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Page 1: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

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

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)

Answer Sheet of Assignment

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.

Erwin Schrödinger

Chapter 2

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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.

EElleeccttrroonn ((––11eeoo))

(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

( 210 − to )10 3 Hgmm− . Blue rays were emerged from the cathode. These rays were termed as 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 910 − to 1110 −

cm/sec)

(v) Cathode rays can cause fluorescence.

Gas at low pressure

Cathode rays

Cathode Anode

TC Vaccum pump

High voltage + –

Discharge tube experiment for production of cathode rays

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(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 −XCu, 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 81076.1( ×− 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.

q The 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 .10602.1 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 gm28101.9 −× 1837/1000549.0 == amu of the mass of hydrogen atom.

(7) According to Einstein’s theory of relativity, mass of electron in motion is, m ′ ])/(1[

)electron(mof mass Rest2cu−

=

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 = .10483.5 4−×

(9) 1.1 2710× electrons =1gram.

(10) 1 mole electron = 5483.0 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 .1028.4 12 cm−×

(15) The density of the electron is = mLg /1017.2 17−× .

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Example : 1 The momentum of electron moving with 1/3rd velocity of light is (in g cm sec–1)

(a) 81069.9 −× (b) 101001.8 × (c) 1810652.9 −× (d) None

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

Where m′ is mass of electron in motion( )2/1 cu

m

−= ; Also 3/cu =

∴ Momentum 118102810

2

28

sec10652.9394.0

10310108.9

3

103

31

10108.9 −−−−

×=×

×××=

××

×−

×= cmg

c

c

Example: 2 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) 8109.2 × (b)

81001.8 × (c) 810652.9 × (d) None

Solution: (a) Mass of electron in motion amu931

2= (1 amu = 931 MeV)

kg271066.1931

2 −××= kg301056.3 −×= (1 kgamu 271066.1 −×= )

Let the speed of the electron be u.

( )2

/1 cu

mm

−=′ or

2

8

30

2

8

27

30

1031

10911.0

1031

1066.1931

511.0

1056.3

×−

×=

×−

××=×

−−

uu

or 06548.0103

12

8=

×−

u or 93452.0109 162 ××=u or mu 8109.2 ×=

Example: 3 A electron of rest mass 271067.1 −× kg is moving with a velocity of 0.9c (c = velocity of light). Find its

mass and momentum.

(a) 191034.10 −× (b)

101001.8 × (c) 1810652.9 −× (d) None

Solution: (a) Mass of a moving object can be calculated using Einsten’s theory of relativity :

( )2

/1 cu

mm

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

kg

c

c

m 27

2

27

1083.3

9.01

1067.1 −−

×=

×=′

Momentum ump ×′=''

11927 1034.109.01083.3 −−− ×=××= mskgcp

PPrroottoonn ((11HH11,, HH++,, PP))

Examples based on Einstein’s theory of relativity

+

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(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.

(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 = 1910602.1 −× coulombs = ...1080.4 10 use−×

(5) Mass of proton = Mass of hydrogen atom= amu00728.1 gram2410673.1 −×= 1837= of the mass of

electron.

(6) Molar mass of proton = mass of proton × Avogadro number 008.1= (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 = 3

3

4rπ ) is ≈ 1.5 .10 338 cm−×

(12) Specific charge of a proton is 41058.9 × Coulomb/gram.

NNeeuuttrroonn ((oonn11,, NN))

Perforated cathode

Cathode rays TC Vaccum pump

High voltage + –

Perforated tube experiment for production of anode rays

Anode rays

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

(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 = gram2410675.1 −× = kg2710675.1 −× = ≈amu00899.1 mass of hydrogen atom.

(5) Specific charge of a neutron is zero.

(6) Density = ../105.1 14 ccgram−×

(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 :

nutrino anti

00

electron

01

proton

11

neutron

10 ν++→ − eHn

(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 Electron(e–) Proton(p+) Neutron(n)

Mass (m)

amu

kg

Relative

0.000546

9.109 × 10–31

1/1837

1.00728

1.673 × 10–27

1

1.00899

1.675 × 10–24

1

Charge(e)

Coulomb (C)

esu

Relative

– 1.602 × 10–19

– 4.8 × 10–10

– 1

+1.602 × 10–19

+4.8 × 10–10

+1

Zero

Zero

Zero

Specific charge (e/m) C/g 1.76 × 108 9.58 × 104 Zero

• The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 126 C , i.e. kg2710660.1 −× .

Other non fundamental particles

Particle Symbol Nature Charge esu

×10–10

Mass

(amu)

Discovered by

Positron ++ β,1, 0ee + + 4.8029 0.0005486

Anderson (1932)

Neutrino ν 0 0 < 0.00002

Pauli (1933) and Fermi (1934)

Anti-proton −p – – 4.8029 1.00787 Chamberlain Sugri (1956) and

Weighland (1955)

Positive mu meson +µ + + 4.8029 0.1152 Yukawa (1935)

Negative mu meson −µ – – 4.8029 0.1152 Anderson (1937)

Positive pi meson +π + + 4.8029 0.1514

Powell (1947) Negative pi meson −π – – 4.8029 0.1514

Neutral pi meson 0π 0 0 0.1454

2.2 Atomic number, Mass number and Atomic species.

1126

42

94 nCHeBe o+→+ or 114

74

211

5 nNHeB o+→+

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(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,

where, −= Xν rays frequency

Z= atomic number of the metal

ba & are constant.

(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,

Note : q 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 Similarities Differences Examples

Isotopes

(Soddy)

(i) Atomic No. (Z)

(ii) No. of protons

(iii) No. of electrons

(iv) Electronic

configuration

(v) Chemical properties

(vi) Position in the periodic

table

(i) Mass No. (A)

(ii) No. of neutrons

(iii) Physical properties

(i) HHH 31

21

11 ,,

(ii) OOO 188

178

168 ,,

(iii) ClCl 3717

3517 ,

1−sν

Z

)( bZa −=ν or abaZ −

A

Z

X

Mass number

Atomic number

Element

e.g. 147

168

199 ,, NOF etc.

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Isobars

(i) Mass No. (A)

(ii) No. of nucleons

(i) Atomic No. (Z)

(ii) No. of protons, electrons

and neutrons

(iii)Electronic configuration

(iv) Chemical properties

(v) Position in the perodic

table.

(i) CaKAr 4020

4019

4018 ,,

(ii) BaXeTe 13056

13054

13052 ,,

Isotones

No. of neutrons (i) Atomic No.

(ii) Mass No., protons and

electrons.

(iii) Electronic

configuration

(iv) Physical and chemical

properties

(v) Position in the periodic

table.

(i) SPSi 32

16

31

15

30

14 ,,

(ii) CaK 4020

3919 ,

(iii) HeH 42

31 ,

(iv) NC 147

136 ,

Isodiaphers

Isotopic No.

(N – Z) or (A – 2Z)

(i) At No., mass No.,

electrons, protons,

neutrons.

(ii) Physical and chemical

properties.

(i) 23190

23592 , ThU

(ii) 199

3919 , FK

(iii) 5524

6529 , CrCu

Isoelectronic

species

(i) No. of electrons

(ii) Electronic

configuration

At. No., mass No. (i) )22(,, 22−− eCNOCOON

(ii) )14(,, 2−− eNCNCO

(iii) )2(,,, 2 −++− eBeLiHeH

(iv)

)18(,,,, 223 −++−−− eCaandKArClSP

Isosters

(i) No. of atoms

(ii) No. of electrons

(iii) Same physical and

chemical properties.

(i) 2N and CO

(ii) 2CO and ON 2

(iii) HCl and 2F

(iv) CaO and MgS

(v) 66 HC and 633 HNB

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

q Average atomic weight/ The average isotopic weight

100

isotope of 2nd mass relative isotope of 2nd % isotope 1stof mass relative isotope 1stof % ×+×=

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Example : 4 The characteristics X- ray wavelength for the lines of the αk series in elements X and Y are 9.87Å and

2.29Å respectively. If Moseley’s equation )75.0(109.4 7 −×= Zν is followed, the atomic numbers of X and

Y are

(a) 12, 24 (b) 10, 12 (c) 6, 12 (d) 8, 10

Solution : (a) λ

νc

=

8

10

8

105132.51087.9

103×=

×

×=

−xν

8

10

8

104457.111029.2

103×=

×

×=

−yν

using Moseley’s equation we get

)75.0(109.4105132.5 78 −×=×∴ xZ …..(i)

and )75.0(1090.4104457.11 78 −×=× yZ ….. (ii)

On solving equation (i) and (ii) .24,12 == yx ZZ

Example : 5 If the straight line is at an angle 45° with intercept, 1 on axis,−ν calculate frequency ν when atomic

number Z is 50.

(a) 20001−s (b) 2010

1−s (c) 24011−s (d) None

Solution : (c) a==°= 145tanν

ab=1

49150 =−=∴ ν

.2401 1−= sν

Example : 6 What is atomic number Z when 12500 −= sν ?

(a) 50 (b) 40 (c) 51 (d) 53

Solution : (c) .51,12500 =−== ZZν

Example : 7 Atomic weight of Ne is 20.2. Ne is a mixutre of 20Ne and 22Ne . Relative abundance of heavier isotope

is

(a) 90 (b) 20 (c) 40 (d) 10

Solution:(d) Average atomic weight/ The average isotopic weight

100

isotope of 2nd mass relative isotope of 2nd % isotope 1stof mass relative isotope 1stof % ×+×=

Examples based on Moseley equation

1−sν

Z

θ

a= θtan ab=intercept

Examples based on Atomic number, Mass number and Atomic species

ZXA

Page 10: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

100

22)100(202.20

×−+×=∴

aa; 90=∴ a ; per cent of heavier isotope 1090100 =−=

Example : 8 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 (b) 85.5 (c) 87.5 (d) 86.0

Solution:(b) Average atomic weight/ The average isotopic weight

100

isotope of 2nd mass relative isotope of 2nd % isotope 1stof mass relative isotope 1stof % ×+×=

5.85100

25877585=

×+×=

Example : 9 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 −3NO

So, in it number of electrons

×= 1 number of electrons of nitrogen ×+3 number of electrons of oxygen +1 (due to negative charge)

3218371 =+×+×=

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 Pa 3115)( Arb 40

18)( Agc 10847)( ?

Solution : 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 Protons Electrons Neutrons

P3115 15 15 31 – 15=16

Ar4018 18 18 40 – 18=22

Ag10847 47 47 108 – 47=61

Example :12 State the number of protons, neutrons and electrons in 12C and .14C

Solution : The atomic number of 12C is 6. So in it number of electrons = 6

Number of protons =6; Number of neutrons =12 – 6=6

The atomic number of 14C is 6. So in it number of electrons = 6

Number of protons = 6; Number of neutrons =14 – 6=8

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Wavelength Crest

Energy

Trough

Vibrating source

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. .2412 Mg 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. 2612 Mg

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 )

or nanometre (nm).

mcmÅ 108 10101 −− ==

mcm 64 10101 −− ==µ

mcmnm 97 10101 −− ==

nmÅcm 748 1010101 === µ

(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

λν

c=

(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., .sec/103 10 cm×

sec/103 10 cmc ×== λν

Thus, a wave of higher frequency has a shorter wavelength while a wave of lower frequency has a

longer wavelength.

Page 12: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

(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 11 or −− mcm .

λν

1=

(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 714 103103 ×−× 95 101101 ×−× Alternating current of high

frequency

Microwave 67 106103 ×−× 119 105101 ×−× Klystron tube

Infrared (IR) 7600106 6 −× 1611 1095.3105 ×−× Incandescent objects

Visible 38007600 − 1416 109.71095.3 ×−× Electric bulbs, sun rays

Ultraviolet (UV) 1503800 − 1614 102107.9 ×−× Sun rays, arc lamps with

mercury vapours

X-Rays 1.0150 − 1916 103102 ×−× Cathode rays striking metal

plate

−γ Rays 01.01.0 − 2019 103103 ×−× Secondary effect of radioactive

decay

Cosmic Rays 0.01- zero −× 20103 infinity Outer space

2.4 Atomic spectrum - Hydrogen spectrum.

AAttoommiicc ssppeeccttrruumm

(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

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Positively charged sphere

Electron –

– –

+ +

+ +

+ +

+

Positive charge spreaded throughout the sphere

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

absorbed.

HHyyddrrooggeenn ssppeeccttrruumm

(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,

−===

2

2

2

1

111

nnR

c

ν

λν

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

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

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.

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 .

2sin

1

4

θ

b

θ

r0 Nucleus

α-particle (energy E eV)

Scattering of α -particle

Page 14: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

(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.

(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 .105.1 13 cm−× to .105.6 13 cm−× i.e. 5.1 to 5.6 Fermi.

Generally the radius of the nucleus ( )nr is given by the following relation,

This exhibited that nucleus is 510 − times small in size as compared to the total size of atom.

(vi) The Volume of the nucleus is about 3910 − 3cm and that of atom is ,10 324 cm− i.e., volume of the

nucleus is 1510 − times that of an atom.

(vii) The density of the nucleus is of the order of 31510 −cmg or 810 tonnes 3−cm or cckg /10 12 . If

nucleus is spherical than,

(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.

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

Nucleus +

10–15 m

10–10 m

Planetry electron

3/113 )104.1( Acmrr on ××== −

Density = = nucleus of the volume

nucleus of the mass

323

3

410023.6

number mass

rπ××

Unstability of atom

e–

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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,

3/113 )104.1( Ar −×= 3/113 19104.1 ××= − cm131073.3 −×= (A for F=19)

Volume of a fluorine atom 3

3

4rπ= 313 )1073.3(14.3

3

4 −×××= 3371018.2 cm−×=

number sAvogadro'

nucleusof mol oneof Mass nucleus singleof Mass = g

2310023.6

19

×=

Thus nucleus singleof Volume

nucleus singleof Mass nucleusof Density =

3723 1018.2

1

10023.6

10−×

××

= 11310616.7 −== cmg

Example: 15 Atomic radius is the order of ,10 8 cm− and nuclear radius is the order of .10 13 cm− Calculate what

fraction of atom is occupied by nucleus.

Solution : Volume of nucleus 3)3/4( pr= 3313 )10()3/4( cmp −×=

Volume of atom 3383 )10()3/4()3/4( cmppr −×==

∴ 15

24

39

1010

10 −

==atom

nucleus

V

V or atomnucleus VV ×= −1510

2.7 Planck's Quantum theory and Photoelectric effect.

PPllaanncckk''ss QQuuaannttuumm tthheeoorryy

(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.

ν∝E or λ

νhc

hE ==

where, =h Planck's constant = 6.62×10–27 erg. sec. or .sec1062.6 34 Joules−×

(iii) The total amount of energy emitted or absorbed by a body will be some whole number quanta.

Hence ,νnhE = where n is an integer.

(iv) The greater the frequency (i.e. shorter the wavelength) the greater is the energy of the radiation.

thus, 1

2

2

1

2

1

λ

λ

ν

ν==

E

E

Examples based on Properties of the nucleus

+

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(v) Also ,21 EEE += hence, 21 λλλ

hchchc+= or

21

111

λλλ+= .

Example: 16 Suppose J1710 − of energy is needed by the interior of human eye to see an object. How many photons of

green light )550( nm=λ 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

1710 −=λ

hcn ; 286.27

10310626.6

105501010834

91717

==×××

××=

×=

−−−

hcn

λ photons

Example: 17 Assuming that a 25 watt bulb emits monochromatic yellow light of wave length .57.0 µ The rate of

emission of quanta per sec. will be

(a) 113 sec1089.5 −× (b) 117 sec1028.7 −× (c) 110 sec105 −× (d) 119 sec1018.7 −×

Solution: (d) Let n quanta are evolved per sec.

1sec25 −=

J

hcn

λ; 25

1057.0

10310626.66

834

×××−

n ; 119 sec1018.7 −×=n

PPhhoottooeelleeccttrriicc eeffffeecctt

(1) When radiations with certain minimum frequency )( 0ν 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 )( 0ν 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

−=−=

0

0

2

max

11

2

1

λλνν hchhmv

where, 0ν and 0λ 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.

Examples based on Planck's Quantum theory

λ

hcE =

Visible light

Visible light

U.V. light Metal

Metal other than alkali

metals

Alkali metals

Photo electrons

No photo electrons

Photo electrons

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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

electron is an integral multiple of .2/ πh Thus, π2

hmvr n=

Where, m = mass of the electron, r = radius of the electronic orbit, v = velocity of the electron in its

orbit.

(v) The angular momentum can be .2

,......2

3,

2

2,

2 ππππ

nhhhh This principal is known as quantization of

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.

(vi) The emission or absorption of radiation by the atom takes place when an electron jumps from one

stationary orbit to another.

(vii) The radiation is emitted or absorbed as a single quantum (photon) whose energy νh is equal to

the difference in energy E∆ of the electron in the two orbits involved. Thus, Eh ∆=ν

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, ++ 2, LiHe etc.

E1

E2 E1 – E2 = hν

Emission

E1

E2 E1 – E2 = hν

Absorption

Page 18: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

(ii) Calculation of radius of Bohr’s orbit : According to Bohr, radius of orbit in which electron moves

is

where, n =Orbit number, m =Mass number [ ],101.9 31 kg−× e =Charge on the electron [ ]19106.1 −×

Z =Atomic number of element, k = Coulombic constant [ ]229109 −× cNm

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

nmZ

nrorÅ

Z

nr nn 529.0529.0

22

×=×=

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

2nr ∝ [Z = constant]

Thus we have 22

21

2

1

n

n

r

r= i.e., .......9:4:1::...........:: 321 rrr 321 rrr <<

(b) For particular orbit of different species

Z

r1

∝ [Z =constant] Considering A and B species, we have A

B

B

A

Z

Z

r

r=

Thus, radius of the first orbit H, 2, ++ LiHe and 3+Be follows the order: 32 +++ >>> BeLiHeH

(iii) Calculation of velocity of electron

2/122

,2

==

mr

ZeV

nh

ZKeV nn

π

For H atom, 18

sec.10188.2 −×

= cmn

Vn

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

n

1V ∝ [Z = constant] Thus, we have,

1

2

2

1

n

n

V

V=

The order of velocity is .........321 VVV >> or ........3

1:

2

1:1::..........:: 321 VVV

(b) For a particular orbit of different species

ZV ∝ [n =constant] Thus, we have 2++ << LiHeH

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

;1:2: 21 =VV ;1:3: 31 =VV 1:4: 41 =VV

(iv) Calculation of energy of electron in Bohr’s orbit

Total energy of electron = K.E. + P.E. of electron r

kZe

r

kZe

r

kZe

22

222

−=−=

Substituting of r, gives us 22

24222

hn

kemZE

π−= Where, n=1, 2, 3………. ∞

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

Z

n

kme

hr

2

22

2

.4

=

π

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atomperergn

ZE

2

212108.21 ××= − )101108.21 7

2

219 ergJ(atomperJ

n

Z=××−= −

molekcaln

ZJatompereV

n

ZE - /.6.313)101.6(1eV6.13

2

219

2

2

×−=×=×−= (1 cal = 4.18J)

or 12

2

1312 −−kJmolZ

n

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

2

1

nE −∝ [Z =constant] Thus, we have

21

22

2

1

n

n

E

E=

The energy increase as the value of n increases

(b) For a particular orbit of different species

2ZE −∝ [n =constant] Thus, we have 2

2

B

A

B

A

Z

Z

E

E=

For the system H, He+ , Li+2, Be+3 (n-same) the energy order is 32 +++ >>> BeLiHeH

The energy decreases as the value of atomic number Z increases.

When an electron jumps from an outer orbit (higher energy) 2n to an inner orbit (lower

energy) ,1n then the energy emitted in form of radiation is given by

atomeVnn

ZEnnh

ZmekEEE nn /

116.13

1122

2

2

1

2

2

2

2

1

2

2422

12

−=∆⇒

−=−=∆

π

As we know that ,νhE = νλ=c and λ

ν1

= ,hc

E∆=

−=

22

21

3

2422 112

nnch

Zmekπ

This can be represented as

−==

2

2

2

1

2 111

nnRZν

λwhere, R

ch

mekR

3

4222π= is known as Rydberg

constant. Its value to be used is .109678 1−cm

(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.

.6.131 eVE −=∴

(ii) In excited state : Energy levels greater than 1n are excited state. i.e. for H- atom 432 ,, nnn are

excited state. For H- atom first excitation state is 2n=

(iii) Excitation potential : Energy required to excite electron from ground state to any excited state.

Ground state → Excited state

Ist excitation potential = =+−=− 6.134.312 EE 10.2 eV.

IInd excitation potential = .1.126.135.113 eVEE =+−=−

(iv) Ionisation energy : The minimum energy required to relieve the electron from the binding of nucleus.

.6.132

2

eff.ionisation eV

n

ZEEE n +=−= ∞

(v) Ionisation potential : e

EV ionisation= ionisation

Page 20: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

(vi) Separation energy : Energy required to excite an electron from excited state to infinity.

S.E. = .excitedEE −∞

(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' = .).(

6.13

EB.

(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,

where, R is = =3

422

ch

meπ Rydberg's constant

It's theoritical value = 109,737 cm–1 and It's experimental value = 1581.677,109 −cm

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.

(iv) Comparative study of important spectral series of Hydrogen

Lyman series

Pfund

series

Brackett

series

Paschen

series

Balmer

series

En

ergy

lev

el

n=7 n=6

n=5

n=4

n=3

n=2

n=1

n=8

Humphrey series

−===

2

2

2

1

111

nnR

c

ν

λν

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S.No. Spectral

series

Lies in

the region

Transitio

n

12 nn >

Rnn

nn

)( 21

22

22

21

max−

=λ R

n 21

min =λ 21

22

22

min

max

nn

n

−=

λ

λ

(1) Lymen

series

Ultraviolet

region

11 =n

∞= ....4,3,22n

2 and 1 21 == nn

R3

4max =λ

∞== 21 and 1 nn

R

1min =λ 3

4

(2) Balmer

series

Visible

region

21 =n

∞= ....5,4,32n

3 and 2 21 == nn

R5

36max =λ

∞== 21 and 2 nn

R

4min =λ

5

9

(3) Paschen

series

Infra red

region

n1 = 3

∞= ....6,5,42n

4 and 3 21 == nn

R7

144max =λ

∞== 21 and 3 nn

R

9min =λ 7

16

(4) Brackett

series

Infra red

region

41 =n

∞= ....7,6,52n

5 and 4 21 == nn

R9

2516max

×=λ

∞== 21 and 4 nn

R

16min =λ 9

25

(5) Pfund series Infra red

region

51 =n

∞= ....8,7,62n

6 and 5 21 == nn

R11

3625max

×=λ

∞== 21 and 5 nn

R

25min =λ 11

36

(6) Humphrey

series

Far

infrared

region

61 =n

∞= ....8,72n

7 and 6 21 == nn

R13

4936max

×=λ

∞== 21 and 6 nn

R

36min =λ 13

49

(v) If an electron from nth excited state comes to various energy states, the maximum spectral lines

obtained will be = .2

)1( −nn n= principal quantum number.

as n=6 than total number of spectral lines = .152

30

2

)16(6==

(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.

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Example: 18 If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be

(a) 29

4r (b) 24r (c) 3

4

9r (d) 29r

Solution : (c) 22

22

4 mZe

hnr

π= ∴

2

2

3

2

3

2=

r

r ∴ 23

4

9rr =

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 = 32 rπ

According to Bohr angular momentum of electron in 3rd orbit is

mvr3 = π2

3h

or 3

2 3r

mv

h π=

by De-Broglie equation, mv

h=λ

3

2 3rπλ =∴ λπ 32 3 =∴ r

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.

Example: 20 The degeneracy of the level of hydrogen atom that has energy 16

11R− is

(a) 16 (b) 4 (c) 2 (d) 1

Solution : (a) 2n

RE H

n −= ∴ 162

HH R

n

R−=−

i.e. for th4 sub-shell

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 18 −× ms Its velocity in the second orbit would be

(a) 181009.1 −× ms (b) 181038.4 −× ms (c) 15105.5 −× ms (d) 181076.8 −× ms

Solution : (a) We know that velocity of electron in nth Bohr's orbit is given by

smn

Zv /1018.2 6×=

for 1, =ZH

Q smv /1

1018.2 6

1

×=

Q smsmv /1009.1/2

1018.2 66

2 ×=×

=

Example: 22 The ionization energy of the ground state hydrogen atom is .1018.2 18 J−× The energy of an electron in its second orbit would be

+1 0 +1

three p

n=4

1=0

m=0

one s

1

five d

+2 –1 0 +1 +2

2

–3 –2 –1 0 +1 +2 +3 seven f

3

Examples based on Bohr’s atomic model and Hydrogen spectrum

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(a) J181009.1 −×− (b) J181018.2 −×− (c) J181036.4 −×− (d) J191045.5 −×−

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

Jn

E2

181018.2 −×−= Q( ionization energy of H = J181018.2 −× )

JJE 19

2

18

2 1045.52

1018.2 −−

×−=×−

=

Example: 23 The wave number of first line of Balmer series of hydrogen atom is 15200 1−cm . What is the wave number

of first line of Balmer series of +3Li ion.

(a) 115200 −cm (b) 6080 1−cm (c) 176000 −cm (d) 136800,1 −cm

Solution : (d) For vvLi =+3 for 2zH × =15200 ×9= 1,36800 1−cm

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 = ÅZ

n120.2

1

)2(530.0530.0

22

=×=×

Example: 25 How many chlorine atoms can you ionize in the process ,−+ +→ eClCl by the energy liberated from the

following process :

−− →+ CleCl for 23106 × atoms

Given electron affinity of ,61.3 eVCl = and IP of eVCl 422.17=

(a) 1.24 2310× atoms (b) 201082.9 × atoms (c) 151002.2 × atoms (d) None of these

Solution : (a) Energy released in conversion of 23106 × atoms of −Cl ions = 23106 × × electron affinity

= 6× 2310 2410166.261.3 ×=× eV.

Let x Cl atoms are converted to +Cl ion

Energy absorbed ×= x ionization energy

2410166.2422.17 ×=×x ; 2310243.1 ×=x 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

Solution : (c) Ionization energy of He 6.132

2

×=n

Z 6.13

1

22

2

×= eV4.54=

Energy required to remove both the electrons

= binding energy + ionization energy

= 4.546.24 + = 79eV

Example: 27 The wave number of the shortest wavelength transition in Balmer series of atomic hydrogen will be

(a) 4215 Å (b) 1437Å (c) 3942Å (d) 3647Å

Solution : (d)

−=

22

21

2

shortest

111

nnRZ

λ

∞−××=

22

2 1

2

11109678

cm510647.3 −×=λ Å3647=

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

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Solution : (b) According to Bohr's model for hydrogen and hydrogen like atoms the velocity of an electron in an atom is

quantised and is given by nh

Zev

22π∝ so

nv

1∝ in this cass 3=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) 3=n to 8=n (c) 2=n to 1=n (d) 8=n to 3=n

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) Hz1410268.1 × and Hz1610864.2 × (b) Hz1010214.3 × and Hz1210124.1 ×

(c) Hz1210806.1 × and Hz1510204.6 × (d) Hz1410568.4 × and Hz1510924.2 ×

Solution : (d) If an electron is in 3rd orbit, two spectral lines are possible

(a) When it falls from 3rd orbit to 2nd orbit.

In equation

−×=

22

21

15 1110289.3

nnν

289.31 =ν36

510289.3

3

1

2

110 15

22

15 ××=

−× = Hz1414568.4 ×

(b) When it falls from 3rd orbit to 1st orbit :

Hz1515

2

152 10924.2

9

810289.3

3

1

1

110289.3 ×=××=

−××=ν

Example: 31 If the first ionisation energy of hydrogen is J1810179.2 −× per atom, the second ionisation energy of helium per atom is

(a) J1810716.8 −× (b) kJ5250.5 (c) J1810616.7 −× (d) J1310016.8 −×

Solution : (a) For Bohrs systems : energy of the electron 2

2

n

Z∝

Ionisation energy is the difference of energies of an electron ),( ∞E when taken to infinite distance and rE

when present in any Bohr orbit and αE is taken as zero so ionisation energy becomes equal to the energy

of electron in any Bohr orbit.

2

2

H

HH

n

ZE ∝ ;

2

2

He

HeHe

n

ZE ∝ or

22

1

×=

He

H

E

E [as ]1,1,2,1 ==== HeHHeH nnZZ

or 1818 10716.8410179.24 −− ×=××=×= HHe EE 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 26.13 ZeVHe ×=+ eVeV 4.5446.13 =×

Example: 33 The ionization energy of +He is 18106.19 −× J atom–1. Calculate the energy of the first stationary state of 2+Li

(a) -118 atom106.19 J−× (b) J181041.4 −× atom–1

(c) -119 atom106.19 J−× (d) 117 atom1041.4 −−× J

Solution : (d) I.E. of ZEHe (2 2×=+ for )2=He

I.E. of 32 3×=+ ELi (Z for Li=3)

9

4

).(.

).(.2

=∴+

+

LiEI

HeEI or I.E. ).(.

4

9)( 2 ++ ×= HeEILi 18106.19

4

9 −××= = 171041.4 −× J atom–1

2.9 Bohr – Sommerfeld’s model.

(1) In 1915, Sommerfield introduced a new atomic model to explain the fine spectrum of hydrogen atom.

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(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.

(3) For circular orbit, the angular momentum = π2

nh where n= principal quantum number only one

component i.e. only angle changes.

(4) For elliptical orbit, angular momentum = vector sum of 2 components. In elliptical orbit two components are,

(i) Radial component (along the radius) = π2

hnr

Where, n r = radial quantum number

(ii) Azimuthal component = n φπ2

h

Where, n φ = azimuthal quantum number

So angular momentum of elliptical orbit =ππ

φ22

hn

hnr +

Angular momentum = π

φ2

)(h

nnr +

(5) Shape of elliptical orbit depends on,

φn

n=

axisminorofLength

axismajorofLength

φ

φ

n

nnr +=

(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 n φ n r Nature of path

3 1 3 elliptical

2 1 elliptical

3 0 circular

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 π2

nh for angular momentum. This model could

not explain Zeeman effect and Stark effect.

• Nuclear

K= 1

K= 2

K= 3

φ = change

r = change

r

φ1 r φ2 φ2 r

φ1 r

φ = change

r = constant

φ1 r

r

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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

,mv

h=λ where h = Planck’s constant.

(4) This can be derived as follows according to Planck’s equation, λ

νch

hE.

==

=

λν

cQ

energy of photon (on the basis of Einstein’s mass energy relationship), 2mcE =

equating both mc

hormc

hc== λ

λ2 which is same as de-Broglie relation. ( )pmc =Q

(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

eVmv =2

2

1; eVmvm 222 =

PeVmmv == 2 ; eVm

h

2=λ

(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

n

rornr

πλλπ

22 ==

From de-Broglie equation, mv

h=λ . Thus

n

r

mv

h π2= or

π2

nhmvr =

Note : q 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.

(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.

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 × m710 − (b) 7.28 × m710 − (c) 2 × 10 m10− (d) 3 × m510 −

Solution : (b) KE 252 1055.42

1 −×== mv J

6

31

252 101

101.9

1055.42×=

×

××=

v ; smv /10 3=

De-Broglie wavelength mmv

h 7

331

34

1028.710101.9

10626.6 −

×=××

×==λ

Example: 35 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, sec10626.6 27 ergh −×=

Examples based on de-Broglie’s equation

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(a) 3.31 × Å310 − (b) 1.33 × Å310 − (c) 3.13 × 10 Å2− (d) 1.31 × Å210 −

Solution : (b) gm2310023.6

1

×=

23

18

27

10023.6sec1031

10626.6××

××

×==

cmmv

hλ = cm111033.1 −×

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 , π4

.h

px ≥∆∆

Where =∆x uncertainty is position of the particle, =∆p uncertainty in the momentum of the particle

Now since vmp ∆=∆

So equation becomes, π4

.h

vmx ≥∆∆ or m

hvx

π4≥∆×∆

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 .4π

hIf x∆ is made small, p∆ increases and vice versa.

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

π4

.h

tE ≥∆∆

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

velocity is 0 and position can be measured accurately.

Example: 36 What is the maximum precision with which the momentum of an electron can be known if the

uncertainty in the position of electron is ?001.0 ű Will there be any problem in describing the

momentum if it has a value of ,2 0a

h

πwhere 0a is Bohr’s radius of first orbit, i.e., 0.529Å?

Solution : π4

.h

px =∆∆

Q mÅx 1310001.0 −==∆

∴ 22

13

34

1027.51014.34

10625.6 −

×=××

×=∆p

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

Examples based on uncertainty principle

π4.

hpx ≥∆∆

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m

hxv

π4. ≈∆∆

xm

hv

∆≈∆

.4π

1031

34

1010108.97

224

10625.6

−−

××××

×= 15 sec108.5 −×= m

Example: 38 A dust particle having mass equal to ,10 11 g− diameter of cm410 − and velocity .sec10 14 −− cm The

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

Solution : 174

sec101100

101.0 −−−

×=×

=∆ cmv

Q m

hxv

π4. =∆∆

∴ cmx 10

711

27

1027.51011014.34

10625.6 −

−−

×=××××

×=∆

The uncertainty in position as compared to particle size.

cmdiameter

x 6

4

10

1027.510

1027.5 −

×=×

=∆

=

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,

0)(8

2

2

2

2

2

2

2

2

=Ψ−+∂

Ψ∂+

Ψ∂+

Ψ∂VE

h

m

zyx

π

Where yx, 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 :

0)(8

2

22 =Ψ−+Ψ∇ VE

h

Where ∇ = laplacian operator.

(5) Physical Significance of Ψ and 2Ψ

(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. )......,,( timeszyxΨ=Ψ

(ii) For a single particle, the square of the wave function )( 2Ψ at any point is proportional to the

probability of finding the particle at that point.

(iii) If 2Ψ 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.

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(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.

QQuuaannttuumm nnuummbbeerrss

(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.

ÅZ

nr 529.0

2

×=

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

moleperKcaln

ZE 3.313

2

2

×−=

(iv) The maximum number of an electron in an orbit represented by this quantum number as .2 2n 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

π2

nhmvr =

(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 fdps <<< (increasing energy).

(v) The value of )1( −= nl 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 )1(2

+llh

π

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(viii) The maximum number of electrons in subshell )12(2 += l

electrons2subshell →−s electrons10subshell →−d

electrons6subshell →−p electrons.14subshell →−f

(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 .2n

(vi) For a given value of ‘l’ the total value of ‘m’ is equal to ).12( +l

(vii) Degenerate orbitals : Orbitals having the same energy are known as degenerate orbitals. e.g. for p

subshell zyx ppp

(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 1/2,-and1/2 is'' +s 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 .)1(2

+ssh

π

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

(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

N

+1/2

S

–1/2

Magnetic field

N S

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2 (L shell) 0

1

0

+1

0

–1 2/1,2/1

2/1,2/1

2/1,2/1

2/1,2/1

−+

−+

−+

−+

2s

2p

6

2

8

3 (M shell)

4(N shell)

0

1

2

0

1

2

3

0

+1

0

–1

+2

+1

0

–1

–2

0

+1

0

–1

+2

+1

0

–1

–2

+3

+2

+1

+0

–1

–2

–3

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

−+

−+

−+

−+

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

−+

−+

−+

−+

−+

−+

−+

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

3s

3p

3d

4s

4p

4d

4f

10

6

2

14

10

6

2

18

32

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Z

Y

X

Nucleus

SShhaappee ooff oorrbbiittaallss

(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.

(ii) The size and energy of ‘s’ orbital with increasing ‘n’ will be

.4321 ssss <<<

(iii)It does not possess any directional property. s orbital has spherical

shape.

(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 .,, zyx ppp

(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.

(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 .222 ,,,,

zyxzxyzxy ddddd−

(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.

(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 .and,,,, 233222222 ),()()( xzyzzxyzyxzyxyyxxfffffff

−−−

(ii) The ‘f’ orbital is complicated in shape.

Y

X

dZ2

Z

Y

X

dX2–Y

2

Z Y

X

dYZ

Z Y

X

dXY

Z Y

X

dZX

Nodal Plane

Z Y

X

Pz orbital Nodal Plane

Nodal Plane

Z Y

X

Py orbital

Nodal Plane

Nodal Plane

Z Y

Px orbital

X

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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:

dfspdspspspss 5665454433221 <<<<<<<<<<<<

(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 22n 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 .2n

(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

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(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 21s 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.

Diagram of multi-electron atoms reveals the following points :

Energy level diagram of one electron system

En

erg

y

1s

5

4

3

2

2s 2p

3s 3p 3d

4s 4p 4d 4f 6 p 5 d 4 f 6 s 5 p 4 d 5 s

4 p 3 d 4 s 3 p

3 s 2 p

2 s 1 s

En

erg

y

Energy level diagram of multiple electron system

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(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.

For 3d n = 3 l= 2 ∴ n + l = 5

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 753 ,, fdp 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 4d and 5d configuraton among electrons

with parallel spins.

To number of possible exchanges = 3 + 2 + 1 =6

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

3 exchanges by 1st e– 2 exchanges by 2nd e– Only 1 exchange by 3rd e–

d4 (1) (2) (3)

2 exchange by 3rd e– 3 exchanges by 2nd e– 4 exchanges by 1st e–

(2) (3) d5 (1)

1 exchange by 4th e–

(4)

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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

Element Atomic

Number 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f

H 1 1

He 2 2

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

F 9 2 2 5

Ne 10 2 2 6

Na 11 2 2 6 1

Mg 12 2

Al 13 2 1

Si 14 10

electrons

2 2

P 15 2 3

S 16 2 4

Cl 17 2 5

Ar 18 2 2 6 2 6

K 19 2 2 6 2 6 1

Ca 20 2

Sc 21 1 2

Ti 22 2 2

V 23 3 2

Cr 24 5 1

Mn 25 5 2

Fe 26 6 2

Co 27 18

electrons

7 2

Ni 28 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

Br 35 10 2 5

Kr 36 2 2 6 2 6 10 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.

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(3) (i) Elements with atomic number 24(Cr), 42(Mo) and 74(W) have 51 )1( dnns − configuration and not 42 )1( dnns − due to extra stability of these atoms.

(ii) Elements with atomic number 29(Cu), 47(Ag) and 79(Au) have 101 )1( dnns − configuration instead

of 92 )1( dnns − due to extra stability of these atoms.

(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 ),24,26(2 −+ = eZ it will not be similar to Cr (with −e24 ) but quite different.

[ ][ ]

+ 62

62

34

34

dsArFe

dsArFeo

outer most orbit is 4th shell hence, electrons from 4s have been removed to make +2Fe .

(5) Ion/atom will be paramagnetic if there are unpaired electrons. Magnetic moment (spin only) is

)2( += nnµ BM (Bohr Magneton). )/1027.91( 24 TJBM −×= 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

[ ] 103dAr is colourless and +2Cu with electronic configuration [ ] 93dAr (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 f-

subshell 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 21−ns ), 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 612 −npns ), 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 d-

block elements having the configuration (n –1) 21101 −− nsd ), the group number is equal to the total

number of electrons present in the (n –1)d orbital and ns orbital.

nl

NUMBER OF PRINCIPAL SHELL

NUMBER OF ELECTRONS PRESENT

SYMBOL OF SUBSHELL

x

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

3d5

4s1

Cr (24) [Ar] 3d10

4s1

Cu (29) [Ar]

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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 (d) None of these

3. Cathode rays are [JIPMER 1991; NCERT 1976]

(a) Protons (b) Electrons (c) Neutrons (d) α-particles

4. Cathode rays have [CPMT 1982]

(a) Mass only (b) Charge only (c) No mass and charge (d) Mass and charge both

5. Cathode rays are made up of [AMU 1983]

(a) Positively charged particles (b) Negatively charged particles

(c) Neutral particles (d) None of these

6. Cathode rays are produced when the pressure in the discharge tube is of the order of

(a) 76 cm of Hg (b) cm610 − of Hg (c) 1 cm of Hg (d) 210 − to 310 − 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

(d) The ratio charge/mass of the particles is much greater than that of anode rays

9. 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 (d) They are electromagnetic waves

10. The electron is [Delhi PMT 1982; MADT Bihar 1980]

(a) α-rays particle (b) β-ray particle (c) Hydrogen ion (d) Positron

11. The charge on an electron is

(a) 10108.4 −×− esu (b) 19106.1 −×− C (c) Unit negative (d) All

12. Mass of an electron is

(a) g28101.9 −× (b) g25101.9 −× (c) g10101.9 −× (d) g18101.9 −×

13. Which of the following has the same mass as that of an electron [AFMC 2002]

(a) Photon (b) Neutron (c) Positron (d) Proton

14. Density of the electron is

(a) mLg /1017.2 17−× (b) mLg /1038.4 17−× (c) mLg /1017.2 14−× (d) None of these

15. A strong argument for the particle nature of cathode rays is that they [CPMT 1986; MLNR 1986]

(a) Produce fluorescence (b) Travel through vacuum

(c) Get deflected by electric and magnetic fields (d) Cast shadow

16. In the discharge tube emission of cathode rays requires

(a) Low potential and low pressure (b) Low potential and high pressure

BBaassiicc LLeevveell

Composition of Atom (Electron, Proton and Neutron)

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(c) High potential and high pressure (d) High potential and low pressure

17. The minimum real charge on any particle which can exist is [Rajasthan PMT 2000]

(a) Coulomb19106.1 −× (b) Coulomb10106.1 −× (c) Coulomb10108.4 −× (d) Zero

18. Which of the following statement is incorrect [CPMT 1973 ; BHU 1985]

(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

19. Ratio of masses of proton and electron is [BHU 1998]

(a) Infinite (b) 3108.1 +× (c) 1.8 (d) None of these

20. The mass of a mole of proton and electron is

(a) g2310023.6 × (b) g008.1 and mg55.0 (c) kg28101.9 −× (d) 2 gm

21. A mass spectrograph is an instrument which is capable of differentiating and identifying particles [NCERT 1977]

(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

22. Anode rays were discovered by [DPMT 1985]

(a) Goldstein (b) J. Stoney (c) Rutherford (d) J.J. Thomson

23. The nature of anode rays depends on [CPMT 1987]

(a) Nature of electrode (b) Nature of discharging tube (c) Nature of residual gas (d) All of these

24. Proton is [NCERT 1976 ; CPMT 1971]

(a) An ionized hydrogen molecule (b) An α-ray particle

(c) A fundamental particle (d) Nucleus of heavy hydrogen

25. Penetration power of proton is [BHU 1985 ; CPMT 1982, 88]

(a) More than electron (b) Less than electron (c) More than neutron (d) None of these

26. 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

27. The e/m for positive rays in comparison to cathode rays is

(a) Very low (b) High (c) Same (d) None of these

28. 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

29. 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

30. The proton and neutron are collectively called as [MP PMT 2001]

(a) Deutron (b) Positron (c) Meson (d) Nucleon

31. Which is correct statement about proton [CPMT 1979 ; MP PMT 1985; NCERT 1985 ; MP PMT 1999]

(a) Proton is nucleus of deuterium (b) Proton is ionized hydrogen molecule

(c) Proton is ionized hydrogen atom (d) Proton is α-particle

32. 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

33. Which of the following reactions led to the discovery of the neutron

(a) nNpC 10

147

11

146 +→+ (b) nCDB 1

0126

21

115 +→+ (c) nCHeBe 1

0126

42

94 +→+ (d) nCHeBe 1

0116

42

84 +→+

34. Heaviest particle is [Delhi PMT 1983 ; MP PET 1999]

(a) Meson (b) Neutron (c) Proton (d) Electron

35. The density of neutrons is of the order [NCERT 1980]

(a) cckg /10 3− (b) cckg /10 6− (c) cckg /10 9− (d) cckg /10 12−

36. The mass of neutron is nearly [MLNR 1988 ; UPSEAT 1999, 2000, 02]

Page 40: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

(a) kg2310 − (b) kg2410 − (c) kg2610 − (d) kg2710 −

37. 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

38. The discovery of neutron becomes very late because [CPMT 1987 ; AIIMS 1998]

(a) Neutrons are present in nucleus (b) Neutrons are highly unstable particles

(c) Neutrons are chargeless (d) Neutrons do not move

39. Which one of the following pairs is not correctly matched [MP PET 2002]

(a) Rutherford-Proton (b) J. J. Thomson-Electron (c) J. H. Chadwick-Neutron (d) Bohr-Isotope

40. An elementary fundamental particle is [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

41. The charge of an electron is C19106.1 −×− . The value of free charge on +Li ion will be

[AFMC 2002 ; Karnataka CET (Engg.) 2002]

(a) C19106.3 −× (b) C19101 −× (c) C19106.1 −× (d) C19106.2 −×

42. The charge on an electron is esu10108.4 −× . What is the value of charge in +Li ion [CPMT 1997]

(a) esu10108.4 −× (b) esu10106.9 −× (c) esu91044.1 −× (d) esu10104.2 −×

43. 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 [IIT 1984]

(a) e, p, n, α (b) n, p, e, α (c) n, p, α, e (d) n, α, p, e

45. The specific charge of proton is 17106.9 −× kgC then for an α-particle it will be [MH CET 1999]

(a) 17104.38 −× kgC (b) 17102.19 −× kgC (c) 17104.2 −× kgC (d) 17108.4 −× kgC

46. The number of atoms in 0.004 g of magnesium are [AFMC 2000]

(a) 20104 × (b) 20108 × (c) 2010 (d) 201002.6 ×

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 (b) 16 (c) 32 (d) 64

48. The number of electrons in −Cl ion is [MP PMT 2003]

(a) 19 (b) 20 (c) 18 (d) 35

49. The number of neutron in tritium is [CPMT 2003]

(a) 1 (b) 2 (c) 3 (d) 0

50. The total number of protons in one molecule of nitrogen dioxide

Advance LLeevveell

Basic LLeevveell

Atomic number, Mass number and Atomic species

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(a) 23 (b) 46 (c) 69 (d) 92

51. Number of neutrons in heavy hydrogen atom is [MP PMT 1986]

(a) 0 (b) 1 (c) 2 (d) 3

52. The nucleus of helium contains [CPMT 1972; Delhi PMT 1982]

(a) Four protons (b) Four neutrons

(c) Two neutrons and two protons (d) Four protons and two electrons

53. Sodium atom differs from sodium ion in the number of [CPMT 1976]

(a) Electron (b) Protons (c) Neutrons (d) Does not differ

54. An atom has 26 electrons and its atomic weight is 56. The number of neutrons in the nucleus of the atom will be [CPMT 1980]

(a) 26 (b) 30 (c) 36 (d) 56

55. The atomic number of an element represents [CPMT 1983; CBSE 1990; NCERT 1973; AMU 1984]

(a) Number of neutrons in the nucleus (b) Number of protons in the nucleus

(c) Atomic weight of element (d) Valency of element

56. The mass of an atom is constituted mainly by [Delhi PMT 1984. 91; AFMC 1990]

(a) Neutron and neutrino (b) Neutron and electron (c) Neutron and proton (d) Proton and electron

57. Which of the following is always a whole number [CPMT 1976, 81, 86]

(a) Atomic weight (b) Atomic radii (c) Equivalent weight (d) Atomic number

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

(a) 30 (b) 32 (c) 34 (d) 42

59. The total number of unpaired electrons in d-orbitals of atoms of element of atomic number 29 is [CPMT 1983]

(a) 10 (b) 1 (c) 0 (d) 5

60. Chlorine atom differs from chloride ion in the number of [NCERT 1972; MP PMT 1995]

(a) Proton (b) Neutron (c) Electrons (d) Protons and electrons

61. The number of electrons in one molecule of 2CO are [IIT 1979; MP PMT 1994; Rajasthan PMT 1999]

(a) 22 (b) 44 (c) 66 (d) 88

62. The nitrogen atom has 7 protons and 7 electrons, the nitride ion )( 3−N will have [NCERT 1977]

(a) 7 protons and 10 electrons (b) 4 protons and 7 electrons

(c) 4 protons and 10 electrons (d) 10 protons and 7 electrons

63. The total number of neutrons in dipositive zinc ions with mass number 70 is [IIT 1979; Bihar MEE 1997]

(a) 34 (b) 40 (c) 36 (d) 38

64. If W is atomic weight and N is the atomic number of an element, then [CPMT 1971, 80, 89]

(a) Number of NWe −=−1 (b) Number of NWn −=10

(c) Number of NWH −=11 (d) Number of Nn =1

0

65. The number of electrons in the atom which has 20 protons in the nucleus is [CPMT 1981, 93; CBSE 1989]

(a) 20 (b) 10 (c) 30 (d) 40

66. Six protons are found in the nucleus of [CPMT 1977, 80, 81; NCERT 1975, 78]

(a) Boron (b) Lithium (c) Carbon (d) Helium

67. A sodium cation has different number of electrons from

(a) −2O (b) −F (c) +Li (d) +++Al

68. An atom which has lost one electron would be [CPMT 1986]

(a) Negatively charged (b) Positively charged

(c) Electrically neutral (d) Carry double positive charge

69. 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

70. 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

71. The nucleus of the atom consists of [CPMT 1973; 74, 78, 83, 84; MADT Bihar 1980; Delhi PMT 1982, 85, MP PMT 1999]

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(a) Proton and neutron (b) Proton and electron

(c) Neutron and electron (d) Proton, neutron and electron

72. The number of electrons in an atom of an element is equal to its [BHU 1979]

(a) Atomic weight (b) Atomic number (c) Equivalent weight (d) Electron affinity

73. Neutrons are found in atoms of all elements except in [MP PMT 1997]

(a) Chlorine (b) Oxygen (c) Argon (d) Hydrogen

74. A transition metal X has a configuration 43][ dAr in its + 3 oxidation state. Its atomic number is [EAMCET 1990]

(a) 25 (b) 26 (c) 22 (d) 19

75. Number of electrons in 2CONH− is [AMU 1988]

(a) 22 (b) 23 (c) 20 (d) 28

76. 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

77. Which of the following atom has more electrons than neutrons

(a) C (b) −F (c) −2O (d) +3Al

78. The present atomic weight scale is based on [EAMCET 1988; MP PMT 2002]

(a) 12C (b) 16O (c) 1H (d) 13C

79. The nucleus of the element 4521 E contains

(a) 45 protons and 21 neutrons (b) 21 protons and 24 neutrons

(c) 21 protons and 45 neutrons (d) 24 protons and 21 neutrons

80. The number of electrons in the nucleus of 12C is [AFMC 1995]

(a) 6 (b) 12 (c) 0 (d) 3

81. 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

82. 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

83. 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]

(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

84. 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

85. 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 (b) 37 (c) 2 (d) 38

86. The number of electrons in 14019 ][ −K is [CPMT 1997; AFMC 1999]

(a) 19 (b) 20 (c) 18 (d) 40

87. In the nucleus of 4020 Ca 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

88. The atomic weight of an element is 39. The number of neutrons in its nucleus is one more than the number of protons. The number of protons, neutrons and electrons respectively in its atom would be [MP PMT 1997]

(a) 19, 20, 19 (b) 19, 19, 20 (c) 20, 19, 19 (d) 20, 19, 20

89. CO has same electrons as or the ion that is iso-electronic with CO is [CPMT 1984; IIT 1982; EAMCET 1990; CBSE 1997]

(a) +2N (b) −CN (c) +

2O (d) −2O

90. +Na ion is iso-electronic with [CPMT 1990]

(a) +Li (b) +2Mg (c) +2Ca (d) +2Ba

91. Which of the following oxides of nitrogen is iso-electronic with 2CO [CBSE 1990]

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(a) 2NO (b) ON2 (c) NO (d) 22ON

92. Which one of the following is not iso-electronic with −2O [CBSE 1994]

(a) −3N (b) −F (c) +Tl (d) +Na

93. Pick out the iso-electronic structures from the following, IV

3III

3II

3I

3−++ CHNHOHCH [IIT 1993]

(a) I and II (b) I and IV (c) I and III (d) II, III and IV

94. The hydride ions )( −H are iso-electronic with [AFMC 1995; Bihar MEE 1997]

(a) Li (b) +He (c) He (d) Be

95. Iso-electronic species are [EAMCET 1989]

(a) −+ ClK , (b) −+ ClNa , (c) ArNa, (d) ArMg ,+

96. Which one of the following grouping represents a collection of iso-electronic species [AIEEE 2003]

(a) +++ 22 ,, MgCaNa (b) +−− NaFN ,,3 (c) −+ ClAlBe ,, 3 (d) BrCsCa ,,2 ++

97. Which of the following are iso-electronic and isostructual 33233 ,,, SOClOCONO −−− [IIT Screening 2003]

(a) −− 233 , CONO (b) −

33, NOSO (c) −− 233 , COClO (d) 3

23 .SOCO −

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]

(a) −F (b) Oxygen atom (c) Mg (d) −N

99. Which of the following is iso-electronic with carbon atom [MP PMT 1994; UPSEAT 2000]

(a) +Na (b) +3Al (c) −2O (d) +N

100. Which of the following is not iso-electronic with Ne [MP PET 2002]

(a) +Na (b) +2Mg (c) −2O (d) −Cl

101. Which of the following is iso-electronic with +2Ca

(a) Kr (b) +K (c) +2Mg (d) Ca

102. Iso-electronic species is [Rajasthan PMT 2002]

(a) 2, −− OF (b) OF ,− (c) +− OF , (d) 2, +− OF

103. Which pair of ions is iso-electronic [DCE 1999]

(a) −F and −Cl (b) −F and −O (c) +Na and +K (d) +Na and 2+Mg

104. Tritium is the isotope of [CPMT 2003]

(a) Hydrogen (b) Oxygen (c) Carbon (d) Sulphur

105. An isostere is [UPSEAT 1999]

(a) −2NO and 3O (b) −

2NO and −34PO (c) −

322 ,, NOONCO (d) −4ClO and −OCN

106. Which of the following pair has same electronic structure [CPMT 1992]

(a) Ca, Ar (b) Mg, +Na (c) Ag, Sn (d) Ar, −Cl

107. Which of the following are iso-electronic with one another [NCERT 1983; EAMCET 1989]

(a) +Na and Ne (b) +K and O (c) Ne and O (d) +Na and +K

108. +2Be is iso-electronic with [EAMCET 1998]

(a) +2Mg (b) +Na (c) +Li (d) +H

109. The nitride ion in lithium nitride is composed of [Karnataka CET 2000]

(a) 7 protons + 10 electrons (b) 10 protons + 10 electrons (c) 7 protons + 7 protons (d) 10 protons + 7 electrons

110. Number of protons, neutrons and electrons in the element 23189 γ is [AFMC 1997]

(a) 89, 231, 89 (b) 89, 89, 242 (c) 89, 142, 89 (d) 89, 71, 89

111. 2CO is isostructural with [IIT 1986; MP PMT 1986, 94, 95]

(a) 2SnCl (b) 2SO (c) 2HgCl (d) All the above

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

Advance LLeevveell

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(a) ν against Z (b) ν

1 against Z (c) ν against Z (d) ν against Z

113. In Moseley's equation )([ bZa −=ν ], 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 114. If molecular mass and atomic mass of sulphur are 256 and 32 respectively, its atomicity is [Rajasthan PET 2000]

(a) 2 (b) 8 (c) 4 (d) 16 115. Assertion (A) : The atoms of different elements having same mass number but different atomic number are known as isobars

Reason (R) : The sum of protons and neutrons, in the isobars is always different [AIIMS 2000] (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

116. The mass number of an anion, 3−X , is 14. If there are ten electrons in the anion, the number of neutrons in the nucleus of atom,

2X of the element will be [MP PMT 1999]

(a) 10 (b) 14 (c) 7 (d) 5 117. 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 126 C [NCERT 1982]

(a) Will remain approximately the same (b) Will become approximately two times (c) Will remain approximately half (d) Will be reduced by 25%

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118. A neutral atom (Atomic no. > 1) consists of [CPMT 1982] (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 (b) Contains more electrons (c) Contains more protons (d) Is a different element

120. Assertion (A) : Nuclide 1330 Al is less stable than 20

40 Ca

Reason (R): Nuclides having odd number of protons and neutrons are generally unstable [IIT 1998] (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. Which of the following are iso-electronic species 3423 IV,III,II,I NHNHNHCH −−−− +−+ [CPMT 1999]

(a) I, II, III (b) II, III, IV (c) I, II, IV (d) I and II

122. The charge on the atom containing 17 protons, 18 neutrons and 18 electrons is [AIIMS 1996]

(a) + 1 (b) – 2 (c) –1 (d) Zero

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 [EAMCET 1998]

(a) 2/10 ARR = (b) ARR .0= (c) 2

0.ARR = (d) 3/10.ARR =

130. The volume of the nucleus is

(a) 410 − times smaller than the volume of an atom (b) 810 − times smaller than the volume of an atom

(c) 1210 − times smaller than the volume of an atom (d) Two-third the volume of the nucleus

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

BBaassiicc Level

Atomic Models and Planck’s Quantum Theory

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

(a) Electrons (b) Protons (c) Nucleus (d) Neutrons

132. The size of nucleus is measured in [EAMCET 1988; CPMT 1994]

(a) amu (b) Angstrom (c) Fermi (d) cm

133. The average distance of an electron in an atom from its nucleus is of the order of [MP PET 1996]

(a) m610 (b) m610 − (c) m1010 − (d) m1510 −

134. Nucleus of an atom is [MNR 1977]

(a) Neutral (b) Negatively charged (c) Positively charged (d) None of them

135. Rutherford’s scatting experiment is related to the size of the [IIT 1983; MADT Bihar 1995; BHU 1995]

(a) Nucleus (b) Atom (c) Electron (d) Neutron

136. The positive charge of an atom is [AFMC 2002]

(a) Spread all over the atom (b) Distributed around the nucleus

(c) Concentrated at the nucleus (d) All of these

137. Atoms have diameters of the order of [NCERT 1971; CPMT 1977]

(a) cm810 − (b) cm1010 − (c) cm1310 − (d) cm1510 −

138. Remaining part of atom except outer orbit is called [CPMT 1982]

(a) Kernel (b) Core (c) Empty space (d) None

139. The radius of an atom is of the order of [AMU 1982; IIT 1985; MP PMT 1995]

(a) cm1010 − (b) cm1310 − (c) cm1510 − (d) cm810 −

140. Discovery of the nucleus of an atom was due to the experiment carried out by [CPMT 1983; MP PET 1983]

(a) Bohr (b) Mosley (c) Rutherford (d) Thomson

141. The order of density in nucleus is [NCERT 1981; CPMT 1981, 2003]

(a) cckg /10 8 (b) cckg /10 8− (c) cckg /10 9− (d) cckg /1012

142. Existence of positively charged nucleus was established by [CBSE 1991]

(a) Positive ray analysis (b) α-ray scattering experiments (c) X-ray analysis (d)

143. The size of nucleus is of the order of [CPMT 1982; MP PMT 1991]

(a) m1210 − (b) m810 − (c) m1510 − (d) m1010 −

144. Bohr’s model can explain [IIT 1985]

(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

(d) Angular momentum of electron is an integral multiple of π2

h

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

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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

153. When an electron revolves in a stationary orbit then [MP PET 1994]

(a) It absorbs energy (b) It gains kinetic energy (c) It emits radiation (d) Its energy remains constant

154. If the radius of first Bohr orbit be 0a , then the radius of third orbit would be [MP PET 1997]

(a) 03 a× (b) 06 a× (c) 09 a× (d) 09

1a

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]

(a) 22

22422

hn

zemEn

π−= (b)

22

2222

hn

zmeEn

π−= (c)

22

2422

hn

zmeEn

π−= (d)

22

4222

hn

zemEn

π−=

159. The expression for Bohr’s radius of an atom is [MP PMT 1999]

(a) 242

22

4 zme

hnr

π= (b)

zme

hnr

22

22

4π= (c)

222

22

4 zme

hnr

π= (d)

2222

22

4 zem

hnr

π=

160. Visible range of hydrogen spectrum will contain the following series [Rajasthan PET 2000]

(a) Pfund (b) Lyman (c) Balmer (d) Brackett

161. Wavelength of spectral line emitted is inversely proportional to [CPMT 2001]

(a) Radius (b) Energy (c) Velocity (d) Quantum number

162. In hydrogen spectrum the different lines of Lyman series are present in [UPSEAT 1999]

(a) UV field (b) IR field (c) Visible field (d) Far IR field

163. In an element going away from nucleus, the energy of particle [Rajasthan PMT 1997]

(a) Decreases (b) Not changing (c) Increases (d) None of these

164. When an electron jumps from lower to higher orbit, its energy [MADT Bihar 1982]

(a) Increases (b) Decreases (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 [MP PET 2003]

(a) Hz101066.15 × (b) Hz141066.24 × (c) Hz141057.30 × (d) Hz241057.40 ×

166. When an electron drops from a higher energy level to a low energy level, then [AMU 1985]

(a) Energy is emitted (b) Energy is absorbed (c) Atomic number increases (d) Atomic number decreases

167. When an electron jumps from L to K shell [CPMT 1983]

(a) Energy is absorbed (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 [AMU (Engg.) 2002]

(a) 12

38.131 −−= molkJn

En (b) 133.131 −−= molkJn

En (c) 12

3.1313 −−= molkJn

En (d) 12

13.313 −−= molkJn

En

170. The spectrum of He is expected to be similar to [AIIMS 1980, 91; Delhi PMT 1983; MP PMT 2002]

(a) H (b) +Li (c) Na (d) +He

171. The series limit for Balmer series of H-spectra is [AMU (Engg.) 1999]

Page 48: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

(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 [DCE 2001]

(a) Heisenberg’s uncertainty principle (b) Bohr’s theory

(c) Ohm’s law (d) Rutherford’s atomic model

175. Zeeman effect refers to the [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

177. Which one of the following is not the characteristic of Planck’s quantum theory of radiation [AIIMS 1991]

(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

179. Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon [IIT 1984; CPMT 1997]

(a) 3 s (b) 2 p (c) 2 s (d) 1 s

180. The frequency of yellow light having wavelength 600 nm is [MP PET 2002]

(a) Hz14100.5 × (b) Hz7105.2 × (c) Hz7100.5 × (d) Hz14105.2 ×

181. The energy of a photon is calculated by [Pb. PMT 2000]

(a) νhE = (b) νEh = (c) V

Eh = (d)

V

hE =

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

AAddvvaannccee Level

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(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 Al2713 will be

(a) m15102.1 −× (b) m151027 −× (c) m15108.10 −× (d) m15106.3 −×

187. The nucleus of an atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by

cmA 3/1131025.1 ×× − . 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) 3100.1 −× (b) 5100.5 −× (c) 2105.2 −× (d) 131025.1 −×

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) ergs111015.2 −× (b) ergs10101911.0 −× (c) ergs1210389.2 −× (d) ergs1010239.0 −×

189. The radius of first Bohr’s orbit for hydrogen is 0.53 Å. The radius of third Bohr’s orbit would be [MP PET 1994]

(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 electrons in Bohr orbits to hydrogen is (are) [IIT 1998]

(a) – 3.4 eV (b) – 4.2 eV (c) – 6.8 eV (d) + 6.8 eV

191. Energy of electron of hydrogen atom in second Bohr orbit is [MP PMT 2000]

(a) J191044.5 −×− (b) kJ191044.5 −×− (c) cal191044.5 −×− (d) eV191044.5 −×−

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

193. The energy of an electron in nth orbit of hydrogen atom is [MP PET 1999]

(a) eVn4

6.13 (b) eV

n3

6.13 (c) eV

n2

6.13 (d) eV

n

6.13

194. As electron moves away from the nucleus, its potential energy [UPSEAT 2003]

(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 account for phenomenon [AIIMS 1983]

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 Hz1410340.2 × . The quantum number 2n which

produces this transition is [Delhi PMT 2001]

(a) 6 (b) 5 (c) 4 (d) 3

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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.

(a) 4/∞R (b) 2/∞R (c) ∞R2 (d) ∞R

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 [ =0a the radius of the first Bohr orbit]

(a) 05.1 a and 0a (b) 0a and 05a (c) 05.1 a and 05.0 a (d) 0a and 05.0 a

202. Choose the correct relations on the basis of Bohr theory

(a) Velocity of electron n

1∝ (b) Frequency of revolution

3

1

n∝

(c) Radius of orbit Zn2∝ (d) Force on electron 4

1

n∝

203. For a hydrogen atom, what is the orbital degeneracy of the level that has energy 9

∞−=

hcR, where ∞R is the Rydberg constant for

the hydrogen atom

(a) 1 (b) 9 (c) 36 (d) 3

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 [AIEEE 2002]

(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 )10022.6 23×= [BHU 1999]

(a) J201069.1 −× (b) J231069.1 −× (c) J231069.1 × (d) J251069.1 ×

206. The value of the energy for the first excited state of hydrogen atom will be [MP PET 2002]

(a) – 13.6 eV (b) – 3.40 eV (c) – 1.51 eV (d) – 0.85 eV

207. 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 element is [CPMT 1989]

(a) 6 (b) 5 (c) 7 (d) 10

208. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [AIEEE 2003]

(a) 3 → 2 (b) 5 → 2 (c) 4 → 1 (d) 2 → 5

209. 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

210. The emission spectrum of hydrogen is found to satisfy the expression for the energy change. ∆E (in joules) such that

Jnn

E

−×=∆

22

21

111018.2 where ....3,2,11 =n and ....4,3,22 =n The spectral lines correspond to Paschen series to [UPSEAT 2002]

(a) 11 =n and 4,3,22 =n (b) 31 =n and 6,5,42 =n (c) 11 =n and 5,4,32 =n (d) 11 =n and =2n infinity

211. The energy required to dislodge electron from excited isolated H-atom., eVIE 6.131 = is [DCE 2000]

(a) = 13.6 eV (b) > 13.6 eV (c) < 13.6 and > 3.4 eV (d) eV4.3≤

212. If change in energy, sJhJE −×=×=∆ −− 348 1064.6,103)( and ,/103 8 smc ×= then wavelength of the light is

[CBSE 2000]

(a) 31036.6 × Å (b) 51036.6 × Å (c) 181064.6 −× Å (d) 181036.6 × Å

213. The value of Planck’s constant is Js341063.6 −× . The velocity of light is 18100.3 −× ms . Which value is closest to the wavelength

in nanometres of a quantum of light with frequency of 115108 −× s [CBSE 2003]

(a) 7103 × (b) 25102 −× (c) 18105 −× (d) 1104 ×

214. The wavelength of a spectral line for an electronic transition is inversely related to [IIT 1988]

(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

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215. If wavelength of photon is m11102.2 −× , Jsh 34106.6 −×= , then momentum of photon is [MP PET 1999]

(a) 123103 −−× mskg (b) 1221033.3 −× mskg (c) 14410452.1 −−× mskg (d) 1431089.6 −× mskg

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 (c) 1/2 (d) 2

217. Wavelength associated with electron motion [BHU 1998]

(a) Increases with increase in speed of electron (b) Remains same irrespective of speed of electron

(c) Decreases with increase in speed of −e (d) Is zero

218. A 200 g golf ball is moving with a speed of 5 m per hour. The associated wave length is sec)10625.6( 34 −×= − Jh [MP PET 2003]

(a) m101038.2 −× (b) m201038.2 −× (c) m301038.2 −× (d) m401038.2 −×

219. The frequency of a wave of light is 1141012 −× s . The wave number associated with this light is [Pb. PMT 1999]

(a) m7105 −× (b) 18104 −−× cm (c) 17102 −−× m (d) 14104 −× cm

220. The energy of a 700 – nm photon is

(a) 1.77 eV (b) 2.47 eV (c) 700 eV (d) 3.57 eV

221. The wave nature of an electron was first given by [CMC Vellore 1991; Punjab PMT 1998]

(a) De–Broglie (b) Heisenberg (c) Mosley (d) Sommerfeld

222. Dual nature of particle is given by [BHU 2003]

(a) Bohr theory (b) Thomson model (c) Heisenberg principle (d) De–Broglie equation

223. Among the following for which one mathematical expression p

h=λ stands

(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) p

h=λ (c) Interference (d) Photoelectric effect

227. Which is the correct relationship between wavelength and momentum of particles [Pb. PMT 2000]

(a) P

h=λ (b)

P

h=π (c)

λ

Ph = (d) None of these

228. Which of the following expressions gives the de–Broglie relationship [MP PMT 1996; MP PET/ PMT 1998]

(a) mv

= (b) mv

h=λ (c)

hv

m=λ (d)

mh

v=λ

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 [Rajasthan PMT 1999]

(a) Electron (b) Proton (c) 2CO molecule (d) 2SO molecule

BBaassiiss Level

Dual nature of electron (de-Broglie equation)

Page 52: ATOMIC STRUCTURE STUDY MATERIAL & ASSIGNMENT

231. The de–Broglie wavelength associated with a particle of mass kg610 − moving with a velocity of 110 −ms , is [AIIMS 2001]

(a) m221063.6 −× (b) m291063.6 −× (c) m311063.6 −× (d) m341063.6 −×

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

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) m331063.6 −× (b) m341063.6 −× (c) m351063.6 −× (d) m351065.6 −×

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 [AMU (Engg.) 2002]

(a) cm101096.9 −× (b) cm81096.9 −× (c) cm41096.9 × (d) cm81096.9 ×

236. What will be de–Broglie wavelength of an electron moving with a velocity of 15102.1 −× ms [MP PET 2000]

(a) 910068.6 −× (b) 3710133.3 −× (c) 910626.6 −× (d) 710018.6 −×

237. An electron has kinetic energy J23108.2 −× . de–Broglie wavelength will be nearly )101.9( 31 kgme−×= [MP PET 2000]

(a) m41028.9 −× (b) m71028.9 −× (c) m81028.9 −× (d) m101028.9 −×

238. Calculate the de–Broglie wavelength for a particle of mass kg3010 − , travelling with a speed of 1710 −ms .

)10626.6( 1234 −−×= smkgh

(a) m410626.6 −× (b) m410509.1 −× (c) m1110626.6 −× (d) m1010509.1 ×

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) metres3310 − (b) metres3110 − (c) metres1610 − (d) metres2510 −

240. De-Broglie wavelength is related to applied voltage as

(a) Åh

3.12=λ (b) Å

V

3.12=λ (c) Å

E

3.12=λ (d) Å

m

3.12λ

241. The possibility of finding an electron in an orbital was conceived by [MP PMT 1994]

(a) Rutherford (b) Bohr (c) Heisenberg (d) Schrodinger

242. The uncertainty principle and the concept of wave nature of matter was proposed by …… and ……. respectively [MP PET 1997]

(a) Heisenberg, de–Broglie (b) De–Brolgie, Heisenberg (c) Heisenberg, Planck (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]

AAddvvaannccee Level

BBaassiicc Level

Uncertainty principle and Schrodinger wave equation

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(a) Heisenberg uncertainty principle (b) Principle of de–Broglie’s wave nature of electron

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

244. In Heisenberg’s uncertainty equation ph

px ∆≥∆×∆ ,4π

stands for

(a) Uncertainty in energy (b) Uncertainty in velocity (c) Uncertainty in momentum (d) Uncertainty in mass

245. According to uncertainty principle [AMU 1990]

(a) 2mcE = (b) π4

hpx ≥∆×∆ (c)

p

h=λ (d)

π6

hpx =∆×∆

246. The uncertainty principle was enunciated by [NCERT 1975; Bihar MEE 1997]

(a) Einstein (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 π4

.h

px >∆∆ shows [MP PET 2000]

(a) De–Brolgie relation (b) Heisenberg’s uncertainty principle

(c) Aufbau principle (d) Hund’s rule

249. Uncertainty principle gave the concept of

(a) Probability (b) An orbital

(c) Physical meaning of ψ, the ψ2 (d) All the above

250. The uncertainty in momentum of an electron is smkg /101 5 −× − . The uncertainty in its position will be

)/1062.6( 234 smkgh −×= −

[AFMC 1998; CBSE 1999; JIPMER 2002]

(a) m281005.1 −× (b) m261005.1 −× (c) m301027.5 −× (d) m281025.5 −×

251. Uncertainty in position of a 0.25 g particle is m510 − . Uncertainty of velocity is )106.6( 34 sJh −×= − [AIEEE 2002]

(a) 34102.1 × (b) 32101.2 −× (c) 20106.1 −× (d) 9107.1 −×

252. If uncertainty in the position of an electron is zero, the uncertainty in its momentum would be [CPMT 1988]

(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 1261050 −−× mskg . The minimum uncertainty in the measurement of the momentum of the helium atom is

[CBSE 1998; AIIMS 2001]

(a) 150 −mskg (b) 160 −mskg (c) 1261080 −−× mskg (d) 1261050 −−× mskg

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

AAddvvaannccee Level

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255. The uncertainty in the position of an electron (mass = )101.9 28 g−× moving with a velocity of 14100.3 −× scm accurate upto

0.001% will be (Use π4

h in the uncertainty expression, where sergh −×= −2710626.6 ) [CBSE 1995]

(a) 51092.1 −× cm (b) 7.68 cm (c) 5.76 cm (d) 3.84 cm

256. The uncertainty in the position of a moving bullet of mass 10 gm is m510 − . Calculate the uncertainty in its velocity [DCE 1999]

(a) sec/102.5 28 m−× (b) sec/100.3 28 m−× (c) sec/102.5 22 m−× (d) sec/103 22 m−×

257. The shape of s-orbital is [NCERT 1978]

(a) Pyramidal (b) Spherical (c) Tetrahedral (d) Dumb-bell shaped

258. The shape of 2p orbital is [CPMT 1983; NCERT 1979]

(a) Spherical (b) Ellipsoidal (c) Dumb-bell (d) Pyramidal

259. Which orbital is dumb-bell shaped [MP PMT 1986; MP PET /PMT 1999]

(a) s-orbital (b) p-orbital (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 [Rajasthan PET 2003]

(a) 1 (b) 0 (c) 4 (d) 3

262. Which of the following pair of orbitals posses two nodal planes [Rajasthan PMT 2000]

(a) 22,yxxy dp

− (b) zxxy dd , (c) zxxy dp , (d) 222 ,

yxzdd

263. Which orbital does not have a spherical node [Kurukshetra CEE 2002]

(a) n = 2, l = 0 (b) n = 3, l = 0 (c) n = 2, l = 1 (d) n = 1, l = 0

264. The number of electrons which can be accommodated in an orbital is [Delhi PMT 1981; AFMC 1988]

(a) One (b) Two (c) Three (d) Four

265. The number of nodal planes in a xp orbital is [IIT Screening 2000]

(a) One (b) Two (c) Three (d) Zero

266. One orbital consists maximum electrons [AFMC 1988]

(a) 2 (b) 1 (c) 8 (d) 18

267. Which of the following orbitals will be dumb-bell shaped [MP PET 1986]

(a) 1s (b) 2s (c) xp2 (d) xyd3

268. The number of orbitals in d sub-shell is [MNR 1981]

(a) 1 (b) 3 (c) 5 (d) 7

269. The shape of p-orbital is [MP PMT 1993]

(a) Elliptical (b) Spherical (c) Dumb-bell (d) Complex geometrical

270. Number of orbitals in h sub-shell is [BHU 2003]

(a) 11 (b) 15 (c) 17 (d) 19

271. Azimuthal quantum number for last electron of Na atom is [BHU 1995]

(a) 1 (b) 2 (c) 3 (d) 0

272. xp orbital can accommodate [MLNR 1990: IIT 1983; MADT Bihar 1995]

(a) 4 electrons (b) 6 electrons

(c) 2 electrons with parallel spins (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]

BBaassiicc Level

Quantum numbers, Shape of Orbitals and Electronic Configuration of elements

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(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 (c) 14 (d) 18

275. Which of the following orbitals does not make sense [Rajasthan PMT 2000]

(a) 7s (b) 5p (c) 2d (d) 4f

276. There is no difference between a 2p and a 3p orbital regarding [BHU 1981]

(a) Shape (b) Size (c) Energy (d) Value of n

277. Which of the sub-shell from the following is dumb-bell

(a) 5s (b) 5p (c) 4d (d) 5f

278. The type of orbitals present in Fe is

(a) s (b) s and p (c) s, p and d (d) s, p, d and f

279. For the dumb-bell shaped orbital, the value of l is [CPMT 1987, 2003]

(a) 3 (b) 1 (c) 0 (d) 2

280. Which of the following orbital is not possible [Rajasthan PMT 1999]

(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 [BHU 1987, 95]

(a) Size (b) Shape (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

286. Principal quantum number of an atom represents [EAMCET 1979; IIT 1983; MLNR 1990; UPSEAT 2000, 02]

(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

293. The angular momentum of an electron depends on [BHU 1978, NCERT 1981]

(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

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295. Which of the following set of quantum number is not valid [AIIMS 2001]

(a) n = 1, l = 2 (b) n = 2, m = 1 (c) n = 3, l = 0 (d) n = 2, l = 0

296. 2p orbital have [NCERT 1981; MP PMT 1993, 97]

(a) n = 1, l = 2 (b) n = 1, l = 0 (c) n = 2, l = 1 (d) n = 2, l = 0

297. The maximum number of electrons which each sub-shell can occupy is [PU CET 1989]

(a) 22n (b) 2n (c) 2 ( 2l + 1) (d) 2l + 1

298. Which of the following represent the correct sets of the four quantum numbers of a 4d electron [MLNR 1992; UPSEAT 2001]

(a) 4, 3, 2, 2

1+ (b) 4, 2, 1, 0 (c) 4, 3, – 2,

2

1+ (d) 4, 2, 1,

2

1−

299. For the n = 2 energy level, how many orbitals of all kinds are possible [Bihar CEE 1995]

(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 [AIIMS 1997]

(a) 2n (b) 22 n (c) n (d) 2n

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

1+=s . The atoms is

[EAMCET 1978]

(a) Lithium (b) Beryllium (c) Hydrogen (d) Boron

302. The maximum number of electrons in an atom with l = 2 and n = 3 is [MP PET / PMT 1998]

(a) 2 (b) 6 (c) 12 (d) 10

303. Correct set of four quantum numbers for valence electron of rubidium (Z = 37) is [IIT 1984; JIPMER 1999; UPSEAT 2003]

(a) 5, 0, 0, 2

1+ (b) 5, 1, 0,

2

1+ (c) 5, 1, 1,

2

1+ (d) 6, 0, 0,

2

1+

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

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306. Orbital angular momentum for a d-electron is [MP PET 2003]

(a) π2

6h (b)

π2

6h (c)

π2

12h (d)

π2

12h

307. The number of quantum numbers required to describe an electron in an atom completely is [CET Pune 1998]

(a) 1 (b) 2 (c) 3 (d) 4

308. An −e has magnetic quantum number as – 3, what is its principal quantum number [BHU 1998]

(a) 1 (b) 2 (c) 3 (d) 4

309. The quantum number which is designated by letters s, p, d and f instead of number is [BHU 1980]

(a) n (b) l (c) lm (d) sm

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 number value is 4 [MP PMT 2000]

(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 2

1+ and a magnetic quantum number of – 1, it cannot be presented in an

[CBSE 1989; UPSEAT 2001]

(a) d-orbital (b) f-orbital (c) p-orbital (d) s-orbital

317. Which statement is not correct for n = 5, m = 3 [CPMT 1996]

(a) 4=l (b) 2/1;2,1,0 +== Sl (c) 3=l (d) All are correct

318. Values of the four quantum numbers for the last electron in the atom are n = 4, l = 1, m = +1 and 2

1−=s . Atomic number of the

atom will be

(a) 22 (b) 32 (c) 33 (d) 36

319. 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

320. For d electrons, the azimuthal quantum number is [MNR 1983; CPMT 1984]

(a) 0 (b) 1 (c) 2 (d) 3

321. 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

322. The magnetic quantum number for d-orbtial is given by [Orissa JEE 2002]

(a) 2 (b) 2,1,0 ±± (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 (d) 16

324. For sodium atom the number of electrons with m = 0 will be [Rajasthan PMT 1999]

(a) 2 (b) 7 (c) 9 (d) 8

325. The quantum numbers n = 2, l = 1 represent [AFMC 2002]

(a) 1s orbital (b) 2s orbital (c) 2p orbital (d) 3d orbital

326. For azimuthal quantum number l = 3, the maximum number of electrons will be [CBSE 1991; EAMCET 1991; Rajasthan PMT 2002; CBSE 2002]

(a) 2 (b) 6 (c) 0 (d) 14

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327. All electrons on the 4p sub-shell must be characterized by the quantum number (s) [MP PET 1996]

(a) n = 4, m = 0, 2

1±=s (b) l = 1 (c) l = 0,

2

1±=s (d)

2

1±=s

328. The magnetic quantum number for valency electrons of sodium is [CPMT 1988; MH CET 1999]

(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 [Roorkee 2000]

(a) Cr (b) Mn (c) +3Fe (d) +3Co

336. How many unpaired electrons are present in cobalt |Co| metal [Rajasthan PMT 2002]

(a) 2 (b) 3 (c) 4 (d) 7

337. Aufbau principle is not satisfied by [MP PMT 1997]

(a) Cr and Cl (b) Cu and Ag (c) Cr and Mg (d) Cu and Na

338. When 3d orbital is complete, the new electron will enter the [EAMCET 1980; MP PMT 1995]

(a) 4p orbital (b) 4f orbital (c) 4s orbital (d) 4d orbital

339. 1622 3221 spss shows configuration of [CPMT 1996]

(a) 3+Al in ground state (b) Ne in excited state (c) 1+Mg in excited state (d) None of these

340. The electronic configuration (outermost) of 2+Mn ion (atomic number of Mn = 25) in its ground state is [MP PET 1993]

(a) 05 4,3 sd (b) 14 4,3 sd (c) 23 4,3 sd (d) 222 44,3 psd

341. The structure of external most shell of inert gases is [JIPMER 1991]

(a) 32 ps (b) 62 ps (c) 21ps (d) 210 sd

342. In a potassium atom, electronic energy levels are in the following order [EAMCET 1979; Delhi PMT 1991]

(a) 4s > 3d (b) 4s > 4p (c) 4s < 3d (d) 4s < 3p

343. Which one is the correct outer configuration of chromium [AIIMS 1980, 91; BHU 1995]

(a) (b)

(c) (d)

344. Which of the following represents the electronic configuration of an element with atomic number 17 [AMU 1982]

(a) 61622 33,22,1 pspss (b) 142622 4,33,22,1 spspss (c) 52622 33,22,1 pspss (d) 241622 4,33,22,1 spspss

345. Which one is the electronic configuration of 2+Fe [MADT Bihar 1982; AIIMS 1989]

(a) 662622 333,22,1 dpspss (b) 2462622 4,333,22,1 sdpspss

↑ ↑ ↑ ↑ ↑↓

↑ ↑ ↑ ↑ ↑ ↑

↑↓ ↑↓ ↑↓

↑↓ ↑↓ ↑ ↑

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(c) 1562622 4,333,22,1 sdpspss (d) None of these

346. Total number of unpaired electrons in an atom of atomic number 29 is [CPMT 1984, 93]

(a) 1 (b) 3 (c) 4 (d) 2

347. Maximum number of electrons present in ‘N’ shell is [EAMCET 1984]

(a) 18 (b) 32 (c) 2 (d) 8

348. Correct configuration of 3+Fe [26] is [CPMT 1994; BHU 1995; Karnataka CET 1992]

(a) 562622 333,22,1 dpspss (b) 2362622 4,333,22,1 sdpspss

(c) 2662622 4,333,22,1 sdpspss (d) 1562622 4,333,22,1 sdpspss

349. According to Aufbau’s principle, which of the three 4d, 5p and 5s will be filled with electrons first [MADT Bihar 1984]

(a) 4d (b) 5p

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

350. The number of unpaired electrons in )26(3 =+ ZFe are [Karnataka CET 2000]

(a) 5 (b) 6 (c) 3 (d) 4

351. The electronic configuration of copper )(29 Cu is [Delhi PMT 1983; BHU 1980; AFMC 1981; CBSE 1991; MP PMT 1995]

(a) 2962622 4,333,22,1 sdpspss (b) 11062622 4,333,22,1 sdpspss

(c) 6262622 44,33,22,1 pspspss (d) 1062622 333,22,1 dpspss

352. The number of electrons in the valence shell of calcium is [IIT 1975]

(a) 6 (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 1522 3221 spss 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 −2O

355. The number of d electrons in +2Fe (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 [IIT 1999]

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

357. The atomic number of an element having the valency shell electronic configuration 62 44 ps is [MP PMT 1991]

(a) 35 (b) 36 (c) 37 (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 (b) C (c) −Cl (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]

(a) 15 43][ sdAr (b) 24 43][ sdAr (c) 06 43][ sdAr (d) 15 44][ sdAr

360. Aufbau principle is obeyed in which of the following electronic configurations [AFMC 1999]

↑↓ ↑↓ ↑ ↑ ↑ ↑↓ ↑↓ ↓ ↑ ↑ ↑↓ ↑↓ ↓ ↑ ↓ ↑↓ ↑↓ ↓ ↓ ↓

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(a) 622 221 pss (b) 232 331 sps (c) 622 331 pss (d) 222 321 sss

361. Nitrogen has the electronic configuration 11122 2222,1 zyx pppss and 01222 2222,1 zyx pppss 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

362. Maximum electron in a d-orbital are [CPMT 1999]

(a) 2 (b) 10 (c) 6 (d) 14

363. A completely filled d-orbital )( 10d [MLNR 1987]

(a) Spherically symmetrical (b) Has octahedral symmetry (c) Has tetrahedral symmetry (d) Depends on the atom

364. Electronic configuration 1562622 4,3,3,3,2,2,1 sdpspss represents [CPMT 2003]

(a) Ground state (b) Excited state (c) Anionic state (d) All of these

365. The correct electronic configuration of )22( =ZTi atom is [MP PMT 1999]

(a) 2262622 3,4,33,22,1 dspspss (b) 2262622 44,33,22,1 pspspss

(c) 462622 333,22,1 dpspss (d) 3162622 34,33,22,1 dspspss

366. Electronic configuration of −H is [CPMT 1985]

(a) 01 s (b) 11 s (c) 21 s (d) 11 21 ss

367. Which electronic configuration is not observing the (n + l) rule

(a) 2162622 4,333,22,1 sdpspss (b) 2762622 4,333,22,1 sdpspss

(c) 1562622 4,333,22,1 sdpspss (d) 2862622 4,333,22,1 sdpspss

368. The electronic configuration of silver atom in ground state is [CPMT 1984, 93]

(a) 110 43][ sdKr (b) 11014 654][ sdfXe (c) 110 54][ sdKr (d) 29 54][ sdKr

369. The order of filling of electrons in the orbitals of an atom will be [CBSE 1991]

(a) 3d, 4s, 4p, 4d, 5s (b) 4s, 3d, 4p, 5s, 4d (c) 5s, 4p, 3d, 4d, 5s (d) 3d, 4p, 4s, 4d, 5s

370. Which of the following has more unpaired d-electrons [CBSE 1999]

(a) +Zn (b) +2Fe (c) +3N (d) +Cu

371. The number of unpaired electrons in 422 2,2,1 pss is [NCERT 1984; CPMT 1991; MP PMT 1996, 2002]

(a) 4 (b) 2 (c) 0 (d) 1

372. Which of the following configuration is correct for iron [CBSE 1999]

(a) 562622 333,22,1 dpspss (b) 5262622 3433221 dspspss

(c) 7262622 34,33,22,1 dspspss (d) 6262622 34,33,22,1 dspspss

373. Which of the following has the maximum number of unpaired electrons [IIT 1996]

(a) +2Mg (b) +3Ti (c) +3V (d) +2Fe

374. An ion has 18 electrons in the outermost shell it is [CBSE 1990]

(a) +Cu (b) +4Th (c) +Cs (d) +K

375. Number of unpaired electrons in inert gas is [CPMT 1996]

(a) Zero (b) 8 (c) 4 (d) 18

376. Which of the following is not correct for electron distribution in the ground state [AIIMS 1982]

4s 3d

(a) Co (Ar) ↑↓ ↑↓ ↑↓ ↑ ↑ ↑

(b) Ni (Ar) ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑

(c) Cu (Ar) ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑

(d) Zn (Ar) ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

377. Electronic configuration of )21(Sc is [BHU 1997]

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(a) 1262622 3433221 dspspss (b) 2162622 3433221 dspspss

(c) 3062622 3433221 dspspss (d) 2262622 3433221 dspspss

378. The electronic configuration 11122 22221 zyx pppss 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

381. The electronic configuration of gadolinium (atomic no. 64) is [CBSE 1997]

(a) 298 654][ sdfXe (b) 217 654][ sdfXe (c) 253 654][ sdfXe (d) 226 654][ sdfXe

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

383. The atomic number of an element is 17. The number of orbitals containing electron pairs in its valence shell is [CPMT 2001]

(a) Eight (b) Six (c) Three (d) Two

384. The electronic configuration of chromium is [MP PMT 1993; MP PET 1995; BHU 2001]

(a) 2462 4333][ sdpsNe (b) 1562 4333][ sdpsNe (c) 4262 44,33][ pspsNe (d) 32162 44,333][ psdpsNe

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 22622 33,22,1 pspss . Its valency electrons are [NCERT 1973]

(a) 6 (b) 2 (c) 3 (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. +2Cu will have the following electronic configuration [MP PMT 1985]

(a) 1062622 333,22,1 dpspss (b) 1962622 4,333,22,1 sdpspss

(c) 962622 333,22,1 dpspss (d) 11062622 4,333,22,1 sdpspss

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 atom of that element [EAMCET (Engg.) 2003]

(a) 6 (b) 11 (c) 17 (d) 23

390. After np orbitals are filled, the next orbital filled will be

(a) sn )1( + (b) pn )2( + (c) dn )1( + (d) sn )2( +

391. The number of unpaired electrons in an 2O molecule is [MNR 1983]

(a) 0 (b) 1 (c) 2 (d) 3

392. How many unpaired electrons are present in +2Ni (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 )( 2+Ca is [CMC Vellore 1991]

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(a) 262622 4,33,22,1 spspss (b) 162622 4,33,22,1 spspss (c) 262622 333,22,1 dpspss (d) 562622 333,22,1 dpspss

(e) 062622 4,33,22,1 spspss

394. The number of unpaired electrons in carbon atom in excited state is [MNR 1987]

(a) One (b) Two (c) Three (d) Four

395. Which of the following electronic configurations is not possible [CPMT 2000]

(a) 22 21 ss (b) 622 221 pss (c) 2210 443 psd (d) 1222 3221 spss

396. The outer electronic structure 52 33 ps is possessed by [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) 2622 3,22,1 spss (b) 1622 3,22,1 spss (c) 622 22,1 pss (d) 262622 4,33,22.1 spspss

399. The electronic configuration of an element is 1562622 4333221 sdpspss . This represents its [IIT Screening 2000]

(a) Excited state (b) Ground state (c) Cationic form (d) Anionic form

400. Electronic configuration of C is [CPMT 1975]

(a) 222 22,1 pss (b) 322 22,1 pss (c) 22 2,1 ss (d) 622 22,1 pss

401. The number of unpaired electrons in the +2Fe ion is [MP PET 1989; Karnataka CET 2000]

(a) 0 (b) 4 (c) 6 (d) 3

402. The electronic configuration of an element with atomic number 7 i.e. nitrogen atom is [CPMT 1982. 84, 87]

(a) 322 22,1 xpss (b) 1222 222,1 yx ppss (c) 11122 2222,1 zyx pppss (d) 2122 222,1 yx ppss

403. The number of orbitals in 2p sub-shell is [NCERT 1973; MP PMT 1996]

(a) 6 (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 [Rajasthan PET 2000]

(a) 53][ dAr (b) 73][ dAr (c) 33][ dAr (d) 83][ dAr

407. Which of the following metal ions will have maximum number of unpaired electrons [CPMT 1996]

(a) 2+Fe (b) 2+Co (c) 2+Ni (d) 2+Mn

408. Number of unpaired electrons in 322 221 pss is

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

(a) 2 (b) 0 (c) 3 (d) 1

409. Energy of atomic orbitals in a particular shell is in the order [AFMC 1990]

(a) s < p < d < f (b) s > p > d > f (c) p < d < f < s (d) f > d > s > p

410. Which of the following ions is not having the configuration of neon

(a) −F (b) ++Mg (c) +Na (d) −Cl

AAddvvaannccee Level

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411. The electron density between 1s and 2s orbital is

(a) High (b) Low (c) Zero (d) None of these

412. p-orbitals of an atom in presence of magnetic field are [Pb. PMT 2002]

(a) Two fold degenerate (b) Non degenerate (c) Three fold degenerate (d) None of these

413. The energy of an electron of yp2 orbital is [AMU 1984]

(a) Greater than of xp2 orbital (b) Less than that of xp2 orbital (c) Equal to that of 2s orbital(d)

414. Assertion (A): A special line will be seen for a yx pp 22 − transition

Reason (R): Energy is released in the form of wave of light when the electron drops from xp2 to yp2 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 [BHU 1981; CPMT 1985; MP PMT 1995]

(a) 1 (b) 3 (c) 4 (d) 9

416. The correct set of quantum numbers for the unpaired electron of chlorine atom is [IIT 1989]

n l m

(a) 2 1 0

(b) 2 1 1

(c) 3 1 1

(d) 3 0 0

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

(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) π2

)1(h

ssS += (b) π2

hsS = (c)

π22

3 hS ×= (d)

π22

1 hS ×±=

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) )1( += sssz (b) π22

3 hsz ×= (c)

π4

hms sz = (d)

π22

1 hsz ×±=

420. The number of electrons that can be accommodated in 2dz orbital is [Kurukshetra CEE 2002]

(a) 10 (b) 1 (c) 4 (d) 2

421. The quantum number ‘m’ of a free gaseous atom is associated with [AIIMS 2003]

(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 [MP PET 1993]

(a) Unsymmetrical (b) Spherically symmetrical (c) Dumb-bell (d) Complicated

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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 [MLNR 1994]

(a) 1=n , 1=l , 1=lm , 2

1+=sm (b) 1=n , 0=l , 0=lm ,

2

1+=sm

(c) 1=n , 0=l , 0=lm , 2

1−=sm (d) 2=n , 0=l , 0=lm ,

2

1+=sm

425. Which of the following statements is not correct for an electron that has the quantum numbers n = 4 and m = 2 [MLNR 1993]

(a) The electron may have the quantum number 2

1+=s (b) The electron may have the quantum number l = 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 in order of increasing energy from the lowest to highest, as [IIT 1999]

(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii) (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii)

427. Which of the following sets of quantum numbers is not allowed [Orissa JEE 1997]

(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, hh 2,6 (b) 1,1, hh 2,4 (c) hh 3,6,1,0 (d) hh 6,20,0,0

429. In an excited state, a calcium atom has the electronic configuration 22 21 ss .44332 626 dspsp What is the angular momentum of

this state.

(a) h4 (b) h16 (c) h20 (d) h10

430. The four quantum number for the valence shell electron or last electron of sodium (Z = 11) is [MP PMT 1999]

(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 [CBSE 1994]

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 [IIT 1986; MP PET 1995]

n l m sm

(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 [DCE 2001]

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

[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 [EAMCET 1998]

(a) n = 4; l = 1; m = – 1; 2

1+=sm (b) n = 4; l = 1; m = 0,

2

1+=sm

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

1+=sm (d) n = 4; l = 1; m = – 1;

2

1+=sm

436. The Pauli exclusion principle is not applicable to

(a) Electrons (b) Positrons (c) Photons (d) Protons

437. If =m magnetic quantum number and l= azimuthal quantum number then

(a) 2+= lm (b) 12 2 += lm (c) 2

1−=

ml (d) 12 += ml

438. Which of the following pairs have identical values of magnetic moment

(a) +2Zn and +Cu (b) +2Co and +2Ni (c) +4Mn and +2Co (d) +2Mg and +Sc

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 [IIT Screening 1994]

(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 [Delhi PMT 2001]

(a) + 1, 0, – 1 (b) + 2, +1, 0, – 1, – 2

(c) – 3, – 2, – 1, – 0, + 1, + 2, + 3 (d) + 1, – 1

442. The four quantum numbers of the outermost orbital of K (atomic no. = 19) are [MP PET 1993, 94]

(a) n = 2, l = 0, m = 0, 2

1+=s (b) n = 4, l = 0, m = 0,

2

1+=s

(c) n = 3, l = 1, m = 1, 2

1+=s (d) n = 4, l = 2, m = – 1,

2

1+=s

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 (b) Na (c) Ne (d) F

445. Be’s 4th electron will have four quantum numbers [MNR 1985]

n l m sm

(a) 1 0 0 + 1/ 2

(b) 1 1 + 1 + 1/2

(c) 2 0 0 – 1/ 2

(d) 2 1 0 + 1/ 2

446. The value of the magnetic moment of particular ion is 2.83 Bohr magneton. The ion is

(a) +2Fe (b) +2Ni (c) +2Mn (d) +3Co

447. Which of the following ions are diamagnetic

(a) +2He (b) +3Sc (c) +2Mg (d) −2

2O

448. Which of the following sets is possible for quantum numbers [Rajasthan PET 2003]

(a) n = 4, l = 3, m = – 2, s = 0 (b) n = 4, l = 4, m = + 2, 2

1−=s

(c) n = 4, l = 4, m = – 2, 2

1+=s (d) n = 4, l = 3, m = – 2,

2

1+=s

449. If the value of azimuthal quantum number is 3, the possible values of magnetic quantum number would be

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[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

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

452. The quantum numbers + 1/ 2 and – 1/ 2 for the electron spin represent [IIT Screening 2001]

(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

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

455. Which of the following electronic configurations have the highest exchange energy

(a) (b)

(c) (d)

456. Which of the following set of quantum numbers is permissible [AIIMS 2001]

(a) n = 3; l = 2; m = 2 and 2

1+=s (b) n = 3; l = 4; m = 0; and

2

1−=s

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

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

2

1+=s

457. Which of the following sets of quantum number is not possible [MP PET 2001]

(a) n = 3; l = + 2; m = 0;2

1+=s (b) n = 3; l = 0; m = 0;

2

1−=s

AB X Y Z 3s 3p

↑↓ ↑↓ ↑↓ ↑ ↓ ↑ ↑↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑

2s 2p

↑↓ ↑↓

2s 2p

↑↓ ↑↓ ↑↓ ↑

2s 2p

↑↑ ↑↓ ↑↓ ↑

2s 2p

↑ ↑↓ ↑ ↑

3d 4s

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑

3d 4s

↑ ↑ ↑ ↑ ↑ ↑ 3d 4s

↑ ↑ ↑ ↑

3d 4s

↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑

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(c) n = 3; l = 0; m = – 1;2

1+=s (d) n = 3; l = 1; m = 0;

2

1−=s

458. Which of the following set of quantum numbers belong to highest energy [CPMT 1999]

(a) n = 4, l = 0, m = 0,2

1+=s (b) n = 3, l = 0, m = 0,

2

1+=s

(c) n = 3, l = 1, m = 1,2

1+=s (d) n = 3, l = 2, m = 1,

2

1+=s

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 [Pb. PMT 2002]

n l 1m 2m

(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 [AIIMS 1983]

(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 [IIT 1988; AMU 1999]

2s xp2 yp2 zp2

(a) ↑↓ ↑↓ ↑

(b) ↑ ↑↓ ↑ ↑

(c) ↑↓ ↑ ↑ ↑

(d) ↑↓ ↑↓ ↑↓ ↑

463. The maximum probability of finding an electron in the xyd orbital is [MP PET 1999]

(a) Along the x-axis (b) Along the y-axis

(c) At an angle of 450 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 610218 434)( pdsAr . 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)

↑↓

↑ ↑

↑↓

↑ ↑ ↑↓

↑ ↑ ↑

↑ ↑

↑↓

↑ ↑ ↑

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466. Correct statement is [BHU 2003]

(a) 210241 43,43,4 sdCusdCrsK === (b) 210242 43,43,4 sdCusdCrsK ===

(c) 210152 43,43,4 sdCusdCrsK === (d) 110151 43,43,4 sdCusdCrsK ===

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 (c) 8, 22, 24 (d) 12, 20, 22

468. Which of the following has maximum energy [AIIMS 2002]

(a) (b)

(c) (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) 2414 7,6,5][ sdfRn (b) 32114 77,6,5][ psdfRn (c) 0614 7,6,5][ sdfRn (d) 1514 7,6,5][ sdfRn

470. Which element is represented by the following electronic configuration [MP PMT 1987]

(a) Nitrogen (b) Oxygen (c) Fluorine (d) Neon

471. Which of the following statements (s) is (are) correct [IIT 1998]

(a) The electronic configuration of Cr is 15 43][ sdAr (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 3HN is – 3

472. The orbital angular momentum of an electron in 2s orbital is [IIT 1996; AIEEE 2003]

(a) π2

.2

1 h+ (b) Zero (c)

π2

h (d)

π2.2

h

473. Energy of orbit [Delhi PMT 1984, 91]

(a) Increases as we move away from nucleus (b) Decreases as we move away from nucleus

(c) Remains same as we move away from nucleus (d) None of these

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 [CPMT 1988, 93]

(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 [EAMCET 1983]

3d 3p 3s

3d 3p 3s

3d 3p 3s

3d 3p 3s

↑↓

↑↓

↑↓ ↑↓ ↑

1s

2s

2p

BBaassiicc Level

Miscellaneous Questions

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(a) β-particles are much heavier than electrons (b) β-particles are positively charged

(c) Most part of the atom is empty space (d) β-particles move with high velocity

478. The atom of the element having atomic number 14 should have [AMU 1984]

(a) One unpaired electron (b) Two unpaired electrons (c) Three unpaired electrons (d) Four unpaired electrons

479. An electronic transition from s1 orbital of an atom causes [JIPMER 1997]

(a) Absorption of energy (b) Release of energy

(c) Both release or absorption of energy (d) Unpredictable

480. Which one pair of atoms or ions will have same configuration [JIPMER 2001]

(a) +F and Ne (b) +Li and −He (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 (c) 34 (d) 33

483. Which of the following cannot be formed [AFMC 1997]

(a) +2He (b) +He (c) He (d) 2He

484. An atom has the electronic configuration of 521062622 4,4,3,3,3,2,2,1 psdpspss . Its atomic weight is 80. Its atomic number and

the number of neutrons in its nucleus shall be [MP PMT 1987]

(a) 35 and 45 (b) 45 and 35 (c) 40 and 40 (d) 30 and 50

485. Which of the following has maximum number of unpaired electron (atomic number of Fe 26) [MP PMT 2001]

(a) Fe (b) Fe (II) (c) Fe (III) (d) Fe (IV)

486. The following has zero valency [DPMT 1991]

(a) Sodium (b) Beryllium (c) Aluminum (d) Krypton

487. What is the electronic configuration of )29(2 =+ ZCu of least position [MP PET /PMT 1998; MP PET 2001]

(a) 81 34][ dsAr (b) 1102 434][ pdsAr (c) 101 34][ dsAr (d) 93][ dAr

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

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) The force of repulsion on the moving alpha particle is small

(b) The force of attraction on the alpha particle to the oppositely charged electrons is very small

(c) There is only one nucleus and large number of electrons

(d) 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 −3N ion is ( AN is the Avogadro’s number) [CBSE 1994]

AAddvvaannccee Level

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(a) AN6.1 (b) AN2.3 (c) AN1.2 (d) AN2.4

495. In neutral atom, which particles are equivalent [Rajasthan PMT 1997]

(a) ++ ep , (b) +− ee , (c) +− pe , (d) 0, np+

496. An element have atomic weight 40 and it’s electronic configuration is 62622 33221 pspss . 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 )(4)( 2,

2, rRrrp lnln π= .

(d) 2Ψ represents the atomic orbital

(e) All the above

500. Which of the following electronic configurations have zero spin multiplicity

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

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 c a b a b b c b b a a,b,c,d b a b a d d a

181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 20

0

BBaassiicc aanndd AAddvvaannccee Level

Answer Sheet

↑ ↑ ↑ ↑ ↑ ↓ ↑ ↓ ↓ ↓ ↓ ↓

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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 a c d d c c a a a d a c a b a e c