ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

Chapter

2 Erwin Schrödinger

CONTENTS 2.1

Composition of atom

2.2

Atomic number, Mass number and Atomic species

2.3

Electromagnetic radiation’s

2.4

Atomic spectrum- Hydrogen spectrum

2.5

Thomson's model

2.6

Rutherford's nuclear model

2.7

Planck's Quantum theory and Photoelectric effect

2.8

Bohr’s atomic model

2.9

Bohr – Sommerfeld’s model

2.10

Dual nature of electron

2.11

Heisenberg’s uncertainty principle

2.12

Schrödinger wave equation

2.13

Quantum numbers and Shapes of orbitals

2.14

Electronic configuration principles

2.15

Electronic configurations of elements Assignment (Basic and Advance Level)

Science

has produced a microscopic structure of the atom, but it’s structure is so detailed and so subtle of something which is far removed from our immediate experience that it is difficult to see how many of its features were constructed. Yet among all the experiments used to form the theory of atomic structure, there stand a few which have been most in-fluential in shaping its major features.

Answer Sheet of Assignment

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John Dalton 1808, believed that matter is made up of extremely minute indivisible particles, called atom which can takes part in chemical reactions. These can neither be created nor be destroyed. However, modern researches have conclusively proved that atom is no longer an indivisible particle. Modern structure of atom is based on Rutherford’s scattering experiment on atoms and on the concepts of quantization of energy.

2.1 Composition of atom. The works of J.J. Thomson and Ernst Rutherford actually laid the foundation of the modern picture of the atom. It is now believed that the atom consists of several sub-atomic particles like electron, proton, neutron, positron, neutrino, meson etc. Out of these particles, the electron, proton and the neutron are called fundamental subatomic particles and others are non-fundamental particles.

Electron (–1eo) (1) It was discovered by J.J. Thomson (1897) and is negatively charged particle. Electron is a component particle of cathode rays. (2) Cathode rays were discovered by William Crooke's & J.J. Thomson (1880) using a cylindrical hard glass tube fitted with two metallic electrodes. The tube has a side tube with a stop cock. This tube was known as discharge tube. They passed electricity (10,000V) through a discharge tube at very low pressure ( 10 −2 to 10 −3 mm Hg ) . Blue rays were emerged from the cathode. These rays were termed as Cathode rays. Cathode rays

Gas at low pressure

Cathode

TC Vaccum pump Anode

High voltage + –

Discharge tube experiment for production of cathode rays

(3) Properties of Cathode rays (i) Cathode rays travel in straight line. (ii) Cathode rays produce mechanical effect, as they can rotate the wheel placed in their path. (iii)Cathode rays consist of negatively charged particles known as electron. (iv) Cathode rays travel with high speed approaching that of light (ranging between 10 −9 to 10 −11 cm/sec) (v) Cathode rays can cause fluorescence.

(vi) Cathode rays heat the object on which they fall due to transfer of kinetic energy to the object. (vii) When cathode rays fall on solids such as Cu, X − rays are produced. (viii) Cathode rays possess ionizing power i.e., they ionize the gas through which they pass. (ix) The cathode rays produce scintillation the photographic plates. (x) They can penetrate through thin metallic sheets. (xi) The nature of these rays does not depend upon the nature of gas or the cathode material used in discharge tube. (xii) The e/m (charge to mass ratio) for cathode rays was found to be the same as that for an e −

(−1 . 76 × 10 8 coloumb per gm). Thus, the cathode rays are a stream of electrons.

Note

:q

When the gas pressure in the discharge tube is 1 atmosphere no electric current flows

through the tube. This is because the gases are poor conductor of electricity. qThe television picture tube is a cathode ray tube in which a picture is produced due to fluorescence on the television screen coated with suitable material. Similarly, fluorescent light tubes are also cathode rays tubes coated inside with suitable materials which produce visible light on being hit with cathode rays. (4) R.S. Mullikan measured the charge on an electron by oil drop experiment. The charge on each electron is − 1 . 602 × 10 −19 C. (5) Name of electron was suggested by J.S. Stoney. The specific charge (e/m) on electron was first determined by J.J. Thomson. (6) Rest mass of electron is 9 . 1 × 10 −28 gm = 0 . 000549 amu = 1 / 1837 of the mass of hydrogen atom. (7) According to Einstein’s theory of relativity, mass of electron in motion is, m ′ =

Rest mass of electron(m ) [1 − (u / c) 2 ]

Where u = velocity of electron, c= velocity of light. When u=c than mass of moving electron =∞. (8) Molar mass of electron = Mass of electron × Avogadro number = 5 . 483 × 10 −4 . (9) 1.1 × 10 27 electrons =1gram. (10) 1 mole electron = 0 . 5483 mili gram. (11) Energy of free electron is ≈ 0. The minus sign on the electron in an orbit, represents attraction between the positively charged nucleus and negatively charged electron. (12) Electron is universal component of matter and takes part in chemical combinations. (13) The physical and chemical properties of an element depend upon the distribution of electrons in outer shells. (14) The radius of electron is 4 .28 × 10 −12 cm . (15) The density of the electron is = 2 . 17 × 10 −17 g / mL .


+

•

Examples based on Einstein’s theory of relativity

Example : 1 The momentum of electron moving with 1/3rd velocity of light is (in g cm sec–1) (a) 9 . 69 × 10 −8

(c) 9 . 652 × 10 −18

(b) 8 . 01 × 10 10

(d) None

Solution: (c) Momentum of electron, ‘p’ = m ′ × u

Where m ′ is mass of electron in motion =

∴ Momentum =

9. 108 × 10 −28  c  1−  3×c

Example: 2

2

×

m 1 − (u / c )2

; Also u = c / 3

3 × 10 10 9 .108 × 10 −28 × 3 × 10 10 = = 9 .652 × 10 −18 g cm sec −1 3 0 . 94 × 3

An electron has a total energy of 2 MeV. Calculate the effective mass of the electron in kg and its speed. Assume rest mass of electron 0.511 MeV. (a) 2 . 9 × 10

(b) 8 . 01 × 10

8

Mass of electron in motion =

Solution: (a) =

(c) 9 . 652 × 10

8

2 amu 931

2 × 1 . 66 × 10 − 27 kg = 3 .56 × 10 −30 kg 931

8

(d) None

(1 amu = 931 MeV)

(1 amu = 1. 66 × 10 −27 kg )

Let the speed of the electron be u. m′ =

m 1 − (u / c )

or 3 .56 × 10

2

− 30

0 . 511 × 1. 66 × 10 − 27 0 .911 × 10 − 30 931 = = 2 2  u   u  1− 1 −     3 × 10 8   3 × 10 8 

2

 u  or 1 −   = 0 .06548  3 × 10 8 

Example: 3

or u 2 = 9 × 10 16 × 0 .93452 or u = 2 .9 × 10 8 m

A electron of rest mass 1. 67 × 10 −27 kg is moving with a velocity of 0.9c (c = velocity of light). Find its mass and momentum. (a) 10 . 34 × 10

Solution: (a)

−19

10 (b) 8 . 01 × 10

−18 (c) 9 . 652 × 10

(d) None

Mass of a moving object can be calculated using Einsten’s theory of relativity : m′ =

m′ =

m 1 − (u / c )

m = rest mass (given), u = velocity (given), c = velocity of light

2

1. 67 × 10 −27  0. 9 c  1−   c 

2

= 3 .83 × 10 − 27 kg

Momentum ' p ' = m ′ × u p = 3 . 83 × 10 −27 × 0 .9 c = 10 .34 × 10 −19 kg ms −1

Proton (1H1, H+, P)

(1) Proton was discovered by Goldstein and is positively charged particle. It is a component particle of anode rays. (2) Goldstein (1886) used perforated cathode in the discharge tube and repeated Thomson's experiment and observed the formation of anode rays. These rays also termed as positive or canal rays. Anode rays

Cathode rays TC Vaccum pump

Perforated cathode

High voltage + –

Perforated tube experiment for production of anode rays

(3) Properties of anode rays (i) Anode rays travel in straight line. (ii) Anode rays are material particles. (iii) Anode rays are positively charged. (iv) Anode rays may get deflected by external magnetic field. (v) Anode rays also affect the photographic plate. (vi) The e/m ratio of these rays is smaller than that of electrons. (vii) Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the tube. It is maximum when gas present in the tube is hydrogen. (viii) These rays produce flashes of light on ZnS screen. (4) Charge on proton = 1 . 602 × 10 −19 coulombs = 4 .80 × 10 −10 e .s. u. (5) Mass of proton = Mass of hydrogen atom= 1 . 00728 amu = 1 . 673 × 10 −24 gram = 1837 of the mass of electron. (6) Molar mass of proton = mass of proton × Avogadro number = 1 . 008 (approx). (7) Proton is ionized hydrogen atom (H + ) i.e., hydrogen atom minus electron is proton. (8) Proton is present in the nucleus of the atom and it's number is equal to the number of electron. (9) Mass of 1 mole of protons is ≈ 1.007 gram. (10) Charge on 1 mole of protons is ≈ 96500 coulombs. (11) The volume of a proton (volume =

4 3 πr ) is ≈ 1.5 × 10 −38 cm 3 . 3

(12) Specific charge of a proton is 9 . 58 × 10 4 Coulomb/gram.

Neutron (on1, N) APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

(1) Neutron was discovered by James Chadwick (1932) according to the following nuclear reaction, 4

Be 9 + 2 He 4 → 6 C 12 + o n 1 or 5 B 11 + 2 He 4 → 7 N 14 + o n 1

(2) The reason for the late discovery of neutron was its neutral nature. (3) Neutron is slightly heavier (0.18%) than proton. (4) Mass of neutron = 1 . 675 × 10 −24 gram = 1 . 675 × 10 −27 kg = 1 . 00899 amu ≈ mass of hydrogen atom. (5) Specific charge of a neutron is zero. (6) Density = 1 . 5 × 10 −14 gram / c.c. (7) 1 mole of neutrons is ≈ 1.008 gram. (8) Neutron is heaviest among all the fundamental particles present in an atom. (9) Neutron is an unstable particle. It decays as follows : 1 0n neutron

→ 1 H 1 + proton

0 0 0ν −1 e + electron anti nutrino

(10) Neutron is fundamental particle of all the atomic nucleus, except hydrogen or protium. Comparison of mass, charge and specific charge of electron, proton and neutron Name of constant

Unit amu kg Relative

Mass (m)

Coulomb (C) esu Relative C/g

Charge(e) Specific charge (e/m)

Electron(e–) 0.000546 9.109 × 10–31 1/1837

Proton(p+) 1.00728 1.673 × 10–27 1

Neutron(n) 1.00899 1.675 × 10–24 1

– 1.602 × 10–19 – 4.8 × 10–10 –1 1.76 × 108

+1.602 × 10–19 +4.8 × 10–10 +1 9.58 × 104

Zero Zero Zero Zero

• The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C 12 , i.e. 1 . 660 × 10 −27 kg . Other non fundamental particles Particle

Symbol

Nature

Charge esu

Positron

e + , 1e 0 , β +

+

×10–10 + 4.8029

Neutrino

ν

0

0

Anti-proton

p−

–

Positive mu meson

µ+

Negative mu meson

Mass (amu)

Discovered by

0.000548 6

Anderson (1932) Pauli (1933) and Fermi (1934)

– 4.8029

< 0.00002 1.00787

+

+ 4.8029

0.1152

Chamberlain Sugri (1956) and Weighland (1955) Yukawa (1935)

µ−

–

– 4.8029

0.1152

Anderson (1937)

Positive pi meson

π+

+

+ 4.8029

0.1514

Negative pi meson

π−

–

– 4.8029

0.1514

Neutral pi meson

π0

0

0

0.1454

2.2 Atomic number, Mass number and Atomic species.

Powell (1947)

(1) Atomic number or Nuclear charge (i) The number of protons present in the nucleus of the atom is called atomic number (Z). (ii) It was determined by Moseley as, ν = a( Z − b) or aZ − ab where, ν = X − rays frequency

ν s −1

Z= atomic number of the metal a & b are constant.

Z

(iii) Atomic number = Number of positive charge on nucleus = Number of protons in nucleus = Number of electrons in nutral atom. (iv) Two different elements can never have identical atomic number. (2) Mass number (i) The sum of proton and neutrons present in the nucleus is called mass number. Mass number (A) = Number of protons + Number of neutrons or Atomic number (Z) or Number of neutrons = A – Z . (ii) Since mass of a proton or a neutron is not a whole number (on atomic weight scale), weight is not necessarily a whole number. (iii) The atom of an element X having mass number (A) and atomic number (Z) may be represented by a symbol, Element Mass number

A

X

Atomic number

e.g.

Note

:q

Z 9

F 19 , 8 O 16 , 7 N 14 etc.

A part of an atom up to penultimate shell is a kernel or atomic core.

q Negative ion is formed by gaining electrons and positive ion by the loss of electrons. q Number of lost or gained electrons in positive or negative ion =Number of protons ± charge on ion. (3) Different Types of Atomic Species Atomic species

Isotopes (Soddy)

Similarities

Differences

(i) Atomic No. (Z)

(i) Mass No. (A)

(ii) No. of protons

(ii) No. of neutrons

(iii) No. of electrons

(iii) Physical properties

(iv) Electronic configuration

Examples (i) 11 H , 12 H , 13 H (ii) (iii)

16 8

18 O, 17 8 O, 8 O

35 17

(v) Chemical properties (vi) Position in the periodic table


37 Cl , 17 Cl

(i) Mass No. (A)

(i) Atomic No. (Z)

(ii) No. of nucleons

(ii) No. of protons, electrons (ii) and neutrons

Isobars

(i)

40 18

40 Ar, 19 K , 40 20 Ca

130 52

130 Te , 130 54 Xe , 56 Ba

(iii)Electronic configuration (iv) Chemical properties (v) Position in the perodic table. No. of neutrons

(i) Atomic No.

(i)

30 14

Si , 1531 P, 1632 S

(ii) Mass No., protons and 39 K , 40 (ii) 19 20 Ca electrons. 3 4 (iii) Electronic (iii) 1 H , 2 He Isotones

configuration

(iv)

(iv) Physical and chemical properties

13 6

C , 14 7 N

(v) Position in the periodic table. Isotopic No. (N – Z) or (A – 2Z) Isodiaphers

(i) At No., electrons, neutrons.

mass No., (i) U 235 , Th 231 92 90 protons, (ii) 19 K 39 , 9 F 19

(ii) Physical and chemical (iii) properties. (i) No. of electrons (ii) configuration

Isoelectronic species

At. No., mass No.

Electronic

29

Cu 65 , 24 Cr 55

(i) N 2 O, CO 2 , CNO − (22 e − ) (ii) CO , CN − , N 2 (14 e − ) (iii) H − , He , Li + , Be 2 + (2 e − ) (iv) P 3 − , S 2 − , Cl − , Ar , K + and Ca 2 + (18 e − )

Isosters

(i) No. of atoms

(i) N 2 and CO

(ii) No. of electrons

(ii) CO 2 and N 2 O

(iii) Same physical and chemical properties.

(iii) HCl and F2 (iv) CaO and MgS (v) C 6 H 6 and B 3 N 3 H 6

Note

:

q In all the elements, tin has maximum number of stable isotopes (ten).

q Average atomic weight/ The average isotopic weight =

% of 1st isotope × relative mass of 1st isotope + % of 2nd isotope × relative mass of 2nd isotope 100

Examples based on Moseley equation Example : 4

The characteristics X- ray wavelength for the lines of the k α series in elements X and Y are 9.87Å and 2.29Å respectively. If Moseley’s equation ν = 4 .9 × 10 7 (Z − 0. 75 ) is followed, the atomic numbers of X and Y are (a) 12, 24

Solution : (a) ν =

(b) 10, 12

(c) 6, 12

(d) 8, 10

c λ

νx =

νy =

3 × 10 8 = 5 .5132 × 10 8 9 . 87 × 10 −10

3 × 10 8 2. 29 × 10 −10

= 11 . 4457 × 10 8

using Moseley’s equation we get ∴ 5 . 5132 × 10 8 = 4 . 9 × 10 7 (Z x − 0 . 75 )

…..(i)

and 11 . 4457 × 10 8 = 4 .90 × 10 7 (Z y − 0 . 75 )

….. (ii)

On solving equation (i) and (ii) Z x = 12, Z y = 24 . Example : 5

If the straight line is at an angle 45° with intercept, 1 on

ν − axis, calculate frequency ν when atomic

number Z is 50. (a) 2000 s

Solution : (c)

−1

(b) 2010 s

−1

(c) 2401 s

−1

(d) None

ν = tan 45 ° = 1 = a

ab=1

νs

∴ ν = 50 − 1 = 49 −1

ν = 2401 s .

Example : 6

What is atomic number Z when ν = 2500 s −1 ?

ν = A ZX

Example : 7

(b) 40

2500 = Z − 1,

(c) 51

(d) 53

Z = 51 .

Examples based on Atomic number, Mass number and Atomic species

Atomic weight of Ne is 20.2. Ne is a mixutre of Ne 20 and Ne 22 . Relative abundance of heavier isotope is (a) 90

Solution:(d)

θ Z

(a) 50

Solution : (c)

a= tan θ ab=intercept

−1

(b) 20

(c) 40

(d) 10

Average atomic weight/ The average isotopic weight =



∴ 20 .2 =

Example : 8

a × 20 + (100 − a) × 22 ; ∴ a = 90 ; per cent of heavier isotope = 100 − 90 = 10 100

The relative abundance of two isotopes of atomic weight 85 and 87 is 75% and 25% respectively. The average atomic weight of element is (a) 75.5

Solution:(b)

Example : 9

(b) 85.5

(c) 87.5

(d) 86.0

Average atomic weight/ The average isotopic weight =


=

85 × 75 + 87 × 25 = 85 . 5 100

Nitrogen atom has an atomic number of 7 and oxygen has an atomic number of 8. The total number of electrons in a nitrate ion is (a) 30

(b) 35

(c) 32

(d) None

Solution : (c) Number of electrons in an element = Its atomic number So number of electrons in N=7 and number of electrons in O=8. Formula of nitrate ion is NO 3− So, in it number of electrons = 1 × number of electrons of nitrogen +3 × number of electrons of oxygen +1 (due to negative charge) = 1 × 7 + 3 × 8 + 1 = 32

Example :10

An atom of an element contains 11 electrons. Its nucleus has 13 neutrons. Find out the atomic number and approximate atomic weight. (a) 11, 25

(b) 12, 34

(c) 10, 25

(d) 11, 24

Solution : (d) Number of electrons =11

∴ Number of protons = Number of electron =11 Number of neutrons = 13 Atomic number of element = Number of proton = Number of electrons =11 Further, Atomic weight = Number of protons + Number of neutrons =11 + 13=24 Example : 11 How many protons, neutrons and electrons are present in (a) Solution :

40 18

Ar (c)

108 47

Ag ?

Protons

Electrons

Neutrons

P

15

15

31 – 15=16

40 18

Ar

18

18

40 – 18=22

108 47

Ag

47

47

108 – 47=61

31 15

Solution :

P (b)

The atomic number subscript gives the number of positive nuclear charges or protons. The neutral atom contains an equal number of negative electrons. The remainder of the mass is supplied by neutrons. Atom

Example :12

31 15

State the number of protons, neutrons and electrons in C 12 and C 14 . The atomic number of C 12 is 6. So in it number of electrons = 6 Number of protons =6; Number of neutrons =12 – 6=6 The atomic number of C 14 is 6. So in it number of electrons = 6 Number of protons = 6; Number of neutrons =14 – 6=8

Example :13

Predict the number of electrons, protons and neutrons in the two isotopes of magnesium with atomic number 12 and atomic weights 24 and 26.

Solution :

Isotope of the atomic weight 24, i.e.

12

Mg 24 . We know that

Number of protons = Number of electrons =12 Further, Number of neutrons = Atomic weight – Atomic number =24 – 12 =12 Similarly, In isotope of the atomic weight 26, i.e.

12

Mg 26

Number of protons = Number of electrons =12 Number of neutrons = 26 – 12 = 14

2.3 Electromagnetic Radiations. (1) Light and other forms of radiant energy propagate without any medium in the space in the form of waves are known as electromagnetic radiations. These waves can be produced by a charged body moving in a magnetic field or a magnet in a electric field. e.g. α − rays, γ − rays, cosmic rays, ordinary light rays etc. (2) Characteristics : (i) All electromagnetic radiations travel with the velocity of light. (ii) These consist of electric and magnetic fields components that oscillate in directions perpendicular to each other and perpendicular to the direction in which the wave is travelling. (3) A wave is always characterized by the following five characteristics: (i) Wavelength : The distance between two nearest crests or nearest troughs is called the wavelength. It is denoted by λ (lambda) and is measured is terms of centimeter(cm), angstrom(Å), micron( µ )

Crest

Wavelength

Vibrating source Energy

or nanometre (nm). 1 Å = 10 −8 cm = 10 −10 m

Trough

1µ = 10 −4 cm = 10 −6 m

1nm = 10 −7 cm = 10 −9 m 1cm = 10 8 Å = 10 4 µ = 10 7 nm

(ii) Frequency : It is defined as the number of waves which pass through a point in one second. It is denoted by the symbol ν (nu) and is expressed in terms of cycles (or waves) per second (cps) or hertz (Hz). λν = distance travelled in one second = velocity =c

ν=

c λ

(iii) Velocity : It is defined as the distance covered in one second by the wave. It is denoted by the letter ‘c’. All electromagnetic waves travel with the same velocity, i.e., 3 × 10 10 cm / sec . c = λν = 3 × 10 10 cm / sec

Thus, a wave of higher frequency has a shorter wavelength while a wave of lower frequency has a longer wavelength. APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

(iv) Wave number : This is the reciprocal of wavelength, i.e., the number of wavelengths per centimetre. It is denoted by the symbol ν (nu bar). It is expressed in cm −1 or m −1 . ν =

1 λ

(v) Amplitude : It is defined as the height of the crest or depth of the trough of a wave. It is denoted by the letter ‘A’. It determines the intensity of the radiation. The arrangement of various types of electromagnetic radiations in the order of their increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum. Name

Wavelength (Å)

Frequency (Hz)

Source

Radio wave

3 × 10 14 − 3 × 10 7

1 × 10 5 − 1 × 10 9

Alternating frequency

current

of

high

Microwave

3 × 10 7 − 6 × 10 6

1 × 10 9 − 5 × 10 11

Klystron tube

Infrared (IR)

6 × 10 6 − 7600

5 × 10 11 − 3 . 95 × 10 16

Incandescent objects

Visible

7600 − 3800

3 . 95 × 10 16 − 7 . 9 × 10 14

Electric bulbs, sun rays

Ultraviolet (UV)

3800 − 150

7.9 × 10 14 − 2 × 10 16

Sun rays, arc mercury vapours

lamps

with

X-Rays

150 − 0 . 1

2 × 10 16 − 3 × 10 19

Cathode rays striking metal plate

γ − Rays

0 .1 − 0 . 01

3 × 10 19 − 3 × 10 20

Secondary effect of radioactive decay

Cosmic Rays

0.01- zero

3 × 10 20 − infinity

Outer space

2.4 Atomic spectrum - Hydrogen spectrum.

Atomic spectrum (1) Spectrum is the impression produced on a photographic film when the radiation (s) of particular wavelength (s) is (are) analysed through a prism or diffraction grating. It is of two types, emission and absorption. (2) Emission spectrum : A substance gets excited on heating at a very high temperature or by giving energy and radiations are emitted. These radiations when analysed with the help of spectroscope, spectral lines are obtained. A substance may be excited, by heating at a higher temperature, by passing electric current at a very low pressure in a discharge tube filled with gas and passing electric current into metallic filament. Emission spectra is of two types, (i) Continuous spectrum : When sunlight is passed through a prism, it gets dispersed into continuous bands of different colours. If the light of an incandescent object resolved through prism or spectroscope, it also gives continuous spectrum of colours. (ii) Line spectrum : If the radiations obtained by the excitation of a substance are analysed with help of a spectroscope a series of thin bright lines of specific colours are obtained. There is dark space in between two consecutive lines. This type of spectrum is called line spectrum or atomic spectrum.. (3) Absorption spectrum : When the white light of an incandescent substance is passed through any substance, this substance absorbs the radiations of certain wavelength from the white light. On analysing the transmitted light we obtain a spectrum in which dark lines of specific wavelengths are observed. These lines

constitute the absorption spectrum. The wavelength of the dark lines correspond to the wavelength of light absorbed.

Hydrogen spectrum (1) Hydrogen spectrum is an example of line emission spectrum or atomic emission spectrum. (2) When an electric discharge is passed through hydrogen gas at low pressure, a bluish light is emitted. (3) This light shows discontinuous line spectrum of several isolated sharp lines through prism. (4) All these lines of H-spectrum have Lyman, Balmer, Paschen, Barckett, Pfund and Humphrey series. These spectral series were named by the name of scientist discovered them. (5) To evaluate wavelength of various H-lines Ritz introduced the following expression, ν =

1 1 ν 1  = = R 2 − 2  λ c  n1 n 2 

Where R is universal constant known as Rydberg’s constant its value is 109, 678 cm −1 .

2.5 Thomson's model. (1) Thomson regarded atom to be composed of positively charged protons and negatively charged electrons. The two types of particles are equal in number thereby making atom Positively charged sphere + – + electrically neutral. –

(2) He regarded the atom as a positively charged sphere in which negative electrons are uniformly distributed like the seeds in a water melon. (3) This model failed to explain the line spectrum of an element and the scattering experiment of Rutherford.

–

+ – + –

+ –

+

Electron

+

Positive charge spreaded throughout the sphere

2.6 Rutherford's nuclear model. (1) Rutherford carried out experiment on the bombardment of thin (10–4 mm) Au foil with high speed positively charged α − particles emitted from Ra and gave the following observations, based on this experiment : (i) Most of the α − particles passed without any deflection. (ii) Some of them were deflected away from their path. (iii) Only a few (one in about 10,000) were returned back to their original direction of propagation. (iv) The scattering of α − particles ∝

1

.  sin   2 Scattering of α -particle 4θ

θ b r0

Nucleus

α-particle (energy E eV)


(2) From the above observations he concluded that, an atom consists of (i) Nucleus which is small in size but carries the entire mass i.e. contains all the neutrons and protons. (ii) Extra nuclear part which contains electrons. This model was similar to the solar system. Planetry electron – +

Nucleus

10–15 m

10 –10 m Size of the nucleus = 1 Fermi = 10–15 m Size of the atom 1 Å = 10–10 m

(3) Properties of the Nucleus (i) Nucleus is a small, heavy, positively charged portion of the atom and located at the centre of the atom. (ii) All the positive charge of atom (i.e. protons) are present in nucleus. (iii) Nucleus contains neutrons and protons, and hence these particles collectively are also referred to as nucleons. (iv) The size of nucleus is measured in Fermi (1 Fermi = 10–13 cm). (v) The radius of nucleus is of the order of 1 . 5 × 10 −13 cm . to 6 . 5 × 10 −13 cm . i.e. 1 . 5 to 6 . 5 Fermi. Generally the radius of the nucleus ( rn ) is given by the following relation,

rn = ro ( = 1 . 4 × 10

− 13

cm ) × A 1 / 3

This exhibited that nucleus is 10 −5 times small in size as compared to the total size of atom. (vi) The Volume of the nucleus is about 10 −39 cm 3 and that of atom is 10 −24 cm 3 , i.e., volume of the nucleus is 10 −15 times that of an atom. (vii) The density of the nucleus is of the order of 10 15 g cm −3 or 10 8 tonnes cm −3 or 10 12 kg / cc . If nucleus is spherical than, Density =

mass of the nucleus volume of the nucleus

=

mass number 4 6 . 023 × 10 23 × πr 3 3

(4) Drawbacks of Rutherford's model (i) It does not obey the Maxwell theory of electrodynamics, according to it “A small charged particle moving around an oppositely charged centre continuously loses its energy”. If an electron does so, it should also continuously lose its energy and should set up spiral motion ultimately failing into the nucleus. (ii) It could not explain the line spectra of H − atom and discontinuous spectrum nature.

e–

Unstability of atom

+

Examples based on Properties of the nucleus

Example:14

Assuming a spherical shape for fluorine nucleus, calculate the radius and the nuclear density of fluorine nucleus of mass number 19.

Solution :

We know that, r = (1. 4 × 10 −13 ) A 1 / 3 = 1 .4 × 10 −13 × 19 1 / 3 = 3 .73 × 10 −13 cm

Volume of a fluorine atom = Mass of single nucleus =

(A for F=19)

4 3 4 πr = × 3 .14 × (3 . 73 × 10 −13 ) 3 = 2 . 18 × 10 −37 cm 3 3 3

Mass of one mol of nucleus 19 = g Avogadro' s number 6 . 023 × 10 23

Thus Density of nucleus =

Mass of single nucleus 10 1 = × 23 Volume of single nucleus 6 . 023 × 10 2 .18 × 10 −37

= 7 . 616 = 10 13 g cm −1

Atomic radius is the order of 10 −8 cm, and nuclear radius is the order of 10 −13 cm . Calculate what

Example: 15

fraction of atom is occupied by nucleus. Solution :

Volume of nucleus = (4 / 3)pr 3 = (4 / 3)p × (10 −13 )3 cm 3 Volume of atom = (4 / 3)pr 3 = (4 / 3)p × (10 −8 )3 cm 3

∴

Vnucleus 10 −39 = − 24 = 10 −15 or Vnucleus = 10 −15 × Vatom Vatom 10

2.7 Planck's Quantum theory and Photoelectric effect.

Planck's Quantum theory (1) Max Planck (1900) to explain the phenomena of 'Black body radiation' and 'Photoelectric effect' gave quantum theory. This theory extended by Einstein (1905). (2) If the substance being heated is a black body (which is a perfect absorber and perfect radiator of energy) the radiation emitted is called black body radiation. (3) Main points (i) The radiant energy which is emitted or absorbed by the black body is not continuous but discontinuous in the form of small discrete packets of energy, each such packet of energy is called a 'quantum'. In case of light, the quantum of energy is called a 'photon'. (ii) The energy of each quantum is directly proportional to the frequency (ν ) of the radiation, i.e. hc E ∝ ν or E = h ν = λ where, h = Planck's constant = 6.62×10–27 erg. sec. or 6 . 62 × 10 −34 Joules sec . (iii) The total amount of energy emitted or absorbed by a body will be some whole number quanta. Hence E = nh ν , where n is an integer. (iv) The greater the frequency (i.e. shorter the wavelength) the greater is the energy of the radiation. E1 ν 1 λ 2 = = thus, E 2 ν 2 λ1


(v)

Also E = E1 + E 2 , hence, E=

hc λ

hc hc hc 1 1 1 or = . = + + λ λ1 λ 2 λ λ1 λ 2

Examples based on Planck's Quantum theory

Example: 16 Suppose 10 −17 J of energy is needed by the interior of human eye to see an object. How many photons of green light (λ = 550 nm ) are needed to generate this minimum amount of energy (a) 14 (b) 28 (c) 39 (d) 42 Solution : (b) Let the number of photons required =n 10 −17 × λ 10 −17 × 550 × 10 −9 hc = = 27 . 6 = 28 photons n = 10 −17 ; n = hc λ 6. 626 × 10 − 34 × 3 × 10 8 Example: 17 Assuming that a 25 watt bulb emits monochromatic yellow light of wave length 0 .57 µ. The rate of emission of quanta per sec. will be (a) 5 .89 × 10 13 sec −1 (b) 7. 28 × 10 17 sec −1 (c) 5 × 10 10 sec −1 Solution: (d) Let n quanta are evolved per sec. 6 .626 × 10 −34 × 3 × 10 8  hc  = 25 ; n = 7. 18 × 10 19 sec −1 n   = 25 J sec −1 ; n 0 . 57 × 10 −6 λ

(d) 7 .18 × 10 19 sec −1

Photoelectric effect (1) When radiations with certain minimum frequency (ν 0 ) strike the surface of a metal, the electrons are ejected from the surface of the metal. This phenomenon is called photoelectric effect and the electrons emitted are called photo-electrons. The current constituted by photoelectrons is known as photoelectric current. (2) The electrons are ejected only if the radiation striking the surface of the metal has at least a minimum frequency (ν 0 ) called Threshold frequency. The minimum potential at which the plate photoelectric current becomes zero is called stopping potential. (3)The velocity or kinetic energy of the electron ejected depend upon the frequency of the incident radiation and is independent of its intensity. (4) The number of photoelectrons ejected is proportional to the intensity of incident radiation. (5) Einstein’s photoelectric effect equation : According to Einstein, Maximum kinetic energy of the ejected electron = absorbed energy – threshold energy 1 1  1 2 mv max = h ν − h ν 0 = hc  −  2  λ λ0  where, ν 0 and λ0 are threshold frequency and threshold wavelength.

Note

:q Nearly all metals emit photoelectrons when exposed to u.v. light. But alkali metals like lithium, sodium, potassium, rubidium and caesium emit photoelectrons even when exposed to visible light. U.V. light

Visible light

Visible light

Metal

Metal other than alkali metals

Alkali metals

Photo electrons

No photo electrons

Photo electrons

q Caesium (Cs) with lowest ionisation energy among alkali metals is used in photoelectric cell.

2.8 Bohr’s atomic model. (1) This model was based on the quantum theory of radiation and the classical law of physics. It gave new idea of atomic structure in order to explain the stability of the atom and emission of sharp spectral lines. (2) Postulates of this theory are : (i) The atom has a central massive core nucleus where all the protons and neutrons are present. The size of the nucleus is very small. (ii) The electron in an atom revolve around the nucleus in certain discrete orbits. Such orbits are known as stable orbits or non – radiating or stationary orbits. (iii) The force of attraction between the nucleus and the electron is equal to centrifugal force of the moving electron. Force of attraction towards nucleus = centrifugal force

(iv) An electron can move only in those permissive orbits in which the angular momentum (mvr) of the h electron is an integral multiple of h / 2π . Thus, mvr = n 2π Where, m = mass of the electron, r = radius of the electronic orbit, v = velocity of the electron in its orbit. h 2h 3 h nh ,...... . This principal is known as quantization of , , 2π 2π 2π 2π angular momentum. In the above equation ‘n’ is any integer which has been called as principal quantum number. It can have the values n=1,2,3, ------- (from the nucleus). Various energy levels are designed as K(n=1), L(n=2), M(n=3) ------- etc. Since the electron present in these orbits is associated with some energy, these orbits are called energy levels.

(v) The angular momentum can be

(vi) The emission or absorption of radiation by the atom takes place when an electron jumps from one stationary orbit to another. E1

E1

E1 – E2 = hν

E1 – E2 = hν E2

E2 Emission

Absorption

(vii) The radiation is emitted or absorbed as a single quantum (photon) whose energy h ν is equal to the difference in energy ∆E of the electron in the two orbits involved. Thus, h ν = ∆E Where ‘h’ =Planck’s constant, ν = frequency of the radiant energy. Hence the spectrum of the atom will have certain fixed frequency. (viii) The lowest energy state (n=1) is called the ground state. When an electron absorbs energy, it gets excited and jumps to an outer orbit. It has to fall back to a lower orbit with the release of energy. (3) Advantages of Bohr’s theory (i) Bohr’s theory satisfactorily explains the spectra of species having one electron, viz. hydrogen atom, He + , Li 2 + etc. APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

(ii) Calculation of radius of Bohr’s orbit : According to Bohr, radius of orbit in which electron moves is   n2 h2 r= 2 . 2   4 π me k  Z

[

]

[

where, n =Orbit number, m =Mass number 9 . 1 × 10 −31 kg , e =Charge on the electron 1 . 6 × 10 −19

[

Z =Atomic number of element, k = Coulombic constant 9 × 10 Nm c 9

2

−2

]

After putting the values of m,e,k,h, we get. rn =

n2 n2 × 0 . 529 Å or rn = × 0 . 529 nm Z Z

(a) For a particular system [e.g., H, He+ or Li+2]

r ∝ n 2 [Z = constant] Thus we have

r1 n12 = i.e., r1 : r2 : r3 .......... . :: 1 : 4 : 9 ....... r1 < r2 < r3 r2 n 22

(b) For particular orbit of different species r∝

r Z 1 [Z =constant] Considering A and B species, we have A = B Z ZA rB

Thus, radius of the first orbit H, He + , Li +2 and Be +3 follows the order: H > He + > Li +2 > Be +3 (iii) Calculation of velocity of electron Vn =

 Ze 2  2πe 2 ZK , Vn =   nh  mr 

For H atom, Vn =

1/ 2

2 . 188 × 10 8 cm . sec −1 n

(a) For a particular system [H, He+ or Li+2] V∝

V n 1 [Z = constant] Thus, we have, 1 = 2 n V2 n 1

The order of velocity is V1 > V2 > V3 ......... or V1 : V2 : V3 .......... :: 1 :

1 1 : ........ 2 3

(b) For a particular orbit of different species V ∝ Z [n =constant] Thus, we have H < He + < Li +2

(c) For H or He+ or Li+2, we have

V1 : V2 = 2 : 1 ; V1 : V3 = 3 : 1 ; V1 : V4 = 4 : 1 (iv) Calculation of energy of electron in Bohr’s orbit Total energy of electron = K.E. + P.E. of electron = Substituting of r, gives us E =

− 2π 2 mZ 2 e 4 k 2

n 2h2 Putting the value of m, e, k, h, π we get

kZe 2 kZe 2 kZe 2 − =− 2r r 2r

Where, n=1, 2, 3………. ∞

]

E = 21 . 8 × 10 −12 ×

E = −13 .6 ×

Z2 Z2 −19 erg per atom = − 21 . 8 × 10 × J per atom (1 J = 10 7 erg ) n2 n2

Z2 Z2 -19 (1eV = 1.6 × 10 ) = − 313 . 6 × kcal . / mole (1 cal = 4.18J) eV per atom J n2 n2

− 1312 2 Z kJmol −1 n2 (a) For a particular system[H, He+ or Li+2]

or

E∝−

E1 n 22 1 = [Z =constant] Thus, we have E 2 n12 n2

The energy increase as the value of n increases (b) For a particular orbit of different species

E ∝ − Z 2 [n =constant] Thus, we have

EA Z2 = A2 EB ZB

For the system H, He+ , Li+2, Be+3 (n-same) the energy order is H > He + > Li +2 > Be +3 The energy decreases as the value of atomic number Z increases. When an electron jumps from an outer orbit (higher energy) n 2 to an inner orbit (lower energy) n1 , then the energy emitted in form of radiation is given by ∆E = E n 2 − E n1 =

2π 2 k 2 me 4 Z 2 h2

 1  1 1  1   2 − 2  ⇒ ∆E = 13 . 6 Z 2  2 − 2  eV / atom  n1 n 2   n1 n 2 

As we know that E = h ν , c = νλ and ν =

2π 2 k 2 me 4 Z 2 1 ∆E = , = λ hc ch 3

 1 1    n2 − n2  2   1

 1 1 1  2π 2 k 2 me 4 = ν = RZ 2  2 − 2  where, R = R is known as Rydberg λ ch 3  n1 n 2  constant. Its value to be used is 109678 cm −1 . (4) Quantisation of energy of electron (i) In ground state : No energy emission. In ground state energy of atom is minimum and for 1st orbit of H-atom, n=1. ∴ E1 = −13 . 6 eV .

This can be represented as

(ii)

In excited state : Energy levels greater than n 1 are excited state. i.e. for H- atom n 2 , n 3 , n 4 are

excited state. For H- atom first excitation state is = n 2 (iii) Excitation potential : Energy required to excite electron from ground state to any excited state. Ist

Ground state → Excited state excitation potential = E 2 − E1 = −3 . 4 + 13 . 6 = 10.2 eV.

IInd excitation potential = E 3 − E1 = −1 . 5 + 13 .6 = 12 . 1 eV . (iv) Ionisation energy : The minimum energy required to relieve the electron from the binding of nucleus. E ionisation = E ∞ − E n = +13 . 6

(v) Ionisation potential : Vionisation =

2 Z eff. eV . n2

E ionisation e


(vi) Separation energy : Energy required to excite an electron from excited state to infinity. S.E. = E ∞ − E excited . (vii) Binding energy : Energy released in bringing the electron from infinite to any orbit is called its binding energy (B.E.).

Note : q

Principal Quantum Number 'n' =

13 . 6 . (B. E.)

(5) Spectral evidence for quantisation (Explanation for hydrogen spectrum on the basisof bohr atomic model) (i) The light absorbed or emitted as a result of an electron changing orbits produces characteristic absorption or emission spectra which can be recorded on the photographic plates as a series of lines, the optical spectrum of hydrogen consists of several series of lines called Lyman, Balmar, Paschen, Brackett, Pfund and Humphrey. These spectral series were named by the name of scientist who discovered them. (ii) To evaluate wavelength of various H-lines Ritz introduced the following expression, ν =

where, R is =

2π 2 me 4 ch 3

1 1 ν 1 = = R 2 − 2  λ c  n1 n 2 

= Rydberg's constant

It's theoritical value = 109,737 cm–1 and It's experimental value = 109 ,677 . 581 cm −1 This remarkable agreement between the theoretical and experimental value was great achievment of the Bohr model. (iii) Although H- atom consists only one electron yet it's spectra consist of many spectral lines as shown in fig. n=8 n=7 n=6 n=5

Humphrey series Pfund

series

Energy level

n=4

Bracket t series

n=3 Paschen

series n=2 Balmer series

n=1

Lyman series

(iv) Comparative study of important spectral series of Hydrogen

S.No.

Spectral series

Lies in the region

Transitio n

λmax =

n12 n 22 − n12 )R

(n 22

λ min =

n12 R

n 2 > n1

(1)

(2)

(3)

(4)

(5)

(6)

Lymen series

Balmer series

Paschen series

Brackett series

Pfund series

Humphrey series

Ultraviolet region

Visible region

Infra red region

Infra red region

Infra red region

Far infrared region

n1 = 1 n 2 = 2,3,4 .... ∞ n1 = 2 n 2 = 3,4 ,5 .... ∞

n1 = 3 n 2 = 4 ,5,6 .... ∞

n1 = 4 n 2 = 5 ,6,7 .... ∞ n1 = 5 n 2 = 6 ,7,8 .... ∞ n1 = 6 n 2 = 7 ,8 .... ∞

n1 = 1 and n 2 = 2

λ max =

4 3R

n1 = 2 and n 2 = 3

λ max =

36 5R

n 1 = 3 and n 2 = 4

λ max

144 = 7R

n1 = 4 and n2 = 5

λmax =

16 × 25 9R

n 1 = 5 and n 2 = 6

λ max

25 × 36 = 11 R

n 1 = 6 and n 2 = 7

λ max

36 × 49 = 13 R

λ max n2 = 2 2 2 λ min n 2 − n1

n 1 = 1 and n 2 = ∞

λ min =

1 R

4 3

n1 = 2 and n2 = ∞

λ min =

4 R

9 5

n 1 = 3 and n 2 = ∞

λ min

9 = R

16 7

n 1 = 4 and n 2 = ∞

λ min =

16 R

25 9

n 1 = 5 and n 2 = ∞

λ min

25 = R

36 11

n 1 = 6 and n 2 = ∞

λ min

36 = R

49 13

(v) If an electron from nth excited state comes to various energy states, the maximum spectral lines n(n − 1) . n= principal quantum number. obtained will be = 2 6(6 − 1) 30 = = 15 . 2 2 (vi) Thus, at least for the hydrogen atom, the Bohr theory accurately describes the origin of atomic spectral lines. (6) Failure of Bohr Model (i) Bohr theory was very successful in predicting and accounting the energies of line spectra of hydrogen i.e. one electron system. It could not explain the line spectra of atoms containing more than one electron. (ii) This theory could not explain the presence of multiple spectral lines. (iii) This theory could not explain the splitting of spectral lines in magnetic field (Zeeman effect) and in electric field (Stark effect). The intensity of these spectral lines was also not explained by the Bohr atomic model. (iv) This theory was unable to explain of dual nature of matter as explained on the basis of De broglies concept. (v) This theory could not explain uncertainty principle. (vi) No conclusion was given for the concept of quantisation of energy.

as n=6 than total number of spectral lines =


Examples based on Bohr’s atomic model and Hydrogen spectrum Example: 18 If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be 4 9 (a) (b) 4r2 (c) r3 (d) 9r2 r2 4 9 Solution : (c) r =

n2h2 4π 2 mZe 2

∴

r2 22 9 = 2 ∴ r3 = r2 r3 4 3

Example: 19 Number of waves made by a Bohr electron in one complete revolution in 3rd orbit is (a) 2 (b) 3 (c) 4 (d) 1 Solution : (b) Circumference of 3rd orbit = 2πr3 According to Bohr angular momentum of electron in 3rd orbit is 2πr3 h h mvr3 = 3 or = 2π mv 3 h by De-Broglie equation, λ = mv 2πr3 ∴2πr3 = 3λ ∴λ = 3 i.e. circumference of 3rd orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in 3rd orbit is three. R Example: 20 The degeneracy of the level of hydrogen atom that has energy − 11 is 16 (a) 16 (b) 4 (c) 2 (d) 1 RH RH RH Solution : (a) En = − 2 ∴ − 2 = − 16 n n i.e. for 4 th sub-shell 1 n=4

2

3

1=0 m=0

+1

0

one s

three p

+1

+2

–1

0

+1

+2

five d

–3

–2

–1

0

+1

+2

+3

seven f

i.e. 1+3+5+7=16, ∴ degeneracy is 16 Example: 21 The velocity of electron in the ground state hydrogen atom is 2.18 × 10 8 ms −1 . Its velocity in the second orbit would be (a) 1 . 09 × 10 8 ms −1

(b) 4 . 38 × 10 8 ms −1

(c) 5 .5 × 10 5 ms −1

(d) 8 . 76 × 10 8 ms −1

Solution : (a) We know that velocity of electron in nth Bohr's orbit is given by v = 2 . 18 × 10 6

Z m /s n

for H , Z = 1

Q v1 =

2 .18 × 10 6 m/s 1

Q v2 =

2 . 18 × 10 6 m / s = 1. 09 × 10 6 m / s 2

Example: 22 The ionization energy of the ground state hydrogen atom is 2 . 18 × 10 −18 J . The energy of an electron in its second orbit would be

(a) − 1. 09 × 10 −18 J

(b) − 2 . 18 × 10 −18 J

(c) − 4 . 36 × 10 −18 J

(d) − 5 . 45 × 10 −19 J

Solution : (d) Energy of electron in first Bohr's orbit of H-atom E=

− 2. 18 × 10 −18 J (Q ionization energy of H = 2 . 18 × 10 −18 J ) n2

E2 =

− 2 .18 × 10 −18 J = −5. 45 × 10 −19 J 22

Example: 23 The wave number of first line of Balmer series of hydrogen atom is 15200 cm −1 . What is the wave number of first line of Balmer series of Li 3 + ion. (a) 15200 cm −1

(b) 6080 cm −1

(d) 1,36800 cm −1

(c) 76000 cm −1

Solution : (d) For Li 3 + v = v for H × z 2 =15200 ×9= 1,36800 cm −1 Example: 24

The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530Å. The radius for the first excited state (n = 2) orbit is (in Å) (a) 0.13

(b) 1.06

(c) 4.77

(d) 2.12

Solution : (d) The Bohr radius for hydrogen atom (n = 1) = 0.530Å The radius of first excited state (n = 2) will be = 0 .530 ×

n2 (2)2 = 0 . 530 × = 2 .120 Å Z 1

Example: 25 How many chlorine atoms can you ionize in the process Cl → Cl + + e − , by the energy liberated from the following process : Cl + e − → Cl − for 6 × 10 23 atoms Given electron affinity of Cl = 3 .61 eV , and IP of Cl = 17 . 422 eV (a) 1.24 × 10 23 atoms

(b) 9 .82 × 10 20 atoms (c) 2 . 02 × 10 15 atoms

Solution : (a) Energy released in conversion of 6 × 10 = 6× 10

23

× 3 . 61 = 2 . 166 × 10

24

23

−

atoms of Cl ions = 6 × 10

23

(d) None of these

× electron affinity

eV.

Let x Cl atoms are converted to Cl + ion Energy absorbed = x × ionization energy x × 17 . 422 = 2 . 166 × 10 24 ; x = 1 .243 × 10 23 atoms Example: 26 The binding energy of an electron in the ground state of the He atom is equal to 24eV. The energy required to remove both the electrons from the atom will be (a) 59eV (b) 81eV (c) 79eV (d) None of these Z2 22 × 13 . 6 = 2 × 13 . 6 = 54 .4 eV 2 n 1 Energy required to remove both the electrons = binding energy + ionization energy

Solution : (c) Ionization energy of He =

= 24 . 6 + 54 . 4 = 79eV Example: 27 The wave number of the shortest wavelength transition in Balmer series of atomic hydrogen will be (a) 4215 Å Solution : (d)

1 λ shortest

(b) 1437Å

(c) 3942Å

(d) 3647Å

 1 1  1   1 = RZ 2  2 − 2  = 109678 × 1 2 ×  2 − 2  n n 2 ∞   2   1

λ = 3 . 647 × 10 −5 cm = 3647 Å Example: 28 If the speed of electron in the Bohr's first orbit of hydrogen atom is x, the speed of the electron in the third Bohr's orbit is (a) x/9 (b) x/3 (c) 3x (d) 9x APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

Solution : (b) According to Bohr's model for hydrogen and hydrogen like atoms the velocity of an electron in an atom is 1 2πZe 2 quantised and is given by v ∝ so v ∝ in this cass n = 3 nh n Example: 29 Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is (a) n=1 to n=2 (b) n = 3 to n = 8 (c) n = 2 to n = 1 (d) n = 8 to n = 3 Solution : (b) Absorption line in the spectra arise when energy is absorbed i.e., electron shifts from lower to higher orbit, out of a & b, b will have the lowest frequency as this falls in the Paschen series. Example: 30 The frequency of the line in the emission spectrum of hydrogen when the atoms of the gas contain electrons in the third energy level are (a) 1 . 268 × 10 14 Hz and 2 . 864 × 10 16 Hz

(b) 3 .214 × 10 10 Hz and 1 . 124 × 10 12 Hz

(c) 1 . 806 × 10 12 Hz and 6 . 204 × 10 15 Hz (d) 4 .568 × 10 14 Hz and 2 . 924 × 10 15 Hz Solution : (d) If an electron is in 3rd orbit, two spectral lines are possible (a) When it falls from 3rd orbit to 2nd orbit. 1 1  In equation ν = 3. 289 × 10 15  2 − 2   n1 n 2  1  5 1 = 4 .568 × 14 14 Hz ν 1 = 3. 289 × 10 15  2 − 2  = 3 . 289 × 10 15 × 36 2 3   (b) When it falls from 3rd orbit to 1st orbit : 8 1 1  ν 2 = 3 . 289 × 10 15 ×  − 2  = 3 . 289 × 10 15 × = 2 . 924 × 10 15 Hz 1 9 3  

Example: 31 If the first ionisation energy of hydrogen is 2 . 179 × 10 −18 J per atom, the second ionisation energy of helium per atom is (a) 8 . 716 × 10 −18 J (b) 5 .5250 kJ (c) 7 .616 × 10 −18 J (d) 8 . 016 × 10 −13 J Z2 n2 Ionisation energy is the difference of energies of an electron (E ∞ ), when taken to infinite distance and Er

Solution : (a) For Bohrs systems : energy of the electron ∝

when present in any Bohr orbit and Eα is taken as zero so ionisation energy becomes equal to the energy of electron in any Bohr orbit. E Z2 Z2 1 [as Z H = 1, Z He = 2, n H = 1, n He = 1] E H ∝ 2H ; E He ∝ 2He or H = E He 2×2 n He nH or E He = E H × 4 = 2 .179 × 10 −18 × 4 = 8 . 716 × 10 −18 Joule per atom. Example: 32 The ionization energy of hydrogen atom is 13.6eV. What will be the ionization energy of He + (a) 13.6eV (b) 54.4eV (c) 122.4eV (d) Zero Solution : (b) I.E. of He + = 13 . 6 eV × Z 2 13 . 6 eV × 4 = 54 . 4 eV Example: 33 The ionization energy of He + is 19 . 6 × 10 −18 J atom–1. Calculate the energy of the first stationary state of Li +2 (b) 4 .41 × 10 −18 J atom–1 (a) 19 . 6 × 10 −18 J atom -1 (c) 19 . 6 × 10 −19 J atom -1

(d) 4. 41 × 10 −17 J atom −1

Solution : (d) I.E. of He + = E × 2 2 (Z for He = 2) I.E. of Li 2 + = E × 3 3 (Z for Li=3) I. E.(He + ) 4 9 9 or I.E. (Li 2 + ) = × I. E.(He + ) = × 19 .6 × 10 −18 = 4 . 41 × 10 −17 J atom–1 ∴ = 2+ 4 4 I. E.(Li ) 9

2.9 Bohr – Sommerfeld’s model. (1) In 1915, Sommerfield introduced a new atomic model to explain the fine spectrum of hydrogen atom.

(2) He gave concept that electron revolve round the nucleus in elliptical orbit. Circular orbits are formed in special conditions only when major axis and minor axis of orbit are equal. nh (3) For circular orbit, the angular momentum = where n= principal quantum number only one 2π component i.e. only angle changes. (4) For elliptical orbit, angular momentum = vector sum of 2 components. In elliptical orbit two components are, h (i) Radial component (along the radius) = nr 2π Where, n r = radial quantum number h 2π Where, n φ = azimuthal quantum number

(ii) Azimuthal component = n φ

So angular momentum of elliptical orbit = nr Angular momentum = (n r + n φ )

h h + nφ 2π 2π r

h 2π

(5) Shape of elliptical orbit depends on,

φ2

r

rr

φ1

φ = change

n r + nφ Length of major axis n = = Length of minor axis nφ nφ

φ2

rr φ11

φ = change

r = change

r = constant

(6) n φ can take all integral values from l to ‘n’ values of n r depend on the value of n φ . For n = 3, n φ can have values 1,2,3 and n r can have (n –1) to zero i.e. 2,1 and zero respectively. Thus for n = 3, we have 3 paths n

nφ

nr

Nature of path

3

1

3

elliptical

2 3

1 0

elliptical circular

K= 3 K= 2 K= 1 • Nuclear

The possible orbits for n = 3 are shown in figure. Thus Sommerfield showed that Bohr’s each major level was composed of several sub-levels. therefore it provides the basis for existance of subshells in Bohr's shells (orbits). (7) Limitation of Bohr sommerfield model : (i) This model could not account for, why electrons does not absorb or emit energy when they are moving in stationary orbits. (ii) When electron jumps from inner orbit to outer orbit or vice –versa, then electron run entire distance but absorption or emission of energy is discontinuous. (iii) It could not explain the attainment of expression of

nh for angular momentum. This model could 2π

not explain Zeeman effect and Stark effect.


2.10 Dual nature of electron. (1) In 1924, the french physicist, Louis de Broglie suggested that if light has both particle and wave like nature, the similar duality must be true for matter. Thus an electron, behaves both as a material particle and as a wave. (2) This presented a new wave mechanical theory of matter. According to this theory, small particles like electrons when in motion possess wave properties. (3) According to de-broglie, the wavelength associated with a particle of mass m , moving with velocity v is given by the relation h , where h = Planck’s constant. λ= mv h.c  c (4) This can be derived as follows according to Planck’s equation, E = h ν = Q ν =  λ λ  2 energy of photon (on the basis of Einstein’s mass energy relationship), E = mc hc h equating both which is same as de-Broglie relation. (Q mc = p ) = mc 2 or λ = λ mc (5) This was experimentally verified by Davisson and Germer by observing diffraction effects with an electron beam. Let the electron is accelerated with a potential of V than the Kinetic energy is 1 mv 2 = eV ; m 2 v 2 = 2eVm 2 h mv = 2eVm = P ; λ = 2eVm (6) If Bohr’s theory is associated with de-Broglie’s equation then wave length of an electron can be determined in bohr’s orbit and relate it with circumference and multiply with a whole number 2πr 2πr = n λ or λ = n h h 2πr nh . Thus or mvr = From de-Broglie equation, λ = = mv mv n 2π

Note

For a proton, electron and an α -particle moving with the same velocity have de-broglie wavelength in the following order : Electron > Proton > α - particle. :q

(7) The de-Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles. Since, we come across macroscopic objects in our everyday life, de-broglie relationship has no significance in everyday life. Examples based on de-Broglie’s equation Example: 34

An electron is moving with a kinetic energy of 4.55 × 10 −25 J. What will be de-Broglie wavelength for this electron (a) 5.28 × 10 −7 m KE =

Solution : (b) v2 =

(b) 7.28 × 10 −7 m

(d) 3 × 10 −5 m

1 mv 2 = 4 .55 × 10 − 25 J 2

2 × 4. 55 × 10 −25 = 1 × 10 6 ; v = 10 3 m / s − 31 9 . 1 × 10

De-Broglie wavelength λ = Example: 35

(c) 2 × 10 −10 m

6 . 626 × 10 −34 h = 7 .28 × 10 −7 m = mv 9 .1 × 10 − 31 × 10 3

The speed of the proton is one hundredth of the speed of light in vacuum. What is the de Broglie wavelength? Assume that one mole of protons has a mass equal to one gram, h = 6 .626 × 10 −27 erg sec

(a) 3.31 × 10 −3 Å m=

Solution : (b) λ=

(b) 1.33 × 10 −3 Å

(c) 3.13 × 10 −2 Å

(d) 1.31 × 10 −2 Å

1 g 6. 023 × 10 23

6 .626 × 10 −27 h = × 6 .023 × 10 23 = 1 . 33 × 10 −11 cm mv 1 × 3 × 10 8 cm sec −1

2.11 Heisenberg’s uncertainty principle. (1) One of the important consequences of the dual nature of an electron is the uncertainty principle, developed by Warner Heisenberg. (2) According to uncertainty principle “It is impossible to specify at any given moment both the position and momentum (velocity) of an electron”. Mathematically it is represented as , ∆x . ∆p ≥

h 4π

Where ∆x = uncertainty is position of the particle, ∆p = uncertainty in the momentum of the particle Now since ∆p = m ∆v So equation becomes, ∆x . m ∆v ≥

h h or ∆x × ∆v ≥ 4π 4πm

The sign ≥ means that the product of ∆x and ∆p (or of ∆x and ∆v ) can be greater than, or equal to but never smaller than

h . If ∆x is made small, ∆p increases and vice versa. 4π

(3) In terms of uncertainty in energy, ∆E and uncertainty in time ∆t, this principle is written as, ∆E . ∆t ≥

h 4π

Note

:q Heisenberg’s uncertainty principle cannot we apply to a stationary electron because its

velocity is 0 and position can be measured accurately. ∆x . ∆p ≥

Example: 36

h 4π

Examples based on uncertainty principle

What is the maximum precision with which the momentum of an electron can be known if the uncertainty in the position of electron is ± 0 .001 Å ? Will there be any problem in describing the h , where a 0 is Bohr’s radius of first orbit, i.e., 0.529Å? momentum if it has a value of 2πa 0

Solution :

∆x . ∆p =

h 4π

Q ∆x = 0 .001 Å = 10 −13 m ∴ ∆p =

6. 625 × 10 −34 = 5 .27 × 10 − 22 4 × 3 . 14 × 10 −13

Example: 37

Calculate the uncertainty in velocity of an electron if the uncertainty in its position is of the order of a 1Å.

Solution :

According to Heisenberg’s uncertainty principle


∆v . ∆x ≈ ∆v ≈

h 4 πm

6. 625 × 10 −34 h = 5. 8 × 10 5 m sec −1 = 22 4πm .∆x − 31 −10 4× × 9 .108 × 10 × 10 7

A dust particle having mass equal to 10 −11 g, diameter of 10 −4 cm and velocity 10 −4 cm sec −1 . The

Example: 38

error in measurement of velocity is 0.1%. Calculate uncertainty in its positions. Comment on the result . Solution :

∆v =

0. 1 × 10 −4 = 1 × 10 −7 cm sec −1 100

Q ∆v . ∆x = ∴ ∆x =

h 4 πm

6 . 625 × 10 −27 = 5 . 27 × 10 −10 cm 4 × 3. 14 × 10 −11 × 1 × 10 −7

The uncertainty in position as compared to particle size. =

5 .27 × 10 −10 ∆x = = 5 . 27 × 10 −6 cm diameter 10 −4

The factor being small and almost being negligible for microscope particles.

2.12 Schrödinger wave equation. (1) Schrodinger wave equation is given by Erwin Schrödinger in 1926 and based on dual nature of electron. (2) In it electron is described as a three dimensional wave in the electric field of a positively charged nucleus. (3) The probability of finding an electron at any point around the nucleus can be determined by the help of Schrodinger wave equation which is, ∂ 2 Ψ ∂ 2 Ψ ∂ 2 Ψ 8π 2m + 2 + 2 + (E − V ) Ψ = 0 h2 ∂x 2 ∂y ∂z

Where x, y and z are the 3 space co-ordinates, m = mass of electron, h = Planck’s constant, E = Total energy, V = potential energy of electron, Ψ = amplitude of wave also called as wave function. ∂ = stands for an infinitesimal change.

(4) The Schrodinger wave equation can also be written as : ∇2 Ψ +

8π 2m (E − V ) Ψ = 0 h2

Where ∇ = laplacian operator. (5) Physical Significance of Ψ and Ψ 2 (i) The wave function Ψ represents the amplitude of the electron wave. The amplitude Ψ is thus a function of space co-ordinates and time i.e. Ψ = Ψ( x , y, z ...... times ) (ii) For a single particle, the square of the wave function (Ψ 2 ) at any point is proportional to the probability of finding the particle at that point. (iii) If Ψ 2 is maximum than probability of finding e − is maximum around nucleus. And the place where probability of finding e − is maximum is called electron density, electron cloud or an atomic orbital. It is different from the Bohr’s orbit.

(iv) The solution of this equation provides a set of number called quantum numbers which describe specific or definite energy state of the electron in atom and information about the shapes and orientations of the most probable distribution of electrons around the nucleus.

2.13 Quantum numbers and Shapes of orbitals.

Quantum numbers (1) Each orbital in an atom is specified by a set of three quantum numbers (n, l, m) and each electron is designated by a set of four quantum numbers (n, l, m and s). (2) Principle quantum number (n) (i) It was proposed by Bohr’s and denoted by ‘n’. (ii) It determines the average distance between electron and nucleus, means it is denoted the size of atom. r=

n2 × 0 . 529 Å Z

(iii) It determine the energy of the electron in an orbit where electron is present. E=−

Z2 × 313 .3 Kcal per mole n2

(iv) The maximum number of an electron in an orbit represented by this quantum number as 2n 2 . No energy shell in atoms of known elements possess more than 32 electrons. (v) It gives the information of orbit K, L, M, N------------. (vi) The value of energy increases with the increasing value of n. (vii) It represents the major energy shell or orbit to which the electron belongs. (viii) Angular momentum can also be calculated using principle quantum number mvr =

nh 2π

(3) Azimuthal quantum number (l) (i) Azimuthal quantum number is also known as angular quantum number. Proposed by Sommerfield and denoted by ‘l’. (ii) It determines the number of sub shells or sublevels to which the electron belongs. (iii) It tells about the shape of subshells. (iv) It also expresses the energies of subshells s < p < d < f (increasing energy). (v) The value of l = (n − 1) always where ‘n’ is the number of principle shell. (vi) Value of l

=

0

1

2

3………..(n-1)

Name of subshell

=

s

p

d

f

Shape of subshell

=

Spherical

Dumbbell

Double dumbbell

Complex

(vii) It represent the orbital angular momentum. Which is equal to

h 2π

l(l + 1)


(viii) The maximum number of electrons in subshell = 2(2l + 1)

s − subshell → 2 electrons d − subshell → 10 electrons p − subshell → 6 electrons

f − subshell → 14 electrons.

(ix) For a given value of ‘n’ the total value of ‘l’ is always equal to the value of ‘n’. (x) The energy of any electron is depend on the value of n & l because total energy = (n + l). The electron enters in that sub orbit whose (n + l) value or the value of energy is less. (4) Magnetic quantum number (m) (i) It was proposed by Zeeman and denoted by ‘m’. (ii) It gives the number of permitted orientation of subshells. (iii) The value of m varies from –l to +l through zero. (iv) It tells about the splitting of spectral lines in the magnetic field i.e. this quantum number proved the Zeeman effect. (v) For a given value of ‘n’ the total value of ’m’ is equal to n 2 . (vi) For a given value of ‘l’ the total value of ‘m’ is equal to (2l + 1). (vii) Degenerate orbitals : Orbitals having the same energy are known as degenerate orbitals. e.g. for p subshell p x p y p z (viii) The number of degenerate orbitals of s subshell =0. (5) Spin quantum numbers (s) (i) It was proposed by Goldshmidt & Ulen Back and denoted by the symbol of ‘s’. (ii) The value of ' s' is + 1/2 and - 1/2, which is signifies the spin or rotation or direction of electron on it’s axis during movement. (iii) The spin may be clockwise or anticlockwise. (iv) It represents the value of spin angular momentum is equal to

h 2π

s(s + 1).

(v) Maximum spin of an atom = 1 / 2 × number of unpaired electron. Magnetic field N

S

+1/2

–1/2

N

S

(vi) This quantum number is not the result of solution of schrodinger equation as solved for H-atom. Distribution of electrons among the quantum levels n

l

m

s

Designation of orbitals

Electrons present

Total no. of electrons

1 (K shell)

0

0

+1/2, –1/2

1s

2

2

2 (L shell)

0

+1 / 2, − 1 / 2 + 1 / 2, − 1 / 2

2s

+ 1 / 2, − 1 / 2 + 1 / 2, − 1 / 2

2p

3s

0

+1 / 2,−1 / 2 + 1 / 2,−1 / 2 + 1 / 2,−1 / 2

–1

+1 / 2,−1 / 2

0 +1

1

0 –1

3 (M shell)

0

0 +1

1

+2 +1 2

0 –1

+ 1 / 2,−1 / 2 + 1 / 2,−1 / 2 + 1 / 2,−1 / 2  + 1 / 2,−1 / 2 + 1 / 2,−1 / 2

3p

3d

–2 0

0 +1

1

0 –1

+2 +1 2

0 –1 –2

4(N shell)

+1 / 2,−1 / 2 + 1 / 2,−1 / 2  + 1 / 2,−1 / 2 + 1 / 2,−1 / 2

+ 1 / 2,−1 / 2 + 1 / 2,−1 / 2 + 1 / 2,−1 / 2  + 1 / 2,−1 / 2 + 1 / 2,−1 / 2

+1 3

+0 –1 –2 –3

+ 1 / 2,−1 / 2 + 1 / 2,−1 / 2 + 1 / 2,−1 / 2  + 1 / 2,−1 / 2 + 1 / 2,−1 / 2  + 1 / 2,−1 / 2 + 1 / 2,−1 / 2

2    6              10 

8

18

4s

4p

4d

+3 +2

2   6 

4f

2    6            10            14 


32

Shape of orbitals (1) Shape of ‘s’ orbital (i) For ‘s’ orbital l=0 & m=0 so ‘s’ orbital have only one unidirectional orientation i.e. the probability of finding the electrons is same in all directions.

Z Y

(ii) The size and energy of ‘s’ orbital with increasing ‘n’ will be 1s < 2 s < 3 s < 4 s. (iii)It does not possess any directional property. s orbital has spherical shape.

X Nucleus

(2) Shape of ‘p’ orbitals (i) For ‘p’ orbital l=1, & m=+1,0,–1 means there are three ‘p’ orbitals, which is symbolised as p x , p y , p z . (ii) Shape of ‘p’ orbital is dumb bell in which the two lobes on opposite side separated by the nodal plane. (iii) p-orbital has directional properties. Z

Y Nodal Plane X

Z

Z

Y

Y

Nodal Plane X

Nodal Plane

Px orbital

Nodal Plane

X Nodal Plane

Py orbital

Pz orbital

(3) Shape of ‘d’ orbital (i) For the ‘d’ orbital l =2 then the values of ‘m’ are –2,–1,0,+1,+2. It shows that the ‘d’ orbitals has five orbitals as d xy , d yz , d zx , d x 2 − y 2 , d z 2 . (ii) Each ‘d’ orbital identical in shape, size and energy. (iii) The shape of d orbital is double dumb bell . (iv) It has directional properties. Z

Z

Y

Z

Y

Y

Y

Y Z

X

X

dZX

dXY

X

X

dYZ

dX2–Y2

X

dZ2

(4) Shape of ‘f’ orbital (i) For the ‘f’ orbital l=3 then the values of ‘m’ are –3, –2, –1,0,+1,+2,+3. It shows that the ‘f’ orbitals have seven orientation as fx (x 2 − y 2 ) , fy ( x 2 − y 2 ), fz ( x 2 − y 2 ), fxyz , fz 3 , fyz 3 and fxz 2 . (ii) The ‘f’ orbital is complicated in shape.

2.14 Electronic configuration principles. The distribution of electrons in different orbitals of atom is known as electronic configuration of the atoms. Filling up of orbitals in the ground state of atom is governed by the following rules: (1) Aufbau principle (i) Auf bau is a German word, meaning ‘building up’. (ii) According to this principle, “In the ground state, the atomic orbitals are filled in order of increasing energies i.e. in the ground state the electrons first occupy the lowest energy orbitals available”. (iii)In fact the energy of an orbital is determined by the quantum number n and l with the help of (n+l) rule or Bohr Bury rule. (iv) According to this rule (a) Lower the value of n + l, lower is the energy of the orbital and such an orbital will be filled up first. (b) When two orbitals have same value of (n+l) the orbital having lower value of “n” has lower energy and such an orbital will be filled up first . Thus, order of filling up of orbitals is as follows:

1s < 2 s < 2 p < 3 s < 3 p < 4 s < 4 p < 5 s < 4 d < 5 p < 6 s < 6 f < 5 d (2) Pauli’s exclusion principle (i) According to this principle, “No two electrons in an atom can have same set of all the four quantum numbers n, l, m and s . (ii) In an atom any two electrons may have three quantum numbers identical but fourth quantum number must be different. (iii)Since this principle excludes certain possible combinations of quantum numbers for any two electrons in an atom, it was given the name exclusion principle. Its results are as follows : (a) The maximum capacity of a main energy shell is equal to 2n 2 electron. (b) The maximum capacity of a subshell is equal to 2(2l+1) electron. (c) Number of sub-shells in a main energy shell is equal to the value of n. (d) Number of orbitals in a main energy shell is equal to n 2 . (e) One orbital cannot have more than two electrons. (iv) According to this principle an orbital can accomodate at the most two electrons with spins opposite to each other. It means that an orbital can have 0, 1, or 2 electron. (v) If an orbital has two electrons they must be of opposite spin. Correct

Incorrect


(3) Hund’s Rule of maximum multiplicity (i) This rule provides the basis for filling up of degenerate orbitals of the same sub-shell. (ii) According to this rule “Electron filling will not take place in orbitals of same energy until all the available orbitals of a given subshell contain one electron each with parallel spin”. (iii)This implies that electron pairing begins with fourth, sixth and eighth electron in p, d and f orbitals of the same subshell respectively. (iv) The reason behind this rule is related to repulsion between identical charged electron present in the same orbital. (v) They can minimise the repulsive force between them serves by occupying different orbitals. (vi) Moreover, according to this principle, the electron entering the different orbitals of subshell have parallel spins. This keep them farther apart and lowers the energy through electron exchange or resonance. (vii) The term maximum multiplicity means that the total spin of unpaired e − is maximum in case of correct filling of orbitals as per this rule. Energy level diagram The representation of relative energy levels of various atomic orbital is made in the terms of energy level diagrams. One electron system : In this system 1s 2 level and all orbital of same principal quantum number have same energy, which is independent of (l). In this system l only determines the shape of the orbital. Multiple electron system : The energy levels of such system not only depend upon the nuclear charge but also upon the another electron present in them. 5

4s

4p

4d

3s

3p

3d

4f

6p 5d 4f 6s 5p

Energy

4

3 2s

2p

3d 4s 3p 3s 2p

2 Energy

4d 5s 4p

2s 1s

1s

Energy level diagram of one electron system

Diagram of multi-electron atoms reveals the following points :

Energy level diagram of multiple electron system

(i) As the distance of the shell increases from the nucleus, the energy level increases. For example energy level of 2 > 1. (ii) The different sub shells have different energy levels which possess definite energy. For a definite shell, the subshell having higher value of l possesses higher energy level. For example in 4th shell. Energy level order 4f > 4d > 4p > 4s l= 3 l=2 l=1 l= 0 (iii)The relative energy of sub shells of different energy shell can be explained in the terms of the (n+l) rule. (a) The sub-shell with lower values of (n + l) possess lower energy. ∴ n+l=5 For 3d n=3 l= 2 For 4s n=4 l=0 n+l=4 (b) If the value of (n + l) for two orbitals is same, one with lower values of ‘n’ possess lower energy level. Extra stability of half filled and completely filled orbitals Half-filled and completely filled sub-shell have extra stability due to the following reasons : (i) Symmetry of orbitals (a) It is a well kown fact that symmetry leads to stability. (b) Thus, if the shift of an electron from one orbital to another orbital differing slightly in energy results in the symmetrical electronic configuration. It becomes more stable. (c) For example p 3 , d 5 , f 7 configurations are more stable than their near ones. (ii) Exchange energy (a) The electron in various subshells can exchange their positions, since electron in the same subshell have equal energies. (b) The energy is released during the exchange process with in the same subshell. (c) In case of half filled and completely filled orbitals, the exchange energy is maximum and is greater than the loss of orbital energy due to the transfer of electron from a higher to a lower sublevel e.g. from 4s to 3d orbitals in case of Cu and Cr . (d) The greater the number of possible exchanges between the electrons of parallel spins present in the degenerate orbitals, the higher would be the amount of energy released and more will be the stability. (e) Let us count the number of exchange that are possible in d 4 and d 5 configuraton among electrons with parallel spins. d4 (1)

(2) 3 exchanges by 1st e–

(3) 2 exchanges by 2nd e–

Only 1 exchange by 3rd e–

To number of possible exchanges = 3 + 2 + 1 =6 d5 (1)

(2) 4 exchanges by 1st e–

(3) 3 exchanges by 2nd

e–

2 exchange by 3rd e–

(4) 1 exchange by 4th e–

To number of possible exchanges = 4 + 3 + 2 +1 = 10


2.15 Electronic configurations of Elements. (1) On the basis of the elecronic configuration priciples the electronic configuration of various elements are given in the following table : Electronic Configuration (E.C.) of Elements Z=1 to 36 Atomic Number

1s

H

1

1

He

2

2

Element

2s

2p

Li

3

2

1

Be

4

2

2

B

5

2

2

1

C

6

2

2

2

N

7

2

2

3

O

8

2

2

4

3s

3p

F

9

2

2

5

Ne

10

2

2

6

Na

11

2

2

6

Mg

12

Al

13

2

1

Si

14

10

2

2

P

15

electrons

2

3

3d

4s

4p

2

S

16

2

4

17

2

5

Ar

18

2

2

6

2

6

K

19

2

2

6

2

6

Ca

20

Sc

21

1

2

Ti

22

2

2

1 2

V

23

3

2

Cr

24

5

1

Mn

25

5

2

Fe

26

6

2

Co

27

18

7

2

Ni

28

electrons

8

2

Cu

29

10

1

Zn

30

10

2

Ga

31

10

2

1

Ge

32

10

2

2

As

33

10

2

3

Se

34

10

2

4

10

2

5

10

2

6

35

Kr

36

4f

1

Cl

Br

4d

2

2

6

2

6

(2) The above method of writing the electronic configurations is quite cumbersome. Hence, usually the electronic configuration of the atom of any element is simply represented by the notation.

nl NUMBER OF PRINCIPAL SHELL

NUMBER OF ELECTRONS PRESENT

x

SYMBOL OF SUBSHELL

e.g. 1s2 means 2 electrons are present in the s- subshell of the 1st main shell.

(3) (i) Elements with atomic number 24(Cr), 42(Mo) and 74(W) have ns 1 (n − 1) d 5 configuration and not ns 2 (n − 1) d 4 due to extra stability of these atoms.

(ii) Elements with atomic number 29(Cu), 47(Ag) and 79(Au) have ns 1 (n − 1) d 10 configuration instead of ns 2 (n − 1) d 9 due to extra stability of these atoms. Cr (24)

3d5

[Ar] 4s1

Cu (29)

3d10

[Ar] 4s1

(4) In the formation of ion, electrons of the outer most orbit are lost. Hence, whenever you are required to write electronic configuration of the ion, first write electronic configuration of its atom and take electron from outermost orbit. If we write electronic configuration of Fe 2 + ( Z = 26 , 24 e − ), it will not be similar to Cr (with 24 e − ) but quite different. Fe [Ar ] 4 s 2 3 d 6  2+  outer most orbit is 4th shell hence, electrons from 4s have been removed to make Fe . 2+ o 6  Fe [ Ar ] 4 s 3 d  (5) Ion/atom will be paramagnetic if there are unpaired electrons. Magnetic moment (spin only) is

µ = n(n + 2) BM (Bohr Magneton). (1 BM = 9 . 27 × 10 −24 J / T ) where n is the number of unpaired electrons.

(6) Ion with unpaired electron in d or f orbital will be coloured. Thus, Cu + with electronic configuration

[Ar ] 3d 10

is colourless and Cu 2 + with electronic configuration [Ar ] 3d 9 (one unpaired electron in 3d) is coloured (blue). (7) Position of the element in periodic table on the basis of electronic configuration can be determined as, (i) If last electron enters into s-subshell, p-subshell, penultimate d-subshell and anti penultimate fsubshell then the element belongs to s, p, d and f – block respectively. (ii) Principle quantum number (n) of outermost shell gives the number of period of the element. (iii)If the last shell contains 1 or 2 electrons (i.e. for s-block elements having the configuration ns 1− 2 ), the group number is 1 in the first case and 2 in the second case. (iv) If the last shell contains 3 or more than 3 electrons (i.e. for p-block elements having the configuration ns 2 np 1 −6 ), the group number is the total number of electrons in the last shell plus 10. (v) If the electrons are present in the (n –1)d orbital in addition to those in the ns orbital (i.e. for dblock elements having the configuration (n –1) d 1−10 ns 1− 2 ), the group number is equal to the total number of electrons present in the (n –1)d orbital and ns orbital.


Composition of Atom (Electron, Proton and Neutron)

Basic Level 1.

The fundamental particles present in the nucleus of an atom are

[CPMT 1983, 84]

(a) Alpha particles and electrons

(b) Neutrons and protons

(c) Neutrons and electrons

(d) Electrons, neutrons and protons

2.

Cathode rays were discovered by (a) William Crookes (b) J. Stoney

(c) Rutherford

3.

Cathode rays are (a) Protons

[JIPMER 1991; NCERT 1976]

(b) Electrons

4.

Cathode rays have

5.

(a) Mass only (b) Charge only Cathode rays are made up of (a) Positively charged particles

6.

(d) None of these

(c) Neutrons

(d) α-particles [CPMT 1982]

(c) No mass and charge

(d) Mass and charge both [AMU 1983]

(b)

Negatively charged particles

(c) Neutral particles (d) None of these Cathode rays are produced when the pressure in the discharge tube is of the order of (a) 76 cm of Hg

(b) 10 −6 cm of Hg

(c) 1 cm of Hg

(d) 10 −2 to 10 −3 mm of Hg

7.

Cathode-ray tube is used in (a) Compound microscope

(b) A ratio receiver

(c) A television set

(d) A Van de Graff generator

8.

Which of the following statement is not correct regarding cathode rays (a) Cathode rays originate from the cathode (b) Charge and mass of the particles constituting cathode rays depends upon the nature of the gas (c) Charge and mass of the particles present does not depend upon the material of the cathode

9.

(d) The ratio charge/mass of the particles is much greater than that of anode rays Which one is not true for the cathode rays (a) They have kinetic energy (b) They cause certain substances to show fluorescence (c) They travel in straight line

10.

The electron is (a) α-rays particle

11.

[Delhi PMT 1982; MADT Bihar 1980]

(b) β-ray particle

(c) Hydrogen ion

(d) Positron

(b) − 1 . 6 × 10 −19 C

(c) Unit negative

(d) All

(b) 9 .1 × 10 −25 g

(c) 9 .1 × 10 −10 g

(d) 9 .1 × 10 −18 g

The charge on an electron is (a) − 4 . 8 × 10 −10 esu

12.

(d) They are electromagnetic waves

Mass of an electron is (a) 9 .1 × 10 −28 g

13.

Which of the following has the same mass as that of an electron (b) Neutron

(c) Positron

(d) Proton

14.

(a) Photon Density of the electron is (a) 2 .17 × 10 −17 g / mL

(b) 4 . 38 × 10 −17 g / mL

(c) 2 .17 × 10 −14 g / mL

(d) None of these

15.

16.

[AFMC 2002]

[CPMT 1986; MLNR 1986] A strong argument for the particle nature of cathode rays is that they (a) Produce fluorescence (b) Travel through vacuum (c) Get deflected by electric and magnetic fields (d) Cast shadow In the discharge tube emission of cathode rays requires (a) Low potential and low pressure (b) Low potential and high pressure

17.

(c) High potential and high pressure pressure The minimum real charge on any particle which can exist is (a) 1 .6 × 10

18.

19.

−19

(b) 1 . 6 × 10

Coulomb

−10

Coulomb

21.

22. 23. 24.

25. 26. 27. 28.

29.

30. 31.

(c) 4 .8 × 10

Coulomb

low

(d) Zero [CPMT 1973 ; BHU 1985]

[BHU 1998]

(d) None of these

(a) 6 .023 × 10 23 g

(c) 9 .1 × 10 −28 kg

(d) 2 gm

[NCERT 1977] A mass spectrograph is an instrument which is capable of differentiating and identifying particles (a) Of different masses (b) Bearing different magnitude of charge (c) Bearing positive and negative charges respectively (d) Of different values of charge and mass ratio Anode rays were discovered by [DPMT 1985] (a) Goldstein (b) J. Stoney (c) Rutherford (d) J.J. Thomson [CPMT 1987] The nature of anode rays depends on (a) Nature of electrode (b) Nature of discharging tube (c) Nature of residual gas (d) All of these [NCERT 1976 ; CPMT 1971] Proton is (a) An ionized hydrogen molecule (b) An α-ray particle (c) A fundamental particle (d) Nucleus of heavy hydrogen [BHU 1985 ; CPMT 1982, 88] Penetration power of proton is (a) More than electron (b) Less than electron (c) More than neutron (d) None of these The ratio of specific charge of a proton and an α-particle is [MP PET 1999] (a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 1 : 1 The e/m for positive rays in comparison to cathode rays is (a) Very low (b) High (c) Same (d) None of these What is false to say about anode rays (a) Their e/m ratio depends upon the nature of residual gas (b) They are deflected by electrical and magnetic field (c) Their e/m ratio is constant (d) These are produced by ionization of molecules of the residual gas Nuclei tend to have more neutrons than protons at high mass numbers because [Roorkee Qualifying 1998] (a) Neutrons are neutral particles (b) Neutrons have more mass than protons (c) More neutrons minimize the coulomb repulsion (d) Neutrons decrease the binding energy [MP PMT 2001] The proton and neutron are collectively called as (a) Deutron (b) Positron (c) Meson (d) Nucleon Which is correct statement about proton [CPMT 1979 ; MP PMT 1985; NCERT 1985 ; MP PMT 1999]

(b) Proton is ionized hydrogen molecule Proton is α-particle

(d)

Who discovered neutron

[IIT 1982 ; BITS 1988 ; CPMT 1977 ; NCERT 1974 ; MP PMT 1992 ; MP PET 2002]

(a) James Chadwick

(b) William Crooks

(c) J. J. Thomson

(d) Rutherford

Which of the following reactions led to the discovery of the neutron (a)

14 6 C

1 +11 p → 14 7 N + 0n

(b)

11 5 B

34.

Heaviest particle is

35.

The density of neutrons is of the order

(a) Meson (a) 10 −3 kg / cc 36.

−10

(c) 1.8

(b) 1 . 008 g and 0 .55 mg

and

[Rajasthan PMT 2000]

(c) Proton is ionized hydrogen atom

33.

potential

(a) Infinite (b) 1 . 8 × 10 The mass of a mole of proton and electron is

(a) Proton is nucleus of deuterium 32.

High

Which of the following statement is incorrect (a) The charge on an electron and on a proton are equal and opposite (b) Neutrons has no charge (c) Electrons and protons have the same weight (d) The mass of a proton and a neutron are nearly identical Ratio of masses of proton and electron is +3

20.

(d)

The mass of neutron is nearly

1 + 12 D → 12 6 C + 0n

(c)

9 4 Be

1 + 42 He → 12 6 C +0 n

(d)

8 4 Be

1 + 42 He → 11 6 C + 0n

[Delhi PMT 1983 ; MP PET 1999]

(b) Neutron (b) 10 −6 kg / cc

(c) Proton

(d) Electron [NCERT 1980]

(c) 10 −9 kg / cc

(d) 10 −12 kg / cc [MLNR 1988 ; UPSEAT 1999, 2000, 02]


(a) 10 −23 kg 37.

38.

(b) 10 −24 kg

(c) 10 −26 kg

(d) 10 −27 kg

Neutron is a fundamental particle carrying

[CPMT 1990]

(a) A charge of +1 unit and a mass of 1 unit

(b) No charge and a mass of 1 unit

(c) No charge and no mass

(d) A charge of –1 and a mass of 1 unit

The discovery of neutron becomes very late because

[CPMT 1987 ; AIIMS 1998]

(a) Neutrons are present in nucleus particles

(b)

(c) Neutrons are chargeless

(d) Neutrons do not move

39.

Which one of the following pairs is not correctly matched

40.

An elementary fundamental particle is

(a) Rutherford-Proton

Neutrons are highly unstable

(b) J. J. Thomson-Electron

[MP PET 2002]

(c) J. H. Chadwick-Neutron

(d) Bohr-Isotope [CPMT 1973]

(a) An element present in a compound

(b) An atom present in an element

(c) A sub-atomic particle

(d) A fragment of an atom

Advance Level

41.

The charge of an electron is − 1 . 6 × 10 −19 C . The value of free charge on Li + ion will be [AFMC 2002 ; Karnataka CET (Engg.) 2002]

(a) 3 . 6 × 10 42.

−19

C

(c) 1 . 6 × 10

C

−19

C

(d) 2 .6 × 10 −19 C

The charge on an electron is 4 . 8 × 10 −10 esu . What is the value of charge in Li + ion (a) 4 . 8 × 10 −10 esu

43.

(b) 1 × 10

−19

(b) 9 .6 × 10 −10 esu

(c) 1 . 44 × 10 −9 esu

[CPMT 1997]

(d) 2 .4 × 10 −10 esu

The specific charge for positive rays is much less than the specific charge for cathode rays. This is because

[CPMT 1990]

(a) Positive rays are positively charged (b) Charge on positive rays is less (c) Positive rays comprise ionised atoms whose mass is much higher (d) Experimental method for determination is wrong 44.

The increasing order (lowest first) for the values of e/m (charge/mass) for (a) e, p, n, α

45.

The specific charge of proton is 9 .6 × 10 C kg 7

(a) 38 .4 × 10 7 C kg −1 46.

(b) n, p, e, α

(c) n, p, α, e −1

[IIT 1984]

(d) n, α, p, e

then for an α-particle it will be

(b) 19 .2 × 10 7 C kg −1

(c) 2 .4 × 10 7 C kg −1

[MH CET 1999]

(d) 4 . 8 × 10 7 C kg −1

The number of atoms in 0.004 g of magnesium are (a) 4 × 10 20

(b) 8 × 10 20

[AFMC 2000]

(c) 10 20

(d) 6 .02 × 10 20

Atomic number, Mass number and Atomic species

Basic Level 47.

Nitrogen atom has an atomic number of 7 and oxygen has an atomic number 8. The total number of electrons in a nitrate ion will be [Pb. PMT 2000]

(a) 8 48. 49. 50.

(b) 16

(c) 32

The number of electrons in Cl − ion is (a) 19 (b) 20 (c) 18 The number of neutron in tritium is (a) 1 (b) 2 (c) 3 The total number of protons in one molecule of nitrogen dioxide

(d) 64 [MP PMT 2003]

(d) 35 [CPMT 2003]

(d) 0

51. 52.

(a) 23 (b) 46 Number of neutrons in heavy hydrogen atom is (a) 0 (b) 1 The nucleus of helium contains (a) Four protons (c) Two neutrons and two protons

(c) 69

(d) 92

(c) 2

(d) 3

[MP PMT 1986] [CPMT 1972; Delhi PMT 1982]

(b) Four neutrons (d) Four protons and two electrons

53.

Sodium atom differs from sodium ion in the number of

54.

(a) Electron (b) Protons (c) Neutrons (d) Does not differ [CPMT 1980] An atom has 26 electrons and its atomic weight is 56. The number of neutrons in the nucleus of the atom will be (a) 26 (b) 30 The atomic number of an element represents

(c) 36

55.

(a) Number of neutrons in the nucleus

(b) Number of protons in the nucleus

(c) Atomic weight of element

(d) Valency of element

56.

(b) Atomic radii

(b) 32 (b) 1 (b) Neutron

(d) Atomic number

[CPMT 1976, 81, 86]

(c) 34

(c) Electrons

(b) 44

(c) 66

The nitrogen atom has 7 protons and 7 electrons, the nitride ion ( N

3−

(b) 4 protons and 7 electrons

(c) 4 protons and 10 electrons

(d) 10 protons and 7 electrons

(a) 34 (a) Number of e

(b) 40 −1

=W −N

(c) Number of 1 H 1 = W − N

69.

70.

71.

[NCERT 1977]

(a) 7 protons and 10 electrons

If W is atomic weight and N is the atomic number of an element, then

68.

(d) 88

) will have

64.

67.

(d) Protons and electrons [IIT 1979; MP PMT 1994; Rajasthan PMT 1999]

The total number of neutrons in dipositive zinc ions with mass number 70 is

66.

[CPMT 1983]

(d) 5 [NCERT 1972; MP PMT 1995]

63.

65.

(d) 42

(c) 0

The number of electrons in one molecule of CO 2 are (a) 22

62.

(c) Equivalent weight

Chlorine atom differs from chloride ion in the number of (a) Proton

61.

(d) Proton and electron

The total number of unpaired electrons in d-orbitals of atoms of element of atomic number 29 is (a) 10

60.

(c) Neutron and proton

The electronic configuration of a dipositive metal M 2 + is 2, 8, 14 and its atomic weight is 56 a.m.u. The number of neutrons in its [MNR 1984, 89; Kerala PMT 1999] nuclei would be (a) 30

59.

(b) Neutron and electron

[Delhi PMT 1984. 91; AFMC 1990]

Which of the following is always a whole number (a) Atomic weight

58.

(d) 56 [CPMT 1983; CBSE 1990; NCERT 1973; AMU 1984]

The mass of an atom is constituted mainly by (a) Neutron and neutrino

57.

[CPMT 1976]

[IIT 1979; Bihar MEE 1997]

(c) 36

(d) 38 [CPMT 1971, 80, 89]

(b) Number of 0 n = W − N 1

(d) Number of 0 n1 = N

The number of electrons in the atom which has 20 protons in the nucleus is (a) 20 (b) 10 (c) 30 Six protons are found in the nucleus of (a) Boron (b) Lithium (c) Carbon A sodium cation has different number of electrons from

[CPMT 1981, 93; CBSE 1989]

(d) 40 [CPMT 1977, 80, 81; NCERT 1975, 78]

(d) Helium

(a) O 2 − (b) F − (c) Li + (d) Al + + + [CPMT 1986] An atom which has lost one electron would be (a) Negatively charged (b) Positively charged (c) Electrically neutral (d) Carry double positive charge The nucleus of the element having atomic number 25 and atomic weight 55 will contain [CPMT 1986; MP PMT 1987] (a) 25 protons and 30 neutrons (b) 25 neutrons and 30 protons (c) 55 protons (d) 55 neutrons Positive ions are formed from the neutral atom by the [CPMT 1976] (a) Increase of nuclear charge (b) Gain of protons (c) Loss of electrons (d) Loss of protons The nucleus of the atom consists of [CPMT 1973; 74, 78, 83, 84; MADT Bihar 1980; Delhi PMT 1982, 85, MP PMT 1999]


72. 73.

(a) Proton and neutron (c) Neutron and electron The number of electrons in an atom of an element is equal to its (a) Atomic weight (b) Atomic number Neutrons are found in atoms of all elements except in (a) Chlorine (b) Oxygen

(b) Proton and electron (d) Proton, neutron and electron [BHU 1979]

(c) Equivalent weight

(d) Electron affinity

(c) Argon

(d) Hydrogen

[MP PMT 1997]

74.

A transition metal X has a configuration [ Ar ] 3 d 4 in its + 3 oxidation state. Its atomic number is

75.

(a) 25 (b) 26 Number of electrons in −CONH 2 is

(c) 22

[EAMCET 1990]

(d) 19 [AMU 1988]

77.

(a) 22 (b) 23 (c) 20 (d) 28 Ca has atomic number 20 and atomic weight 40. Which of the following statements is not correct about Ca atom [MP PET 1993] (a) The number of electrons is same as the number of neutrons (b) The number of nucleons is double of the number of electrons (c) The number of protons is half of the number of neutrons (d) None of these Which of the following atom has more electrons than neutrons

78.

(a) C (b) F − The present atomic weight scale is based on

76.

(a) C 79.

12

(b) O

The nucleus of the element

21 E

45

16

81.

82. 83.

(d) Al 3 + [EAMCET 1988; MP PMT 2002]

(c)

H

1

(d) C13

contains

(a) 45 protons and 21 neutrons (c) 21 protons and 45 neutrons 80.

(c) O 2 −

(b) 21 protons and 24 neutrons (d) 24 protons and 21 neutrons

The number of electrons in the nucleus of C12 is [AFMC 1995] (a) 6 (b) 12 (c) 0 (d) 3 The atomic number of an element is always equal to [MP PMT 1994] (a) Atomic weight divided by 2 (b) Number of neutrons in the nucleus (c) Weight of the nucleus (d) Electrical charge of the nucleus The ratio between the neutrons in C and Si with respect to atomic masses 12 and 28 is [EAMCET 1990] (a) 2 : 3 (b) 3 : 2 (c) 3 : 7 (d) 7 : 3 If the atomic weight of an element is 23 times that of the lightest element and it has 11 protons, then it contains [EAMCET 1986; AFMC 1989]

84. 85.

(a) 11 protons, 23 neutrons, 11 electrons (b) 11 protons, 11 neutrons, 11 electrons (c) 11 protons, 12 neutrons, 11 electrons (d) 11 protons, 11 neutrons, 23 electrons The nucleus of tritium contains [MP PMT 2002] (a) 1 proton + 1 neutron (b) 1 proton + 3 neutron (c) 1 proton + 0 neutron (d) 1 proton + 2 neutron The number of electrons and neutrons of an element is 18 and 20 respectively. Its mass number is [CPMT 1997; Pb. PMT 1999; MP PMT 1999]

(a) 17 86.

(b) 37

The number of electrons in

40 [19

(a) 19 87.

88.

89.

In the nucleus of

20 Ca

(d) 38

is

[CPMT 1997; AFMC 1999]

(c) 18

(d) 40

there are

[CPMT 1990; EAMCET 1991]

(a) 40 protons and 20 electrons (b) 20 protons and 40 electrons (c) 20 protons and 20 neutrons (d) 20 protons and 40 neutrons The atomic weight of an element is 39. The number of neutrons in its nucleus is one more than the number of protons. The [MP PMT 1997] number of protons, neutrons and electrons respectively in its atom would be (a) 19, 20, 19 (b) 19, 19, 20 (c) 20, 19, 19 (d) 20, 19, 20 CO has same electrons as or the ion that is iso-electronic with CO is [CPMT 1984; IIT 1982; EAMCET 1990; CBSE 1997] (b) CN −

(c) O 2+

(d) O 2−

Na + ion is iso-electronic with

(a) Li 91.

K]

(b) 20 40

(a) N 2+ 90.

−1

(c) 2

+

[CPMT 1990]

(b) Mg

2+

(c) Ca

Which of the following oxides of nitrogen is iso-electronic with CO 2

2+

(d) Ba

2+

[CBSE 1990]

(a) NO 2 92.

3−

(b) F

95.

(c) Tl CH 3+ I

H3O

+

+

NH 3

II

(b) I and IV

(d) Na

(a) Li Iso-electronic species are −

(b) He

+

CH 3− IV

III

(c) I and III

[IIT 1993]

(d) II, III and IV

The hydride ions (H ) are iso-electronic with

(a) K , Cl

[AFMC 1995; Bihar MEE 1997]

+

(c) He

(d) Be [EAMCET 1989]

+

(b) Na , Cl

−

(c)

+

(d) Mg , Ar

Na, Ar

Which one of the following grouping represents a collection of iso-electronic species +

(a) Na , Ca 97.

[CBSE 1994]

−

−

+

96.

(d) N 2 O2

2−

Pick out the iso-electronic structures from the following, (a) I and II

94.

(c) NO

Which one of the following is not iso-electronic with O (a) N

93.

(b) N 2 O

2+

, Mg

2+

(b) N

3−

−

, F , Na

+

(c)

Be , Al

3+

, Cl

−

[AIEEE 2003]

(d) Ca

2+

Which of the following are iso-electronic and isostructual NO 3− , CO 32 − , ClO3− , SO 3 (a)

NO 3− , CO 32 −

(b)

SO 3 , NO 3−

(c)

+

, Cs , Br [IIT Screening 2003]

ClO 3− , CO 32 −

(d)

CO 32 − .SO 3

98.

Which of the following atoms and ions are iso-electronic i.e. have the same number of electrons with the neon atom [NCERT 1978]

99.

(a) F − (b) Oxygen atom Which of the following is iso-electronic with carbon atom +

[MP PMT 1994; UPSEAT 2000]

3+

(a) Na (b) Al 100. Which of the following is not iso-electronic with Ne (a) Na

+

(b) Mg

(d) N −

(c) Mg (c) O

2−

(d) N +

2−

−

[MP PET 2002]

2+

(c) O

(d) Cl

101. Which of the following is iso-electronic with Ca 2 + (a) Kr 102. Iso-electronic species is −

(b) K +

(c)

−

(b) F , O (a) F , O 103. Which pair of ions is iso-electronic −

(a) F and Cl 104. Tritium is the isotope of (a) Hydrogen 105. An isostere is (a)

NO 2−

and O3

(d) Ca [Rajasthan PMT 2002]

−2

−

Mg 2 +

(c)

−

F ,O

+

−

(d) F , O +2 [DCE 1999]

−

(b) F and O

−

(c)

Na

+

and K

+

(d) Na

+

and Mg

[CPMT 2003]

(b) Oxygen

(c) Carbon

(d) Sulphur [UPSEAT 1999]

(b)

NO 2−

and

PO43 −

(c) CO 2 , N 2 O,

NO 3−

(d)

ClO 4−

and OCN −

106. Which of the following pair has same electronic structure

[CPMT 1992]

+

(c) Ag, Sn

(a) Ca, Ar (b) Mg, Na 107. Which of the following are iso-electronic with one another (a) Na 108.

+

and Ne

(d) Ar, Cl

−

[NCERT 1983; EAMCET 1989]

+

(b) K and O

(c) Ne and O

(d) Na + and K +

Be 2 + is iso-electronic with 2+

[EAMCET 1998] +

(a) Mg (b) Na 109. The nitride ion in lithium nitride is composed of (a) 7 protons + 10 electrons (b) 10 protons + 10 electrons 110. Number of protons, neutrons and electrons in the element (a) 89, 231, 89 111. CO 2 is isostructural with (a) SnCl 2

+2

(b) 89, 89, 242

89 γ

(c)

Li

+

(d) H

+

[Karnataka CET 2000]

(c) 7 protons + 7 protons 231

(d) 10 protons + 7 electrons

is

[AFMC 1997]

(c) 89, 142, 89

(d) 89, 71, 89 [IIT 1986; MP PMT 1986, 94, 95]

(b) SO 2

(c)

HgCl 2

(d) All the above

Advance Level 112.

In an X-ray experiment, different metals are used as the target. In each case, the frequency (ν) of the radiation produced is measured. If Z= atomic number, which of the following plots will be a straight line


(a) ν against Z 113.

114. 115.

116.

117.

(b)

1 against Z ν

(c)

ν against Z

(d) ν against

Z

In Moseley's equation [ ν = a( Z − b ) ], which was derived from the observations made during the bombardment of metal targets with X-rays, (a) a is independent but b depends on the metal (b) Both a and b depend on the metal (c) Both a and b are independent of the metal and are constant (d) b is independent but a depends on the metal [Rajasthan PET 2000] If molecular mass and atomic mass of sulphur are 256 and 32 respectively, its atomicity is (a) 2 (b) 8 (c) 4 (d) 16 Assertion (A) : The atoms of different elements having same mass number but different atomic number are known as isobars [AIIMS 2000] Reason (R) : The sum of protons and neutrons, in the isobars is always different (a) Both A and R are true and R is a correct explanation of A (b) Both A and R are true but R is not a correct explanation of A (c) A is true but the R is false (d) A is false but R is true The mass number of an anion, X 3− , is 14. If there are ten electrons in the anion, the number of neutrons in the nucleus of atom, X 2 of the element will be [MP PMT 1999] (a) 10 (b) 14 (c) 7 (d) 5 Atoms consists of protons, neutrons and electrons. If the mass of neutrons and electrons were made half and two times respectively to their actual masses, then the atomic mass of 6 C 12 (a) Will remain approximately the same (c) Will remain approximately half

[NCERT 1982]

(b) Will become approximately two times (d) Will be reduced by 25%

[CPMT 1982] 118. A neutral atom (Atomic no. > 1) consists of (a) Only protons (b) Neutrons + protons (c) Neutrons + electrons (d) Neutrons +proton + electron 119. Compared with an atom of atomic weight 12 and atomic number 6, the atom of atomic weight 13 and atomic number 6 [NCERT 1971]

(a) Contains more neutrons (c) Contains more protons 120. Assertion (A) : Nuclide

30

(b) Contains more electrons (d) Is a different element Al13 is less stable than

40

Ca 20

Reason (R): Nuclides having odd number of protons and neutrons are generally unstable (a) Both A and R are correct and R is the correct explanation of A (b) Both A and R are correct but R is not the correct explanation of A (c) A is correct but R is incorrect (d) A is incorrect but R is correct 121.

[IIT 1998]

Which of the following are iso-electronic species I − CH 3+ , II − NH 2− , III − NH 4+ , IV − NH 3

(a) I, II, III (b) II, III, IV (c) I, II, IV 122. The charge on the atom containing 17 protons, 18 neutrons and 18 electrons is (a) + 1 (b) – 2 (c) –1

[CPMT 1999]

(d) I and II [AIIMS 1996]

(d) Zero

Atomic Models and Planck’s Quantum Theory

Basic Level 123. Rutherford’s α-particle scattering experiment proved that atom has

[MP PMT 2001]

(a) Electrons (b) Neutron (c) Nucleus (d) Orbitals 124. Rutherford’s alpha particle scattering experiment eventually led to the conclusion that [IIT 1986; Rajasthan PMT 2002] (a) Mass and energy are related (b) Electrons occupy space around the nucleus (c) Neutrons are buried deep in the nucleus (d) The point of impact with matter can be precisely determined 125. The element used by Rutherford in his famous scattering experiment was [Karnataka CET 1998] (a) Gold (b) Tin (c) Silver (d) Lead 126. The α-particle scattering experiment of Rutherford concluded that

[Orissa JEE 1997]

(a) The nucleus is made up of protons and neutrons (b) The number of electrons is exactly equal to number of protons in atom (c) The positive charge of the atom is concentrated in a very small space (d) Electrons occupy discrete energy levels 127. Experimental evidence for the existence of the atomic nucleus comes from

[CBSE 1989]

(a) Millikan’s oil drop experiment

(b) Atomic emission spectroscopy

(c) The magnetic bending of cathode rays

(d) Alpha scattering by a thin metal foil

128. Which of the following is not true in Rutherford’s nuclear model of atom

[Orissa JEE 2002]

(a) Protons and neutrons are present inside nucleus (b) Volume of nucleus is very small as compared to volume of atom (c) The number of protons and neutrons are always equal (d) The number of electrons and protons are always equal 129. The radius of the nucleus is related to the mass number A by (a) R = R0 A

1/2

(b) R = R0 . A

[EAMCET 1998]

(c)

R = R0 .A

2

(d) R = R0 .A

1/3

130. The volume of the nucleus is

131.

(a) 10 −4 times smaller than the volume of an atom

(b) 10 −8 times smaller than the volume of an atom

(c) 10 −12 times smaller than the volume of an atom

(d) Two-third the volume of the nucleus

Rutherford’s experiment on scattering of particles showed for the first time that the atom has


[IIT 1981, NCERT 1981; CMC Vellore 1991; CPMT 1984; Kurukshetra CEE 1998]

(a) Electrons (b) Protons (c) Nucleus 132. The size of nucleus is measured in (a) amu (b) Angstrom (c) Fermi 133. The average distance of an electron in an atom from its nucleus is of the order of 6

(a) 10 m

(b) 10

−6

m

134. Nucleus of an atom is (a) Neutral (b) Negatively charged 135. Rutherford’s scatting experiment is related to the size of the (a) Nucleus (b) Atom 136. The positive charge of an atom is (a) Spread all over the atom (c) Concentrated at the nucleus 137. Atoms have diameters of the order of −8

−10

(a) 10 cm (b) 10 cm 138. Remaining part of atom except outer orbit is called (a) Kernel (b) Core 139. The radius of an atom is of the order of

(c) 10

−10

(a) 10 kg / cc

(b) 10

−8

kg / cc

142. Existence of positively charged nucleus was established by (a) Positive ray analysis (b) α-ray scattering experiments 143. The size of nucleus is of the order of (a) 10 −12 m

(b) 10 −8 m

[EAMCET 1988; CPMT 1994]

(d) cm [MP PET 1996]

(d) 10

m

−15

m [MNR 1977]

(c) Positively charged

(d) None of them

[IIT 1983; MADT Bihar 1995; BHU 1995]

(c) Electron

(d) Neutron [AFMC 2002]

(b) Distributed around the nucleus (d) All of these [NCERT 1971; CPMT 1977]

(c) 10

−13

cm

(d) 10 −15 cm [CPMT 1982]

(c) Empty space

(d) None [AMU 1982; IIT 1985; MP PMT 1995]

(b) 10 −13 cm (c) 10 −15 cm (a) 10 −10 cm 140. Discovery of the nucleus of an atom was due to the experiment carried out by (a) Bohr (b) Mosley (c) Rutherford 141. The order of density in nucleus is 8

(d) Neutrons

(c) 10

−9

kg / cc

(d) 10 −8 cm [CPMT 1983; MP PET 1983]

(d) Thomson [NCERT 1981; CPMT 1981, 2003]

(d) 10 12 kg / cc [CBSE 1991]

(c)

X-ray analysis

(d)

[CPMT 1982; MP PMT 1991]

(c) 10 −15 m

(d) 10 −10 m

[IIT 1985] 144. Bohr’s model can explain (a) The spectrum of hydrogen atom only (b) Spectrum of atom or ion containing one electron only (c) The spectrum of hydrogen molecule (d) The solar spectrum 145. Which one of the following is considered as the main postulate of Bohr’s model of atom [AMU 2000] (a) Protons are present in the nucleus (b) Electrons are revolving around the nucleus (c) Centrifugal force produced due to the revolving electrons balances the force of attraction between the electron and the protons h (d) Angular momentum of electron is an integral multiple of 2π 146. The electronic energy levels of the hydrogen atom in the Bohr’s theory are called [AMU 2000] (a) Rydberg levels (b) Orbits (c) Ground states (d) Orbitals 147. Which of the following statements does not form part of Bohr’s model of the hydrogen atom [CBSE 1989] (a) Energy of the electrons in the orbit is quantized (b) The electron in the orbit nearest the nucleus has the lowest energy (c) Electrons revolve in different orbits around the nucleus (d) The position and velocity of the electrons in the orbit cannot be determined simultaneously 148. Bohr model of atom is contradicted by [MP PMT 2002] (a) Pauli’s exclusion principle (b) Planck quantum theory (c) Heisenberg uncertainty principle (d) All of these 149. Bohr’s radius can have [Delhi PMT 1996] (a) Discrete values (b) + ve values (c) – ve values (d) Fractional values 150. Who modified Bohr’s theory by introducing elliptical orbits for electron path [CBSE 1999; AFMC 2003] (a) Hund (b) Thomson (c) Rutherford (d) Sommerfeld

151.

Bohr model of an atom could not account for (a) Emission spectrum (b) Absorption spectrum (c) Line spectrum of hydrogen (d) Fine spectrum 152. Radius of the first Bohr’s orbit of hydrogen atom is [Rajasthan PET 2000] (a) 1.06 Å (b) 0.22 Å (c) 0.28 Å (d) 0.53 Å [MP PET 1994] 153. When an electron revolves in a stationary orbit then (a) It absorbs energy (b) It gains kinetic energy (c) It emits radiation (d) Its energy remains constant [MP PET 1997] 154. If the radius of first Bohr orbit be a0 , then the radius of third orbit would be (a) 3 × a0

(b) 6 × a0

(c) 9 × a0

(d)

1 a0 9

155. The ratio between kinetic energy and the total energy of the electrons of hydrogen atom according to Bohr’s model is [Pb. PMT 2002] (a) 2 : 1 (b) 1 : 1 (c) 1 : – 1 (d) 1 : 2 156. The postulate of Bohr theory that electrons jump from one orbit to the other, rather than flow is according to (a) The quantisation concept (b) The wave nature of electron (c) The probability expression for electron (d) Heisenberg uncertainty principle 157. Ratio of radii of second and first Bohr orbits of H atom [BHU 2003] (a) 2 (b) 4 (c) 3 (d) 5 158. The energy of an electron revolving in nth Bohr’s orbit of an atom is given by the expression [MP PMT 1999] 2π 2 m 4 e 2 z 2

(a) En = −

n 2h 2

(b) En = −

2π 2 me 2 z 2 n 2h 2

(c)

2π 2 me 4 z 2

En = −

n 2h 2

2πm 2 e 2 z 4

(d) En = −

n 2h2

159. The expression for Bohr’s radius of an atom is (a) r =

n 2h 2

(b) r =

4 π me z 2

4 2

[MP PMT 1999]

n 2h 2 2

2

4π me z

(c) r =

n 2h 2

(d) r =

4 π me z 2

2 2

160. Visible range of hydrogen spectrum will contain the following series 161.

(a) Pfund (b) Lyman Wavelength of spectral line emitted is inversely proportional to

n 2h 2 4π m 2e 2 z 2 2

[Rajasthan PET 2000]

(c) Balmer

(d) Brackett [CPMT 2001]

(a) Radius (b) Energy (c) Velocity 162. In hydrogen spectrum the different lines of Lyman series are present in

(d) Quantum number

(a) UV field (b) IR field 163. In an element going away from nucleus, the energy of particle (a) Decreases (b) Not changing

(d) Far IR field

164. When an electron jumps from lower to higher orbit, its energy (a) Increases (b) Decreases

[UPSEAT 1999]

(c) Visible field


(c) Increases

(d) None of these [MADT Bihar 1982]

(c) Remains the same

(d) None of these

165. The frequency corresponding to transition of electron n = 2 to n =1 in hydrogen atom is (a) 15 .66 × 10 10 Hz

(b) 24 . 66 × 10 14 Hz

(c) 30 .57 × 10 14 Hz

[MP PET 2003]

(d) 40 .57 × 10 24 Hz

166. When an electron drops from a higher energy level to a low energy level, then (a) Energy is emitted (b) Energy is absorbed 167. When an electron jumps from L to K shell (a) Energy is absorbed

[AMU 1985]

(c) Atomic number increases (d) Atomic number decreases [CPMT 1983]

(b) Energy is released

(c) Energy is sometimes absorbed and sometimes released (d) Energy is neither absorbed nor released 168. The third line in Balmer series corresponds to an electronic transition between which Bohr’s orbits in hydrogen [MP PMT 2001] (a) 5 → 3

(b) 5 → 2

(c) 4 → 3

(d) 4 → 2

169. Energy of the electron in hydrogen atom is given by (a) En = −

131 .38 n

2

kJ mol −1

131 .33 kJ mol −1 (b) E n = − n

[AMU (Engg.) 2002]

(c)

En = −

170. The spectrum of He is expected to be similar to +

171.

(a) H (b) Li The series limit for Balmer series of H-spectra is

1313 . 3 n

2

kJ mol −1

(d) En = −

313 .13 n2

kJ mol −1

[AIIMS 1980, 91; Delhi PMT 1983; MP PMT 2002]

(c)

Na

(d) He +


[AMU (Engg.) 1999]

(a) 3800Å (b) 4200Å (c) 3664Å (d) 4000Å 172. Which of the following electron transition in a hydrogen atom will require the largest amount of energy [UPSEAT 1999, 2000, 01] (a) From n = 1 to n = 2 (b) From n = 2 to n = 3 (c) From n = ∞ to n = 1 (d) n = 3 to n = 5 173. Which of the following transitions are allowed in the normal electronic emission spectrum of an atom (a) 2s → 1s (b) 2p → 1s (c) 3d → 2p (d) 5d → 2s 174. The formation of energy bonds in solids are in accordance with (a) Heisenberg’s uncertainty principle (c) Ohm’s law 175. Zeeman effect refers to the

[DCE 2001]

(b) Bohr’s theory (d) Rutherford’s atomic model [AFMC 1995]

(a) Splitting up of the lines in an emission spectrum in a magnetic field (b) Splitting up of the lines in an emission spectrum in the presence of an external electrostatic field (c) Emission of electrons from metals when light falls upon them (d) Random scattering of light by colloidal particles 176. The first use of quantum theory to explain the structure of atom was made by [IIT 1997; CPMT 2001] (a) Heisenberg (b) Bohr (c) Planck (d) Einstein [AIIMS 1991] 177. Which one of the following is not the characteristic of Planck’s quantum theory of radiation (a) The energy is not absorbed or emitted in whole number multiple of quantum (b) Radiation is associated with energy (c) Radiation energy is not emitted or absorbed continuously but in the form of small packets called quanta (d) This magnitude of energy associated with a quantum is proportional to the frequency 178. The Planck constant has the dimension of (a) Length (b) Energy (c) Momentum (d) Angular momentum [IIT 1984; CPMT 1997] 179. Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon (a) 3 s (b) 2 p (c) 2 s (d) 1 s [MP PET 2002] 180. The frequency of yellow light having wavelength 600 nm is (a) 5 .0 × 10 14 Hz 181.

(b) 2 .5 × 10 7 Hz

(c) 5 .0 × 10 7 Hz

(d) 2 .5 × 10 14 Hz

E (c) h = V

h (d) E = V

The energy of a photon is calculated by (a) E = h ν

(b) h = Eν

[Pb. PMT 2000]

Advance Level

182. Which of the following is true for Thomson's model of the atom (a) The radius of an electron can be calculated using Thomson's model. (b) In an undisturbed atom, the electrons will be at their equilibrium positions, where the attraction between the cloud of positive charge and the electrons balances their mutual repulsion (c) When the electrons are disturbed by collision, they will vibrate around their equilibrium positions and emit electromagnetic radiation whose frequency is of the order of magnitude of the frequency of electromagnetic radiation of a vibrating electron. (d) It can explain the existence of protons. 183. When a gold sheet is bombarded by a beam of α–particles, only a few of them get deflected whereas most go straight, undeflected. This is because (a) The force of attraction exerted on the α–particles by the oppositely charged electrons is not sufficient. (b) A nucleus has a much smaller volume than that of an atom. (c) The force of repulsion acting on the fast moving α–particles is very small. (d) The neutrons in the nucleus do not have any effect on the α–particles. 184. From the α–particle scattering experiment, Rutherford concluded that (a) α–particles can come within a distance of the order of 10–14m of the nucleus (b) The radius of the nucleus is less than 10–14m (c) Scattering follows Coulomb's law

(d) The positively charged parts of the atom move with extremely high velocities. 185. Rutherford's scattering formula fails for very small scattering angles because (a) The full nuclear charge of the target atom is partially screened by its electron (b) The impact parameter between the α–particle source and the nucleus of the target is very large compared to the size of the nucleus (c) The kinetic energy of the α–particles is large (d) The gold foil is very thin 186. The radius of

27 13

Al will be

(a) 1 . 2 × 10 −15 m

(b) 27 × 10 −15 m

(c) 10 .8 × 10 −15 m

(d) 3 . 6 × 10 −15 m

187. The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by 1 . 25 × 10 −13 × A1 / 3 cm . Radius of atom is one Å. If the mass number is 64, then the fraction of the atomic volume that is occupied by the nucleus is [NCERT 1983]

(a) 1 . 0 × 10 −3 (b) 5 .0 × 10 −5 (c) 2 .5 × 10 −2 (d) 1 . 25 × 10 −13 188. In a Bohr’s model of atom when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed [CBSE 1996] (a) 2 .15 × 10 −11 ergs

(b) 0 .1911 × 10 −10 ergs

(c) 2 .389 × 10 −12 ergs

(d) 0 .239 × 10 −10 ergs

[MP PET 1994] 189. The radius of first Bohr’s orbit for hydrogen is 0.53 Å. The radius of third Bohr’s orbit would be (a) 0.79 Å (b) 1.59 Å (c) 3.18 Å (d) 4.77 Å 190. The energy of an electron in the first Bohr orbit of H atom is ––13.6 eV. The possible energy value (s) of the excited state (s) for [IIT 1998] electrons in Bohr orbits to hydrogen is (are) (a) – 3.4 eV (b) – 4.2 eV (c) – 6.8 eV (d) + 6.8 eV [MP PMT 2000] 191. Energy of electron of hydrogen atom in second Bohr orbit is

(a) − 5 .44 × 10 −19 J

(b) − 5 .44 × 10 −19 kJ

(c) − 5 .44 × 10 −19 cal

(d) − 5 .44 × 10 −19 eV

192. The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) orbit is [CBSE 1998; BHU 1999]

(a) 0.13 Å (b) 1.06 Å (c) 4.77 Å (d) 2.12 Å [MP PET 1999] 193. The energy of an electron in nth orbit of hydrogen atom is 13 . 6 13 . 6 13 . 6 13 . 6 eV (a) (b) (c) (d) eV eV eV n n4 n3 n2 [UPSEAT 2003] 194. As electron moves away from the nucleus, its potential energy (a) Increases (b) Decreases (c) Remains constant (d) None of these 195. In which one of the following pairs of experimental observations and phenomenon does the experimental observation correctly [AIIMS 1983] account for phenomenon Experimental observation Phenomenon (a) X- ray spectra (a) Charge on the nucleus (b )

α-particle scattering

(b )

Quantized electron orbit

(c)

Emission spectra

(c)

The quantization of energy

(d )

The photoelectric effect

(d )

The nuclear atom

196. When an electron jumps from ‘L’ level to ‘M’ level, there occurs [EAMCET 1979] (a) Emission of energy (b) Emission of X-rays (c) Absorption of energy (d) Emission of γ-rays 197. In Balmer series of hydrogen atom spectrum which electronic transition causes third line [MP PMT 2000] (a) Fifth Bohr orbit to second one (b) Fifth Bohr orbit to first one (c) Fourth Bohr orbit to second one (d) Fourth Bohr orbit to first one 198. In which of the following transitions will the wavelength be minimum (a) n = 6 to n = 4 (b) n = 4 to n = 2 (c) n = 3 to n = 1 (d) n = 2 to n = 1 199. The frequency of one of the lines in Paschen series of hydrogen atom is 2 .340 × 10 14 Hz . The quantum number n2 which produces this transition is (a) 6

[Delhi PMT 2001]

(b) 5

(c) 4

(d) 3


200. Positronium consists of an electron and a positron (a particle which has the same mass as an electron, but opposite charge) orbiting round their common centre of mass. Calculate the value of the Rydberg constant for this system. (b) R ∞ / 2 (c) 2 R∞ (d) R∞ (a) R∞ / 4 201. What are the average distance and the most probable distance of an electron from the nucleus in the 1s orbital of a hydrogen atom [ a 0 = the radius of the first Bohr orbit] (a) 1 . 5 a0 and a 0

(b) a 0 and 5a 0

(c) 1 . 5 a0 and 0 .5 a0

202. Choose the correct relations on the basis of Bohr theory 1 (a) Velocity of electron ∝ n (c) Radius of orbit ∝ n 2 Z

(d) a 0 and 0 .5 a0

(b) Frequency of revolution ∝ (d) Force on electron ∝

203. For a hydrogen atom, what is the orbital degeneracy of the level that has energy =

1 n3

1 n4

− hcR ∞ , where R∞ is the Rydberg constant for 9

the hydrogen atom (a) 1 (b) 9 (c) 36 (d) 3 [AIEEE 2002] 204. In a hydrogen atom, if energy of an electron in ground state is 13.6 eV, then that in the 2nd excited state is (a) – 1.51 eV (b) – 3.4 eV (c) – 6.04 eV (d) – 13.6 eV 205. The ionization energy of hydrogen atom is – 13.6 eV. The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is (Avogadro’s constant = 6 .022 × 10 23 ) 206. 207.

208.

209. 210.

[BHU 1999]

(a) 1 . 69 × 10 −20 J (b) 1 . 69 × 10 −23 J (c) 1 . 69 × 10 23 J (d) 1 . 69 × 10 25 J [MP PET 2002] The value of the energy for the first excited state of hydrogen atom will be (a) – 13.6 eV (b) – 3.40 eV (c) – 1.51 eV (d) – 0.85 eV An atom has 2 electrons in K shell, 8 electrons in L shell and 6 electrons in M shell. The number of s-electrons present in that [CPMT 1989] element is (a) 6 (b) 5 (c) 7 (d) 10 In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter[AIEEE 2003] orbit jumps of the electron for Bohr orbits in an atom of hydrogen (a) 3 → 2 (b) 5 → 2 (c) 4 → 1 (d) 2 → 5 If electron falls from n = 3 to n = 2, then emitted energy is [AFMC 1997; MP PET 2003] (a) 10.2 eV (b) 12.09 eV (c) 1.9 eV (d) 0.65 eV The emission spectrum of hydrogen is found to satisfy the expression for the energy change. ∆E (in joules) such that  1 1  ∆E = 2 .18 × 10  2 − 2  J where n1 = 1, 2, 3 .... and n2 = 2, 3, 4 .... The spectral lines correspond to Paschen series to [UPSEAT 2002]  n  1 n2 

(a) n1 = 1 and n2 = 2, 3, 4 211.

(b) n1 = 3 and n2 = 4 , 5 , 6

(c) n1 = 1 and n2 = 3, 4 , 5

(d) n1 = 1 and n2 = infinity

The energy required to dislodge electron from excited isolated H-atom., IE1 = 13 .6 eV is (a) = 13.6 eV

(b) > 13.6 eV

212. If change in energy, (∆E) = 3 × 10

−8

(c) < 13.6 and > 3.4 eV

J , h = 6 .64 × 10

−34

[DCE 2000]

(d) ≤ 3 .4 eV

J − s and c = 3 × 10 m / s, then wavelength of the light is 8

[CBSE 2000]

(a) 6 .36 × 10 Å 3

(b) 6 .36 × 10 Å 5

(c) 6 .64 × 10

−18

Å

(d) 6 .36 × 10

18

Å

213. The value of Planck’s constant is 6 .63 × 10 −34 Js . The velocity of light is 3 . 0 × 10 8 ms −1 . Which value is closest to the wavelength in nanometres of a quantum of light with frequency of 8 × 10 15 s −1 (b) 2 × 10 −25 (c) 5 × 10 −18 (a) 3 × 10 7 214. The wavelength of a spectral line for an electronic transition is inversely related to (a) The number of electrons undergoing the transition (b) The nuclear charge of the atom (c) The difference in the energy of the energy levels involved in the transition (d) The velocity of the electron undergoing the transition

[CBSE 2003]

(d) 4 × 10 1 [IIT 1988]

215. If wavelength of photon is 2 .2 × 10 −11 m , h = 6 .6 × 10 −34 Js , then momentum of photon is (a) 3 × 10 −23 kg ms −1

(b) 3 . 33 × 10 22 kg ms −1

[MP PET 1999]

(c) 1 . 452 × 10 −44 kg ms −1

(d) 6 .89 × 10 43 kg ms −1

216. The ratio of the energy of a photon of 2000 Å wavelength radiation to that of 4000 Å radiation is [IIT 1986; DCE 2000; JIPMER 2000]

(a) 1/4 (b) 4 217. Wavelength associated with electron motion (a) Increases with increase in speed of electron (c) Decreases with increase in speed of e

(c) 1/2

(d) 2 [BHU 1998]

(b) Remains same irrespective of speed of electron

−

(d) Is zero

218. A 200 g golf ball is moving with a speed of 5 m per hour. The associated wave length is ( h = 6 .625 × 10 −34 J − sec) [MP PET 2003] (a) 2 .38 × 10 −10 m

(b) 2 .38 × 10 −20 m

(c) 2 .38 × 10 −30 m

(d) 2 .38 × 10 −40 m

219. The frequency of a wave of light is 12 × 10 14 s −1 . The wave number associated with this light is (a) 5 × 10 −7 m

(b) 4 × 10 −8 cm −1

220. The energy of a 700 – nm photon is (a) 1.77 eV (b) 2.47 eV

[Pb. PMT 1999]

(c) 2 × 10 −7 m −1

(d) 4 × 10 4 cm −1

(c) 700 eV

(d) 3.57 eV

Dual nature of electron (de-Broglie equation)

Basis Level 221. The wave nature of an electron was first given by (a) De–Broglie (b) Heisenberg 222. Dual nature of particle is given by (a) Bohr theory (b) Thomson model

[CMC Vellore 1991; Punjab PMT 1998]

(c) Mosley

(d) Sommerfeld [BHU 2003]

(c) Heisenberg principle

(d) De–Broglie equation

h 223. Among the following for which one mathematical expression λ = stands p

(a) De–Broglie equation (b) Einstein equation (c) Uncertainty equation (d) Bohr equation 224. De–Broglie equation describes the relationship of wavelength associated with the wave motion of an electron and its [MP PMT 1986] (a) Mass

(b) Energy

(c) Momentum

(d) Charge

225. De–Broglie equation tells about

[MP PMT 1993]

(a) Relation between electrons and nucleus

(b) Relation between electrons and protons

(c) Relation between electrons and neutrons

(d) Electron’s dual nature of wave and particle

226. Which one of the following explains light both as a stream of particles and as wave motion [AIIMS 1983; IIT 1992; UPSEAT 2003] (a) Diffraction

(b) λ =

h p

(c) Interference

(d) Photoelectric effect

227. Which is the correct relationship between wavelength and momentum of particles (a) λ =

h P

(b) π =

h P

(c) h =

P λ

228. Which of the following expressions gives the de–Broglie relationship (a) h =

λ mv

(b) λ =

h mv

(c) λ =

[Pb. PMT 2000]

(d) None of these [MP PMT 1996; MP PET/ PMT 1998]

m hv

(d) λ =

v mh

229. Which particle among the following will have the smallest de Broglie wavelength, assuming that they have the same velocity (a) A positron

(b) A photon

(c) An α -particle

(d) A neutron

230. Minimum de–Broglie wavelength is associated with (a) Electron

(b) Proton


(c) CO 2 molecule

(d) SO 2 molecule


231. The de–Broglie wavelength associated with a particle of mass 10 −6 kg moving with a velocity of 10 ms −1 , is (a) 6 .63 × 10 −22 m

(b) 6 .63 × 10 −29 m

(c) 6 .63 × 10 −31 m

[AIIMS 2001]

(d) 6 .63 × 10 −34 m

232. Davisson and Germer’s experiment showed that

[MADT Bihar 1983]

(a) β-particles are electrons

(b) Electrons come from nucleus

(c) Electrons show wave nature

(d)

None of the above

Advance Level

233. The de–Brolglie wavelength of a particle with mass 1 g and velocity 100 m/s is [CBSE 1999; EAMCET 1997; AFMC 1999; AIIMS 2000]

(a) 6 .63 × 10

−33

m

(b) 6 .63 × 10

−34

(c) 6 .63 × 10 −35 m

m

(d) 6 .65 × 10 −35 m

234. The de–Broglie wavelength associated with a material particle is

[JIPMER 2000]

(a) Directly proportional to its energy

(b) Directly proportional to momentum

(c) Inversely proportional to its energy

(d) Inversely proportional to momentum

235. What is the de–Broglie wavelength associated with the hydrogen electron in its third orbit (a) 9 .96 × 10 −10 cm

(b) 9 .96 × 10 −8 cm

(c) 9 .96 × 10 4 cm

[AMU (Engg.) 2002]

(d) 9 .96 × 10 8 cm

236. What will be de–Broglie wavelength of an electron moving with a velocity of 1 . 2 × 10 5 ms −1 (a) 6 .068 × 10 −9

(b) 3 . 133 × 10 −37

(c) 6 .626 × 10 −9

[MP PET 2000]

(d) 6 .018 × 10 −7

237. An electron has kinetic energy 2 .8 × 10 −23 J . de–Broglie wavelength will be nearly (m e = 9 .1 × 10 −31 kg ) (a) 9 .28 × 10 −4 m

(b) 9 .28 × 10 −7 m

(c) 9 .28 × 10 −8 m

238. Calculate the de–Broglie wavelength for a particle of mass

[MP PET 2000]

(d) 9 .28 × 10 −10 m

10 −30 kg , travelling with a speed of

10 7 ms −1 .

( h = 6 .626 × 10 −34 kg m 2 s −1 )

(a) 6 .626 × 10 −4 m

(b) 1 . 509 × 10 −4 m

(c) 6 .626 × 10 −11 m

(d) 1 . 509 × 10 10 m

239. The de–Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately [AIEEE 2003]

(a) 10

−33

metres

(b) 10

−31

(c) 10

metres

−16

metres

(d) 10

−25

metres

240. De-Broglie wavelength is related to applied voltage as (a) λ =

12 . 3 h

Å

(b) λ =

12 . 3 V

(c)

Å

λ=

12 . 3 E

Å

(d) λ

12 .3 m

Å

Uncertainty principle and Schrodinger wave equation

Basic Level 241. The possibility of finding an electron in an orbital was conceived by (a) Rutherford

(b) Bohr

(c) Heisenberg

[MP PMT 1994]

(d) Schrodinger

242. The uncertainty principle and the concept of wave nature of matter was proposed by …… and ……. respectively (a) Heisenberg, de–Broglie

(b) De–Brolgie, Heisenberg

(c) Heisenberg, Planck

[MP PET 1997]

(d) Planck, Heisenberg

243. “The position and velocity of a small particle like electron cannot be simultaneously determined.” This statement is [NCERT 1979; BHU 1981, 87]

(a) Heisenberg uncertainty principle nature of electron

(b)

(c) Pauli’s exclusion principle

(d) Aufbau’s principle

244. In Heisenberg’s uncertainty equation ∆x × ∆p ≥ (a) Uncertainty in energy

Principle of de–Broglie’s wave

h , ∆p stands for 4π

(b) Uncertainty in velocity

(c) Uncertainty in momentum (d) Uncertainty in mass

245. According to uncertainty principle

[AMU 1990]

h (b) ∆x × ∆p ≥ 4π

(a) E = mc 2

h (d) ∆x × ∆p = 6π

h (c) λ = p

246. The uncertainty principle was enunciated by (a) Einstein

[NCERT 1975; Bihar MEE 1997]

(b) Heisenberg

(c) Rutherford

(d) Pauli

247. Simultaneous determination of exact position and momentum of an electron is

[BHU 1979]

(a) Possible

(b) Impossible

(c) Sometimes possible sometimes impossible

(d) Non of the above

248. The equation ∆x .∆p >

h shows 4π

[MP PET 2000]

(a) De–Brolgie relation

(b) Heisenberg’s uncertainty principle

(c) Aufbau principle

(d) Hund’s rule

Advance Level

249. Uncertainty principle gave the concept of (a) Probability (c) Physical meaning of ψ, the ψ2 250. The uncertainty

(b) An orbital (d) All the above

in momentum of an electron is

1 × 10 −5 kg − m / s .

The uncertainty

in its position will

be

( h = 6 .62 × 10 −34 kg − m 2 / s) [AFMC 1998; CBSE 1999; JIPMER 2002]

(a) 1 . 05 × 10

−28

m

(b) 1 . 05 × 10

−26

m

(c) 5 .27 × 10

−30

(d) 5 .25 × 10 −28 m

m

251. Uncertainty in position of a 0.25 g particle is 10 −5 m . Uncertainty of velocity is ( h = 6 .6 × 10 −34 J − s) (a) 1 . 2 × 10 34

(b) 2 .1 × 10 −32

[AIEEE 2002]

(d) 1. 7 × 10 −9

(c) 1 . 6 × 10 −20

[CPMT 1988] 252. If uncertainty in the position of an electron is zero, the uncertainty in its momentum would be (a) Zero (b) < h / 2λ (c) > h / 2λ (d) Infinite 253. The position of both an electron and a helium atom is known within 1.0 nm and the momentum of the electron is known within

50 × 10 −26 kg ms −1 . The minimum uncertainty in the measurement of the momentum of the helium atom is [CBSE 1998; AIIMS 2001]

(a) 50 kg ms

−1

(b) 60 kg ms

−1

(c) 80 × 10

−26

kg ms

−1

(d) 50 × 10 −26 kg ms −1

254. Assertion (A): The position of an electron can be determined exactly with the help of an electron microscope. Reason (R): The product of uncertainty in the measurement of its momentum and the uncertainty in the measurement of the position cannot be less than a finite limit. [NDA 1999] (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true


255. The uncertainty in the position of an electron (mass = 9 .1 × 10 −28 g ) moving with a velocity of 3 . 0 × 10 4 cm s −1 accurate upto 0.001% will be (Use

h in the uncertainty expression, where h = 6 .626 × 10 −27 erg − s ) 4π

(a) 1 . 92 × 10 −5 cm

(b) 7.68 cm

(c) 5.76 cm

256. The uncertainty in the position of a moving bullet of mass 10 gm is 10 (a) 5 .2 × 10 −28 m / sec

[CBSE 1995]

(b) 3 . 0 × 10 −28 m / sec

−5

(d) 3.84 cm

m . Calculate the uncertainty in its velocity

(c) 5 .2 × 10 −22 m / sec

[DCE 1999]

(d) 3 × 10 −22 m / sec

Quantum numbers, Shape of Orbitals and Electronic Configuration of elements

Basic Level 257. The shape of s-orbital is (a) Pyramidal

[NCERT 1978]

(b) Spherical

(c) Tetrahedral

(d) Dumb-bell shaped

(b) Ellipsoidal

(c) Dumb-bell

(d) Pyramidal

258. The shape of 2p orbital is (a) Spherical

[CPMT 1983; NCERT 1979]

259. Which orbital is dumb-bell shaped (a) s-orbital

(b) p-orbital

[MP PMT 1986; MP PET /PMT 1999]

(c) d-orbital

(d) f-orbital

260. A 3p orbital has

[IIT 1995]

(a) Two spherical nodes

(b) Two non-spherical nodes

(c) One spherical and one non-spherical nodes

(d) One spherical and two non-spherical nodes

261. Number of nodal centres for 2s orbital (a) 1

(b) 0


(c) 4

(d) 3

262. Which of the following pair of orbitals posses two nodal planes (a) p xy , d x 2 −y 2

(b) d xy , d zx


(c)

p xy , d zx

(d) d z 2 , d x 2 −y 2

263. Which orbital does not have a spherical node (a) n = 2, l = 0

(b) n = 3, l = 0

[Kurukshetra CEE 2002]

(c) n = 2, l = 1

264. The number of electrons which can be accommodated in an orbital is (a) One

(b) Two

(c) Three

(d) n = 1, l = 0 [Delhi PMT 1981; AFMC 1988]

(d) Four

265. The number of nodal planes in a p x orbital is (a) One (b) Two 266. One orbital consists maximum electrons (a) 2 (b) 1 267. Which of the following orbitals will be dumb-bell shaped (a) 1s

(b) 2s

268. The number of orbitals in d sub-shell is (a) 1 (b) 3 269. The shape of p-orbital is (a) Elliptical (b) Spherical 270. Number of orbitals in h sub-shell is (a) 11 (b) 15 271. Azimuthal quantum number for last electron of Na atom is (a) 1 272.

(b) 2

p x orbital can accommodate

(a) 4 electrons (c) 2 electrons with parallel spins

[IIT Screening 2000]

(c) Three

(d) Zero

(c) 8

(d) 18

[AFMC 1988] [MP PET 1986]

(c) 2 p x

(d) 3 d xy

(c) 5

(d) 7

(c) Dumb-bell

(d) Complex geometrical

[MNR 1981] [MP PMT 1993] [BHU 2003]

(c) 17

(d) 19

(c) 3

(d) 0

[BHU 1995]

[MLNR 1990: IIT 1983; MADT Bihar 1995]

(b) 6 electrons (d) 2 electrons with opposite spins

273. The maximum number of electrons that can be accommodated in ‘f’ sub shell is [CPMT 1983, 84; MP PET / PMT 1988; BITS 1988]

(a) 2

(b) 8

(c) 32

(d) 14

274. The maximum number of electrons accommodated in 5f orbitals are

[MP PET 1996]

(a) 5 (b) 10 275. Which of the following orbitals does not make sense

(c) 14

(d) 18

(a) 7s (b) 5p 276. There is no difference between a 2p and a 3p orbital regarding

(c) 2d

(d) 4f

(c) Energy

(d) Value of n

(c) 4d

(d) 5f

(c) s, p and d

(d) s, p, d and f

(a) Shape

(b) Size

277. Which of the sub-shell from the following is dumb-bell (a) 5s (b) 5p 278. The type of orbitals present in Fe is (a) s (b) s and p 279. For the dumb-bell shaped orbital, the value of l is (a) 3

(b) 1

[Rajasthan PMT 2000] [BHU 1981]

[CPMT 1987, 2003]

(c) 0

(d) 2

280. Which of the following orbital is not possible (a) 3f

(b) 4f


(c) 5f

(d) 6f

281. Quantum numbers of an atom can be defined on the basis of

[AIIMS 2002]

(a) Hund’s rule

(b) Aufbau’s principle

(c) Pauli’s exclusion principle

(d) Heisenberg’s uncertainty principle

282. Principal, azimuthal and magnetic quantum numbers are respectively related to

[CPMT 1988; AIIMS 1999]

(a) Size, shape and orientation(b) Shape, size and orientation (c) Size, orientation and shape(d) None of the above 283. The magnetic quantum number specifies [MNR 1986; BHU 1982; CPMT 1989, 94; MP PET 1999; AFMC 1999; AMU (Engg.) 1999] (a) Size of orbitals

(b) Shape of orbitals

(c) Orientation of orbitals

(d) Nuclear stability

284. The azimuthal quantum number is related to (a) Size

(b) Shape

[BHU 1987, 95]

(c) Orientation

(d) Spin

285. The principal quantum number represents

[CPMT 1991]

(a) Shape of an orbital

(b) Distance of electron from nucleus

(c) Number of electrons in an orbit

(d) Number of orbitals in an orbit

[EAMCET 1979; IIT 1983; MLNR 1990; UPSEAT 2000, 02] 286. Principal quantum number of an atom represents (a) Size of the orbital (b) Spin angular momentum (c) Orbital angular momentum (d) Space orientation of the orbital 287. Azimuthal quantum number defines [AIIMS 2002] (a) e/m ratio of electron (b) Spin of electron (c) Angular momentum of electron (d) Magnetic momentum of electron 288. Which quantum number will determine the shape of the sub-shell [CPMT 1999; Punjab PMT 1998] (a) Principal quantum number (b) Azimuthal quantum number (c) Magnetic quantum number (d) Spin quantum number 289. Which quantum number is not related with Schrodinger equation [Rajasthan PMT 2002] (a) Principal (b) Azimuthal (c) Magnetic (d) Spin 290. The quantum number which specifies the location of an electron as well as energy is [Delhi PMT 1983] (a) Principal quantum number (b) Azimuthal quantum number (c) Spin quantum number (d) 291. When the azimuthal quantum number has a value of l = 0, the shape of the orbital is [MP PET 1995] (a) Rectangular (b) Spherical (c) Dumb-bell (d) Unsymmetrical 292. If n = 3, then the value of ‘l’ which is incorrect [CPMT 1994] (a) 0 (b) 1 (c) 2 (d) 3 [BHU 1978, NCERT 1981] 293. The angular momentum of an electron depends on (a) Principal quantum number (b) Azimuthal quantum number (c) Magnetic quantum number (d) All of these 294. The shape of an orbital is given by the quantum number [NCERT 1984; MP PMT 1996] (a) n (b) l (c) m (d) s


295. Which of the following set of quantum number is not valid (a) n = 1, l = 2 (b) n = 2, m = 1 (c) n = 3, l = 0 296. 2p orbital have (a) n = 1, l = 2 (b) n = 1, l = 0 (c) n = 2, l = 1 297. The maximum number of electrons which each sub-shell can occupy is 2

[AIIMS 2001]

(d) n = 2, l = 0 [NCERT 1981; MP PMT 1993, 97]

(d) n = 2, l = 0 [PU CET 1989]

(a) 2n (b) 2n (c) 2 ( 2l + 1) 298. Which of the following represent the correct sets of the four quantum numbers of a 4d electron 1 (a) 4, 3, 2, + 2

(b) 4, 2, 1, 0

1 (c) 4, 3, – 2, + 2

(d) 2l + 1 [MLNR 1992; UPSEAT 2001]

(d) 4, 2, 1, −

299. For the n = 2 energy level, how many orbitals of all kinds are possible (a) 2 (b) 3 (c) 4 (d) 5 300. The total number of orbitals in an energy level designated by principal quantum number n, is equal to (a) 2n

(b) 2 n 2

(c) n

1 2 [Bihar CEE 1995] [AIIMS 1997]

(d) n 2

301. The quantum numbers for the outermost electron of an element are given below as n = 2, l = 0, m = 0, s = +

1 . The atoms is 2 [EAMCET 1978]

(a) Lithium (b) Beryllium (c) Hydrogen 302. The maximum number of electrons in an atom with l = 2 and n = 3 is (a) 2 (b) 6 (c) 12 303. Correct set of four quantum numbers for valence electron of rubidium (Z = 37) is 1 (a) 5, 0, 0, + 2

1 (b) 5, 1, 0, + 2

1 (c) 5, 1, 1, + 2

(d) Boron [MP PET / PMT 1998]

(d) 10 [IIT 1984; JIPMER 1999; UPSEAT 2003]

(d) 6, 0, 0, +

1 2

304. If n + l = 6, then total possible number of sub-shells would be [Rajasthan PMT 1997] (a) 3 (b) 4 (c) 2 (d) 5 305. If value of azimuthal quantum number l is 2, then total possible values of magnetic quantum number will be (a) 7 (b) 5 (c) 3 (d) 2

306. Orbital angular momentum for a d-electron is (a)

6h 2π

(b)

[MP PET 2003]

6h 2π

(c)

12h 2π

307. The number of quantum numbers required to describe an electron in an atom completely is (a) 1 (b) 2 (c) 3 308. An e − has magnetic quantum number as – 3, what is its principal quantum number (a) 1 (b) 2 (c) 3 309. The quantum number which is designated by letters s, p, d and f instead of number is (a) n

(b) l

(c) m l

12 h 2π

(d)

[CET Pune 1998]

(d) 4 [BHU 1998]

(d) 4 [BHU 1980]

(d) m s

310. An electron having the quantum numbers n = 4 , l = 3, m = 0, s = – 1/2 would be in the orbital [Orissa JEE 1997] (a) 3s (b) 3p (c) 4d (d) 4f 311. The magnetic quantum number of valence electron of sodium (Na) is [Rajasthan PMT 2002] (a) 3 (b) 2 (c) 1 (d) 0 312. How many electrons can be fit into the orbitals that comprise the 3rd quantum shell n = 3 [MP PMT 1986, 87; Orissa JEE 1997] (a) 2 (b) 8 (c) 18 (d) 32 313. A sub-shell l = 2 can take how many electrons [NCERT 1973, 78] (a) 3 (b) 10 (c) 5 (d) 6 314. What is the maximum number of electrons which can be accommodated in an atom in which the highest principal quantum [MP PMT 2000] number value is 4 (a) 10 (b) 18 (c) 32 (d) 54 315. How many electrons can be accommodated in a sub-shell for which n = 3, l = 1 [CBSE 1990] (a) 8 (b) 6 (c) 18 (d) 32 316. If an electron has spin quantum number of +

1 and a magnetic quantum number of – 1, it cannot be presented in an 2 [CBSE 1989; UPSEAT 2001]

(a) d-orbital (b) f-orbital 317. Which statement is not correct for n = 5, m = 3 (a) l = 4 (b) l = 0, 1, 2; S = +1 / 2

(c) p-orbital

(d) s-orbital

(c) l = 3

(d) All are correct

[CPMT 1996]

318. Values of the four quantum numbers for the last electron in the atom are n = 4, l = 1, m = +1 and s = −

319. 320. 321. 322.

1 . Atomic number of the 2

atom will be (a) 22 (b) 32 (c) 33 (d) 36 An electron has principal quantum number 3. The number of its (i) sub-shells and (ii) orbitals would be respectively [MP PET 1997] (a) 3 and 5 (b) 3 and 7 (c) 3 and 9 (d) 2 and 5 For d electrons, the azimuthal quantum number is [MNR 1983; CPMT 1984] (a) 0 (b) 1 (c) 2 (d) 3 The magnetic quantum number for an electron when the value of principal quantum number is 2 can have [CPMT 1984] (a) 3 values (b) 2 values (c) 9 values (d) 6 values The magnetic quantum number for d-orbtial is given by [Orissa JEE 2002] (a) 2 (b) 0 , ± 1, ± 2 (c) 0, 1, 2 (d) 5

323. The number of orbitals in the fourth principal quantum number will be (a) 4 (b) 8 (c) 12 324. For sodium atom the number of electrons with m = 0 will be (a) 2 (b) 7 (c) 9 325. The quantum numbers n = 2, l = 1 represent (a) 1s orbital (b) 2s orbital (c) 2p orbital 326. For azimuthal quantum number l = 3, the maximum number of electrons will be

(d) 16 [Rajasthan PMT 1999]

(d) 8 [AFMC 2002]

(d) 3d orbital

[CBSE 1991; EAMCET 1991; Rajasthan PMT 2002; CBSE 2002]

(a) 2

(b) 6

(c) 0

(d) 14


327. All electrons on the 4p sub-shell must be characterized by the quantum number (s) [MP PET 1996] 1 1 1 (a) n = 4, m = 0, s = ± (b) l = 1 (c) l = 0, s = ± (d) s = ± 2 2 2 [CPMT 1988; MH CET 1999] 328. The magnetic quantum number for valency electrons of sodium is (a) 3 (b) 2 (c) 1 (d) 0 329. Which set of quantum numbers are not possible from the following (a) n = 3, l = 2, m = 0, s = – 1/2 (b) n = 3, l = 2, m = – 2, s = – 1/2 (c) n = 3, l = 3, m = – 3, s= – 1/2 (d) n = 3, l = 0, m = 0, s = – 1/2 330. The two electrons in K shell will differ in [MLNR 1988; UPSEAT 1999, 2000; Kerala PMT 2003] (a) Principal quantum number (b) Azimuthal quantum number (c) Magnetic quantum number (d) Spin quantum number 331. Electron occupies the available orbital singly before pairing in any one orbital occurs, it is [CBSE 1991] (a) Pauli’s exclusion principle (b) Hund’s rule (c) Heisenberg’s principle (d) Prout’s hypothesis 332. Which of the following explains the sequence of filling the electrons in different shells [AIIMS 1998; BHU 1999] (a) Hund’s rule (b) Octet rule (c) Aufbau principle (d) All of these 333. Following Hund’s rule which element contains six unpaired electron [Rajasthan PET 2000] (a) Fe

(b) Co

(c) Ni

(d) Cr

334. The explanation for the presence of three unpaired electrons in the nitrogen atom can be given by [NCERT 1979; Rajasthan PMT 1999; DCE 1999; CPMT 2001; MP PMT 2002; Pb. PMT 2002]

(a) Pauli’s exclusion principle (b) Hund’s rule

(c) Aufbau’s principle

(d) Uncertainty principle

335. Which of the following have the same number of unpaired electrons in ‘d’ orbitals (a) Cr

(b) Mn

(c)

Fe

[Roorkee 2000]

3+

(d) Co

336. How many unpaired electrons are present in cobalt |Co| metal (a) 2

(b) 3

3+


(c) 4

(d) 7

337. Aufbau principle is not satisfied by (a) Cr and Cl

[MP PMT 1997]

(b) Cu and Ag

(c) Cr and Mg

(d) Cu and Na

338. When 3d orbital is complete, the new electron will enter the (a) 4p orbital 2

2

6

(b) 4f orbital

[EAMCET 1980; MP PMT 1995]

(c) 4s orbital

(d) 4d orbital

1

339. 1 s 2 s 2 p 3 s shows configuration of (a)

Al +3 in ground state

(b) Ne in excited state

[CPMT 1996]

(c)

Mg +1 in excited state

(d) None of these

340. The electronic configuration (outermost) of Mn +2 ion (atomic number of Mn = 25) in its ground state is (a) 3 d 5 ,4 s 0

(b) 3 d 4 ,4 s1

(c) 3 d 3 ,4 s 2

(d) 3 d 2 , 4 s 2 4 p 2

341. The structure of external most shell of inert gases is 2 3

2 6

(a) s p

(b) s p

[JIPMER 1991] 1 2

10 2

(c) s p

(d) d s

342. In a potassium atom, electronic energy levels are in the following order (a) 4s > 3d (b) 4s > 4p 343. Which one is the correct outer configuration of chromium (a)

↑

↑

↑

↑

(c)

↑

↑

↑

↑

↑

[EAMCET 1979; Delhi PMT 1991]

(c) 4s < 3d

(d) 4s < 3p [AIIMS 1980, 91; BHU 1995]

↑↓

(b)

↑↓ ↑↓ ↑↓

↑

(d)

↑↓ ↑↓ ↑

↑

344. Which of the following represents the electronic configuration of an element with atomic number 17 (a) 1 s 2 , 2 s 2 2 p 6 ,3 s1 3 p 6

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 4 ,4 s1

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 5

2

6

2

6

(a) 1 s , 2 s 2 p ,3 s 3 p 3 d

6

[AMU 1982]

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s1 3 p 4 , 4 s 2

345. Which one is the electronic configuration of Fe +2 2

[MP PET 1993]

[MADT Bihar 1982; AIIMS 1989] 2

2

6

2

6

4

(b) 1 s , 2 s 2 p ,3 s 3 p 3 d ,4 s 2


(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 5 ,4 s1

(d) None of these

346. Total number of unpaired electrons in an atom of atomic number 29 is (a) 1

(b) 3

[CPMT 1984, 93]

(c) 4

(d) 2

(c) 2

(d) 8

347. Maximum number of electrons present in ‘N’ shell is (a) 18

[EAMCET 1984]

(b) 32

348. Correct configuration of Fe +3 [26] is 2

2

6

2

6

(a) 1 s , 2 s 2 p ,3 s 3 p 3 d

[CPMT 1994; BHU 1995; Karnataka CET 1992]

5

2

(b) 1 s , 2 s 2 p ,3 s 2 3 p 6 3 d 3 ,4 s 2

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 6 ,4 s 2

2

6

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 5 ,4 s1

349. According to Aufbau’s principle, which of the three 4d, 5p and 5s will be filled with electrons first (a) 4d

[MADT Bihar 1984]

(b) 5p

(c) 5s

(d) 4d and 5s will be filled simultaneously

350. The number of unpaired electrons in Fe (a) 5

3+

(Z = 26 ) are

[Karnataka CET 2000]

(b) 6

(c) 3

351. The electronic configuration of copper (29 Cu ) is

(d) 4

[Delhi PMT 1983; BHU 1980; AFMC 1981; CBSE 1991; MP PMT 1995]

(a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 9 ,4 s 2

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 10 ,4 s1

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 4 p 6

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 10

352. The number of electrons in the valence shell of calcium is (a) 6

[IIT 1975]

(b) 8

(c) 2

(d) 4

353. Pauli’s exclusion principle states that

[MNR 1983; AMU 1984]

(a) Two electrons in the same atoms can have the same energy (b) Two electrons in the same atom cannot have the same spin (c) The electrons tend to occupy different orbitals as far as possible (d) Electrons tend to occupy lower energy orbitals preferentially (e) None of these 354. The configuration 1 s 2 2 s 2 2 p 5 3 s1 shows

[AIIMS 1997; Pb. PMT 2002]

(a) Ground state of fluorine atom

(b) Excited state of fluorine atom

(c) Excited state of neon atom

(d) Excited state of ion O 2−

355. The number of d electrons in Fe 2 + (atomic number of Fe = 26) is not quite equal to that of the

[MLNR 1993]

(a) p-electrons in Ne (At. No. = 10)

(b) s-electrons in Mg (At. No. = 12)

(c) d-electrons in Fe

(d) p-electrons in Cl − (At. No. of Cl = 17)

356. Ground state electronic configuration of nitrogen atom can be represented by (a)

↑↓ ↑↓

↑

↑

↑

(b)

↑↓ ↑↓

↑

↓

↑

(c)

↑↓ ↑↓

[IIT 1999]

↑

↓ 2

357. The atomic number of an element having the valency shell electronic configuration 4 s 4 p (a) 35

(b) 36

(c) 37

↓ 6

(d)

↑↓ ↑↓

is

↓

↓

↓

[MP PMT 1991]

(d) 38

358. A filled or half-filled set of p or d-orbitals is spherically symmetric. Point out the species which has spherical symmetry [NCERT 1983] (a) Na

(c) Cl −

(b) C

(d) Fe

359. The correct ground state electronic configuration of chromium atom is [IIT 1989, 94, MP PMT 1993; EAMCET 1997; MP PAT 1996; AFMC 1997; Bihar MEE 1996; ISM Dhanbad 1994; MP PET 1995, 97; CPMT 1999; Kerala PMT 2003] 5

1

(a) [ Ar ] 3 d 4 s

4

(b) [ Ar ] 3 d 4 s

2

(c) [ Ar ] 3 d 6 4 s 0

(d) [ Ar ] 4 d 5 4 s1

360. Aufbau principle is obeyed in which of the following electronic configurations APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

[AFMC 1999]

(a) 1 s 2 2 s 2 2 p 6

(b) 1 s 2 3 p 3 3 s 2

(c) 1 s 2 3 s 2 3 p 6

(d) 1 s 2 2 s 2 3 s 2

361. Nitrogen has the electronic configuration 1 s 2 , 2 s 2 2 p 1x 2 p 1y 2 p 1z and 1 s 2 , 2 s 2 2 p x2 2 p 1y 2 p z0 which is determined by [Delhi PMT 1982, 83, 89, MP PMT /PET 1988; EAMCET 1988]

(a) Aufbau’s principle

(b) Pauli’s exclusion principle

(c) Hund’s rule

(d) Uncertainty principle

(c) 6

(d) 14

362. Maximum electron in a d-orbital are (a) 2

[CPMT 1999]

(b) 10

363. A completely filled d-orbital (d 10 ) (a) Spherically symmetrical

[MLNR 1987]

(b) Has octahedral symmetry

(c) Has tetrahedral symmetry (d) Depends on the atom

364. Electronic configuration 1 s 2 , 2 s 2 , 2 p 6 ,3 s 2 ,3 p 6 ,3 d 5 ,4 s1 represents (a) Ground state

(b) Excited state

[CPMT 2003]

(c) Anionic state

(d) All of these

365. The correct electronic configuration of Ti ( Z = 22 ) atom is

[MP PMT 1999]

(a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 ,3 d 2

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 4 p 2

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 4

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s1 3 d 3

366. Electronic configuration of H − is (a) 1 s

0

[CPMT 1985]

(b) 1 s

1

(c) 1 s

2

1

(d) 1 s 2 s

1

367. Which electronic configuration is not observing the (n + l) rule (a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 1 ,4 s 2

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 7 ,4 s 2

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 5 ,4 s1

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 8 ,4 s 2

368. The electronic configuration of silver atom in ground state is (a) [Kr ] 3 d

10

1

(b) [ Xe ] 4 f

4s

14

10

1

5d 6 s

[CPMT 1984, 93]

(c) [Kr ] 4 d

10

1

5s

369. The order of filling of electrons in the orbitals of an atom will be (a) 3d, 4s, 4p, 4d, 5s (b) 4s, 3d, 4p, 5s, 4d (c) 5s, 4p, 3d, 4d, 5s 370. Which of the following has more unpaired d-electrons (a) Zn

+

(b) Fe

2+

2

(c) 2

371. The number of unpaired electrons in 1 s , 2 s , 2 p

4

N

3+

is

(a) 4 (b) 2 372. Which of the following configuration is correct for iron

9

(d) [Kr ] 4 d 5 s 2 [CBSE 1991]

(d) 3d, 4p, 4s, 4d, 5s [CBSE 1999]

(d) Cu

+

[NCERT 1984; CPMT 1991; MP PMT 1996, 2002]

(c) 0

(d) 1 [CBSE 1999]

(a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 5

(b) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 5

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 3 d 7

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 3 d 6

373. Which of the following has the maximum number of unpaired electrons (a) Mg

2+

(b) Ti 3 +

(c) V 3 +

[IIT 1996]

(d) Fe 2 +

374. An ion has 18 electrons in the outermost shell it is

[CBSE 1990]

(a) Cu + (b) Th 4 + (c) Cs + 375. Number of unpaired electrons in inert gas is (a) Zero (b) 8 (c) 4 376. Which of the following is not correct for electron distribution in the ground state 4s 3d (a) Co

(Ar)

↑↓

↑↓

↑↓

↑

↑

↑

(b) Ni

(Ar)

↑↓

↑↓

↑↓

↑↓

↑

↑

(c) Cu

(Ar)

↑↓

↑↓

↑↓

↑↓

↑↓

↑

(d) Zn

(Ar)

↑↓

↑↓

↑↓

↑↓

↑↓

↑↓

377. Electronic configuration of Sc

(21 )

(d) K + [CPMT 1996]

(d) 18

is


[AIIMS 1982]

[BHU 1997]

(a) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 1

(b) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s1 3 d 2

(c) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 0 3 d 3

(d) 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 2

378. The electronic configuration 1 s 2 2 s 2 2 p1x 2 p y1 2 p 1z is

[AFMC 1997; Pb. PMT 1999; CBSE 2001; AIIMS 2001]

(a) Oxygen (b) Nitrogen (c) Hydrogen (d) Fluorine 379. The electrons would go to lower energy levels first and then to higher energy levels according to which of the following [BHU 1990; MP PMT 1993]

(a) Aufbau principle (b) Pauli’s exclusion principle (c) Hund’s rule of maximum multiplicity (d) Heisenberg’s uncertainty principle 380. The atomic orbitals are progressively filled in order of increasing energy. This principle is called is [MP PET 2001] (a) Hund’s rule (b) Aufbau principle (c) Exclusion principle (d) De–Broglie rule [CBSE 1997] 381. The electronic configuration of gadolinium (atomic no. 64) is (a) [ Xe ] 4 f 8 5 d 9 6 s 2

(b) [ Xe ] 4 f 7 5 d 1 6 s 2

(c) [ Xe ] 4 f 3 5 d 5 6 s 2

(d) [ Xe ] 4 f 6 5 d 2 6 s 2

382. The correct order of increasing energy of atomic orbitals is [MP PET 2002] (a) 5p < 4f < 6s < 5d (b) 5p < 6s < 4f < 5d (c) 4f < 5p < 5d < 6s (d) 5p < 5d < 4f < 6s [CPMT 2001] 383. The atomic number of an element is 17. The number of orbitals containing electron pairs in its valence shell is (a) Eight (b) Six (c) Three (d) Two 384. The electronic configuration of chromium is [MP PMT 1993; MP PET 1995; BHU 2001] (a) [ Ne ] 3 s 2 3 p 6 3 d 4 4 s 2

(b) [ Ne ] 3 s 2 3 p 6 3 d 5 4 s1

(c) [ Ne ] 3 s 2 3 p 6 ,4 s 2 4 p 4

(d) [ Ne ] 3 s 2 3 p 6 3 d 1 ,4 s 2 4 p 3

385. In a given atom no two electrons can have the same values for all the four quantum numbers. This is called [BHU 1979; AMU 1983; EAMCET 1980, 83; MADT Bihar 1980; CPMT 1986, 90, 92; NCERT 1978, 84; Raj. PMT 1997; CBSE 1991; MP PET 1986, 99]

(a) Hund’s rule

(b) Aufbau’s principle

(c) Uncertainty principle

(d) Pauli’s exclusion principle

386. An element has the electronic configuration 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 2 . Its valency electrons are (a) 6

(b) 2

(c) 3

[NCERT 1973]

(d) 4

387. Pauli’s exclusion principle states that

[CPMT 1983, 84]

(a) Nucleus of an atom contains no negative charge (b) Electrons move in circular orbits around the nucleus (c) Electrons occupy orbitals of lowest energy (d) All the four quantum numbers of two electrons in an atom cannot be equal 388. Cu 2 + will have the following electronic configuration

[MP PMT 1985]

(a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 10

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 9 ,4 s1

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 9

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 10 ,4 s1

389. The atomic number of an element is 35. What is the total number of electrons present in all the p-orbitals of the ground state [EAMCET (Engg.) 2003] atom of that element (a) 6 (b) 11 (c) 17 (d) 23 390. After np orbitals are filled, the next orbital filled will be (a) (n + 1)s

(b) (n + 2)p

(c) (n + 1)d

(d) (n + 2)s

391. The number of unpaired electrons in an O2 molecule is (a) 0

(b) 1

392. How many unpaired electrons are present in Ni

[MNR 1983]

(c) 2 2+

(d) 3

(atomic number = 28) cation [IIT 1981; MNR 1984; MP PAT 1993; MP PMT 1995; Kerala PMT 2003]

(a) 0

(b) 2

(c) 4

(d) 6

393. The electronic configuration of calcium ion (Ca 2+ ) is APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

[CMC Vellore 1991]

(a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s1

(c) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 2

(d) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 3 d 5

(e) 1 s 2 , 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 0 394. The number of unpaired electrons in carbon atom in excited state is (a) One (b) Two (c) Three 395. Which of the following electronic configurations is not possible 2

(a) 1 s 2 s

2

2

2

(b) 1 s 2 s 2 p

6

(c) 3 d

10

[MNR 1987]

(d) Four [CPMT 2000] 2

4s 4 p

2

2

2

(d) 1 s 2 s 2 p 3 s1

396. The outer electronic structure 3 s 2 3 p 5 is possessed by

2

[Pb. PMT 2002]

(a) Cl (b) O (c) Ar (d) Br 397. The total number of electrons present in all the p-orbitals of bromine are [MP PET 1994] (a) Five (b) Eighteen (c) Seventeen (d) Thirty five 398. Which one of the following configuration represents a noble gas [CPMT 1983, 89, 93; NCERT 1973; MP PMT 1989; Delhi PMT 1984] (a) 1 s 2 , 2 s 2 2 p 6 ,3 s 2

(b) 1 s 2 , 2 s 2 2 p 6 ,3 s1

(c) 1 s 2 , 2 s 2 2 p 6

(d) 1 s 2 . 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2

399. The electronic configuration of an element is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 5 4 s1 . This represents its (a) Excited state

(b) Ground state


(c) Cationic form

(d) Anionic form

400. Electronic configuration of C is (a) 1 s 2 , 2 s 2 2 p 2

[CPMT 1975]

(b) 1 s 2 , 2 s 2 2 p 3

(c) 1 s 2 , 2 s 2

(d) 1 s 2 , 2 s 2 2 p 6

401. The number of unpaired electrons in the Fe 2 + ion is (a) 0

(b) 4

[MP PET 1989; Karnataka CET 2000]

(c) 6

(d) 3

402. The electronic configuration of an element with atomic number 7 i.e. nitrogen atom is (a) 1 s 2 , 2 s 2 2 p 3x

(b) 1 s 2 , 2 s 2 2 p x2 2 p 1y

(c) 1 s 2 , 2 s 2 2 p1x 2 p 1y 2 p1z

[CPMT 1982. 84, 87]

(d) 1 s 2 , 2 s 2 2 p 1x 2 p y2

403. The number of orbitals in 2p sub-shell is (a) 6

[NCERT 1973; MP PMT 1996]

(b) 2

(c) 3

(d) 4

404. The statements

[AIIMS 1982]

(i) In filling a group of orbitals of equal energy, it is energetically preferable to assign electrons to empty orbitals rather than pair them into a particular orbital. (ii) When two electrons are placed in two different orbitals, energy is lower if the spins are parallel are valid for (a) Aufbau principle

(b) Hund’s rule

(c) Pauli’s exclusion principle (d) Uncertainty principle

405. Which of the following principles/rules limits the maximum number of electrons in an orbital to two [CBSE 1989; MP PAT 1993] (a) Aufbau principle

(b) Pauli’s exclusion principle

(c) Hund’s rule of maximum multiplicity

(d) Heisenberg’s uncertainty principle

406. Electronic configuration of ferric ion is (a) [ Ar ] 3 d

5


(b) [ Ar ] 3 d

7

(c) [ Ar ] 3 d

3

(d) [ Ar ] 3 d 8

407. Which of the following metal ions will have maximum number of unpaired electrons (a) Fe

+2

(b) Co 2

+2

2

408. Number of unpaired electrons in 1 s 2 s 2 p

(c) 3

Ni

+2

[CPMT 1996]

(d) Mn

+2

is

[CPMT 1982; MP PMT 1987; BHU 1987; CBSE 1990; CET Pune 1998; AIIMS 2000]

(a) 2 (b) 0 (c) 3 409. Energy of atomic orbitals in a particular shell is in the order (a) s < p < d < f (b) s > p > d > f (c) p < d < f < s 410. Which of the following ions is not having the configuration of neon (a) F −

(b) Mg + +

(c)

Na +

(d) 1 [AFMC 1990]

(d) f > d > s > p (d) Cl −

Advance Level APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

411.

The electron density between 1s and 2s orbital is (a) High

(b) Low

(c) Zero

(d) None of these

(c) Three fold degenerate

(d) None of these

412. p-orbitals of an atom in presence of magnetic field are (a) Two fold degenerate

(b) Non degenerate

[Pb. PMT 2002]

413. The energy of an electron of 2 p y orbital is (a) Greater than of 2 p x orbital

[AMU 1984]

(b)

Less than that of 2 p x orbital

(c) Equal to that of 2s orbital(d)

414. Assertion (A): A special line will be seen for a 2 p x − 2 py transition Reason (R): Energy is released in the form of wave of light when the electron drops from 2 p x to 2 p y orbital

[AIIMS 1996]

(a) Both A and R are true statements and R is the correct explanation of A (b) Both A and R are true statements and R is not the correct explanation of A (c) A is true but R is a false statement (d) Both A and R are false statements 415. For n = 3 energy level, the number of possible orbitals (all kinds) are (a) 1

(b) 3

[BHU 1981; CPMT 1985; MP PMT 1995]

(c) 4

(d) 9

416. The correct set of quantum numbers for the unpaired electron of chlorine atom is n

l

m

(a) 2

1

0

(b) 2

1

1

(c) 3

1

1

(d) 3

0

0

[IIT 1989]

417. Assertion (A) : Two electrons in an atom can have the same values of four quantum numbers. Reason (R) : Two electrons in an atom can be present in the same shell, sub-shell and orbital and have the same spin [AIIMS 2001] (a) Both A and R are true and R is a correct explanation of A A

(b) Both A and R are true but R is not a correct explanation of

(c) A is true but R are false

(d) Both A and R are false

(e) A is false but R is true 418. The magnitude of spin angular momentum of an electron is given by (a) S = s(s + 1)

h 2π

(b) S = s

h 2π

(c) S =

3 h × 2 2π

(d) S = ±

1 h × 2 2π

419. If a magnetic field is applied to the electron of a hydrogen atom in the z-direction, the z- component of the spin angular momentum is given by (a) s z = s(s + 1)

(b) s z =

3 h × 2 2π

(c)

sz = m s

h 4π

420. The number of electrons that can be accommodated in dz 2 orbital is (a) 10 (b) 1 (c) 4 421. The quantum number ‘m’ of a free gaseous atom is associated with (a) The effective volume of the orbital (b) The shape of the orbital (c) The spatial orientation of the orbital (d) The energy of the orbital in the absence of a magnetic field 422. When the azimuthal quantum number has a value of l = 1, the shape of the orbital is (a) Unsymmetrical (b) Spherically symmetrical (c) Dumb-bell

(d) s z = ±

1 h × 2 2π

[Kurukshetra CEE 2002]

(d) 2 [AIIMS 2003]

[MP PET 1993]

(d) Complicated


423. For a given value of quantum number l, the number of allowed values of m is given by (a) l + 2 (b) 2l + 2 (c) 2l + 1 (d) l + 1 424. The set of quantum numbers not applicable for an electron in an atom is 1 1 (a) n = 1 , l = 1 , m l = 1 , m s = + (b) n = 1 , l = 0 , m l = 0 , m s = + 2 2 (c) n = 1 , l = 0 , m l = 0 , m s = −

1 2

(d) n = 2 , l = 0 , m l = 0 , m s = +

[MLNR 1994]

1 2

425. Which of the following statements is not correct for an electron that has the quantum numbers n = 4 and m = 2 [MLNR 1993] 1 (a) The electron may have the quantum number s = + (b) The electron may have the quantum number l = 2 2 (c) The electron may have the quantum number l = 3 (d) The electron may have the quantum number l = 0, 1, 2, 3 426. The electrons identified by quantum numbers n and l (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l = 2 (iv) n = 3, l = 1 can be placed [IIT 1999] in order of increasing energy from the lowest to highest, as (a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii) (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii) [Orissa JEE 1997] 427. Which of the following sets of quantum numbers is not allowed (a) n = 1, l = 0, m = 0, s = + 1/2 (b) n = 1, l = 1, m = 0, s = – 1/2 (c) n = 2, l = 1, m = 1, s = + 1/2 (d) n = 2, l = 0, m =0, s = – ½ 428. What are the values of the orbital angular momentum of an electron in the orbitals 1s, 3s, 3d and 2p (a) 0, 0,

6 h, 2 h

(b) 1,1, 4 h, 2 h

(c) 0 ,1, 6 h, 3 h

(d) 0 , 0 , 20 h, 6 h

429. In an excited state, a calcium atom has the electronic configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 4 d . What is the angular momentum of this state. (a)

4h

16 h

(b)

(c)

20 h

(d)

430. The four quantum number for the valence shell electron or last electron of sodium (Z = 11) is (a) n = 2, l = 1, m = – 1, s = – 1/2 (b) n = 3, l = 0, m = 0, s = + 1/2 (c) n = 3, l = 2, m = – 2, s = – 1/2 (d) n = 3, l = 2, m = 2, s = + 1/2 431. For which of the following sets four quantum numbers, an electron will have the highest energy n l m s (a) 3 2 1 + 1/ 2 (b) 4 2 – 1 + 1/2 (c) 4 1 0 – 1/ 2 (d) 5 0 0 – 1/ 2 432. Which of the following sets of quantum numbers represent an impossible arrangement ms n l m

10 h [MP PMT 1999]

[CBSE 1994]

[IIT 1986; MP PET 1995]

(a) 3 2 – 2 1/2 (b) 4 0 0 1/2 (c) 3 2 – 3 1/2 (d) 5 3 0 1/2 433. Which of the following set of quantum numbers is correct for the 19th electron of chromium n l m s (a) 3 0 0 1/2 (b) 3 2 –2 1/2 (c) 4 0 0 1/2 (d) 4 1 –1 1/2 434. When the principal quantum number (n) = 3, the possible values of azimuthal quantum number (l) is

[DCE 2001]

[Bihar MEE 1996; Karnataka CET 2000]

(a) 0, 1, 2, 3

(b) 0, 1, 2

(c) – 2, –1, 0, 1, 2

(d) 1, 2, 3

435. Which one of the following set of quantum numbers is not possible for 4p electron (a) n = 4; l = 1; m = – 1; m s = +

1 2

(b)

n = 4; l = 1; m = 0, m s = +

[EAMCET 1998]

1 2


(c) n = 4; l = 1; m = 2; m s = +

1 2

(d) n = 4; l = 1; m = – 1; m s = +

436. The Pauli exclusion principle is not applicable to (a) Electrons (b) Positrons (c) Photons 437. If m = magnetic quantum number and l= azimuthal quantum number then (a) m = l + 2

(b) m = 2 l 2 + 1

(c) l =

1 2

(d) Protons

m −1 2

(d) l = 2m + 1

438. Which of the following pairs have identical values of magnetic moment (a) Zn 2 + and Cu +

(b) Co 2 + and Ni 2 +

(c)

(d) Mg 2 + and Sc +

Mn 4 + and Co 2 +

439. Which set of quantum numbers for an electron of an atom is not possible

[Rajasthan PMT; DCE 1999]

(a) n = 1, l = 0, m = 0, s = + 1/2

(b) n = 1, l = 1, m = 1, s = +1/2

(c) n = 1, l = 0, m = 0, s = – 1/2

(d) n = 2, l = 1, m = –1, s = + 1/2

440. From the given sets of quantum numbers the one that is inconsistent with the theory is


(a) n = 3; l = 2; m = – 3; s = + 1/2 (b)

n = 4; l = 3; m = 3; s = + 1/2

(c) n = 2; l = 1; m = 0; s = – 1/2

(d) n = 4; l = 3; m = 2; s = + 1/2

441. When the value of azimuthal quantum number is 3, magnetic quantum number can have values (a) + 1, 0, – 1

(b) + 2, +1, 0, – 1, – 2

(c) – 3, – 2, – 1, – 0, + 1, + 2, + 3

(d) + 1, – 1

[Delhi PMT 2001]

442. The four quantum numbers of the outermost orbital of K (atomic no. = 19) are (a) n = 2, l = 0, m = 0, s = (c) n = 3, l = 1, m = 1, s =

+1 2 +1 2

[MP PET 1993, 94]

+1 2

(b)

n = 4, l = 0, m = 0, s =

(d)

n = 4, l = 2, m = – 1, s =

+1 2

443. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 is [CPMT 1971, 89, 91] (a) 2

(b) 4

(c) 6

(d) 8

444. The set of quantum numbers n = 3, l = 0, m = 0 s = – 1/ 2 belongs to the element (a) Mg 445. Be’s

4th

(b) Na

(c) Ne

(d) F

electron will have four quantum numbers

n

l

m

ms

(a) 1

0

0

+ 1/ 2

(b) 1

1

(c) 2

0

0

– 1/ 2

(d) 2

1

0

+ 1/ 2

[MNR 1985]

+ 1 + 1/2

446. The value of the magnetic moment of particular ion is 2.83 Bohr magneton. The ion is (a) Fe 2 + (b) Ni 2 + 447. Which of the following ions are diamagnetic (a) He 2+

(b) Sc 3 +

(c)

Mn 2 +

(d) Co 3 +

(c)

Mg 2 +

(d) O22 −

448. Which of the following sets is possible for quantum numbers (a) n = 4, l = 3, m = – 2, s = 0 (c) n = 4, l = 4, m = – 2, s = +

(b) 1 2


n = 4, l = 4, m = + 2, s = −

1 2

(d) n = 4, l = 3, m = – 2, s = +

1 2

449. If the value of azimuthal quantum number is 3, the possible values of magnetic quantum number would be APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

[MP PMT 1987; Rajasthan PMT 1999; AFMC 2002; Karnataka CET 2002]

(a) 0, 1, 2, 3

(b) 0, – 1, – 2, – 3

(c) 0, ± 1 ± 2, ± 3

(d) ± 1, ± 2, ± 3

450. When the value of the principal quantum number n is 3, the permitted values of the azimuthal quantum number l and the magnetic quantum numbers m, are l m (a) 0 0 1 + 1, 0, – 1 2 + 2,+ 1, 0, – 1, – 2 (b) 1 1 2 + 2, 1, – 2 3 + 3,+ 2, 1, – 2, – 3 (c) 0 0 1 1, 2, 3 2 + 3, + 2, 1, – 2, – 3 (d) 1 0, 1 2 0, 1, 2 3 0, 1, 2, 3 451. Five valence electrons of 15 P are labelled as AB

X

3s

Y

Z

3p

If the spin quantum of B and Z is + 1 /2, the group of electrons with three of the quantum number same are [JIPMER 1997] (a) AB, XYZ, BY (b) AB (c) XYZ, AZ (d) AB, XYZ [IIT Screening 2001] 452. The quantum numbers + 1/ 2 and – 1/ 2 for the electron spin represent (a) Rotation of the electron in clockwise and anticlockwise direction respectively (b) Rotation of the electron in anticlockwise and clockwise direction respectively (c) Magnetic moment of the electron pointing up and down respectively (d) Two quantum mechanical spin states with have no classical analogue 453. Which of the following violates the Pauli exclusion principle (a)

↑↓ ↑↓

(b)

↑↓

↑ ↓ ↑

(c)

↑↑

↑ ↑ ↑

(d)

↑

↑ ↑ ↑

454. Which of the following violates the Aufbau principle 2p

2s

(a)

2s

↑↓ ↑↓

↑↓ ↑↓ ↑↓ ↑

(b)

2s

2p

(c)

2p

2s

↑↑ ↑↓ ↑↓ ↑

(d)

2p

↑ ↑↓ ↑ ↑

455. Which of the following electronic configurations have the highest exchange energy (a) (c)

3d

4s

↑ ↑ ↑

↑

3d

4s

↑↓ ↑↓ ↑↓ ↑↓ ↑

↑

(b) (d)

3d

4s

↑ ↑ ↑ ↑ ↑

↑

3d

4s

↑↓ ↑↓ ↑↓ ↑↓ ↑↓

↑

456. Which of the following set of quantum numbers is permissible (a) n = 3; l = 2; m = 2 and s = +

[AIIMS 2001]

1 (b) 2

n = 3; l = 4; m = 0; and s = −

1 2

1 (d) 2

n = 4; l = 4; m = 3; and s = +

1 2

(c) n = 4; l = 0; m = 2; and s = +

457. Which of the following sets of quantum number is not possible (a) n = 3; l = + 2; m = 0; s = +

1 2

[MP PET 2001]

(b) n = 3; l = 0; m = 0; s = −

1 2


(c) n = 3; l = 0; m = – 1; s = +

1 2

(d) n = 3; l = 1; m = 0; s = −

1 2

458. Which of the following set of quantum numbers belong to highest energy 1 (a) n = 4, l = 0, m = 0, s = + 2

(c) n = 3, l = 1, m = 1, s = +

1 2

[CPMT 1999]

(b)

1 n = 3, l = 0, m = 0, s = + 2

(d)

n = 3, l = 2, m = 1, s = +

1 2

459. Assertion (A) : The cation energy of an electron is largely determined by its principal quantum number Reason (R) : The principal quantum number n is a measure of the most probable distance of finding the electron around the nucleus [AIIMS 1996]

(a) Both A and R are true statements and R is the correct explanation of A (b) Both A and R are true statements and R is not the correct explanation of A (c) A is true but R is a false statement (d) Both A and R are false statements 460. Which of the following set of quantum number is not possible n l m1 m2

[Pb. PMT 2002]

(a) 3 2 1 + 1/2 (b) 3 2 1 – 1/2 (c) 3 2 1 0 (d) 5 2 –1 + 1/2 461. For the energy levels in an atom, which one of the following statements is correct (a) There are seven principal electron energy levels (b) The second principal energy level can have four orbitals and contains a maximum of eight electrons (c) The M energy level can have maximum of 32 electrons (d) The 4s sub-energy level is at a higher energy than the 3d sub-energy level 462. The orbital diagram in which the Aufbau’s principle is violated is 2s

2 px

2 py

(a) ↑↓

↑↓

↑

(b) ↑

↑↓

↑

↑

(c) ↑↓

↑

↑

↑

(d) ↑↓

↑↓

↑↓

↑

[AIIMS 1983]

[IIT 1988; AMU 1999]

2 pz

463. The maximum probability of finding an electron in the d xy orbital is (a) Along the x-axis (c) At an angle of

450

[MP PET 1999]

(b) Along the y-axis from the x and y-axes

(d) At an angle of 900 from the x and y-axes

464. Krypton (36 Kr ) has the electronic configuration (18 Ar ) 4 s 2 3 d 10 4 p 6 . The 37th electron will go into which one of the following sub-levels

[CBSE 1989; CPMT 1989; EAMCET 1991]

(a) 4f

(b) 4d

(c) 3p

(d) 5s

465. Which one is in the ground state

[Delhi PMT 1996]

↑ ↑

(a)

↑

↑

(b)

↑↓ ↑ ↑

↑

↑↓ ↑

(c)

↑↓ ↑

↑

↑

↑

↑ (d)

↑

↑

↑

↑↓


466. Correct statement is

[BHU 2003]

(a) K = 4 s , Cr = 3 d 4 s , Cu = 3 d 1

(c)

4

2

10

4s

(b) K = 4 s , Cr = 3 d 4 s , Cu = 3 d

2

2

K = 4 s 2 , Cr = 3 d 5 4 s1 , Cu = 3 d 10 4 s 2

4

2

10

4s

2

(d) K = 4 s1 , Cr = 3 d 5 4 s1 , Cu = 3 d 10 4 s1

467. The total number of electrons present in all the s-orbitals, all the p-orbitals and all the d-orbitals of cesium ion are respectively [EAMCET 2003]

(a) 8, 26, 10 (b) 10, 24, 20 468. Which of the following has maximum energy 3s

3p

(c) 8, 22, 24

(d) 12, 20, 22 [AIIMS 2002]

3s

3d

(a)

3p

3d

(b) 3s

3p

3s

3d

(c)

3p

3d

(d)

469. Elements upto atomic number 103 have been synthesized and studied. If a newly discovered element is found to have an atomic number 106, its electronic configuration will be [AIIMS 1980] (a) [Rn ] 5 f 14 ,6 d 4 ,7 s 2

(b) [Rn ] 5 f 14 ,6 d 1 ,7 s 2 7 p 3

(c) [Rn ] 5 f 14 ,6 d 6 ,7 s 0

(d) [Rn ] 5 f 14 ,6 d 5 ,7 s1

470. Which element is represented by the following electronic configuration

[MP PMT 1987]

2p 2s 1s

↑↓ ↑↓ ↑

↑↓

↑↓ (a) Nitrogen (b) Oxygen 471. Which of the following statements (s) is (are) correct 5

(c) Fluorine

(d) Neon [IIT 1998]

1

(a) The electronic configuration of Cr is [ Ar ] 3 d 4 s (Atomic no. of Cr = 24) (b) The magnetic quantum number may have a negative value (c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type (Atomic no. of Ag = 47) (d) The oxidation state of nitrogen in HN 3 is – 3 472. The orbital angular momentum of an electron in 2s orbital is 1 h (a) + . 2 2π

(b) Zero

473. Energy of orbit (a) Increases as we move away from nucleus (c) Remains same as we move away from nucleus

[IIT 1996; AIEEE 2003]

(c)

h 2π

(d)

2.

h 2π [Delhi PMT 1984, 91]

(b) Decreases as we move away from nucleus (d) None of these

Miscellaneous Questions

Basic Level 474. When beryllium is bombarded with α-particles, extremely penetrating radiations which cannot be deflected by electrical or magnetic field are given out. These are [CPMT 1983] (a) A beam of protons (b) α-rays (c) A beam of neutrons (d) X -rays 475. When α-particles are sent through a thin metal foil, most of them go straight through the foil because (one or move are correct) [IIT 1984]

(a) Alpha particles are much heavier than electrons (b) Alpha particles are positively charged (c) Most part of the atom is empty space (d) Alpha particles move with high velocity 476. Number of electrons in the outermost orbit of the element of atomic number 15 is (a) 1 (b) 3 (c) 5 (d) 7 477. When β-particles are sent through a tin metal foil, most of them go straight through the foil as APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

[CPMT 1988, 93] [EAMCET 1983]

(a) β-particles are much heavier than electrons

(b) β-particles are positively charged (d) β-particles move with high velocity

(c) Most part of the atom is empty space 478. The atom of the element having atomic number 14 should have (a) One unpaired electron (b) Two unpaired electrons 479. An electronic transition from 1 s orbital of an atom causes (a) Absorption of energy (c) Both release or absorption of energy 480. Which one pair of atoms or ions will have same configuration +

(a) F and Ne

+

(b) Li and He

[AMU 1984]

(c) Three unpaired electrons (d) Four unpaired electrons [JIPMER 1997]

(b) Release of energy (d) Unpredictable [JIPMER 2001]

−

−

(c) Cl and Ar

(d) Na and K

481. Fe (atomic number = 26) atom has the electronic arrangement [NCERT 1974; MNR 1980] (a) 2, 8, 8, 8 (b) 2, 8, 16 (c) 2, 8, 14, 2 (d) 2, 8, 12, 4 482. An element has electronic configuration 2, 8, 18, 1. If its atomic weight is 63, then how many neutrons will be present in its nucleus [MP PAT 1996]

(a) 30 (b) 32 483. Which of the following cannot be formed (a) He

2+

(b) He

(c) 34

(d) 33 [AFMC 1997]

+

(c) He 2

2

6

2

6

10

(d) He 2 2

5

484. An atom has the electronic configuration of 1 s , 2 s , 2 p ,3 s ,3 p ,3 d ,4 s ,4 p . Its atomic weight is the number of neutrons in its nucleus shall be (a) 35 and 45 (b) 45 and 35 (c) 40 and 40 (d) 485. Which of the following has maximum number of unpaired electron (atomic number of Fe 26) (a) Fe (b) Fe (II) (c) Fe (III) (d) 486. The following has zero valency (a) Sodium (b) Beryllium (c) Aluminum (d) 487. What is the electronic configuration of Cu 2 + ( Z = 29 ) of least position (a) [ Ar ] 4 s1 3 d 8

(b) [ Ar ] 4 s 2 3 d 10 4 p 1

80. Its atomic number and [MP PMT 1987]

30 and 50 [MP PMT 2001]

Fe (IV) [DPMT 1991]

Krypton

[MP PET /PMT 1998; MP PET 2001]

(c) [ Ar ] 4 s1 3 d 10

(d) [ Ar ] 3 d 9

488. Fertile nuclides are [CPMT 2000] (a) Isotopes (b) Fissionable (c) Not fissionable (d) None of these 489. The valence electron in the carbon atom are [MLNR 1982] (a) 0 (b) 2 (c) 4 (d) 6 490. The atomic number of an element is 35 and mass number is 81. The number of electrons in the outer most shell is [UPSEAT 2001] (a) 7 (b) 6 (c) 5 (d) 3 491. The atomic weight of an element is double its atomic number. If there are four electrons in 2p orbital, the element is [AMU 1983] (a) C (b) N (c) O (d) Ca

Advance Level

492. If electron, hydrogen, helium and neon nuclei are all moving with the velocity of light, then the wavelengths associated with these particles are in the order [MP PET 1993] (a) Electron > hydrogen > helium > neon (b) Electron > helium > hydrogen > neon (c) Electron < hydrogen < helium < neon (d) Neon < hydrogen < helium < electron 493. When atoms are bombarded with alpha particles, only a few in million suffer deflection, others pass out undeflected. This is because [MNR 1979; NCERT 1980; AFMC 1995]

(a) (b) (c) (d)

The force of repulsion on the moving alpha particle is small The force of attraction on the alpha particle to the oppositely charged electrons is very small There is only one nucleus and large number of electrons The nucleus occupies much smaller volume compared to the volume of the atom

494. The total number of valence electrons in 4.2 gm of N 3− ion is ( N A is the Avogadro’s number) APEX INSTITUTE FOR IIT-JEE /AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 www.apexiit.co.in/

[CBSE 1994]

(a) 1 . 6 N A

(b) 3 . 2 N A

(c) 2 .1 N A

(d) 4 . 2 N A

495. In neutral atom, which particles are equivalent +

(a) p , e

+

−

(b) e , e


+

−

(c) e , p

+

+

(d) p , n 0

496. An element have atomic weight 40 and it’s electronic configuration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 . Then its atomic number and number of neutrons will be [Rajasthan PMT 2002] (a) 18 and 22 (b) 22 and 18 (c) 26 and 20 (d) 40 and 18 497. Which phrase would be incorrect to use [AMU (Engg.) 1997] (a) A molecule of a compound (b)A molecule of an element (c)An atom of an element (d) None of these 498. Splitting of signals is caused by [Pb. PMT 2000] (a) Proton (b) Neutron (c) Positron (d) Electron 499. Choose the correct statement (a) A node is a point in space where the wave function (Ψ ) has zero amplitude. (b) The number of peaks in radial distribution is n – l (c) Radial probability density p n,l (r) = 4 πr 2 Rn2,l (r) . (d) Ψ 2 represents the atomic orbital (e) All the above 500. Which of the following electronic configurations have zero spin multiplicity (a)

(b)

↑ ↑ ↑

(c)

↑ ↑ ↓

↑ ↓ ↓

(d)

↓ ↓ ↓ Answer Sheet

Basic and Advance Level

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

b

a

b

d

b

d

c

b

d

b

d

a

c

a

c

d

a

c

b

b

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

d

a

c

c

b

a

a

c

b

d

c

a

c

b

d

d

b

c

d

c

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

c

a

c

d

d

c

c

c

b

a

b

c

a

b

b

c

d

a

c

c

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

a

a

b

b

a

c

c

b

a

c

a

b

d

a

b

c

c

a

b

c

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

d

c

c

d

d

a

c

a

b

b

b

c

d

c

a

b

a

a

d

d

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

b

a

d

a

a

d

a

c

a

c

b

c

c

b

c

c

d

d

a

a

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

b

c

c

b

a

c

d

c

d

c

c

c

b

c

a

c

a

a

d

c

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

d

b

c

b

d

b

d

c

b

d

d

d

d

c

c

a

b

c

b

c

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

b

a

b

a

d

d

a

194

195

196

197

198

199

20 0

b

a

c

a

b

a

b

b

c

b

b

a

a,b,c,d

181

182

183

184

185

186

187

188

189

190

191

192

193


a

b,c

b

a,b,c

a,b

d

d

b

d

a

a

d

c

a

c

c

a

c

c

b

201

202

203

204

205

206

207

20 8

209

210

211

212

213

214

215

216

217

218

219

220

a

a

b

a

b

b

a

b

c

b

d

c

d

c

a

d

c

c

d

a

221

222

223

224

225

226

227

228

229

230

231

232

233

234

235

236

237

238

239

240

a

d

a

c

d

b

a

b

c

d

b

c

a

d

b

a

c

c

a

b

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

256

257

258

259

260

d

a

a

c

b

b

b

b

a

c

b

d

d

b

a

a

b

c

b

c

261

262

263

264

265

266

267

268

269

270

271

272

273

274

275

276

277

278

279

28 0

a

b

c

b

a

a

c

c

c

a

d

d

d

c

c

a

b

c

b

a

281

282

283

284

285

286

287

288

289

290

291

292

293

294

295

296

297

298

299

30 0

c

a

c

b

b

a

c

b

d

a

b

d

b

b

a

c

c

d

c

d

301

302

303

304

305

306

307

30 8

309

310

311

312

313

314

315

316

317

318

319

320

a

d

a

a

b

b

d

d

b

d

d

c

b

c

b

d

a

d

a

c

321

322

323

324

325

326

327

328

329

330

331

332

333

334

335

336

337

338

339

340

a

d

d

b

c

d

b

d

c

d

b

c

d

b

a,b,c

b

b

a

c

a

341

342

343

344

345

346

347

348

349

350

351

352

353

354

355

356

357

358

359

360

b

c

c

c

a

a

b

a

c

a

b

c

e

c

b

a,d

b

c

a

a

361

362

363

364

365

366

367

368

369

370

371

372

373

374

375

376

377

378

379

38 0

c

b

a

a

a

c

c

c

b

b

b

d

d

a

a

c

a

b

a

b

381

382

383

384

385

386

387

388

389

390

391

392

393

394

395

396

397

398

399

40 0

b

b

c

b

d

d

d

c

c

a

c

b

e

d

d

a

c

c

b

a

401

402

403

404

405

406

407

40 8

40 9

410

411

412

413

414

415

416

417

418

419

420

b

c

c

b

b

a

d

c

a

d

c

c

d

d

d

c

d

a,c

d

d

421

422

423

424

425

426

427

428

429

430

431

432

433

434

435

436

437

438

439

440

c

c

c

a

d

a

b

a

c

b

b

c

c

b

c

c

c

a,c

b

a

441

442

443

444

445

446

447

448

449

450

451

452

453

454

455

456

457

458

459

460

c

b

c

a

c

b

b,c,d

d

c

a

b

a,d

c

d

d

a

c

d

a

c

461

462

463

464

465

466

467

468

469

470

471

472

473

474

475

476

477

478

479

48 0

b

a,b

c

d

b

d

b

b

d

c

a,b,c

b

a

d

c

c

c

b

a

c

481

482

483

484

485

486

487

48 8

489

490

491

492

493

494

495

496

497

498

499

50 0

c

c

d

c

d

d

c

c

a

a

a

d

a

c

a

b

a

e

c

a


ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

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