H 2 NH 3 H 2 O CH 4 (c) NH 3 H 2 O CH 4 C 2 H 2 (d) H 2 O NH 3 CH 4 C 2 H 2 [CET 1994; Karnatka CEE 1992] 272. The hybrid states of carbon in diamond, graphite and acetylene are respectively
CHEMICAL BONDING
(a) sp 2 , sp, sp 3
(b) sp, sp 2 , sp 3
(c) sp 3 , sp 2 , sp
(d) sp 2 , sp 3 , sp
273. The compound 1, 2-butadiene has
(a) Only sp hybrid carbon atoms
(b) Only sp 2 hybrid carbon atoms
(c) Both sp and sp 2 hybrid carbon atoms
(d) sp, sp 2 and sp 3 hybrid carbon atoms
274. The hybridization in methane, ethene & ethyne respectively is
(a) sp 3 , sp 2 and sp 275. The structure of H 2O2 is
(b) sp 3 , sp, sp 2
(c) sp 2 , sp 3 and sp
(d) sp 2 , sp, sp 3 [AFMC 2003]
(a) Planar (b) Non-planar 276. Increasing order (lower first) of size of hybrid orbitals is (a) sp, sp 2 , sp 3
[CPMT 2003]
(b) sp 3 , sp 2 , sp
(c) Spherical
(d) Tetrahedral [CPMT 1988]
(c) sp 2 , sp 3 , sp
(d) sp 2 , sp, sp 3
277. Both sp 2 and sp 3 hybrid carbons are present in which one of the following compounds
(a) CH 3 CH 3
(b) CH 2 CH 2
(c) CH CH
(d) CH 3 CH CH CH 3
[CPMT 1992]
278. Which of the following has a bond formed by the overlap of sp sp 3 hybrid orbitals
(a) CH 3 C C H
(b) CH 3 CH CH CH 3
(c) CH 2 CH CH CH 2
[UPSEAT 2002]
(d) HC CH
279. The hybridization of carbon atoms in C C single bond of HC C CH CH 2 is
(a) sp 3 sp 3
(b) sp 2 sp 3
(c) sp sp 2
[IIT 1991]
(d) sp 3 sp
280. The type of hybridization present in SO 2 and SO 3 is respectively
(a) sp, sp 2 (b) sp 2 , sp 2 (c) sp 2 , sp 3 (d) sp, sp 3 281. Match list I (Type of hybridization) with list II (shape of hybridized orbitals) and select the correct answer using the codes given below the lists List I List II I. sp 3 d 2
A. Linear
II. sp 3
B. Triangular planar
III. sp 2 IV. sp
C. Octahedral D. Tetrahedral E. Pyramidal [SCRA 1998] (a) I-E, II-C, III-D, IV-A (b) I-C, II-D, III-B, IV-A (c) I-C, II-D, III-A, IV-B (d) I-E, II-C, III-A, IV-D [KCET 2003] 282. The percentage s -character of the hybrid orbitals in methane, ethene & ethyne are respectively (a) 25, 33, 50 (b) 25, 50, 75 (c) 50, 75, 100 (d) 10, 20, 40 283. The correct order of hybridization of the central atom in the following species NH 3 , [ PtCl 4 ]2 , PCl 5 and BCl 3 is [IIT Screening test 2001]
(a) dsp 2 , dsp 3 , sp 2 and sp 3
(b) sp 3 , dsp 2 , dsp3 , sp 2
(c) dsp 2 , sp 2 , sp 3 , dsp 3 (d) dsp 2 , sp 3 , sp 2 , dsp 3 284. The bond between carbon atom (1) and carbon atom (2) in compound N C C H CH 2 involves the hybrid orbitals
[IIT 1987]
(1) (2)
(a) sp and sp 2 (b) sp 2 and sp 3 285. Carbon atoms in C 2 (CN )4 are
(c) sp and sp 3
(d) sp and sp [Roorkee 1999]
(a) sp -hybridized
(b) sp 2 -hybridized
(c) sp -and sp 2 -hybridized
(d) sp, sp 2 and sp 3 -hybridized
286. Both sp 3 and sp 2 hybrid carbons are present in which one of the following compounds
[CPMT 1993]
CHEMICAL BONDING
(a) CH 3 C CH 3
(b) CH 2 CH 2
(c) CH CH
(d) CH 3 CH CH CH 3
287. What type of hybridization is involved in [ Fe(CN )6 ]4
[CPMT 1995]
(a) sp (b) sp 2 (c) sp 3 d 2 288. Hybridization state of sulphur and % d -character in SF 6 will be respectively (a) sp 3 d 2 , 33 .3%
(b) sp 3 d , 20 %
(c) sp 2 d , 25 %
(d) d 2 sp 3 (d) sp 3 d , 75 %
289. Which of the following will be octahedral
[AIIMS 1999]
(b) BF 4
(a) SF 6
(c) PCl 5
(d) BO 33
290. The geometry of the molecule with sp 3 d 2 hybridized central atom is
(a) Square planar
(b) Trigonal bipyramidal
[RPMT 2000]
(c) Octahedral
(d) Square pyramidal IV. Resonance
Basic Level 291. The actual structure of a compound which lies some where in between the canonical structure is called
(a)
292.
293.
294.
295.
Resonance hybrid (b) Resonating structure (c) Lewis structure None of these Resonance arises due to the (a) Migration of H atoms (b) Migration of protons (c) Delocalisation of sigma electrons (d) Delocalisation of pi electrons Resonating structures have different (a) Atomic arrangements (b) Electronic arrangements (c) Functional (d) Alkyl groups As a result of resonance (a) Bond length decreases (b) Energy of the molecule decreases (c) Stability of the molecule increases (d) All are correct The compound which represents the maximum number of canonical forms (a) CO
(b) CO 2 (c) C 6 H 6
(d)
groups
(d) CH 3 COO ion
296. Point out incorrect statement about resonance
[MP PET 1997]
(a) Resonance structures should have equal energy (b) In resonance structures, the constituent atoms should be in the same position (c) In resonance structures, there should not be the same number of electron pairs (d) Resonance structures should differ only in the location of electron around the constituent atoms. 297. Resonance structures can be written for (a) (c) CH 4 (d) H 2 O O 3 (b) NH 3 298. Which of the following will not show resonance
(a)
H 2 O 2
(b) Allene
[AFMC 2000]
(c) Ozone
(d) Oxygen
299. The number of possible resonance structures for CO 32 is
(a) 2 (b) 3 (c) 6 300. Which of the following resonating structures is not correct for CO 2
[MP PMT 2000]
(d)
9
CHEMICAL BONDING
O
(a)
O
C O (b)
O C O
(c)
O C O
(d)
C O
301. The carbon monoxide molecule may be represented by the following structures except C
(a)
O (b)
C
O
(c)
C O
(d)
C
302. Resonance hybrid of nitrate ion is 1 / 2
O
N
O 1 / 2
2 / 3
1 / 3
O
O
N O 2 / 3
O 1 / 3
2 / 3
(c)
O
O 2 / 3
(b)
1 / 2
N
[Raj. PET 2000]
(a) O
O
O
N
O 2 / 3
(d)
1 / 3
O 2 / 3
303. Resonance in not shown by
(a)
C 6 H 6
(b) CO 2 (c) CO 3 2
(d) SiO 2
Ad Ad vance Level 304. Which of the following conditions is not correct for resonating structures
(a) The contributing structures must have the same number of unpaired electrons (b) The contributing structures should have similar energies (c) The contributing structures should be so written that unlike charges reside on atoms that are far apart (d) The positive charge should be present on the electropositive element and the negative charge on the electronegative element 305. A molecule may be represented by three structures having energies E 1 , E 2 and E 3 , respectively. The energies of these structures follow the order E 3 E 2 E 1 , respectively. If the experimental bond energy of the molecule is E 0 , the resonance energy is (a)
( E 1 E 2 E 3 ) E 0
(b) E 0 E 3
(c) E 0 E 1 (d)
306. Which of the following is the correct electron dot structure of N 2 O molecule
(a)
N N
O
(b)
N N
E 0 E 2
[MP CEE 1990] O
(c) N N O
(d)
N N
O
307. In which case the bond length is minimum between carbon and nitrogen
(a)
CH 3 NH 2 (b) C 6 H 5 CH NOH
(c) CH 3 CONH 2
(d) CH 3 CN
308. The C – H bond distance is the longest in
(a) C 2 H 2
(b) C 2 H 4
(c)
[MLNR 1990]
C 2 H 4 Br 2
(d) C 6 H 6
309. The distance between the two adjacent carbon atoms is largest in
(a) Benzene
(b) Ethene
[CBSE PMT 1994]
(c) Butane
(d) Ethyne
CHEMICAL BONDING
III. VSEPR Theory
Basic Level 310. VSEPR theory was proposed by
(a) Nyhom and Gillespie (b) Nyhom (c) Gillespie (d) 311. Which of the following statement about repulsion between bond pairs (bp) and lone pairs (lp) is correct (a) lp lp lp bp bp bp
Kossel
(b) lp bp lp lp bp bp (c) bp bp lp bp lp lp (d) Any of the three depending upon the type of molecule 312. Which of the following accounts for the bond angle of 104.5 o around O in water molecule (a) lp-bp repulsions (b) Low IE of oxygen (c) High electron affinity of oxygen (d) Small size of O and H-atoms 313. Which of the following has one lone pair of electrons on the central atom (a)
H 2 (b) CH 4
(c) NH 4
(d) NCl 3
314. The shape of covalent molecule AX 3 is
(a) Triangular (b) T-shaped (c) Pyramidal (d) Any of the above depending upon the number of lone pairs of electron on A 315. Which of the following does not have a tetrahedral structure (a)
BH 4
(b) B2 H 6
(c) NH 4
(d) H 2 O
316. BCl 3 molecule is planar while NCl 3 is pyramidal because
(a) BCl 3 does not have lone pair on B but NCl 3 has (c) N atom is smaller than B 317.The molecule that has linear structure is (a) N 2 O (b) NO 2
(b) B Cl bond is more polar than N Cl bond (d) N Cl bond is more covalent than B Cl bond (c) SO 2
(d) SiO2
(c) H 2 O
(d) NH 3
(c) NO 2
(d) H 2 O
318. Which one of the following molecules is linear
(a)
Hg 2 Cl 2 (b) CH 4
319. Which one of the following molecules is linear
(a)
HgCl 2 (b) SO 2
320. A molecule has seven bond pairs around the central a tom, the shape associated with the molecule is
(a) Heptagonal (b) Octahedral bipyramidal 321. The pair of species with similar shape is (a) PCl 3 , NH 3 (b)
(c) Pentagonal pyramidal
(d)
CF 4 , SF 4
(c) PbCl 2 , CO 2
(c) CO 2
(d) SnCl 2
ClF 3
(d) AlCl 3
PF 3
(d) NH 3
PF 5 , IF 5 322. The following molecule is not linear in shape
(a)
HgCl 2 (b) Hg 2 Cl 2
323. Which of the following will be planar trigonal
(a) PCl 3
(b) NH 3
(c)
324. In which molecule are all atoms coplanar
(a) CH 4
(b) BF 3
(c)
325. The molecule/ion which has pyramidal shape is
Pentagonal
(d)
CHEMICAL BONDING
(a)
PCl 3
(b) SO 3
(c) CO 32
(d) NO 3
(c) SO 2
(d) ClO2 or SiO 2
XeF 2
(d) XeF 6
326. Amongst the following the molecule that is linear is
(a)
CO 2
(b) NO 2
327. Which of the following has pyramidal shape
(a) XeO 3
(b) XeF 4 (c)
328. Among NH 3 , BeCl 2 , CO 2 and H 2 O , the non-linear molecules are
(a) BeCl 2 and H 2 O (b) BeCl 2 and CO 2
(c) NH 3 and H 2 O
(d) NH 3 and CO 2
329. Which one of the following compounds has bond angle as nearly 90o
(a) NH 3
(b) H 2 S
(c)
H 2 O
(d) CH 4
330. The bond angle around the O atom in ethanol (C 2 H 5 OH ) is
(a) 90o (b) 120o (c) 109o 331. From among the following triatomic species, the least angle around the central atom is in (a) O 3
(b) I 3 (c) NO 2
(d) 180o
(d) H 2 S
332. The H O H angle in water molecule is about
(a) 90o (b) 180o(c) 105o 333. Which of the following has the least bond angle (a) BeF 2 (b) H 2 O (c)
(d) 75o NH 3
(d) CH 4
334. Which one of the following has the shortest carbon, carbon bond length
(a) Benzene (b) Ethene 335. Which of the following has the highest bond angle (a) (b) H 2 O BF 3
(c) Ethyne
(d) Ethane
(c) NH 3
(d) CH 4
(c) NH 3
(d) PH 3
336. Which of the following has the lowest bond angle
(a)
H 2 O
(b) H 2 S
337. In which of the following species the angle around the central atom is exactly equal to 109, 28
(a)
SF 4
(b) NH 3
(c) NH 4
(d) None of the above
338. The bond angle in H 2 O is nearly 105o where as bond angle in H 2 S is nearly 92o. This is because
(a) Electronegativity of oxygen is greater than that of sulphur (b) Size of sulphur atom is greater than that of oxygen (c) Sulphur contains d -orbital whereas oxygen does not (d) The number of lone pairs present on oxygen and sulphur is not equal 339. The bond angle in H 2 S is (a)
NH 3 (b) Same as in BeCl 2
(c) H 2 Se H 2 O
(d) Same as in CH 4
340. The bond angle in H 2 O molecule is less than that of NH 3 molecule because
(a) The hybridisation of O in H 2 O and N in NH 3 is different (b) The atome radii of N and O are different (c) There is one lone pair of electrons on O and two lone pairs of electrons on N (d) There are two lone pairs of electrons on O and one lone pair of electrons on N 341. The correct sequence of decrease in the bond angle of the following hydrides is (a) NH 3 PH 3 AsH 3 SbH 3 (b) NH 3 AsH 3 PH 3 SbH 3 (c)
SbH 3 AsH 3 PH 3 NH 3
(d) PH 3 NH 3 AsH 3 SbH 3
342. The bond angle of H 2 X (where X is a sixth group element) as one goes down the group
(a)
Increases (b) Changes irregularly
Decreases
(c) Does not change (d)
CHEMICAL BONDING
343. Which of the following is the correct reducing order of bond-angle
(a) NH 3 CH 4 C 2 H 2 H 2 O
(b) C 2 H 2 NH 3 H 2 O CH 4
(c) NH 3 H 2 O CH 4 C 2 H 2
(d) H 2 O NH 3 CH 4 C 2 H 2
344. The correct increasing bond angle among BF 3 , PF 3 and ClF 3 follows the order
(a)
BF 3 PF 3 ClF 3
(b) PF 3 BF 3 ClF 3
(c) ClF 3 PF 3 BF 3
(d) BF 3 PF 3 ClF 3
345. The correct order of the bond angle is
(a) NH 3 H 2 O PH 3 H 2 S
(b) NH 3 PH 3 H 2 O H 2 S
(c) NH 3 H 2 S PH 3 H 2O
(d) PH 3 H 2 S NH 3 H 2 O
346. The bond angle around the central atom is highest in
(a)
BBr 3
(b) CS 2
(c) SO 2
(d) SF 4
347. Which of the following set contains species having same angle around the central atom
(a)
SF 4 , CH 4 , NH 3
(b) NF 3 , BCl 3 , NH 3
(c) BF 3 , NF 3 , AlCl 3 (d)
BF 3 , BCl 3 , BBr 3 348. Among the following orbital bonds, the angle is minimum between
(a)
sp 3 bonds
(b)
p x and p y orbitals
(c) H O H
in
water
(d) sp bonds 349. The bond angle in Cl 2 O is nearly 111o whereas bond angle in F 2 O is nearly 103o. This is because (a) (b) (c) (d)
Electronegativity of Fluorine is greater than that of oxygen Size of chlorine atom is greater than that of Fluorine Chlorine contains d orbitals whereas Fluorine does not The number of lone pairs present on Fluorine and chlorine is not equal
Ad Ad vance Level 350. In compounds of type ECl 3 , where E B, P . As or Bi , the angles Cl E Cl for different E are in the order
B P As Bi B P As Bi 351. The correct order of bond angles is (a) PF 3 PCl 3 PBr 3 PI 3
(a)
(c) PI 3 PBr 3 PCl 3 PF 3 352. The pair of molecules having identical geometry is (a) BCl 3 , PCl 3 (b)
(b) B P As Bi
(c) B P As Bi
(d)
(b)
PF 3 PBr 3 PCl 3 PI 3
(d)
PF 3 PCl 3 PBr 3 PI 3
BF 2 , NF 3
(c) CCl 4 , CH 4
(d)
(c) CH 4 only
(d)
CHCl 3 , CH 3 Cl 353. Out of CHCl 3 , CH 4 and SF 4 the molecules having regular geometry are
(a)
CHCl 3 only (b) CH 4 and
CHCl 3 and SF 4
SF 4
354. According to VSEPR theory, the most probable shape of the molecule having 4 electron pairs in the outer shell of the central
atom is [MP PET 1996, 2001]
(a) Linear
(b) Tetrahedral
(c) Hexahedral
(d) Octahedral
355. The geometry of ClO 3 ion according to Valence Shell Electron Pair Repulsion (VSEPR) theory will be [KCET 1996; MP PET 1997]
(a) Planar triangular (b) Pyramidal (c) Tetrahedral 356. Which of the following molecules has three fold axis of symmetry (a) NH 3 (b) C 2 H 4 (c) CO 2
(d) Square planar (d) SO 2
CHEMICAL BONDING
357. Of the three molecules XeF 4 , SiF 4 , SF 4 which have tetrahedral structures
(a) All the three
(b) Only SiF 4
(c) Both SF 4 and XeF 4
(d) Only SF 4 and XeF 4
358. A molecule XY 2 contains two , two bonds and one lone pair of electron in the valence shell of X . The arrangement of lone
pairs well as bond pairs is (a) Square pyramidal (b) Linear (c) Trigonal planar 359. A bonded molecule MX 3 is T-shaped. The number of non-bonding pairs of electrons is
(d) Unpredictable
(a) 0 (b) 2 (c) 1 (d) Can be predicted only if atomic number of M is known 360. The molecule ML x is planar with six pairs of electrons around M in the valence shell. The value of x is (a) 6 (b) 2 (c) 4 (d) 3 361. Which of the following molecules does not have a linear arra ngement of atoms (a) C 2 H 2 (b) H 2 S (c) (d) CS 2 BeCl 2
[CBSE 1989]
362. CO 2 has the same geometry as: ( A) HgCl 2 , ( B) NO 2 , (C ) SnCl 4 , ( D) C 2 H 2
(a) A and C (b) B and D (c) 363. The pair having similar geometry is (a) BO 33 and PO 4 3
(b) SO 3 and CO 3
A and D
(d) C and D
(c) NO 3 and SO 4
(d) PO4 3 and ClO 4 II. Molecular Orbital Theory
Basic Level 364. Molecular orbital theory was given by
(a) Mulliken (b) Moseley 365. Which of the following combination of orbitals is correct (a)
+
– –
(c)
+
– +
+ +
–(b)
+
–(d)
+
(c) Wemer
+
+
+
– –
+ – –
+
+
+ –
366. Which among the following molecule /ions is diamagnetic
(a) (c) molecule
(d) Kossel
–
Super oxide ion Hydrogen molecule
–
+ ––
–
+
– – –
(b) Oxygen (d) Unipositive ion of N 2
367. Combination of two AO’s lead to the formation of
(a) (c) Three MO’s Two MO’s (b) One MO 368. When the atomic orbitals combine, the new molecular orbitals formed are such that (a) They have greater energy than the combining orbitals (b) They have lower energy than the combining orbitals (c) They have energy equal to those of the combining orbitals (d) One of them has lower energy and the other has a higher energy 369. The energy of antibonding molecular orbital is (a) Greater than the bonding M.O. (b) (c) Equal to that of bonding M.O. (d) 370. Which of the following molecular orbital has the lowest energy (a)
2 p z
(b) * 2 p z
(c) * 2 p x
371.In the molecular orbital diagram for O 2 ion the highest occupied orbital is
(d) Four MO’s
Smaller than the bonding M.O. None (d) * 2 p y
CHEMICAL BONDING
(a)
MO
orbital (b)
MO orbital
(c) * MO orbital
(d)
* MO orbital
372. For a stable molecule the value of bond order should be
(a) Negative (b) Positive (c) Zero (d) No relationship of stability and bond order 373. Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statements is true about it? The bond order [AIIMS 1980] (a) Cannot be a negative quantity (b) Has always an integral value (c) Can assume any value, positive or negative integral or fractional, including zero (d) Is a non-zero quantity 374. N 2 and O 2 are converted into monoanions, N 2 and O 2 respectively. Which of the following statements is wrong
(a)
In N 2 , N N bond weakens
(b) In O 2 , O O bond order increases
(c)
In O 2 , O O bond order decreases
II.
N 2
B. 1.5
III.
Be 2
C. 1.0
IV.
O2
D. 0
(d) N 2 becomes paramagnetic 375. Match List I (Molecules) with List II (Bond order) and select the correct answer using the codes given below the lists List I List II I. Li 2 A. 3
E. 2 Codes: (a) I–B, II–C, III– A, IV –E (c) I–D, II– A, III–E, IV –C
(b) I–C, II– A, III–D, IV –E (d) I–C, II–B, III–E, IV – A
376. The calculated bond order in the superoxide (O 2 ) ion is
(a)
2.5 (b)
2
(c)
1.5 (d)
1
2
(d)
1.0
N 2 (d)
N 2
377. The bond order in peroxide ion (O 22 ) is
(a) 2.5 (b) 1.5 (c) 378. Which of the following ion has not having bond order of 2.5 (a)
O 2 (b)
O 2 (c)
379. In which of the following pairs the two molecules have identical bond order
(a)
N 2 , O22 (b) N 2 , O2
(c) N 2 , O2
(d) O2 , N 2
380. Out of the following which has smallest bond length
(a)
O2
(b) O2 (c) O2
[RPMT 1997]
(d) O2 2
381. What is correct sequence of bond order
(a)
O 2 O 2 O 2
(b)
O 2 O 21 O 21
[BHU 1998]
(c) O 2 O 2 O 2
(d) O 21 O 21 O 2
Covalent character
(c) Bond length
382. Higher is the bond order greater is
(a) Bond dissociation energy (b) Para magnetism 383. Which of the following has fractional bond order (a)
O 22
(b) O 22 (c) F 22
(d) H 2
384. Bond order of N 2 anion is
(a) 3.0 (b) 2.0 (c) 2.5 (d) 1.5 385. The bond order of CO molecule on the basis of molecular orbital theory is
(d)
CHEMICAL BONDING
(a)
Zero
(b) 2
(c) 3
(d) 1
386. The bond order of C 2 is
(a) 1 (b) 2 (c) 3/2 387. The bond order is maximum in (a)
H 2
(d) 1/2 (d) He 2
(b) H 2 (c) He 2
388. Which has maximum bond order
(a)
H 2
(b) N 2 (c) F 2
(d) O 2
389. What bond order does Li 2 have
(a)
1
(b) 2
(c) Zero
(d) 3
390. In PO43 , the formal charge on each oxygen atom and the P O bond order respectively are
(a)
– 0.75, 0.6 (b) – 0.75, 1.0
(c) – 0.75, 1.25
[CBSE 1998]
(d) – 3, 1.25
391. According to molecular orbital theory, O 2 possesses
(a) (c)
Bond order of 2.5 Diamagnetic character
(b) Three unpaired electrons (d) Stability lower than O 2
392. The bond order for a species with the configuration 1 s 2 , * 1 s 2 , 2 s 2 , * 2 s 2 , 2 p x1 will be
(a)
1
(b)
1 2
(c)
Zero
[Haryana CEET 1991]
(d) 1.
393. The bond order of NO molecule is
(a) 1.5 (b) 2.0 (c) 2.5 (d) 3.0 394. Ground state electronic configuration of valence shell electrons in nitrogen molecule ( 2 s )2 ( *2 s )2 ( 2 p y )4 ( 2 p )2
(a)
2
( N 2 )
is written as
. Hence the bond order of nitrogen molecule is
(b)
3
(c)
0
(d)
1.
395. The common features among the species CN , CO and NO are
(a) Bond order three and isoelectronic (c) Bond order two and -acceptors ligands 396. Which of the following have identical bond order
I. C H 3
II. H 3 O
III. NH 3
IV. CH 3
(a) I and II (b) III and IV 397. The bond order is not three for (a)
N 2 (b)
(b) Bond order three and weak field ligands (d) Isoelectronic and weak field
O 22
[IIT 1992]
(c) I and III
(d) II, III and IV
(c) N 2
(d) NO
398. The bond lengths in the species O 2 , O 2 and O 2 are in the order
(a)
O 2 O 2 O 2 (b)
[Pb. CET 1990]
O 2 O 2 O 2
(c) O 2 O 2 O 2
(d)
O 2 O 2 O 2 399. The bond angle between H O H in ice is closest to
(a) (c) 90o 109 28 (b) 60o 400. The angle between the overlapping of one s-orbital and one p-orbital is (a) 180o(b) 120o (c) 109o 28 401. Which of the following bonds has the highest energy (a) Se-Se (b) Te-Te (c) S-S
(d) 105o (d) 120o 60 [CBSE 1996]
(d) O-O
2
402. The bond order in N ion is
(a)
1
(b)
[BHU 2000]
2
(c)
2.5 (d)
3
CHEMICAL BONDING
403. The correct order of increasing C –O bond length of CO, CO 32 , CO 2 is
(a)
CO 32 CO 2 CO
(d) CO CO 2 CO 32
(b) CO 2 CO 32 CO
(c) CO CO 32 CO 2
404. Which one of the following species has the lowest bond order
(a)
O 2 (b)
O 2 (c)
[AIIMS 1994]
O 2 (d)
O 22
(c) O 2
(d) O 2
405. Molecular species having highest bond order from the following is
(a)
O 22
(b) O 2
406. In the formation of N 2 molecule according to M.O.T. the outermost electron goes to
(a)
MO
(b) sp hybrid orbital
(c) MO
(d) 2 p orbital
407. The true statements from the following is/are
1. PH 5 and BiCl 5 donot exist 2. p d bond is present is SO 2 3. Electrons move with the speed of light 4. SeF 4 and CH 4 have same shape 5. I 3 has bent geometry (a) 1, 3 (b) 1, 2, 5 (c) 1, 3, 5 408. The chemical inertness of N 2 is attributed to
(d) 1, 2, 4
(a) The presence of large no. of bondings electrons in comparison to antibonding electron (b) Its high heat of dissociation (c) Presence of triple bonds between nitrogen atoms which make the molecule quite stable (d) All the statements are correct 409. If Z-axis is taken as the molecular axis, then -orbitals are formed by (a) (b) (c) 2s and 2 p y 2 p x and 2 p y 2 p x and 2 p z
(d) 2s and 2 p z
410. Which of the following has a linear structure
(a)
CCl 4
(b) SO 2 (c) C 2 H 2
[IIT 1986]
(d) C 2 H 4
411. Which of the following molecules is planar
(a)
NH 3
(b) CH 4 (c) C 2 H 4
[Pb. CET 1988]
(d) SiCl 4
412. Which of the following molecules does not have a linear arra ngement of atoms
(a)
H 2 S
(b) C 2 H 2
(c)
BeH 2
[CBSE PMT 1989]
(d) CO 2
413. The number of antibonding electron pairs in O 22 molecular ion on the basis of molecular orbital theory is
(a) 4 (b) 3 (c) 2 (d) 5 414. The number of unpaired electrons in O2 is (a) 1 (b) 2 (c) 3 (d) 0 415. Which of the following molecules or molecular ions have only one unpa ired electron (a)
O 2
(b) O 2 (c) O 22
(d) H 2
416. Which of the following molecules have unpaired electrons in antibonding molecular orbitals
(a)
O2
(b) N 2 (c) C 2
(d) B 2
417. The energy of 2 s is greater than 1* s orbital because
(a)
2 s orbital is bigger than 1 s
orbital
(b)
2 s is a bonding orbital whereas 1* s is an antibonding orbital
(c)
2 s orbital has a greater value of n than 1* s orbital
[MP PET 2000]
CHEMICAL BONDING
(d)
2 s orbital is formed only after 1 s
418. The molecular orbital shown in the diagram can be described as
+
–
(b) +
+
–
(c) +
+
*
(d)
+
+
–
–
–
+
+ +
–
+
–
(a) (b) * (c) 419. Which of the following overlap is correct (a) + – + 1s 1s
–
–
(d) None of above 420. Which is not paramagnetic (a)
O 2 (b)
O 2 (c)
[DCE 1999, 2000]
O 22
(d) O 2
421. Which of the following species is paramagnetic
(a)
CO (b)
NO (c)
[CBSE PMT 1995]
O 22
(d) CN
422. Which one of the following molecules is paramagnetic
(a)
N 2 (b)
O2 (c)
[AIIMS 1994]
Cl 2 (d)
HCl
423. Which one among the following is not paramagnetic (At. No. Be 4 , Ne 10 , Cl 17 , As 33 )
(a) (b) Ne 2 Be 424. Which of the following is not paramagnetic (a) NO
(b) S 2 (c) O 21
(c) Cl
(d) As [AIIMS 1997]
(d) N 2
425. Paramagnetism is exhibited by molecules which
(a) Are not attracted by magnetic field (c) Contain unpaired electrons 426. Which one of the following molecules is paramagnetic CO 2 (a) (b) SO 2
[Manipal (Med.) Ent. 1995]
(b) Contain only paired electrons (d) Carry positive charge [Pb. PMT 1998]
(c) NO
(d) H 2 O
427. Which one of the following is paramagnetic
(a)
NO
(b) O 2
[CPMT 1992]
(c) CN
(d) CO
428. Which of the following is not paramagnetic
(a)
N 2 (b)
CO (c)
[CBSE PMT 2000]
O 2 (d)
NO
429. Which of the following species would be expected parama gnetic
(a)
Copper crystals
[UPSEAT 2000]
(b) Cu
(c) Cu
N 2 (d)
O2
(d)
H 2
430. Which is paramagnetic and has bond order 0.5
(a)
H 2 (b)
F 2 (c)
431. Which of the following is diamagnetic
(a) O2
(b) O 2
[CPMT 1988]
(c) O 2
(d) O 22
CHEMICAL BONDING
Ad Ad vance Level 432. Anti-bonding molecular orbital is formed by
[DPMT 2000]
(a) Addition of wave functions of atomic orbitals (b) Subtraction of wave functions of atomic orbitals (c) Multiplication of wave functions of atomic orbitals (d) None of these [ AIIMS 1989] 433. In a homonuclear molecule which of the following set of orbitals are degenerate (a) (c) 2 px and 2 pz (d) 2 px and 2 py 2 s and 1 s (b) 2 pz
and * 2 px
434. Which of the following molecular orbitals has two nodal pla nes
(a)
2 x
(b) 2 p y
[Kurukshetra CEE 1996]
(c) 2* p y
(d) 2* p x
435. If Z-axis is the molecular axis, then -molecular orbitals are formed by the overlap of
(a)
s p z
(b) p x p y
(c) p z p z
436. In the process O 2 O 22 e the electron lost is from
(a)
Bonding -orbital (b)
Antibonding -orbital
(d) p x p x [Orissa JEE 2002 ]
(c) 2 P x orbital
(d) 2 P x anti bonding orbital
(c) 1* s orbital
(d) Both (b) and (c) are
437. For a homonuclear diatomic molecule the energy of 2 s orbital is
(a)
2* s orbital (b) 2* s orbital
correct 438. Which of the following phenomenon occur when two atom of the element, having same spin of electron approach for bonding [ AFMC 1994]
(a) Orbital overlap will not occur (b) Bonding will not occur (c) Both (a) and (b) are correct (d) None of the above are correct 439. If N x is the number of bonding orbitals of an atom and N y is the number of antibonding orbital, then the molecule/atom will be stable if [DPMT 1996] (a) N x N y (b) N x N y (c) N x N y (d) N x N y 440. The distribution of electrons in the molecular orbitals of the O 2 molecule is as follows 1 s 2 , * 1 s 2 , 2 s 2 , * 2 s 2 , 2 p z2 .... [ISM 1994]
(a)
2 p x2 , 2 p y2 , * 2 p x2 , * .2 p y0
(b) 2 p x2 , 2 p y2 , * 2 p x1 , * 2 p y1
(c)
2 p x2 , 2 p y1 , * 2 p x2 , * .2 p y2
(d) 2 p x2 , 2 p y0 , * 2 p x2 , * 2 p y2
441. The molecular orbital configuration of CN is
(a)
K K (2 s)2 , * (2 s)2 , (2 p x )2 (2 p y )2
(b) K K (2 s)2 , * (2 s)2 , (2 p x )2 (2 p x )1 , (2 p y )1
(c)
K K (2 s)2 , * (2 s)2 , (2 p z )2 (2 p s )2 , (2 p y )
(d) K K (2 s)2 , * (2 s)2 , (2 p z )2 (2 p s )2 , (2 p y )2
442. Which one of the following is a correct electronic configuration for diatomic nitrogen
(a)
(b)
(c)
(d)
CHEMICAL BONDING
*2 p
*2 p
*2 p
*2 p
*2 p
*2 p
*2 p
*2 p
2 p
2 p
2 p
2 p
2 p
2 p
2 p
*2 s
*2 s
2 s
2 s
*1 s
2 p
*2 s
2 s *1 s
*2 s
2 s
*1 s
1 s
1 s
1 s
*1 s
1 s
443. The sequence of energy levels of MO’s formed from the outermost shells of C 2 molecule is
(a)
(2 s) * (2 s) (2 p x ) (2 p y ) (2 p z ) * (2 p x ) * (2 p y ) * (2 p z )
(b)
(2 s) * (2 s) (2 p x ) (2 p y ) (2 p z ) * (2 p z ) * (2 p y ) * (2 p z )
(c)
(2 s) *(2 s) (2 p x ) (2 p y ) (2 p z ) *(2 p x ) *(2 p y ) *(2 p z )
(d)
(2 s) * (2 s) (2 p z ) (2 p x ) (2 p y ) * (2 p z ) * (2 p x ) * (2 px y )
444. Which one of the following does not exhibit paramagnetism
(a)
NO (b)
NO 2
[BHU 1993]
(c) ClO2
(d) ClO 2
445. N 2 and O 2 are converted into monocations, N 2 and O 2 respectively. Which of the following is wrong
(a)
In N 2 , N N bond weakens
(b) In O 2 , the O O bond order increases
(c)
In O 2 , paramagnetism decreases
(d) N 2 becomes diamagnetic
446. The species isoelectronic with C 2 H 4 is
(a)
CN
(b) O 2
(c) O 2
(d) N 2
CN
(d) O 2
447. The molecule having one unpaired electron is
(a)
NO (b)
CO (c)
448. Which of the following combinations is not allowed (assume Z -axis is internuclear axis)
(a)
2s and 2s (b) 2 p x and 2 p x
(c) 2s and 2 p x
(d) 2 p x and 2 p y
449. Which of the following statements is not correct regarding bonding molecular orbitals
(a) Bonding molecular orbitals possess less energy than the atomic orbitals from which they are formed (b) Bonding molecular orbitals have low electron densities between the two nuclei (c) Every electron in bonding molecular orbitals contributes to the attraction between atoms (d) They are formed when the lobes of the combining atomic orbitals have the same sign 450. Which of the following combinations is not allowed in the LCAO method for the formation of a MO (consider the Z-axis as the molecular axis) (a) (b) s p x (c) p x p x (d) p z p z s p z 451. Which sequence correctly describes the relative bond strength of oxygen molecule, superoxide ion, peroxide ion and unipositiv e
oxygen molecule (a)
O 2 O 2 O 22 O 2
(d) O 22 O 2 O 2 O 2
452. The bond energies in NO, NO and NO follow the order
(b) O 2 O 2 O 2 O 22
(c) O 22 O 2 O 2 O 2
CHEMICAL BONDING
(a) NO NO NO (d) NO NO NO 453. Which of the following statements is incorrect (a) He 2 does not exist because its bond order is zero
(b) NO NO NO
(b)
O 2 , O 2 and O 2 are all paramagnetic
(c) (d)
Any two atomic orbitals can combine to form two molecular orbitals (2 p x ) and (2 p y ) are degenerate molecular orbitals
(c) NO NO NO
454. Which of the following statements is correct about N 2 molecule
(a) (b) (c)
It has a bond order of 3 The number of unpaired electrons present in it is zero and hence it is diamagnetic The order of filling of MOs is (2 p x ) (2 p y ), (2 p z )
(d)
All the above three statements are correct
455. In which of the following diatomic molecules is the bond order of each molecule 2
(a) N 2 , NO, O 2
(b) O 2 , NO, CN
1 2
(c) N 2 , CN , O 2
(d) CN , N 2 , N 2 XVI.
Hydrogen Bonding
Basic Level 456. H-Bonding is exhibited by
(a) All the molecules containing H-atoms (b) Molecules in which H is covalently bonded to F, O, or N (c) Molecules in which two H atoms are present (d) Molecules in which H is bonded to atoms with electronegativity greater than 2.1 457. Which of the following hydrogen bond is strongest in vapour phase (a) HF ------------- HF (b) HF ------------ HCl (c) HCl ------------- HCl 458. Which of the following shows hydrogen bonding (a) (b) PH 3 (c) AsH 3 NH 3
(d) HF ----------- HI [CPMT 2000]
(d) SbH 3
459. Among HF , CH 4 , CH 3 OH , and N 2 H 4 intermolecular hydrogen bonding is expected
(a)
In all
(b) In all leaving one
(c) In two
[RPET 1998]
(d) None of these
460. Intramolecular H-bonding is present in
(a)
Meta nitrophenol (b) Salicylaldehyde (c) Hydrogen (d) Benzophenone 461. Out of the two compounds shown below, the vapour pressure of B at a particular temperature is expected to be
chloride
OH
OH
and O2 N
A
B
(a) Higher than that of A (c) Same as that of A (d) 462. The O– H bond distance in water molecule is (a) 1.0 Å (b) 1.33 Å 463. The pair of molecules forming strongest hydrogen bonds are (a) SiH 4 and SiF 6
NO 2
(b) Lower than that of A Can be higher or lower depending upon the size of the vessel (c) 0.96 Å
(d) 1.45 Å [IIT 1981]
(b) CH 3 C CH 3 and CHCl 3 ||
O
CHEMICAL BONDING
(c)
H C OH and CH 3 C OH ||
||
O
O
(d)
H 2 O and H 2 O 2
464. Strongest hydrogen bonding is shown by 465. 466. 467.
468. 469.
[Pb. CET 1991]
(a) Methanol (b) Dimethylamine (c) Acetic acid Hydrogen bonding is maximum in the (a) Carbinol (b) Ethyl fluoride (c) Ethyl chloride Organic compound soluble in water contains (a) C, H (b) C, H, O (c) C, S Density of ice is less than that of water because of (a) Extensive hydrogen bonding (b) Crystal modification of ice (c) Open porous structure of ice due to hydrogen bonding (d) Different physical states of these The hydrogen bond is strongest in (a) (c) O H .... S F H .... O (b) F H .... F Which of the following does not contain any coordinate bond but hydrogen bond is present (a)
H 3 O
(b) BF 4 (c) HF 2
(d) Methyl thioalcohol [DPMT 1992]
(d) Triethyl amine [ AFMC 1992]
(d) C, H, Cl [BHU 1994]
[IIT 1986; Pb. CET 1999]
(d) O H .... N
(d) NH 4
470. Hydrogen fluoride is a liquid unlike other hydrogen halides because
(a) (b)
HF molecules associate due to hydrogen bonding F 2 is highly reactive
(c) HF is the weakest acid of all hydrogen halides (d) Fluorine atom is the smallest of all halogens 471. Which one is appreciably soluble in water CS 2 (a) (b) C 2 H 5 OH
(c) CCl 4
(d) CHCl 3
472. Hydrogen bond formation depends upon
(a) High electronegativity of atoms (c) Both (a) and (b) 473. Incorrect order of decreasing boiling points is (a) HF HI HBr HCl H 2 O H 2 Te H 2 Se H 2 S (c)
Br 2 Cl 2 F 2
(b)
High electropositivity of atoms (d) None of these (b)
(d)
CH 4 GeH 4 SiH 4
474. In which one of the following compounds does hydrogen bonding occur
(a)
SiH 4
(b) LiH
(c) HI
(d) NH 3
475. Between the HF --------- HF , hydrogen bond will be
(a) Compact (b) Strong (c) Weak 476. Which one shows maximum hydrogen bonding (a) (b) H 2 Se H 2 O
[CPMT 1990]
(d) None of these [CBSE PMT 1990]
(c) H 2 Te
(d) HF
477. Hydrogen bonding lies in between
(a)
Ionic bond and covalent bond bond (c) Metallic bond and dipole-dipole bond 478. The intermolecular attractive forces vary in the order (a) Water < Alcohol < Ether (c) Alcohol > Water < Ether 479. Which does not show hydrogen bonding (a) C 2 H 5 OH (b) Liquid NH 3
[Pb. CET 1990]
(b)
(d) Dipole-dipole bond and Vander Waal bond (b) Water > Alcohol > Ether (d) Ether > Water > Alcohol (c) H 2 O
480. NH 3 has abnormally high boiling point because it has
(a)
Alkaline nature (b) Hydrogen bonding
Covalent bond and metallic
(d) Liquid HBr [Pb. CET 1997]
Distorted shape
(c) sp3-Hybridization (d)
CHEMICAL BONDING
481. The boiling point of p–nitrophenol is higher than that of o–nitrophenol because
[CBSE PMT 1994, KCET 2002]
(a) NO 2 group at p–position behaves in a different way from that at o–position (b) Intramolecular hydrogen bonding exists in p–nitrophenol (c) There is intermolecular hydrogen bonding in p–nitrophenol (d) p–nitrophenol has a higher molecular weight than o–nitrophenol 482. Which one of the following hydrogen halides has the lowest boiling point (a) HF (b) HCl (c) HBr (d) 483. NH 3 has a much higher boiling point than PH 3 because (a)
NH 3 has larger molecular weight
(b)
NH 3 undergoes umbrella inversion
(c)
NH 3 forms hydrogen bond
(d)
NH 3 contains ionic bonds whereas PH 3 contains covalent bonds
[ AIIMS 1998]
HI [MLNR 1994]
484. Vaporisation capacity increases due to
(a) Inter molecular hydrogen bonding (c) Both (a) and (b) (d) 485. Which is appreciably soluble in water (a) CH 3 COOH (b) H 2 S (c) CCl 4
(b) Intra molecular hydrogen bonding None of these (d) CHCl 3
486. Which one has the highest boiling point
[MP PMT 2002]
(a) Acetone (b) Ethyl alcohol (c) Diethyl ether 487. Which molecule does not have hydrogen bonding in its molecules (a) HF (b) (c) NH 3 H 2 O
(d) Chloroform (d) HI
488. Which one among the following does not have the hydrogen bond
(a)
Phenol
(b) Liquid NH 3
(c) Water
(d) Liquid HCl
489. There is no hydrogen bonding in
(a) Acetic acid (b) Ammonia (c) Ethyl alcohol (d) Diethyl ether [MP PMT 1994] 490. In which of the following compounds intramolecular hydrogen bond is present (a) Ethyl alcohol (b) Water (c) Salicylaldehyde (d) Hydrogen sulphide 491. Which of the following compounds has the least tendency to form hydrogen bonds between molecules (a) (b) NH 2 OH (c) HF (d) CH 3 F NH 3 492. The pair likely to form the strongest hydrogen bonding
(a)
H 2 O 2 and H 2 O
(b)
(c)
CH 3 COOH and CH 3 COOCH 3
[IIT 1981; DCE 2000]
HCOOH and CH 3 COOH
(d) SiH 4 and SiCl 4
493. The bond that determines the secondary structure of a protein is
[NCERT 1984; MP PET 1996]
(a)
Coordinate bond (b) Covalent bond (c) Hydrogen bond (d) Ionic bond 494. Water ( H 2 O) is liquid while hydrogen sulphide ( H 2 S ) is a gas because (a) Water has higher molecular weight (b) Hydrogen sulphide is a weak acid (c) Sulphur has high electronegativity than oxygen (d) Water molecules associate through hydrogen bonding [CPMT 1989] 495. Methanol and ethanol are miscible in water due to (a) Covalent character (b) Hydrogen bonding character (c) Oxygen bonding character (d) None of the above [NCERT 1978] 496. Which of the following compounds can form hydrogen bond (a) (b) H 2 O (c) NaCl (d) CHCl 3 CH 4 497. Maximum possible number of hydrogen bonds in which a water molecule can participate is
(a)
1
(b)
4
(c)
3
(d)
[CPSE 1994]
2
CHEMICAL BONDING
498. Hydrogen bonding is maximum in
[IIT 1987; MLNR 1995; KCET 2000]
(a) Ethyl chloride (b) ether 499. Strongest hydrogen bond is shown by (a) Water (b) Ammonia 500. Bond nature of hydrogen bond is
(a)
Ionic
Triethylamine
(c) Ethanol (d)
Diethyl [CBSE 1991]
(b) Covalent
(c) Hydrogen fluoride
(d) Hydrogen sulphide
(c) Co-ordinate
(d) None of these
Ad Ad vance Level 501. The boiling points of methanol, water and diethyl ether are respectively 65 oC , 100oC and 34.5oC . Which of the following best
explain these wide variations in b.p. (a) The molecular mass increases from water (18) to methanol (32) to diethyl ether (74) (b) The extent of H-bonding decreases from water to methanol while it is absent in ether (c) The extent of intermolecular H-bonding decreases from ether to methanol to water (d) The number of H atoms per molecule increases from water to methanol to ether 502. Ethanol and methoxymethane have the same molecular weight but methoxymethane boils at a lower temperature because it has [DCE 1994]
(a) (c)
Low density Molecular association
(b) No hydrogen bonding (d) Oxygen atom attached to two methyl groups
503. The dielectric constant of H 2 O is 80. The electrostatic force of attraction between K and Cl will be
(a) (c)
Reduced to 1/40 in water than air Increased 80 times in water than air
(b) Reduced to 1/80 times in water than air (d) Will remain unchanged XV.
Type of Bonding and Forces in solids
Basic Level 504. ZnS is an example of
(a)
Ionic crystal (b) Metallic crystal
Covalent crystal
(c) Molecular crystal (d)
Metallic crystal
(c) Covalent crystal (d)
Molecular crystal
(c) Covalent crystal (d)
505. LiF is an example of
(a)
Ionic crystal (b) Molecular crystal 506. Which one has highest melting point (a) Ionic crystal (b) Metallic crystal 507. Which force is strongest (a) Dipole – dipole forces (c) Ion – Dipole forces 508. The solid NaCl is a bad conductor of electricity since (a) In solid NaCl there are no ions (c) In solid NaCl there is no velocity of ions 509. In crystals of which of the following ionic compounds would anions
(b) Ion – ion force (d) Ion – induced dipole forces [AIIMS 1980]
(b) Solid NaCl is covalent (d) In solid NaCl there are no electrons you expect maximum distance between centres of cations and [CBSE 1998]
(a) CsF (b) CsI (c) LiI (d) LiF [CBSE PMT 1997] 510. For two ionic solids, CaO and KI. Identify the wrong statement among the following (a) Lattice energy of CaO is much higher than that of KI (b) KI is soluble in benzene (c) CaO has higher melting point (d) KI has lower melting point [NCERT 1974; CPMT 1989; MP PMT 1999] 511. When NaCl is dissolved in water the sodium ion be comes (a) oxidised (b) Reduced (c) Hydrolysed (d) Hydrated
CHEMICAL BONDING
512. Silicon dioxide is an example of
(a) Metallic crystal (b) Ionic crystal (c) Covalent crystal (d) None 513. Graphite is an example of (a) Ionic solid (b) Covalent solid (c) Vander Walls’ crystal (d) Metallic crystal 514. Which one of the following is a two dimensional covalent soli d (a) Graphite (b) Quartz (c) Carborundum (d) Pure germanium 515. Which one of the following is a good conductor of electricity (a) Diamond (b) Graphite (c) Silicon (d) Amorphous carbon 516. A solid melts above 3000 K and is a poor conductor of heat and electricity. To which of the following catagories does it below (a) Ionic (b) Metallic (c) Covalent (d) Molecular 517. Which substance will conduct the current in the solid state (a) Diamond (b) Graphite (c) Iodine (d) Sodium chloride [AFMC 1997] 518. Which of the following is true for diamond (a) Diamond is a good conductor of electricity (b) Diamond is soft (c) Diamond is a bad conductor of heat (d) Diamond is made up of C, H and O [BHU 1984] 519. Crystals of covalent compounds always have (a) Atoms as their structural units (b) Molecules as structural units (c) Ions held together by electrostatic forces (d) High melting point 520. Particles of quartz are packed by (a) Electrical attraction forces (b) Covalent bond forces (c) Vander Waal’s forces (d) None of three 521. Which of the following is an example of covalent crystal solid (a) Si (b) Al (c) Ar (d) NaF 522. Which crystal is expected to be soft and have low melting point (a) Covalent (b) Metallic (c) Molecular (d) Ionic In a crystal, all the lattice sites are found to be occupied by covalent molecules. To which kind of solid does it belong 523. (a) Ionic (b) Covalent (c) Molecular (d) None [DPMT 1994] 524. A crystal will be hard and have high melting point (a) Covalent crystal (b) Ionic (c) Metallic (d) Molecular 525. Solid CO 2 is an example of (a)
Molecular crystal Ionic crystal
(b) Covalent crystal
(c) Metallic crystal
(d)
Ionic solid
(c) Molecular solid
(d)
526. Iodine crystals are
(a)
527.
528. 529. 530. 531.
532.
Metallic solid (b) Covalent solid p -and n -type of semi conductors are formed due to (a) Covalent bonds (b) Co-ordinate bonds A molecular crystalline solid (a) Is very hard (b) Is volatile (c) Which of the following is a molecular crystal (a) SiC (b) NaCl (c) Graphite Which of the following is not an example of molecular crystal (a) Hydrogen (b) Iodine (c) Ice In solid argon, the atoms are held together by (a) Ionic bonds (b) (d) Hydrophobic forces Lorentz explanation is for (a) Ionic bonds (b)
[UPSEAT 2002]
Metallic bonds
(c) Ionic bonds
(d)
Has a high melting point
(d) Is a good conductor
(d) Ice (d) Sodium chloride Hydrogen bonds
(c) Vander
Covalent bonds
(c) Metallic bonds
Waal’s
for ces
(d)
Vander Waal’s forces 533. A metallic bond is
[Pb. CET 1998]
CHEMICAL BONDING
534.
535.
536.
537.
(a) Ionic (b) Polar covalent (c) Non-polar covalent (d) Electrostatic in nature In the metallic crystal (a) The valence electrons remain within the field of influence of their own kernels (b) The valence electrons constitute a sea of mobile electrons (c) The valence electrons are localised between the two kernels (d) Both kernels as well as electrons move rapidly With increase in temperature, the conductivity of the metals [PUCET 1986] (a) Increases (b) Decreases (c) Remains unaffected (d) May increase or decrease Metallic sodium is [PUCET 1986] (a) Insulator (b) Semi conductor (c) Conductor of electricity (d) Conductor only in molten state The conductivity of the metal decreases with increase in temperature because (a) The kinetic energy of the electrons increases (b) The movement of electrons becomes haphazard (c) The kernels start vibrating (d) The metal becomes hot and starts emitting radiations
Ad Ad vance Level 538. As it cools, olive oil slowly solidifies and forms a solid over a wide range of temperature. Which term best describes the solid
(a) Ionic (b) Covalent network (c) Metallic (d) Molecular crystal 539. At very low temperature, oxygen O 2 , freezes and forms a crystalline solid. Which term best describes the solid (a) Covalent network (b) Molecular crystals (c) Metallic (d) Ionic [AMU 1985] 540. Which of the following substances has covalent bonding (a) (b) NaCl (c) Solid Ne (d) Cu Ge 541. A solid X melts slightly above 273 K and is a poor conductor of heat and electricity. To which of the following categories does it belong (a) Ionic solid (b) Covalent solid (c) Metallic (d) Molecular 542. Wax is an example of (a) Ionic crystal (b) Covalent crystal (c) Molecular crystal (d) Metallic crystal 543. The interparticle forces in solid hydrogen are (a) Hydrogen bonds (b) Covalent bonds (c) Co-ordinate bonds (d) Vander Waal’s forces 544. Which solid will have weakest intermolecular forces (a) Ice (b) Phosphorus (c) Naphthalene (d) Sodium fluoride 545. Which of the following exhibits the weakest intermolecular forces (a) He (b) HCl (c) NH 3 (d) H 2 O 546. Which of the following is an example of metallic crystal s olid (a) C (b) Si (c) W (d) AgCl 547. Crystals which are good conductor of electricity and heat are known as (a) Ionic crystals (b) Covalent crystals (c) Metallic crystals (d) Molecular crystals 548. The oxide which shows metallic conduction (a) (b) VO (c) CrO 2 (d) All Re O 3 549. An increase in the charge of the positive ions that occupy lattice positions brings in a/an ……in metallic bonding (a) Increase (b) Decrease (c) Neither increase nor decrease (d) Either increase or decrease 550. Metallic bond can explain (a) Ductility (b) Toughness (c) Malleability (d) All 551. Metals are malleable and ductile because of
CHEMICAL BONDING
552.
553.
554.
555. 556.
557.
558.
559. 560. 561.
562. 563.
(a) Presence or mobile electrons (b) Slipping of kernels over each other (c) They non-directional nature of the bonds (d) Their high density and melting point Metals possess luster when freshly cut because (a) They have a hard surface and light is reflected back (b) Their crystal structure contains ordered arrangement of their constituent atoms (c) They contain loosely bound electrons which absorb the photons and then re-emit (d) They are obtained from the minerals on which light has been falling for years The metallic lustre exhibited by sodium is explained by [IIT 1987, Pb. CET 1998] (a) Diffusion of sodium ions (b) Excitation of free protons (c) Oscillation of loose electrons (d) Existence of body centred cubic lattice Iron is tougher than sodium because (a) Iron atom is smaller (b) Iron atoms are more closely packed (c) Metallic bonds are stronger in iron (d) None of these Metallic bonds do not play a role in [AFMC 1993] (a) Brass (b) Copper (c) Germanium (d) Zinc Which of the following does not apply to bonding in metals [CBSE 1989] (a) Non directional bonds (b) Mobility of valence electrons (c) Delocalisation of electrons (d) Highly directional bonds Out of the following which is strongest bond (a) Vander Waal (b) Dipole-dipole (c) Metallic bond (d) Hydrogen bond One would expect the elemental form of Cs at room temperature to be (a) A network solid (b) A metallic solid (c) Non-polar liquid (d) An ionic liquid Which does not apply to metallic bond [CBSE 1989] (a) Overlapping (b) Mobile valency electrons (c) Delocalised electrons (d) None Among the following metals interatomic forces are probably weakest in (a) Copper (b) Silver (c) Zinc (d) Mercury The enhanced force of cohesion in metals is due to (a) The covalent linkages between atoms (b) The electrovalent linkages between atoms (c) The lack of exchange of valency electrons (d) The exchange energy of mobile electrons Which of the following has the highest melting point [CPMT 1994] (a) (b) Diamond (c) Pb Fe (d) Na In metallic crystals, its atoms are located at the position of [AMU 1985] (a) Maximum potential energy (b) Minimum potential energy (c) Zero potential energy (d) Infinite potential energy XIV.
Miscellaneous Questions
Basic Level 564. NH 3 and BF 3 combine readily because of the formation of [AFMC 1982; MP PMT 1985; MLNR 1994; Karmatala CET 2000; MP PET 2001]
(a) A covalent bond (b) A hydrogen bond A co-ordinate bond 565. The ionization of hydrogen atom would give rise to (a) Hybridization (b) Hydronium ion 566. The interionic attraction depends on interaction of (a) Solute-solute (b) Solvent-solvent Molecular properties (e) Atomic particles
(c) An ionic bond
(c) Proton
(d)
(d) Hydroxyl ion [Kerala PMT 2002]
(c) The charges
(d)
CHEMICAL BONDING
567. When an element of very low ionisation potential reacts with an element with very high electron affinity, the bond formed will be
predominantly (a) Ionic (b) Covalent 568. In dry ice there are (a) Ionic bond (b) Covalent bond 569. Bonding in ferric chloride is (a) Covalent (b) Ionic
(c) Co-ordinate
(d) Hydrogen
(c) Hydrogen bond
(d) None of these
(c) Co-ordinate
(d) None of these
[Pb CET 1997]
570. The type of bond formed between H ion and NH 3 in NH 4 ion is
(a) Ionic
(b) Covalent
(c) Dative
(d) Hydrogen
571. NH 3 molecules attaches itself to Hydrogen ion in NH 4OH by
(a) Electrovalency (b) Covalent (c) 572. Sulphuric acid provides a simple example of (a) Co-ordinate bonds (b) Non-covalent compound (c) 573. What is the nature of the bond between B and O in (C 2 H 5 )2 OBH 3 (a) Covalent (b) Co-ordinate bond (c) 574. Which of the followings is ionic as well as covalent (a) H 2 O (b) NaOH (c)
Co-ordinate valency
(d) None of these [Kerala PMT 2002]
Co-valent ion
(d) None of these
Ionic bond
(d) Banana shaped bond [Haryana 1999]
CO 2
(d) H 2 O 2
575. AlCl 3 anhydrous is covalent but AlCl 3 .6 H 2 O is ionic because
(a) AlCl 3 dissolves in CS 2 (c) IE of Al is low 576. Which of the following contains both covalent and ionic bond? (a) NH 4 Cl (b) H 2 O
(b) AlCl 3 has planar structure (d) Hydration energy compensates the high IE of Al [KCET 2000]
(c) CCl 4
(d) CaCl 2
577. Sodium chloride is an ionic compound where as hydrogen chloride gas is mainly covalent because
(a) Electronegativity difference in the case of hydrogen is less than 2.1 (b) Hydrogen chloride is a gas (c) Hydrogen is a non-metal (d) Sodium is reactive 578. Indicate the nature of bonding in CCl 4 and CaH 2 (a) Covalent in CCl 4 and electrovalent in CaH 2
(b) Electrovalent in both CCl 4 and CaH 2
(c) Covalent in both CCl 4 and CaH 2
(d) Electrovalent in CCl 4 and covalent in CaH 2
579. HCl molecule contains a
(a) Covalent bond (c) Co-ordinate bond 580. The bonds in K 4 [ Fe(CN )6 ] are (a) All ionic (c) Ionic, covalent and co-ordinate 581. Which type of bond is not present in HNO 2 molecule (a) Covalent (c) Ionic
[CPMT 1984]
(b) Double bond (d) Electrovalent bond [EAMCET 1991]
(b) All covalent (d) Ionic and covalent (b) Co-ordinate (d) Ionic as well as co-ordinate
582. Zeise’s salt contains which type of bonds
(a) Ionic (b) Ionic and covalent (c) Hydrogen bonds (d) Ionic, covalent and co-ordiante bonds 2 583. Given electronic configuration of four elements as (I) 1s (II) 1s2 2s22 p2, (III) 1s22s22 p5 and (IV) 1s22s22 p6. The one which is capable of forming ionic as well as covalent bond is [MNR 1995] (a) I (b) II (c) III (d) IV 584. Which of the following bond does not exhibit ionic and covalent bonding (a) BaSO 4 (b) CH 3 Cl (c) NH 4 Cl (d) Ca( NO 3 )2 585. The structure of orthophosphoric acid is
[KCET 2003]
CHEMICAL BONDING
O
(a) H O P O H |
O |
H
H
H
|
|
(b) O P O H
(c) O P O H
|
O
O (d) H O P O
|
H
|
H
586. Which of the following compound shows ionic, covalent & co-ordination b onds as well
(a)
CaSO 4 .5 H 2 O
(b) CuSO 4 .5 H 2 O
(c) HCl
[DCE 2001]
(d) NaCl
587. In the compound CH 3 C OCl , which type of orbital have been used by the circle carbon in bond formation 588. 589. 590. 591. 592.
(a) sp3 (b) sp2 (c) sp o Bond angle between two hybrid orbitals is 105 , % s-character of hybrid orbital is (a) Between 20 – 21% (b) Between 19 – 20% (c) Between 21 – 22% The bonds present in N 2O5 are (a) Ionic (b) Covalent and co-ordinate (c) Covalent only The number of covalent and co-ordinate bonds in pyrosulphuric acid ( H 2 S 2O7) are (a) 6, 4 (b) 6, 6 (c) 4, 4 CO is isoelectronic with (a) NH 3 (b) N 2 (c) O2 Which of the following is a non-crystalline s olid (a) CsCl (b) NaCl (c) CaF 2
[MP PET 1994]
(d) p [MP PMT 1986]
(d) Between 22 – 25% (d) Ionic and covalent (d) 4, 6 [AFMC 2003]
(d) NO2 (d) Glass
593. For the various types of interactions the correct order of increasing strength is
594. 595.
596. 597. 598.
599. 600.
601.
(a) Covalent < hydrogen bonding < Vander Waals < dipole-dipole (b) Vander Waals < hydrogen bonding < dipole-dipole < covalent (c) Vander Waals < dipole-dipole < hydrogen bonding < covalent (d) Dipole-dipole < Vander Waals < hydrogen bonding < covalent Crystals of sodium chloride belongs to the system [NCERT 1975; MP PMT 1997] (a) Orthorhombic (b) Cubic (c) Trigonal (d) Monoclinic Reintzer and Gatlermann found that parazoxyanide melts at 389 K to give a turbid non-uniform liquid but at 408K, it melts to give a clear uniform isotropic liquid. What type of crystal is being described ? (a) Molecular crystal (b) Covalent crystals (c) Liquid crystals (d) H-bonded crystals Consider two elements with atomic no. 37 and 53, the bond between their atoms would be [CPMT 1991, 1994] (a) Covalent (b) Ionic (c) Co-ordinate (d) Metallic Two element have electronegativities of 1.2 and 3.0. Bond formed between them would [CPMT 1982] (a) Ionic (b) Polar covalent (c) Co-ordinate (d) Metallic The element with atomic number 14 when reacts with other elements [CPMT 1991, 1994] (a) Generally forms ionic compounds (b) Generally forms covalent compound (c) Forms both covalent and ionic compounds with equal ease (d) None of these Lateral overlapping give rise to (a) -bonds (b) – bonds (c) Metallic bonds (d) None of these Shape of molecules is decided by (a) Sigma bond (b) -bond (c) Both sigma and -bonds (d) Neither sigma nor -bonds The boiling point of water ( H 2 O) is 100o C whereas that of hydrogen sulphide ( H 2 S ) is 42 o C . This can be attributed to
(a) A larger bond angle in H 2O than in H 2 S (b) Smaller size of oxygen atom as compared to sulphur (c) Larger ionization energy of oxygen than sulphur (d) Larger tendency of H 2O to form hydrogen bonds than H 2 S 602. The b.p. of glycerol is more than propanol because of (a) Hydrogen bonding (b) Hybridization (c) Resonance 603. HCl is a gas but HF is a low boiling liquid. This is because (a) H F bond is strong (b) H F bond is weak
[CPMT 1997]
(d) All the above [NCERT 1984]
CHEMICAL BONDING
(c) Molecules aggregate because of hydrogen bonding (d) HF is a weak acid 604. Reason for excessive solubility of alcohol in water is due to (a) Covalent bond (b) Ionic bond (c) Hydrogen bond with water (d) None of these [CPMT 1986] 605. An ionic solid is a poor conductor of electricity because (a) Ionic solids do not conduct electricity (b) Charge on the ions is uniformly distributed (c) Ions occupy fixed positions in solids (d) Ions have uniform field of influence around it [BHU 1996] 606. Which theory shows oxygen is paramagnetic that (a) Valency theory (b) Molecular orbital theory (c) Quantum theory (d) None of these [MP PMT 2002] 607. O-nitrophenol is more volatile than p -nitrophenol due to (a) Intramolecular H -bonding (b) Intermolecular H -bonding (c) Resonance (d) Inductive effect 608. Two ice cubes are pressed over each other until they unite to form one block. Which one of the following forces dominates for holding them together [Kerala CEE 2000] (a) Dipole-dipole interaction (b) Vander Waals’ forces (c) Hydrogen bond formation (d) Covalent attraction 609. The weakest bond among the following is (a) Ionic (b) Covalent (c) Metallic (d) Hydrogen bond [CPMT 1990] 610. The predominant intermolecular forces in hydrogen fluoride is due to (a) Dipole-induced dipole interactions (b) Dipole-dipole interactions (c) Hydrogen bond formation (d) Dispersion interaction [Roorkee 1990] 611. In benzene, all the C C bonds are of equal length because of (a) Hyperconjugation (b) Resonance (c) Electromeric effect (d) Inductive effect XIII.
nswer Sheet
Basic aand A d v ance Level 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
d
b
a
c
a
d
c
c
c
a
d
c
d
a
c
b
b
c
b
d
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
d
c
a
b
a
d
d
c
b
a
a
c
a
c
b
b
b
c
b
a
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
b
c
a
b
b
b
a
b
b
d
d
b
c
a
a
d
d
d
b
c
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
b
c
b
b
b
d
d
b
d
d
c
a
b
a
c
c
a
b
c
a
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
b
d
b
d
d
a
d
d
b
c
c
b
c
a
a
a
a
b
c
a
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
b
b
d
d
c
a
a
b
b
b
c
c
c
c
c
d
c
d
a
d
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
c
a
c
c
d
c
d
c
b
b
b
b
a
b
b
c
d
a
d
c
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
d
c
b
b
d
b
a
c
a
a
c
b
d
b
b
a
d
d
b
b
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
b
c
b
a
a
b
b
a
a
b
d
a
c
d
a
c
a
d
c
c
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
20 0
d
d
c
a
c
c
c
d
c
b
b
d
d
b
b
c
a
c
b
d
201
202
203
204
205
206
207
20 8
209
210
211
212
213
214
215
216
217
218
219
220
b
c
a
d
b
b
b
b
c
b
b
b
b
b
c
d
a
c
c
c
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
c
a
b
c
d
c
a
b
c
a
a
d
d
a
a
c
c
b
c
c
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
CHEMICAL BONDING
c
d
c
c
a
b
a
d
b
a
a
d
d
c
c
c
a
c
b
c
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
28 0
c
a
d
c
b
b
c
b
b
d
b
c
d
a
b
a
d
a
c
b
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
30 0
b
a
b
a
c
d
d
a
a
c
a
d
b
d
c
c
a
a
b
b
301
302
303
304
305
306
307
30 8
309
310
311
312
313
314
315
316
317
318
319
320
d
c
d
c
b
b
d
c
c
a
a
a
d
d
b
a
a
a
a
d
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
a
d
d
b
a
a
a
c
b
c
d
c
b
c
a
b
c
a
c
d
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
a
b
b
c
a
b
d
b
a
b
d
c
c
b
b
a
b
c
b
c
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
38 0
b
c
d
a
b
c
a
d
a
a
c
b
a
b
b
c
d
a
a
b
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
40 0
c
a
d
c
c
c
a
b
a
c
a
b
c
b
a
d
a
d
d
a
401
402
403
404
405
406
407
40 8
40 9
410
411
412
413
414
415
416
417
418
419
420
c
c
d
d
b
c
b
d
a
c
c
a
a
b
a
a
c
c
c
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
b
b
c
b
c
c
b
b
c
a
d
b
b
c
d
b
d
c
a
b
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
b
c
a
d
d
c
a
d
b
b
b
a
c
d
a
b
a
a
b
b
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
48 0
a
c
c
c
a
b
c
b
c
b
a
d
d
b
d
c
b
d
d
481
482
483
484
485
486
487
48 8
489
490
491
492
493
494
495
496
497
498
499
50 0
c
b
c
b
a
b
d
d
d
c
d
b
c
d
b
b
b
c
c
a
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
b
b
b
a
a
a
b
c
b
b
d
c
b
a
b
c
b
c
a
b
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
a
c
c
b
a
c
a
b
d
d
c
c
d
b
b
c
c
d
b
a
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
d
c
d
a
a
c
c
d
a
d
b
c
c
c
a
d
c
b
a
d
561
562
563
564
565
566
567
568
570
571
572
573
574
575
576
577
578
579
580
d
b
b
d
c
c
a
b
a
c
c
d
b
b
d
a
a
a
a
c
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
60 0
a
d
c
b
a
b
b
d
b
a
b
d
c
b
c
b
a
b
b
a
601
602
603
604
605
606
607
60 8
60 9
610
611
d
a
c
c
c
b
a
c
d
c
b
569
a
c
