36 Structure of atom

Chapter

2
Structure of atom

John Dalton 1808, believed that matter is made up of extremely (vi) Cathode rays heat the object on which they fall due to transfer
minute indivisible particles, called atom which can takes part in chemical of kinetic energy to the object.
reactions. These can neither be created nor be destroyed. However, modern (vii) When cathode rays fall on solids such as Cu, X  rays are
researches have conclusively proved that atom is no longer an indivisible
produced.
particle. Modern structure of atom is based on Rutherford’s scattering
experiment on atoms and on the concepts of quantization of energy. (viii) Cathode rays possess ionizing power i.e., they ionize the gas
through which they pass.
Composition of atom (ix) The cathode rays produce scintillation on the photographic
The works of J.J. Thomson and Ernst Rutherford actually laid the plates.
foundation of the modern picture of the atom. It is now believed that the atom (x) They can penetrate through thin metallic sheets.
consists of several sub-atomic particles like electron, proton, neutron, positron, (xi) The nature of these rays does not depend upon the nature of gas
neutrino, meson etc. Out of these particles, the electron, proton and the neutron or the cathode material used in discharge tube.
are called fundamental subatomic particles and others are non-fundamental (xii) The e/m (charge to mass ratio) for cathode rays was found to
particles. 
be the same as that for an e (1.76  10 8 coloumb per
Electron ( e )
–1
o

gm). Thus, the cathode rays are a stream of electrons.

(1) It was discovered by J.J. Thomson (1897) and is negatively (xiii) According to Einstein’s theory of relativity, mass of electron in
charged particle. Electron is a component particle of cathode rays. Rest mass of electron(m)
(2) Cathode rays were discovered by William Crooke's & J.J. motion is, m  
Thomson (1880) using a cylindrical hard glass tube fitted with two metallic [1  (u / c)2 ]
electrodes. The tube has a side tube with a stop cock. This tube was known Where u = velocity of electron, c= velocity of light.
as discharge tube. They passed electricity (10,000V) through a discharge When u=c than mass of moving electron =.
tube at very low pressure ( 10 2 to 10 3 mm Hg) . Blue rays were
Proton ( H , H , P)
1
1 +

emerged from the cathode. These rays were termed as Cathode rays.
(1) Proton was discovered by Goldstein and is positively charged
(3) Properties of Cathode rays particle. It is a component particle of anode rays.
(i) Cathode rays travel in straight line.
(2) Goldstein (1886) used perforated cathode in the discharge tube
(ii) Cathode rays produce mechanical effect, as they can rotate the and repeated Thomson's experiment and observed the formation of anode
wheel placed in their path. rays. These rays also termed as positive or canal rays.
(iii) Cathode rays consist of negatively charged particles known as (3) Properties of anode rays
electron.
(i) Anode rays travel in straight line.
(iv) Cathode rays travel with high speed approaching that of light
(ii) Anode rays are material particles.
(ranging between 10 9 to 10 11 cm/sec)
(iii) Anode rays are positively charged.
(v) Cathode rays can cause fluorescence.
Table : 2.1 Comparison of mass, charge and specific charge of electron, proton and neutron
 Structure of atom 37

Name of constant Unit Electron(e ) –

Proton(p )
+
Neutron(n)
Amu 0.000546 1.00728 1.00899
Mass (m) Kg 9.109 × 10 –31
1.673 × 10 –27
1.675 × 10 –27

Relative 1/1837 1 1

Coulomb (C) – 1.602 × 10 –19

+1.602 × 10 –19
Zero
Charge(e) Esu – 4.8 × 10 –10
+4.8 × 10–10
Zero
Relative –1 +1 Zero
Specific charge (e/m) C/g 1.76 × 10
8
9.58 × 10 4
Zero
Density Gram / cc 2.17  10 17 1.114  1014 1.5  10 14

 The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C 12 , i.e. 1.660  10 27 kg .
Table : 2.2 Other non fundamental particles
Particle Symbol Nature Charge esu Mass Discovered by
10–10 (amu)
Positron + + 4.8029 0.0005486 Anderson (1932)
e  , 1e 0 ,  
Neutrino  0 0 < 0.00002 Pauli (1933) and Fermi (1934)
Anti-proton – – 4.8029 1.00787 Chamberlain Sugri (1956) and Weighland (1955)
p
Positive mu meson + + 4.8029 0.1152 Yukawa (1935)

Negative mu meson – – 4.8029 0.1152 Anderson (1937)

Positive pi meson + + 4.8029 0.1514

Negative pi meson – – 4.8029 0.1514
 Powell (1947)
Neutral pi meson 0 0 0.1454
0

(iv) Anode rays may get deflected by external magnetic field. Atomic number, Mass number and Atomic species
(v) Anode rays also affect the photographic plate. (1) Atomic number or Nuclear charge
(i) The number of protons present in the nucleus of the atom is
(vi) The e/m ratio of these rays is smaller than that of electrons. called atomic number (Z).
(vii) Unlike cathode rays, their e/m value is dependent upon the (ii) It was determined by Moseley as,
nature of the gas taken in the tube. It is maximum when gas present in the   a(Z  b) or aZ  ab  s 1
tube is hydrogen.
Where,   X  ray’s frequency Z Fig. 2.1
(viii) These rays produce flashes of light on ZnS screen.
Z= atomic number of the metal a & b are constant.
Neutron ( n , N)
o
1
(iii) Atomic number = Number of positive charge on nucleus =
Number of protons in nucleus = Number of electrons in nutral atom.
(1) Neutron was discovered by James Chadwick (1932) according to
(iv) Two different elements can never have identical atomic number.
the following nuclear reaction,
(2) Mass number
4 Be 9  2 He 4  6 C 12  o n 1 Mass number (A) = Number of protons or Atomic number (Z) +
Number of neutrons or Number of neutrons = A – Z .
or 5 B 11  2 He 4  7 N 14  o n 1 (i) Since mass of a proton or a neutron is not a whole number (on
atomic weight scale), weight is not necessarily a whole number.
(2) Neutron is an unstable particle. It decays as follows, (ii) The atom of an element X having mass number (A) and atomic
number (Z) may be represented by a symbol, XA.
 1 H 1  1 e  0
1 0 0 Z
0n
neutron Proton electon antinutrino

Table: 2.3 Different types of atomic species

Atomic species Similarities Differences Examples
Isotopes (i) Atomic No. (Z) (i) Mass No. (A) (i) 11 H , 12 H , 13 H
(Soddy) (ii) No. of protons (ii) No. of neutrons
 38 Structure of atom

(iii) No. of electrons (iii) Physical properties (ii) 16 17 18

8 O, 8 O, 8 O
(iv) Electronic configuration
35 37
(v) Chemical properties (iii) 17 Cl, 17 Cl

(vi) Position in the periodic table

(i) Mass No. (A) (i) Atomic No. (Z) (i) 40 40
Ar, 19 K, 40
18 20 Ca
(ii) No. of nucleons (ii) No. of protons, electrons and
130
neutrons (ii) 52 Te , 130 130
54 Xe, 56 Ba
Isobars
(iii)Electronic configuration
(iv) Chemical properties
(v) Position in the perodic table.
No. of neutrons (i) Atomic No. (i) 30 31
Si, 15 32
P, 16 S
14
(ii) Mass No., protons and electrons.
39
(iii) Electronic configuration (ii) 19 K, 40
20 Ca
Isotones
(iv) Physical and chemical properties (iii) 3
H , 42 He
1
(v) Position in the periodic table.
13 14
(iv) 6 C, 7 N
Isotopic No. (i) At No., mass No., electrons, protons, (i) 92 U 235 , 90 Th 231
(N – Z) or (A – 2Z) neutrons.
Isodiaphers (ii) Physical and chemical properties. (ii) 19 K 39 , 9 F19
65
(iii) 29 Cu , 24 Cr 55
(i) No. of electrons At. No., mass No. (i) N 2 O, CO 2 , CNO  (22e  )
(ii) Electronic configuration
(ii) CO , CN  , N 2 (14 e  )
Isoelectronic species
(iii) H  , He, Li  , Be 2  (2e  )
(iv) P 3  , S 2 , Cl  , Ar, K  and Ca 2 (18 e  )

(i) No. of atoms (i) N 2 and CO

(ii) No. of electrons
(ii) CO 2 and N 2 O
(iii) Physical and chemical properties.
Isosters (iii) HCl and F2

(iv) CaO and MgS

(v) C 6 H 6 and B3 N 3 H 6

Electromagnetic radiations
(1) Light and other forms of radiant energy propagate without any measured is terms of centimeter(cm), angstrom(Å), micron(  ) or
medium in the space in the form of waves are known as electromagnetic nanometre (nm).
radiations. These waves can be produced by a charged body moving in a
magnetic field or a magnet in a electric field. e.g.   rays,   rays,
cosmic rays, ordinary light rays etc.

(2) Characteristics
Crest Wavelength
(i) All electromagnetic radiations travel with the velocity of light.
(ii) These consist of electric and magnetic fields components Vibrating
that oscillate in directions perpendicular to each other and source
perpendicular to the direction in which the wave is travelling.
Energy
(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  (lambda) and is Trough
Fig. 2.2
 Structure of atom 39
incandescent object resolved through prism or spectroscope, it also gives
continuous spectrum of colours.
(ii) Line spectrum : If the radiation’s 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
1 Å  10 8 cm  10 10 m ; 1  10 4 cm  10 6 m ; consecutive lines. This type of spectrum is called line spectrum or atomic
spectrum..
1nm  10 7 cm  10 9 m ; 1cm  10 8 Å  10 4   10 7 nm
(2) Absorption spectrum : Spectrum produced by the absorbed
(ii) Frequency : It is defined as the number of waves which pass
radiations is called absorption spectrum.
through a point in one second. It is denoted by the symbol  (nu) and is
expressed in terms of cycles (or waves) per second (cps) or hertz (Hz). Hydrogen spectrum
(1) Hydrogen spectrum is an example of line emission spectrum or
  distance travelled in one second = velocity =c
atomic emission spectrum.
c
  (2) When an electric discharge is passed through hydrogen gas at
 low pressure, a bluish light is emitted.
(iii) Velocity : It is defined as the distance covered in one second by (3) This light shows discontinuous line spectrum of several isolated
the wave. It is denoted by the letter ‘c’. All electromagnetic waves travel sharp lines through prism.
with the same velocity, i.e., 3  1010 cm / sec . (4) All these lines of H-spectrum have Lyman, Balmer, Paschen,
c    3  1010 cm / sec Barckett, Pfund and Humphrey series. These spectral series were named by
the name of scientist discovered them.
(iv) Wave number : This is the reciprocal of wavelength, i.e., the
(5) To evaluate wavelength of various H-lines Ritz introduced the
number of wavelengths per centimetre. It is denoted by the symbol  (nu
following expression,
bar). It is expressed in cm 1 or m 1 .
1  1 1 
1     R 2  2 
   c  n1 n 2 

(v) Amplitude : It is defined as the height of the crest or depth of Where R is universal constant known as Rydberg’s constant its value
the trough of a wave. It is denoted by the letter ‘A’. It determines the is 109, 678 cm 1 .
intensity of the radiation.
The arrangement of various types of electromagnetic radiations in Plum pudding model of Thomson
the order of their increasing or decreasing wavelengths or frequencies is (1) He suggected that atom is a positively charged sphere having
known as electromagnetic spectrum. electrons embedded uniformly giving an overall picture of plum pudding.
Table: 2.4 Positively charged
Name Wavelength (Å) Frequency (Hz) + – + sphere
Radio wave
3  10 14  3  10 7 1  10 5  1  10 9 – + – +
–
Microwave Electron unifromly
3  10  6  10
7 6
1  10  5  10
9 11
+ –
+
embedded
Infrared (IR) –
6  10 6  7600 5  10 11  3.95  10 16 +
Visible 7600  3800 3.95  10 16
 7.9  10 14 Positive charge spreaded throughout the sphere

Ultraviolet (UV) 3800  150 (2) This model failed toFig.explain

2.3 the line spectrum of an element and
7.9  10 14
 2  10 16
the scattering experiment of Rutherford.
X-Rays 150  0.1 2  10 16  3  10 19
  Rays 0.1  0.01 3  10 19  3  10 20
Cosmic Rays 0.01- zero Rutherford's nuclear model
3  10 20  infinity
(1) Rutherford carried out experiment on the bombardment of thin
(10 mm) Au foil with high speed positively charged   particles emitted
–4

from Ra and gave the following observations based on this experiment,

(i) Most of the  particles passed without any deflection.
Atomic spectrum - Hydrogen spectrum
(ii) Some of them were deflected away from their path.
Atomic spectrum (iii) Only a few (one in about 10,000) were returned back to their
Spectrum is the impression produced on a photographic film when original direction of propagation.
the radiation (s) of particular wavelength (s) is (are) analysed through a Deflected
prism or diffraction grating. -particles
Types of spectrum
(1) Emission spectrum : Spectrum produced by the emitted radiation -rays
is known as emission spectrum. This spectrum corresponds to the radiation
+ve Nucleus
emitted (energy evolved) when an excited electron returns back to the
ground state.
(i) Continuous spectrum : When sunlight is passed through a prism, it
gets dispersed into continuous bands of different colours. If the light of an
Fig. 2.4 ZnS screen
 40 Structure of atom
(2) From the above observations he concluded that, an atom (iii) The total amount of energy emitted or absorbed by a body
consists of will be some whole number quanta. Hence E  nh , where n is an
(i) Nucleus which is small in size but carries the entire mass i.e. integer.
contains all the neutrons and protons.
Photoelectric effect
(ii) Extra nuclear part which contains electrons. This model was
similar to the solar system. (1) When radiations with certain minimum frequency ( 0 ) strike
(3) Properties of the nucleus the surface of a metal, the electrons are ejected from the surface of the
(i) Nucleus is a small, heavy, positively charged portion of the atom metal. This phenomenon is called photoelectric effect and the electrons
and located at the centre of the atom. emitted are called photo-electrons. The current constituted by
photoelectrons is known as photoelectric current.
(ii) All the positive charge of atom (i.e. protons) are present in
(2) The electrons are ejected only if the radiation striking the
nucleus.
surface of the metal has at least a minimum frequency ( 0 ) called
(iii) Nucleus contains neutrons and protons, and hence these
particles collectively are also referred to as nucleons. Threshold frequency. The minimum potential at which the plate
photoelectric current becomes zero is called stopping potential.
(iv) The size of nucleus is measured in Fermi (1 Fermi = 10 cm). –13

(3) The velocity or kinetic energy of the electron ejected depend

(v) The radius of nucleus is of the order of 1.5  10 13 cm. to upon the frequency of the incident radiation and is independent of its
6.5  10 13 cm. i.e. 1 .5 to 6 .5 Fermi. Generally the radius of the intensity.
(4) The number of photoelectrons ejected is proportional to the
nucleus ( rn ) is given by the following relation,
intensity of incident radiation.
rn  ro ( 1.4  10 13 cm)  A1 / 3 (5) Einstein’s photoelectric effect equation
According to Einstein,
This exhibited that nucleus is 10 5 times small in size as Maximum kinetic energy of the ejected electron = absorbed energy
compared to the total size of atom. – threshold energy

(vi) The Volume of the nucleus is about 10 39 cm 3 and that of 1 1 1 

2
mv max  h   h  0  hc   
atom is 10 24 3
cm , i.e., volume of the nucleus is 10 15
times that of an 2    0 

atom. Where,  0 and 0 are threshold frequency and threshold

15 3 wavelength.
(vii) The density of the nucleus is of the order of 10 g cm or
10 8 tonnes cm 3 or 10 12 kg / cc . If nucleus is spherical than, Bohr’s atomic model
Bohr retained the essential features of the Rutherford model of the
mass of the nucleus mass number
Density =  atom. However, in order to account for the stability of the atom he
volume of the nucleus 4
6 .023  10 23  r 3 introduced the concept of the stationary orbits. The Bohr postulates are,
3
(1) An atom consists of positively charged nucleus responsible for
almost the entire mass of the atom (This assumption is retention of
(4) Drawbacks of Rutherford's model Rutherford model).
(i) It does not obey the Maxwell theory of electrodynamics,
(2) The electrons revolve around the nucleus in certain permitted
according to it “A small charged particle moving around an oppositely
circular orbits of definite radii.
charged centre continuously loses its energy”. If an electron does so, it
should also continuously lose its energy and should set up spiral motion (3) The permitted orbits are those for which the angular
ultimately failing into the nucleus. momentum of an electron is an intergral multiple of h / 2 where h is
(ii) It could not explain the line spectra of H  atom and the Planck’s constant. If m is the mass and v is the velocity of the
electron in a permitted orbit of radius r, then
discontinuous spectrum nature.
Planck's quantum theory nh
L  mvr  ; n  1 , 2, 3, …… 
2
When black body is heated, it emits thermal radiation’s of different
wavelengths or frequency. To explain these radiations, max planck put Where L is the orbital angular momentum and n is the number
forward a theory known as planck’s quantum theory. of orbit. The integer n is called the principal quantum number. This
(i) The radiant energy which is emitted or absorbed by the black equation is known as the Bohr quantization postulate.
body is not continuous but discontinuous in the form of small discrete
(4) When electrons move in permitted discrete orbits they do not
packets of energy, each such packet of energy is called a 'quantum'. In case
radiate or lose energy. Such orbits are called stationary or non-radiating
of light, the quantum of energy is called a 'photon'.
orbits. In this manner, Bohr overcame Rutherford’s difficulty to account for
(ii) The energy of each quantum is directly proportional to the
the stability of the atom. Greater the distance of energy level from the
frequency (  ) of the radiation, i.e. nucleus, the more is the energy associated with it. The different energy
hc levels were numbered as 1,2,3,4 .. and called as K, L, M , N , …. etc.
E   or E  hv 

(5) Ordinarily an electron continues to move in a particular
Where, h Planck's constant = 6.62×10 –27
erg. sec. or stationary state or orbit. Such a state of atom is called ground state.
6.62  10 34 Joules sec . When energy is given to the electron it jumps to any higher energy
level and is said to be in the excited state. When the electron jumps
from higher to lower energy state, the energy is radiated.
 Structure of atom 41
Advantages of Bohr’s theory  1 
2 2k 2me 4 Z 2  1 
(i) Bohr’s theory satisfactorily explains the spectra of species having   n2  n2 
ch3  1 2 
one electron, viz. hydrogen atom, He  , Li 2  etc.
1  1 1 
(ii) Calculation of radius of Bohr’s orbit : According to Bohr, radius This can be represented as    RZ 2  2  2 
of n orbit in which electron moves is
th
  1
n n 2 

 h2  n2 2 2 k 2 me 4
rn   2 2 . Where, R  ; R is known as Rydberg constant. Its
ch 3
 4 me k  Z
value to be used is 109678cm 1 .
Where, n  Orbit number, m  Mass number
9.1  10 31

kg , e  Charge on the electron 1.6  10 19   Z  Atomic
The negative sign in the above equations shows that the electron
and nucleus form a bound system, i.e., the electron is attracted towards the
number of element, k = Coulombic constant 9  10 9 Nm 2c 2   nucleus. Thus, if electron is to be taken away from the nucleus, energy has
to be supplied. The energy of the electron in n  1 orbit is called the
After putting the values of m,e,k,h, we get. ground state energy; that in the n  2 orbit is called the first excited state
energy, etc. When n   then E  0 which corresponds to ionized atom
n2
rn   0 .529 Å i.e., the electron and nucleus are infinitely separated H  H   e 
Z
(ionization).
(6) Spectral evidence for quantisation (Explanation for hydrogen spectrum
(iii) Calculation of velocity of electron on the basisof bohr atomic model)
1/2
2e 2 ZK  Ze 2  2 .188  10 8 Z (i) The light absorbed or emitted as a result of an electron changing
Vn  , Vn    ; Vn  cm. sec 1 orbits produces characteristic absorption or emission spectra which can be
nh  mr  n
recorded on the photographic plates as a series of lines, the optical
(iv) Calculation of energy of electron in Bohr’s orbit spectrum of hydrogen consists of several series of lines called Lyman,
Balmar, Paschen, Brackett, Pfund and Humphrey. These spectral series were
Total energy of electron = K.E. + P.E. of electron
named by the name of scientist who discovered them.
2 2 2
kZe kZe kZe
   (ii) To evaluate wavelength of various H-lines Ritz introduced the
2r r 2r following expression,
 2 2 mZ 2 e 4 k 2 1  1 1
Substituting of r, gives us E  Where, n=1, 2,     R 2  2 
n 2h2  c  n1 n 2 
3………. 
Putting the value of m, e, k, h,  we get 2 2me 4
Where, R is =  Rydberg's constant
Z2 ch 3
E  21.8  10 12  erg per atom
n2 It's theoritical value = 109,737 cm –1
and It's experimental value =
1
109,677.581cm
Z2
 21.8  10 19  J per atom(1 J  10 7 erg) This remarkable agreement between the theoretical and
n2
2
experimental value was great achievment of the Bohr model.
Z
E  13.6  eV per atom(1eV  1.6  10 -19 J ) (iii) Although H-atom consists of only one electron yet it's spectra
n2 consist of many spectral lines.
Z2 (iv) Comparative study of important spectral series of Hydrogen is
 13.6  k .cal / mole (1 cal = 4.18J)
n2 shown in following table.
(v) If an electron from n excited state comes to various energy
th

1312 2
or Z kJmol 1 n(n  1)
n2 states, the maximum spectral lines obtained will be = . n=
2
When an electron jumps from an outer orbit (higher energy) n 2 to principal quantum number.
an inner orbit (lower energy) n1 , then the energy emitted in form of As n=6 than total number of spectral lines =
radiation is given by 6(6  1) 30
  15.
2 2
2 2 k 2 me 4 Z 2  1 1 
E  En2  En1  
 n2  n2 
(vi) Thus, at least for the hydrogen atom, the Bohr theory accurately
h2  1 2  describes the origin of atomic spectral lines.
(7) Failure of Bohr model
 1 1 
 E  13.6 Z 2  2  2 eV / atom (i) Bohr theory was very successful in predicting and accounting the
 1
n n 2  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.
1 E
As we know that E  h , c   and    , (ii) This theory could not explain the presence of multiple spectral
 hc lines.
 42 Structure of atom
(iii) This theory could not explain the splitting of spectral lines in (iv) This theory was unable to explain of dual nature of matter as
magnetic field (Zeeman effect) and in electric field (Stark effect). The explained on the basis of De broglies concept.
intensity of these spectral lines was also not explained by the Bohr atomic (v) This theory could not explain uncertainty principle. (vi) No
model. conclusion was given for the concept of quantisation of energy.
Table: 2.5

S.No. Spectral series Lies in the Transition n12n22 n12  max n2

region max   min   2 2 2
n2  n1 (n22 n12 )R R  min n 2  n1
(1) Lymen series Ultraviolet n1  1 n1  1 and n 2  2 n1  1 and n 2  
region 4
n 2  2,3,4....  4 1
 max   min  3
3R R

(2) Balmer series Visible region n1  2 n1  2 and n 2  3 n1  2 and n2  

n 2  3,4,5....  36 4 9
 max   min 
5R R 5
(3) Paschen series Infra red n=3
1 n1  3 and n 2  4 n1  3 and n 2  
region n 2  4,5,6....  16
144 9
 max   min  7
7R R
(4) Brackett series Infra red n1  4 n1  4 and n2  5 n1  4 and n 2  
region 25
n 2  5,6,7....  16  25 16
max   min  9
9R R

(5) Pfund series Infra red n1  5 n1  5 and n 2  6 n1  5 and n 2  

region 36
n 2  6,7,8....  25  36 25
 max   min  11
11R R
(6) Humphrey Far infrared n1  6 n1  6 and n 2  7 n1  6 and n 2  
series region 49
n 2  7,8....  36  49 36
 max   min  13
13 R R

Bohr–Sommerfeild’s model energy of photon (on the basis of Einstein’s mass energy

It is an extension of Bohr’s model. The electrons in an atom revolve relationship), E  mc 2

around the nuclei in elliptical orbit. The circular path is a special case of
hc h
ellipse. Association of elliptical orbits with circular orbit explains the fine Equating both  mc 2 or   which is same as de-
line spectrum of atoms.  mc
Broglie relation.  mc  p 
Dual nature of electron
(5) This was experimentally verified by Davisson and Germer by
(1) In 1924, the French physicist, Louis de Broglie suggested that if
observing diffraction effects with an electron beam. Let the electron is
light has both particle and wave like nature, the similar duality must be true
accelerated with a potential of V than the Kinetic energy is
for matter. Thus an electron, behaves both as a material particle and as a
wave. 1
mv 2  eV ; m 2 v 2  2eVm
(2) This presented a new wave mechanical theory of matter. 2
According to this theory, small particles like electrons when in motion
h
possess wave properties. mv  2eVm  P ;  
(3) According to de-broglie, the wavelength associated with a 2eVm
particle of mass m, moving with velocity v is given by the relation (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
h
 , where h = Planck’s constant. with circumference and multiply with a whole number
mv
2r
(4) This can be derived as follows according to Planck’s equation, 2r  n  or  
n
h.c  c
E  h     
   From de-Broglie equation,  
h
.
mv
 Structure of atom 43

h 2r nh (iii) If  2 is maximum than probability of finding e  is maximum

Thus  or mvr 
mv n 2 around nucleus and the place where probability of finding e  is maximum
(7) The de-Broglie equation is applicable to all material objects but is called electron density, electron cloud or an atomic orbital. It is different
it has significance only in case of microscopic particles. Since, we come from the Bohr’s orbit.
across macroscopic objects in our everyday life, de-broglie relationship has
(iv) The solution of this equation provides a set of number called
no significance in everyday life.
quantum numbers which describe specific or definite energy state of the
Heisenberg’s uncertainty principle electron in atom and information about the shapes and orientations of the
most probable distribution of electrons around the nucleus.
This principle states “It is impossible to specify at any given
Radial probability distribution curves : Radial probability is
moment both the position and momentum (velocity) of an electron”.
R  4r 2 dr 2 . The plats of R distance from nucleus as follows
h
Mathematically it is represented as , x . p 
4 14
12
Where x  uncertainty is position of the particle, 10 5 5
p  uncertainty in the momentum of the particle

4r2 dr 2

4r2 dr 2

4r2 dr 2
8 4 4
6 3 3
Now since p  m v 4 2 2
2 1 1
h h
So equation becomes, x . m v  or x  v  0 0 0
4 4m 2 4 6 8 2 4 6 8 2 4 6 8
0.53Å r(Å) 0.53Å 2.7Å r(Å) 2.1Å r(Å)
In terms of uncertainty in energy, E and uncertainty in time 1s 2s 2s
h Fig. 2.5
t, this principle is written as, E . t  Quantum numbers
4
Each orbital in an atom is specified by a set of three quantum
Schrödinger wave equation numbers (n, l, m) and each electron is designated by a set of four quantum
(1) Schrodinger wave equation is given by Erwin Schrödinger in numbers (n, l, m and s).
1926 and based on dual nature of electron. (1) Principle quantum number (n)
(2) In it electron is described as a three dimensional wave in the electric (i) It was proposed by Bohr and denoted by ‘n’.
field of a positively charged nucleus.
(ii) It determines the average distance between electron and nucleus,
(3) The probability of finding an electron at any point around the means it denotes the size of atom.
nucleus can be determined by the help of Schrodinger wave equation which
is, (iii) It determine the energy of the electron in an orbit where
electron is present.
2 2 2 8 2m (iv) The maximum number of an electron in an orbit
   (E  V )   0
x 2
y 2
z 2
h2
represented by this quantum number as 2n 2 . No energy shell in
Where x, y and z are the 3 space co-ordinates, m = mass of atoms of known elements possess more than 32 electrons.
electron, h = Planck’s constant, E = Total energy, V = potential energy of (v) It gives the information of orbit K, L, M, N------------.
electron,  = amplitude of wave also called as wave function,  = for an (vi) Angular momentum can also be calculated using principle
infinitesimal change. quantum number
(4) The Schrodinger wave equation can also be written as, (2) Azimuthal quantum number (l)
8 2m (i) Azimuthal quantum number is also known as angular
2   (E  V )   0
h 2 quantum number. Proposed by Sommerfield and denoted by ‘ l ’.

Where  = laplacian operator. (ii) It determines the number of sub shells or sublevels to which the
electron belongs.
(5) Physical significance of  and  2 (iii) It tells about the shape of subshells.
(i) The wave function  represents the amplitude of the (iv) It also expresses the energies of subshells s  p  d  f
electron wave. The amplitude  is thus a function of space co-ordinates (increasing energy).
and time i.e.   (x , y, z...... times)
(v) The value of l  (n  1) always. Where ‘n’ is the number of
(ii) For a single particle, the square of the wave function ( ) at 2 principle shell.
any point is proportional to the probability of finding the particle at that (vi) Value of l = 0 1 2 3…..(n-1)
point. Name of subshell = s p d f

Shape of subshell = Spheric Dumbbell Double Complex

al dumbbell
 44 Structure of atom
(vii) It represent the orbital angular momentum. Which is equal to (vii) Degenerate orbitals : Orbitals having the same energy are
h known as degenerate orbitals. e.g. for p subshell p x p y p z
l(l  1)
2
(viii) The number of degenerate orbitals of s subshell =0.
(viii) The maximum number of electrons in subshell  2(2l  1)
(4) Spin quantum numbers (s)
s  subshell 2 electrons d  subshell 10 electrons (i) It was proposed by Goldshmidt & Ulen Back and denoted by the
symbol of ‘s’.
p  subshell 6 electrons f  subshell 14 electrons.
(ii) The value of ' s' is  1/2 and - 1/2, which signifies the spin
(ix) For a given value of ‘n’ the total values of ‘l’ is always equal to
or rotation or direction of electron on it’s axis during movement.
the value of ‘n’.
(iii) The spin may be clockwise or anticlockwise.
(3) Magnetic quantum number (m)
(iv) It represents the value of spin angular momentum is equal to
(i) It was proposed by Zeeman and denoted by ‘m’.
h
s(s  1).
(ii) It gives the number of permitted orientation of subshells. 2
(iii) The value of m varies from –l to +l through zero.
(v) Maximum spin of an atom  1 / 2  number of unpaired
(iv) It tells about the splitting of spectral lines in the magnetic field electron.
i.e. this quantum number proves the Zeeman effect.
(vi) This quantum number is not the result of solution of
(v) For a given value of ‘n’ the total value of ’m’ is equal to n 2 . schrodinger equation as solved for H-atom.

(vi) For a given value of ‘l’ the total value of ‘m’ is equal to (2l  1).
Table : 2.6 Distribution of electrons among the quantum levels
Designation of Number of Orbitals in the subshell
n l m
orbitals
1 0 0 1s 1
2 0 0 2s 1

2 1 –1, 0, +1 2p 3
3 0 0 3s 1
3 1 –1, 0, +1 3p 3

3 2 –2, –1, 0, +1, +2 3d 5

4 0 0 4s 1

4 1 –1, 0, +1 4p 3

4 2 –2, –1, 0, +1, +2 4d 5

4 3 –3, –2, –1, 0, +1, +2, +3 4f 7

Shape of orbitals
(1) Shape of ‘s’ orbital (2) Shape of ‘p’ orbitals
(i) For ‘s’ orbital l=0 & m=0 so ‘s’ orbital have only one (i) For ‘p’ orbital l=1, & m=+1,0,–1 means there are three ‘p’ orbitals,
unidirectional orientation i.e. the probability of finding the electrons is same which is symbolised as p x , p y , p z .
in all directions.
(ii) The size and energy of ‘s’ orbital with increasing ‘n’ will be (ii) Shape of ‘p’ orbital is dumb bell in which the two lobes on
opposite side separated by the nodal plane.
1s  2s  3 s  4 s.
(iii) p-orbital has directional properties.
(iii) s-orbitals known as radial node or modal surface. But there is
no radial node for 1s orbital since it is starting from the nucleus. Z Z Z
Y Y Y

X X X

Px PY Pz

Fig. 2.7
1S 2S
Fig. 2.6
 Structure of atom 45

(3) Shape of ‘d’ orbital The orbital diagram does not represent a possible
(i) For the ‘d’ orbital l =2 then the values of ‘m’ are –2, –1, 0, +1, +2. arrangement of electrons
It shows that the ‘d’ orbitals has five orbitals as d xy , d yz , d zx , d x 2 y 2 , d z 2 . 1s
Because there are only two possible values of s, an orbital can hold
(ii) Each ‘d’ orbital identical in shape, size and energy. not more than two electrons.
(iii) The shape of d orbital is double dumb bell . (4) Hund’s Rule of maximum multiplicity
(iv) It has directional properties. This rule deals with the filling of electrons in the orbitals having
Z Z equal energy (degenerate orbitals). According to this rule,
Y Y
“Electron pairing in p, d and f orbitals cannot occur until each
orbitals of a given subshell contains one electron each or is singly occupied”.
X X
This is due to the fact that electrons being identical in charge, repel
each other when present in the same orbital. This repulsion can however be
dZX dXY minimised if two electrons move as far apart as possible by occupying
Z different degenerate orbitals. All the unpaired electrons in a degenerate set
Z Y
Y of orbitals will have same spin.
Z As we now know the Hund’s rule, let us see how the three electrons
X
are arranged in p orbitals.
X Z
Y The important point ot be remembered is that all the singly
occupied orbitals should have electrons with parallel spins i.e in the same
dYZ dX2–Y2 direction either-clockwise or anticlockwise.
2 px 2 py 2 pz 2 px 2 py 2 pz
X or
Electronic configurations of elements
d
z2
On the basis of the elecronic configuration principles the electronic
Fig. 2.8 configuration of various elements are given in the following table :
The above method of writing the electronic configurations is quite
(4) Shape of ‘f’ orbital cumbersome. Hence, usually the electronic configuration of the atom of any
(i) For the ‘f’ orbital l=3 then the values of ‘m’ are –3, –2, – element is simply represented by the notation.
1,0,+1,+2,+3. It shows that the ‘f’ orbitals have seven orientation as Number of
fx ( x 2  y 2 ) , fy( x 2  y 2 ) , fz ( x 2  y 2 ), fxyz , fz 3 , fyz 2 and fxz 2 . electrons
Present
(ii) The ‘f’ orbital is complicated in shape.
number of nlx symbol of
Rules for filling of electrons in various orbitals principal
subshell
shell
The atom is built up by filling electrons in various orbitals according Some Unexpected Electronic Configuration
to the following rules,
Some of the exceptions are important though, because they occur
(1) Aufbau’s principle with common elements, notably chromium and copper.
This principle states that the electrons are added one by one to the Cu has 29 electrons. Its excepted electronic configuration is
various orbitals in order of their increasing energy starting with the orbital
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 9 but in reality the configuration is
of lowest energy. The increasing order of energy of various orbitals is
2 2 6 2 6 1 10
1s 2 s 2 p 3 s 3 p 4 s 3 d as this configuration is more stable. Similarly
1s  2s  2 p  3 s  3 p  4 s  3d  4 p  5 s  4 d  5 p  6 s  4 f
Cr has the configuration of 1s 2 2 s 2 sp 6 3 s 2 3 p 6 4 s 1 3d 5 instead of
 5d  6 p  7 s  5 f  6 d  7 p.........
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 4 .
(2) (n+l) Rule
Factors responsible for the extra stability of half-filled and
In neutral isolated atom, the lower the value of (n + l) for an orbital, completely filled subshells,
lower is its energy. However, if the two different types of orbitals have the
(i) Symmetrical distribution : It is well known fact that symmetry
same value of (n + l), the orbitals with lower value of n has lower energy.
leads to stability. Thus the electronic configuration in which all the orbitals
(3) Pauli’s exclusion principle of the same subshell are either completely filled or are exactly half filled are
According to this principle “no two electrons in an atom will have more stable because of symmetrical distribution of electrons.
same value of all the four quantum numbers”. (ii) Exchange energy : The electrons with parallel spins present in
the degenerate orbitals tend to exchange their position. The energy released
If one electron in an atom has the quantum numbers n  1 , l  0 , during this exchange is called exchange energy. The number of exchanges
m  0 and s  1 / 2 , no other electron can have the same four that can take place is maximum when the degenerate orbtials (orbitals of
quantum numbers. In other words, we cannot place two electrons with the same subshell having equal energy) are exactly half-filled or completely. As a
same value of s in a 1s orbital. result, the exchange energy is maximum and so it the stability.
 46 Structure of atom

 When energy or frequency of scattered ray is lesser than the

incident ray, it is known as Compton effect.
 The instrument used to record solar spectrum is called
spectrometer or spectrograph developed by Bunsen and Kirchoff in 1859.
 The intensities of spectral lines decreases with increase in the value
 All lines in the visible region are of Balmer series but reverse is not of n . For example, the intensity of first Lyman line (2  1) is greater
true i.e., all Balmer lines will not fall in visible region. than second line (3  1) and so on.
 A part of an atom up to penultimate shell is a kernel or atomic
core.  In Balmer series of hydrogen spectrum the first line (3  2) is
 If the energy supplied to hydrogen atom is less than 13.6 eV it will also known as L line. The second line (4  2) is L  line. The line
awpt or absorb only those quanta which can take it to a certain higher
energy level i.e., all those photons having energy less than or more than a from infinity energy shell is called limiting line.
particular energy level will not be absorbed by hydrogen atom, but if
energy supplied to hydrogen atom is more than 13.6eV then all photons
are absorbed and excess energy appear as kinetic energy of emitted
photo electron.
 No of nodes in any orbital  (n  l  1)
 No of nodal planes in an orbitals  l
 The d orbital which does not have four lobes is d z 2 Discovery and Properties of anode, cathode rays
 The d orbital whose lobes lie along the axis is d x 2 y 2 neutron and Nuclear structure
h
 Spin angular momentum  s(s  1) 1. A neutral atom (Atomic no. > 1) consists of
2
[CPMT 1982]
n
 Total spin   ; where n is no of unpaired e  (a) Only protons
2
(b) Neutrons + protons
 Magnetic moment  n(n  2) B.M. (Bohr magnetron) of n (c) Neutrons + electrons
unpaired e  (d) Neutron + proton + electron
 Ion with unpaired electron in d or f orbital will be coloured. 2. The nucleus of the atom consists of
 Exception of E.C. are Cr(24) , Cu(29) , Mo(42) , Ag(47) , [CPMT 1973, 74, 78, 83, 84; MADT Bihar 1980;
W (74 ) , Au(79) . DPMT 1982, 85; MP PMT 1999]

2r h (a) Proton and neutron

 No. of waves n  (where   )
 mv (b) Proton and electron
v (c) Neutron and electron
 No. of revolutions of e  per second is  .
2r (d) Proton, neutron and electron
 The solution of schrodinger wave equation gives principal, 3. The size of nucleus is of the order of
azimuthal and magnetic quantum numbers but not the spin quantum [CPMT 1982; MP PMT 1991]
number. 12 8
(a) 10 m (b) 10 m
 In the Rydberg formula, when n 2   the line produced is called
the limiting line of that series. (c) 10 15 m (d) 10 10 m
4. Positive ions are formed from the neutral atom by the
 Among various forms of visible light, violet colour has shortest
wavelength, highest frequency and highest energy. [CPMT 1976]
(a) Increase of nuclear charge
 Red coloured light has largest wavelength, least frequency and
lowest energy in visible light. (b) Gain of protons
(c) Loss of electrons
 Elements give line spectra. The line spectrum is characteristic of
the excited atom producing it. No two elements have identical line (d) Loss of protons
spectrum. 5. The electron is
 The line spectrum results from the emission of radiations from the [DPMT 1982; MADT Bihar 1980]
atoms of the elements and is therefore called as atomic spectrum.
(a)  -ray particle (b)  -ray particle
 Atoms give line spectra (known as atomic spectrum) and the molecules (c) Hydrogen ion (d) Positron
give band spectra (known as molecular spectrum).
6. Who discovered neutron
 The negative potential at which the photoelectric current becomes [IIT 1982; BITS 1988;CPMT 1977; NCERT 1974;
zero is called cut off potential or stopping potential.
MP PMT 1992; MP PET 2002]
 Structure of atom 47
(a) James Chadwick (b) William Crooks
(c) J.J. Thomson (d) Rutherford
7. The ratio of charge and mass would be greater for
[BHU 2005]
(a) Proton (b) Electron
(c) Neutron (d) Alpha
 48 Structure of atom

8. Magnitude of K.E. in an orbit is equal to [BCECE 2005] (a) Proton is nucleus of deuterium
(a) Half of the potential energy (b) Proton is ionized hy drogen m olecule
(b) Twice of the potential energy (c) Proton is ionized hy drogen atom
(c) One fourth of the potential energy (d) Proton is  -particle
(d) None of these 20. Cathode rays are made up of [A MU 1983]
9. The density of neutrons is of the order [NCERT 1980] (a) Positiv ely charged particles
(a) 10 3 kg / cc (b) 10 6 kg / cc (b) Negativ ely charged particles
(c) Neutral particles
(c) 10 9 kg / cc (d) 1011 kg / cc
(d) None of these
10. The discov ery of neutron becom es v ery late because
[CPMT 1987; AIIMS 1998] 21. Anode rays were discov ered by [DPMT 1985]
(a) Neutrons are present in nucleus (a) Goldstein (b) J. Stoney
(b) Neutrons are highly unstable particles (c) Rutherford (d) J.J. Thom son
(c) Neutrons are chargeless 22. The radius of an atom is of the order of
(d) Neutrons do not m ov e [A MU 1982; IIT 1985; MP PMT 1995]
11. The fundamental particles present in the nucleus of an (a) 10 10
cm (b) 10 13 cm
atom are [CPMT 1983, 84]
(a) Alpha particles and electrons (c) 10 15 cm (d) 10 8 cm
23. Neutron possesses [CPMT 1982]
(b) Neutrons and protons
(a) Positiv e charge (b) Negativ e charge
(c) Neutrons and electrons
(c) No charge (d) All are correct
(d) Electrons, neutrons and protons
24. Neutron is a fundamental particle carrying
12. The order of density in nucleus is
[CPMT 1990]
[NCERT 1981, CPMT 1981, 2003]
(a) A charge of + 1 unit and a m ass of 1 unit
(a) 10 8 kg / cc (b) 10 8 kg / cc (b) No charge and a m ass of 1 unit
(c) 10 9 kg / cc (d) 10 12 kg / cc (c) No charge and no m ass
13. Cathode rays are [JIPMER 1991; NCERT 1976] (d) A charg of –1 and a m ass of 1 unit
(a) Protons (b) Electrons 25. Cathode rays have [CPMT 1982]

(c) Neutrons (d)  -particles (a) Mass only (b) Charge only
(c) No m ass and charge (d) Mass and charge both
14. Number of neutron in C 12 is [BCECE 2005]
26. The size of nucleus is m easured in
(a) 6 (b) 7 [EA MCET 1988; CPMT 1994]
(c) 8 (d) 9 (a) am u (b) Angstrom
15. Heaviest particle is [DPMT 1983; MP PET 1999] (c) Ferm i (d) cm
(a) Meson (b) Neutron 27 . Which phrase would be incorrect to use
(c) Proton (d) Electron [A MU (Engg.) 1999]
16. Penetration power of prot on is (a) A m olecular of a com pound
[BHU 1985; CPMT 1982, 88] (b) A m olecule of an elem ent
(a) More than electron (b) Less than electron (c) An atom of an elem ent
(c) More than neutron (d) None (d) None of these
17. An elementary particle is [CPMT 1973] 28. Which one of the following pairs is not correctly m atched
(a) An elem ent present in a com pound [MP PET 2002]
(b) An atom present in an elem ent (a) Rutherford-Proton
(c) A sub-atom ic particle (b) J.J. Thom som -Electron
(d) A fragm ent of an atom (c) J.H. Chadwick-Neutron
18. The nucleus of helium contains (d) Bohr-Isotope
[CPMT 1972; DPMT 1982] 29. Proton was discov ered by [A FMC 2004]
(a) Four protons (a) Chadwick (b) Thom son
(b) Four neutrons (c) Goldstein (d) Bohr
(c) Two neutrons and two protons 30. The m inimum real charge on any particle which can exist
(d) Four protons and two electrons is
19. Which is correct statem ent about proton [RPMT 2000]
[CPMT 1979; MP PMT 1985; NCERT 1985; MP PET 1999] (a) 1.6  10 19 Coulomb (b) 1.6  10 10 Coulomb
 Structure of atom 49
(a) Atom ic weight (b) Atom ic num ber
(c) 4.8  10 10 Coulomb (d) Zero
(c) Equiv alent weight (d) Electron affinity
31. The nature of anode ray s depends upon 2. The nucleus of the element having atomic number 2 5 and
[MP PET 2004] atom ic weight 55 will contain
(a) Nature of electrode (b) Nature of residual gas [CPMT 1986; MP PMT 1987]
(c) Nature of discharge tube (d) All the abov e (a) 2 5 protons and 3 0 neutrons
32. One would expect proton to hav e v ery large (b) 2 5 neutrons and 3 0 protons
(c) 55 protons
[Pb. CET 2004]
(d) 55 neutrons
(a) Ionization potential (b) Radius
3. If W is atom ic weight and N is the atom ic num ber of an
(c) Charge (d) Hy dration energy elem ent, then [CPMT 1971, 80, 89]
33. The m ass of a m ol of proton and electron is (a) Num ber of e 1  W  N
(a) 6.023  10 23 g (b) 1.008 g and 0.55mg (b) Num ber of 0 n 1  W  N

(c) 9.1  10 28 kg (d) 2 gm (c) Num ber of 1 H 1  W  N

34. The average distance of an electron in an atom from its (d) Num ber of 0 n 1  N
nucleus is of the order of [MP PET 1996]
4. The total number of neutrons in dipositive zinc ions with
(a) 10 6 m (b) 10 6 m mass number 70 is [IIT 1979; Bih a r MEE 1997]

(c) 10 10 m (d) 10 15 m (a) 3 4 (b) 4 0

(c) 3 6 (d) 3 8
35. The mass of 1 mole of electrons is [Pb. CET 2004]
5. Which of the following are isoelectronic with one another
(a) 9.1  10 28 g (b) 1.008 mg [NCERT 1983; EAMCET 1989]

(c) 0.55 mg (d) 9.1  10 27
g (a) Na and Ne (b) K  and O
(c) Ne and O (d) Na  and K 
36. The ratio of specific charge of a proton and an  -particle
6. The num ber of electrons in one m olecule of CO 2 are
is
[IIT 1979; MP PMT 1994; RPMT 1999]
[MP PET 1999]
(a) 2 2 (b) 4 4
(a) 2 : 1 (b) 1 : 2
(c) 6 6 (d) 88
(c) 1 : 4 (d) 1 : 1
7. Chlorine atom differs from chloride ion in the num ber of
37 . Ratio of masses of proton and electron is [BHU 1998] [NCERT 1972; MP PMT 1995]
(a) Infinite (b) 1.8  10 3 (a) Proton (b) Neutron
(c) Electrons (d) Protons and electrons
(c) 1 .8 (d) None of these
8. CO has sam e electrons as or the ion that is isoelectronic
38. Splitting of signals is caused by [Pb. PMT 2000]
with CO is [CPMT 1984; IIT 1982;
(a) Proton (b) Neutron EA MCET 1990; CBSE PMT 1997]
(c) Positron (d) Electron (a) N 2 (b) CN 
39. The proton and neutron are collectiv ely called as
(c) O 2 (d) O 2
[MP PET 2001]
9. The m ass of an atom is constituted m ainly by
(a) Deutron (b) Positron
[DPMT 1984, 91; AFMC 1990]
(c) Meson (d) Nucleon
(a) Neutron and neutrino (b) Neutron and electron
40. Which of the following has the sam e m ass as that of an (c) Neutron and proton (d) Proton and electron
electron [A FMC 2002]
10. The atom ic num ber of an elem ent represents
(a) Photon (b) Neutron [CPMT 1983; CBSE PMT 1990; NCERT 1973; AMU 1984]
(c) Positron (d) Proton (a) Num ber of neutrons in the nucleus
41. What is the ratio of m ass of an electron to the m ass of a (b) Num ber of protons in the nucleus
proton (c) Atom ic weight of elem ent
[UPSEAT 2004] (d) Valency of elem ent
11. An atom has 2 6 electrons and its atomic weight is 56. The
(a) 1 : 2 (b) 1 : 1
num ber of neutrons in the nucleus of the atom will be
(c) 1 : 1 83 7 (d) 1 : 3 [CPMT 1980]
(a) 2 6 (b) 3 0
Atomic number, Mass number, Atomic species
(c) 3 6 (d) 56
12. The m ost probable radius (in pm) for finding the electron
1. The number of electrons in an atom of an element is equal
to its [BHU 1979] in He  is [A IIMS 2005]
 50 Structure of atom
(a) 0.0 (b) 52 .9 (b) The number of nucleons is double of the number of
(c) 2 6 .5 (d) 1 05.8 electrons
(c) The number of protons is half of the number of neutrons
13. The number of unpaired electrons in the Fe 2  ion is
(d) The number of nucleons is double of the atomic num ber
[MP PET 1989; KCET 2000]
(a) 0 (b) 4 24. Pick out the isoelectronic structures from the following
(c) 6 (d) 3 CH 3 H3O NH 3 CH 3 [IIT 1993]
I II III IV
14. A sodium cation has different num ber of electrons from
(a) I and II (b) I and IV
(a) O 2  (b) F 
(c) I and III (d) II, III and IV
(c) Li  (d) Al 3
25. Number of electrons in CONH 2 is [A MU 1988]
15. An atom which has lost one electron would be
[CPMT 1986] (a) 2 2 (b) 2 4
(a) Negativ ely charged (c) 2 0 (d) 2 8
(b) Positiv ely charged 26. The atomic number of an element having the valency shell
(c) Electrically neutral electronic configuration 4 s 2 4 p 6 is [MP PMT 1991]
(d) Carry double positiv e charge (a) 3 5 (b) 3 6
16. Num ber of electrons in the outermost orbit of the element (c) 3 7 (d) 3 8
of atomic num ber 15 is [CPMT 1988, 93]
27 . The present atom ic weight scale is based on
(a) 1 (b) 3 [EA MCET 1988; MP PMT 2002]
(c) 5 (d) 7
(a) C 12
(b) O 16
17. The atom ic weight of an elem ent is double its atom ic
num ber. If there are four electrons in 2 p orbital, the (c) H 1 (d) C 13
elem ent is [A MU 1983] 28. Isoelectronic species are [EA MCET 1989]
(a) C (b) N 
(a) K , Cl 
(b) Na , Cl 


(c) O (d) Ca
(c) Na, Ar (d) Na  , Ar
18. An atom has the electronic configuration of 1s 2 ,2 s 2 2 p 6 ,
29. If the atom ic weight of an element is 2 3 tim es that of the
3 s 2 3 p 6 3d 10 ,4 s 2 4 p 5 . Its atom ic weight is 80. Its atom ic lightest element and it has 1 1 protons, then it contains
num ber and the number of neutrons in its nucleus shall [EA MCET 1986; AFMC 1989]
be
(a) 1 1 protons, 2 3 neutrons, 1 1 electrons
[MP PMT 1987]
(b) 1 1 protons, 1 1 neutrons, 1 1 electrons
(a) 3 5 and 4 5 (b) 4 5 and 3 5
(c) 4 0 and 4 0 (d) 3 0 and 50 (c) 1 1 protons, 1 2 neutrons, 1 1 electrons
19. Which of the following particles has m ore electrons than (d) 1 1 protons, 1 1 neutrons, 2 3 electrons
neutrons 30. Which of the following oxides of nitrogen is isoelectronic
 with CO 2 [CBSE PMT 1990]
(a) C (b) F
(c) O 2
(d) Al 3 (a) NO 2 (b) N 2 O
20. Com pared with an atom of atomic weight 1 2 and atom ic (c) NO (d) N 2 O 2
num ber 6 , the atom of atom ic weight 1 3 and atom ic
31. The ratio between the neutrons in C and Si with respect
num ber 6 [NCERT 1971] to atom ic m asses 1 2 and 2 8 is [
(a) Contains m ore neutrons(b) Contains m ore electrons
(a) 2 : 3 (b) 3 : 2
(c) Contains m ore protons(d) Is a different elem ent
(c) 3 : 7 (d) 7 : 3
21. In the nucleus of 20 Ca 40 there are 32. The atom ic num ber of an elem ent is alway s equal to
[CPMT 1990; EAMCET 1991] [MP PMT 1994]
(a) 40 protons and 2 0 electrons (a) Atom ic weight div ided by 2
(b) 20 protons and 4 0 electrons (b) Num ber of neutrons in the nucleus
(c) 20 protons and 2 0 neutrons (c) Weight of the nucleus
(d) 20 protons and 4 0 neutrons (d) Electrical charge of the nucleus
22. Na  ion is isoelectronic with [CPMT 1990] 33. Which of the following is isoelectronic with carbon atom
2 [MP PMT 1994; UPSEAT 2000]
(a) Li  (b) Mg

(a) Na (b) Al 3 
(c) Ca 2 (d) Ba 2
23. Ca has atomic no. 2 0 and atomic weight 4 0. Which of the (c) O 2  (d) N 
following statem ents is not correct about Ca atom 34. CO 2 is isostructural with
[MP PET 1993] [IIT 1986; MP PMT 1986, 94, 95]
(a) The num ber of electrons is sam e as the num ber of (a) SnCl 2 (b) SO 2
neutrons
 Structure of atom 51
(c) HgCl2 (d) All the abov e 47 . The num ber of electrons in [19
40
K ]1 is
35. The hy dride ions (H  ) are isoelectronic with [CPMT 1997; AFMC 1999]
(a) 1 9 (b) 2 0
[A FMC 1995; Bihar MEE 1997]
(c) 1 8 (d) 4 0
(a) Li (b) He  48. The number of electrons and neutrons of an element is 1 8
(c) He (d) Be and 2 0 respectiv ely . Its m ass num ber is
[CPMT 1997; Pb. PMT 1999; MP PMT 1999]
36. The num ber of electrons in the nucleus of C 12 is
[A FMC 1995] (a) 1 7 (b) 3 7
(c) 2 (d) 3 8
(a) 6 (b) 1 2
49. Num ber of protons, neutrons and electrons in the
(c) 0 (d) 3
37 . An elem ent has electronic configuration 2 , 8, 1 8, 1 . If its elem ent 89
231
Y is [A FMC 1997]
atom ic weight is 6 3 , t hen how m any neutrons will be (a) 89, 2 3 1 , 89 (b) 89, 89, 2 4 2
present in its nucleus (c) 89, 1 4 2 , 89 (d) 89, 7 1 , 89
(a) 3 0 (b) 3 2 50. Be 2  is isoelectronic with [EA MCET 1998]
(c) 3 4 (d) 3 3 2 
(a) Mg (b) Na
38. The nucleus of the elem ent E 45 contains 
21 (c) Li (d) H 
(a) 4 5 protons and 2 1 neutrons 51. An isostere is [UPSEA T 1999]
(b) 2 1 protons and 2 4 neutrons (a) NO 2 and O 3 (b) NO 2 and PO43 
(c) 2 1 protons and 4 5 neutrons
(d) 2 4 protons and 2 1 neutrons (c) CO 2 , N 2 O, NO 3 (d) ClO4 and OCN 
39. Neutrons are found in atom s of all elem ents except in 52. Nitrogen atom has an atomic number of 7 and oxygen has
[MP PMT 1997] an atomic number 8. The total num ber of electrons in a
(a) Chlorine (b) Oxy gen nitrate ion will be [Pb. PMT 2000]
(a) 8 (b) 1 6
(c) Argon (d) Hy drogen
(c) 3 2 (d) 6 4
40. The m ass number of an anion, X 3  , is 14. If there are ten 53. If m olecular mass and atom ic m ass of sulphur are 2 56
electrons in the anion, the num ber of neutron s in the and 32 respectively, its atom icity is [RPET 2000]
nucleus of atom , X 2 of the elem ent will be (a) 2 (b) 8
[MP PMT 1999] (c) 4 (d) 1 6
(a) 1 0 (b) 1 4 54. The nitride ion in lithium nitride is com posed of
(c) 7 (d) 5 [KCET 2000]
41. Which of the following are isoelectronic species (a) 7 protons + 1 0 electrons
I  CH 3 , II  NH 2 , III  NH 4 , IV  NH 3 [CPMT 1999] (b) 1 0 protons + 1 0 electrons
(c) 7 protons + 7 protons
(a) I, II, III (b) II, III, IV (d) 1 0 protons + 7 electrons
(c) I, II, IV (d) I and II 55. The atomic number of an elem ent is 1 7 . The num ber of
42. The charge on the atom containing 1 7 protons, 1 8 orbitals containing electron pairs in its v alence shell is
neutrons and 18 electrons is [A IIMS 1996] [CPMT 2001]
(a) 1 (b) 2 (a) Eight (b) Six
(c) 1 (d) Zero (c) Three (d) Two
43. Number of unpaired electrons in inert gas is[CPMT 1996] 56. The atomic number of an element is 3 5 and mass number
(a) Zero (b) 8 is 81 . The number of electrons in the outer m ost shell is
(c) 4 (d) 1 8 [UPSEAT 2001]
44. In neutral atom , which particles are equiv alent (a) 7 (b) 6
[RPMT 1997] (c) 5 (d) 3
   
(a) p , e (b) e , e 57 . Which of the following is not isoelectronic[MP PET 2002]

(c) e , p 
(d) p , n  o
(a) Na  (b) Mg 2 
45. Nuclei tend to have m ore neutrons than protons at high (c) O 2  (d) Cl 
mass numbers because [Roor kee Qu a l ify in g 1998]
58. The charge of an electron is  1.6  10 19 C. The v alue of
(a) Neutrons are neutral particles 
(b) Neutrons hav e m ore m ass than protons free charge on Li ion will be
(c) More neutrons m inim ize the coulom b repulsion [A FMC 2002; KCET (Engg.) 2002]
19
(d) Neutrons decrease the binding energy (a) 3.6  10 C (b) 1  10 19 C
46. Which one of the following is not isoelectronic with O 2  (c) 1.6  10 19 C (d) 2.6  10 19 C
[CBSE PMT 1994] 59. Iso-electronic species is [RPMT 2002]
(a) N 3  (b) F  
(a) F , O 2 
(b) F , O
(c) Tl  (d) Na 
 52 Structure of atom

(c) F  , O  (d) F  , O 2 (b) 4 protons and 7 electrons

(c) 4 protons and 1 0 electrons
60. An elem ent hav e atom ic weight 4 0 and it’s electronic
(d) 1 0 protons and 7 electrons
configuration is 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 . Then its atom ic
number and number of neutrons will be [RPMT 2002] 73. Num ber of neutrons in heav y hy drogen atom is
(a) 1 8 and 2 2 (b) 2 2 and 1 8 [MP PMT 1986]

(c) 2 6 and 2 0 (d) 4 0 and 1 8 (a) 0 (b) 1

61. The nucleus of tritium contains [MP PMT 2002] (c) 2 (d) 3
(a) 1 proton + 1 neutron (b) 1 proton + 3 neutron 74. Which of the following is alway s a whole num ber
(c) 1 proton + 0 neutron (d) 1 proton + 2 neutron [CPMT 1976, 81, 86]
62. Which one of the following groupings represents a (a) Atom ic weight (b) Atom ic radii
collection of isoelectronic species [A IEEE 2003] (c) Equiv alent weight (d) Atom ic num ber
(a) Na  , Ca 2  , Mg 2  (b) N 3  , F  , Na 
Atomic models and Planck's quantum theory
(c) Be, Al 3  , Cl  (d) Ca 2  , Cs  , Br
63. Which of the following are isoelectronic and isostructural 1. Rutherford's experiment on scattering of particles showed
NO 3 , CO 32  , ClO3 , SO 3 [IIT Scr een in g 2003] for the first tim e that the atom has
(a) NO 3 , CO 32  (b) SO 3 , NO 3 [IIT 1981; NCERT 1981; CMC Vellore 1991;
CPMT 1984; Kurukshetra CEE 1998]
(c) ClO3 , CO 32  (d) CO 32  , SO 3
(a) Electrons (b) Protons
64. The number of electrons in Cl  ion is [MP PMT 2003] (c) Nucleus (d) Neutrons
(a) 1 9 (b) 2 0
2. Rutherford's scattering experiment is related to the size of
(c) 1 8 (d) 3 5 the
65. The number of neutron in tritium is [CPMT 2003]
[IIT 1983; MADT Bihar 1995; BHU 1995]
(a) 1 (b) 2
(c) 3 (d) 0 (a) Nucleus (b) Atom
66. Tritium is the isotope of [CPMT 2003] (c) Electron (d) Neutron
(a) Hy drogen (b) Oxy gen 3. Rutherford's alpha particle scattering experim ent
(c) Carbon (d) Sulpher ev entually led to the conclusion that[IIT 1986; RPMT 2002]
67 . The atomic number of an element is 3 5. What is the total (a) Mass and energy are related
num ber of electrons present in all the p-orbitals of the
ground state atom of that elem ent (b) Electrons occupy space around the nucleus
[EA MCET (Engg.) 2003] (c) Neutrons are buried deep in the nucleus
(a) 6 (b) 1 1 (d) The point of im pact with m atter can be precisely
(c) 1 7 (d) 2 3 determ ined
68. The nucleus of an element contain 9 protons. Its v alency
4. Bohr's m odel can explain [IIT 1985]
would be [MP PET 2004]
(a) 1 (b) 3 (a) The spectrum of hy drogen atom only
(c) 2 (d) 5 (b) Spectrum of atom or ion containing one electron only
69. The compound in which cation is isoelectronic with anion is
[UPSEAT 2004] (c) The spectrum of hy drogen m olecule
(a) NaCl (b) CsF (d) The solar spectrum
(c) NaI (d) K 2 S 5. When atoms are bombarded with alpha particles, only a
70. Which among the following species have the same number of few in m illion suffer deflection, others pass out
electrons in its outerm ost as well as penultim ate shell undeflected. This is because[MNR 1979; NCERT 1980; A FMC 19
[DCE 2004] (a) The force of repulsion on th e m oving alpha particle is
(a) Mg 2
(b) O 2 sm all
(b) The force of attraction on the alpha particle to the
(c) F  (d) Ca 2 
oppositely charged electrons is v ery sm all
7 1. Six protons are found in the nucleus of
[CPMT 1977, 80, 81; NCERT 1975, 78] (c) There is only one nucleus and large num ber of
electrons
(a) Boron (b) Lithium
(c) Carbon (d) Helium (d) The nucleus occupies m uch sm aller v olum e
com pared to the v olum e of the atom
72. The nitrogen atom has 7 protons and 7 electrons, the
3 6. Positronium consists of an electron and a positron (a
nitride ion ( N ) will hav e [NCERT 1977]
particle which has the sam e m ass as an electron, but
(a) 7 protons and 1 0 electrons opposite charge) orbiting round their com m on centre of
 Structure of atom 53
m ass. Calculate the value of the Rydberg constant for this (a) Positiv e ray analy sis
sy stem. (b)  -ray scattering experim ents
(a) R / 4 (b) R / 2 (c) X-ray analy sis
(c) 2 R (d) R (d) Discharge tube experim ents
15. Electron occupies the av ailable orbital singly before
7. When  -particles are sent through a thin m etal foil, most pairing in any one orbital occurs, it is [CBSE PMT 1991]
of them go straight through the foil because (one or m ore (a) Pauli's exclusion principle
are correct) [IIT 1984]
(b) Hund's Rule
(a) Alpha particles are m uch heav ier than electrons
(c) Heisenberg's principle
(b) Alpha particles are positiv ely charged
(b) Prout's hy pothesis
(c) Most part of the atom is em pty space 16. The wav elength of a spectral line for an electronic
(d) Alpha particles m ov e with high v elocity transition is inversely related to [IIT 1988]
8. When an electron jumps from L to K shell (a) The number of electrons undergoing the transition
[CPMT 1983] (b) The nuclear charge of the atom
(a) Energy is absorbed (c) The difference in the energy of the energy lev els
inv olv ed in the transition
(b) Energy is released
(d) The v elocity of the electron undergoing the transition
(c) Energy is som etim es absorbed and som etim es
17. When an electron drops from a higher energy lev el to a
released
low energy level, then [A MU 1985]
(d) Energy is neither absorbed nor released (a) Energy is em itted
9. When bery llium is bom barded with  -particles, (b) Energy is absorbed
extrem ely penetrating radiations which cannot be
(c) Atom ic num ber increases
deflected by electrical or m agnetic field are giv en out.
These are (d) Atom ic num ber decreases
[CPMT 1983] 18. Dav isson and Germ er's experim ent showed that
(a) A beam of protons (b)  -ray s [MA DT Bihar 1983]
(a)  -particles are electrons
(c) A beam of neutrons (d) X-ray s
(b) Electrons com e from nucleu s
10. Which one of the following is not the characteristic of
Planck's quantum theory of radiation [A IIMS 1991] (c) Electrons show wav e nature
(a) The energy is not absorbed or em itted in whole (d) None of the abov e
num ber or m ultiple of quantum 19. When an electron jum ps from lower to higher orbit, its
(b) Radiation is associated with energy energy [MA DT Bi h a r 1982]
(a) Increases (b) Decreases
(c) Radiation energy is not em itted or absorbed conti -
nuously but in the form of sm all packets called (c) Rem ains the sam e (d) None of these
quanta 20. Experimental ev idence for the existence of the a tom ic
nucleus com es from [CBSE PMT 1989]
(d) This m agnitude of energy associated with a quantum
is proportional to the frequency (a) Millikan's oil drop experim ent
(b) Atom ic em ission spectroscopy
11. The spectrum of He is expected to be sim ilar to
(c) The m agnetic bending of cathode ray s
[A IIMS 1980, 91; DPMT 1983; MP PMT 2002]
(d) Alpha scattering by a thin m etal foil
(a) H (b) Li  21. Which of the following statem ents does not form part of
Bohr's m odel of the hydrogen atom [CBSE PMT 1989]
(c) Na (d) He 
(a) Energy of the electrons in the orbit is quantized
12. Energy of orbit [DPMT 1984, 91]
(b) The electron in the orbit nearest the nucleus has the
(a) Increases as we m ov e away from nucleus
lowest energy
(b) Decreases as we m ov e away from nucleus
(c) Electrons rev olv e in different orbits around the
(c) Rem ains sam e as we m ov e away from nucleus nucleus
(d) None of these (d) The position and velocity of the electrons in the orbit
13. Bohr m odel of an atom could not account for cannot be determ ined sim ultaneously
(a) Em ission spectrum 22. When  -particles are sent through a tin m etal foil, m ost
(b) Absorption spectrum of them go straight through the foil as [EA MCET 1983]
(c) Line spectrum of hy drogen (a)  -particles are m uch heav ier than electrons
(d) Fine spectrum
(b)  -particles are positiv ely charged
14. Existence of positively charged nucleus was established by
(c) Most part of the atom is em pty space
[CBSE PMT 1991]
 54 Structure of atom

(d)  -particles m ov e with high v elocity 2 2 m 4 e 2 z 2 2 2 me 2 z 2

(a) E n   (b) E n  
23. The energy of second Bohr orbit of the hydrogen atom is – n2h2 n2h2
3 2 8 kJ mol–1, hence the energy of fourth Bohr orbit would be
2 me z
2 4 2
2m 2 e 2 z 4
[CBSE PMT 2005] (c) E n   (d) E n  
n2h2 n2h2
(a) – 4 1 kJ mol–1 (b) –1 3 1 2 kJ mol–1
33. Who m odified Bohr's theory by introducing ellipt ical
(c) –1 6 4 kJ mol–1 (d) – 82 kJ mol–1
orbits for electron path [CBSE PMT 1999; A FMC 2003]
24. When an electron rev olv es in a stationary orbit then (a) Hund (b) Thom son
[MP PET 1994] (c) Rutherford (d) Som m erfield
(a) It absorbs energy 34. Bohr's radius can have [DPMT 1996]
(b) It gains kinetic energy
(a) Discrete v alues (b) ve v alues
(c) It em its radiation
(c) ve v alues (d) Fractional v alu es
(d) Its energy rem ains constant 35. The first use of quantum theory to explain the structure of
25. A m ov ing particle m ay hav e wav e m otion, if atom was made by[IIT 1997; CPMT 2001; J&K CET 2005]
(a) Its m ass is v ery high (a) Heisenberg (b) Bohr
(b) Its v elocity is negligible (c) Planck (d) Einstein
(c) Its m ass is negligible 36. An electronic transition from 1s orbital of an atom causes
(d) Its m ass is v ery high and v elocity is negligible [JIPMER 1997]
26. The postulate of Bohr theory that electrons jum p from (a) Absorption of energy
one orbit to the other, rather than flow is according to (b) Release of energy
(a) The quantisation concept (c) Both release or absorption of energy
(b) The wav e nature of electron (d) Unpredictable
(c) The probability expression for electron 37 . In an element going away from nucleus, the energy of
(d) Heisenberg uncertainty principle particle [RPMT 1997]
27 . The frequency of an electrom agnetic radiation is (a) Decreases (b) Not changing
2  10 Hz . What is its wav elength in m etres
6
(c) Increases (d) None of these
(Velocity of light  3  10 8 ms 1 ) 38. The  -particle scattering experim ent of Rutherford
concluded that [Or issa JEE 1997]
(a) 6.0  10 14
(b) 1.5  10 4
(a) The nucleus is m ade up of protons and neutrons
(c) 1.5  10 2
(d) 0.66  10 2 (b) The number of electrons is exactly equal to number of
28. What is the packet of energy called [A FMC 2005] protons in atom
(a) Electron (b) Photon (c) The positive charge of the atom is concentrated in a
(c) Positron (d) Proton v ery sm all space
29. The energy of an electron in n th orbit of hydrogen atom is (d) Electrons occupy discrete energy lev els
[MP PET 1999] 39. Wavelength associated with electron m otion [BHU 1998]
13 .6 13 .6 (a) Increases with increase in speed of electron
(a) eV (b) eV (b) Rem ains sam e irrespectiv e of speed of electron
n4 n3
13 .6 13 .6 (c) Decreases with increase in speed of e 
(c) eV (d) eV (d) Is zero
n2 n
40. The elem ent used by Rutherford in his famous scattering
30. If wav elength of photon is 2.2  10 11 m, h  6.6  10 34 J- experim ent was [KCET 1998]
sec, then m om entum of photon is [MP PET 1999] (a) Gold (b) Tin
(a) 3  10 23 kg ms 1 (b) 3.33  10 22 kg ms 1 (c) Silv er (d) Lead
41. If electron falls from n  3 to n  2 , then emitted energy
(c) 1.452  10 44 kg ms 1 (d) 6.89  10 43 kg ms 1 is
31. The expression for Bohr's radius of an atom is [A FMC 1997; MP PET 2003]
[MP PMT 1999] (a) 10.2eV (b) 12.09eV
n2h2 n2h2 (c) 1.9 eV (d) 0.65eV
(a) r  (b) r  42. The radius of the nucleus is related to the m ass num ber
4 me z
2 4 2
4 2 me 2 z
A by [
n2h2 n2h2
(c) r  (d) r  (a) R  R o A 1/2
(b) R  R o A
4 2 me 2 z 2 4 2 m 2 e 2 z 2
(c) R  R o A 2
(d) R  R o A 1 / 3
32. The energy of an electron revolving in n th Bohr's orbit of
an atom is given by the expression [MP PMT 1999] 43. The specific charge of proton is 9.6  10 6 C kg 1 then for an
 -particle it will be [MH CET 1999]
 Structure of atom 55
(a) 38.4  10 7 C kg 1 (b) 19.2  10 7 C kg 1 54. Rutherford’s  -particle scattering experiment proved that
atom has [MP PMT 2001]
(c) 2.4  10 7 C kg 1 (d) 4.8  10 7 C kg 1 (a) Electrons (b) Neutron
44. In hy drogen spectrum the different lines of Ly m an series (c) Nucleus (d) Orbitals
are present is [UPSEA T 1999] 55. Wav elength of spectral line em itted is inv ersely
(a) UV field (b) I R field proportional to [
(a) Radius (b) Energy
(c) Visible field (d) Far I R field
(c) Velocity (d) Quantum num ber
45. Which one of the following is considered as the m ain
postulate of Bohr’s m odel of atom [A MU 2000] 56. The energy of a radiation of wavelength 8000 Å is E1 and
(a) Protons are present in the nucleus energy of a radiation of wavelength 16000 Å is E 2 . What
(b) Electrons are rev olv ing around the nucleus is the relation between these two [Kerala CET 2005]
(c) Centrifugal force produced due to the rev olv ing (a) E1  6 E 2 (b) E1  2E 2
electrons balances the force of attraction between the (c) E1  4 E 2 (d) E1  1 / 2 E2
electron and the protons
(e) E1  E 2
(d) Angular momentum of electron is an integral multiple of
h 57 . The form ation of energy bonds in solids are in accordance
with [DCE 2001]
2
(a) Heisenberg’s uncertainty principle
46. The electronic energy levels of the hydrogen atom in the
(b) Bohr’s theory
Bohr’s theory are called [A MU 2000]
(c) Ohm ’s law
(a) Ry dberg lev els (b) Orbits
(d) Rutherford’s atom ic m odel
(c) Ground states (d) Orbitals 58. The frequency of y ellow light having wavelength 600 nm
47 . The energy of a photon is calculated by [Pb. PMT 2000] is
(a) E  h (b) h  E [MP PET 2002]
E h (a) 5.0  10 Hz14
(b) 2.5  10 Hz
7
(c) h  (d) E 
  (c) 5.0  10 7 Hz (d) 2.5  10 14 Hz
48. Visible range of hy drogen spectrum will contain the 59. The v alue of the energy for the first excited state of
following series [RPET 2000] hy drogen atom will be [MP PET 2002]
(a) Pfund (b) Ly m an (a)  13.6 eV (b)  3.40 eV
(c) Balm er (d) Brackett (c)  1.51 eV (d)  0.85 eV
49. Radius of the first Bohr’s orbit of hy drogen atom is 60. Bohr m odel of atom is contradicted by [MP PMT 2002]
[RPET 2000]
(a) Pauli’s exclusion principle
(a) 1 .06 Å (b) 0 .22 Å (b) Planck quantum theory
(c) Heisenberg uncertainty principle
(c) 0 .28 Å (d) 0 .53 Å
(d) All of these
50. In Balm er ser ies of hy drogen atom spectrum which
61. Which of the following is not true in Rutherford’s nuclear
electronic transition causes third line [MP PMT 2000]
m odel of atom [Or issa JEE 2002]
(a) Fifth Bohr orbit to second one (a) Protons and neutrons are present inside nucleus
(b) Fifth Bohr orbit to first one (b) Volum e of nucleus is v ery sm all as com pared to
(c) Fourth Bohr orbit to second one v olum e of atom
(d) Fourth Bohr orbit to first one (c) The num ber of protons and neutrons are alway s
51. Energy of electron of hydrogen atom in second Bohr orbit equal
is (d) The num ber of electrons and protons are alway s
[MP PMT 2000] equal
(a)  5.44  10 19 J (b)  5.44  10 19 kJ 62. The em ission spectrum of hydrogen is found to satisfy the
expression for the energy change. E (in joules) such
(c)  5.44  10 19 cal (d)  5.44  10 19 eV
 1 1 
52. If change in energy (E)  3  10 8 J , h  6.64  10 34 J - s that E  2 .18  10 2  2  J where n 1 = 1, 2 , 3 ….. and
 n1 n2 
and c  3  10 8 m/s, then wavelength of the light is n 2 = 2 , 3 , 4 ……. The spectral lines correspond to Paschen
[CBSE PMT 2000] series to [UPSEA T 2002]
(a) 6.36  10 Å3
(b) 6.36  10 5 Å (a) n1  1 and n 2  2, 3, 4
8
(c) 6.64  10 Å (d) 6.36  10 18 Å (b) n1  3 and n 2  4, 5, 6
53. The radius of first Bohr’s orbit for hydrogen is 0.53 Å. The
radius of third Bohr’s orbit would be [MP PMT 2001] (c) n1  1 and n 2  3, 4, 5
(a) 0.7 9 Å (b) 1 .59 Å (d) n1  2 and n 2  3, 3, 5
(c) 3 .1 8 Å (d) 4 .7 7 Å
(e) n1  1 and n 2  infinity
 56 Structure of atom
63. The ratio between kinetic energy and the total energy of 74. The wav elength of the radiation em itted, when in a
the electrons of hydrogen atom according to Bohr’s m odel hy drogen atom electron falls from infinity to stationary
is state 1 , would be (Ry dberg constant  1.097 10 7 m 1 )
[Pb. PMT 2002]
[A IEEE 2004]
(a) 2 : 1 (b) 1 : 1 (a) 4 06 nm (b) 1 92 nm
(c) 1 : – 1 (d) 1 : 2
(c) 91 nm (d) 9.110 8 nm
64. Energy of the electron in Hy drogen atom is giv en by
[A MU (Engg.) 2002] 75. In Bohr’s m odel, atomic radius of the first orbit is  , the
131.38 131.33 radius of the 3 rd orbit, is [MP PET 1997; Pb. CET 2001]
(a) En   kJ mol 1 (b) En   kJ mol 1 (a)  / 3 (b) 
n2 n
1313.3 313.13 (c) 3 (d) 9
(c) En   kJ mol 1 (d) En   kJ mol 1
n 2
n 2 76. According to Bohr’s principle, the relation between
65. Ratio of radii of second and first Bohr orbits of H atom principle quantum number (n) and radiu s of orbit is [BHU 2004]
[BHU 2003] (a) r  n (b) r  n 2
(a) 2 (b) 4 1 1
(c) 3 (d) 5 (c) r  (d) r  2
n n
66. The frequency corresponding to transition n  2 to n  1 77. The ionisation potential of a hydrogen atom is –1 3 .6 eV.
in hydrogen atom is [MP PET 2003] What will be the energy of the atom corresponding to
(a) 15.66  10 10 Hz (b) 24.66  10 14 Hz n2
[Pb. CET 2000]
(c) 30.57  10 14 Hz (d) 40.57  10 24 Hz
(a) –3 .4 eV (b) –6.8 eV
67 . The m ass of a photon with a wav elength equal to
(c) –1 .7 eV (d) – 2 .7 eV
1.54  10 8 cm is [Pb. PMT 2004] 7 8. The energy of electron in hy drogen atom in its grounds
(a) 0.8268  10 34 kg (b) 1.2876  10 33 kg state is –13.6 eV. The energy of the level corresponding to
the quantum number equal to 5 is [Pb. CET 2002]
(c) 1.4285  10 32 kg (d) 1.8884  10 32 kg (a) –0.54 eV (b) – 0.85 eV
68. Splitting of spectral lines under the influence of m agnetic (c) – 0.64 eV (d) – 0.4 0 eV
field is called [MP PET 2004] 79. The positive charge of an atom is [A FMC 2002]
(a) Zeem an effect (b) Stark effect (a) Spread all ov er the atom
(c) Photoelectric effect (d) None of these (b) Distributed around the nucleus
69. The radius of electron in the first excited state of (c) Concentrated at the nucleus
hy drogen atom is [MP PMT 2004] (d) All of these
(a) a0 (b) 4a0 80. A metal surface is exposed to solar radiations [DPMT 2005]
(c) 2a0 (d) 8a0 (a) The em itted electrons hav e energy less than a
m aximum value of energy depending upon frequency
70. The ratio of area cov ered by second orbital to the first of incident radiations
orbital is [A FMC 2004] (b) The em itted electrons hav e energy less than
(a) 1 : 2 (b) 1 : 1 6 m aximum value of energy depending upon intensity
(c) 8 : 1 (d) 1 6 : 1 of incident radiation
7 1. Tim e taken for an electron to com plete one rev olution in (c) The em itted electrons hav e zero energy
the Bohr orbit of hydrogen atom is [Ker a l a PMT 2004] (d) The em itted electrons have energy equal to energy of
4 2 mr 2 nh photos of incident light
(a) (b) 81. Which of the following transitions hav e m inim um
nh 4 2 mr wav elength [DPMT 2005]
(c)
nh
(d)
h (a) n4  n1 (b) n 2  n1
4 2 mr 2 2mr
(c) n4  n 2 (d) n3  n1
72. The radius of which of the following orbit is sam e as that
of the first Bohr's orbit of hy drogen atom Dual nature of electron
[IIT Screening 2004]
(a) He  (n  2) (b) Li 2 (n  2) 1. De broglie equation describes the relationship of wavelengt h
2 3
(c) Li (n  3) (d) Be (n  2) associated with the motion of an electron and its[MP PMT
1986]
73. The frequency of radiation em itted when the electron falls
from n  4 to n  1 in a hy drogen atom will be (Giv en (a) Mass (b) Energy
18 1 (c) Mom entum (d) Charge
ionization energy of H  2.18  10 J atom and
34 2. The wav e nature of an electron was first giv en by
h  6.625  10 Js ) [CBSE PMT 2004]
[CMC V ellore 1991; Pb. PMT 1998; CPMT 2004]
(a) 3.08  1015 s 1 (b) 2.00  1015 s 1 (a) De-Broglie (b) Heisenberg
(c) 1.54  1015 s 1 (d) 1.03  1015 s 1 (c) Mosley (d) Som m erfield
 Structure of atom 57
3. Am ong the following for which one m athem atical 13. The de-Broglie wavelength associated with a particle of
expression  
h
stands m ass 10 6 kg m ov ing with a v elocity of 10 ms 1 , is
p [A IIMS 2001]
(a) De Broglie equation (b) Einstein equation 22 29
(a) 6.63  10 m (b) 6.63  10 m
(c) Uncertainty equation (d) Bohr equation
4. Which one of the following explains light both as a stream (c) 6.63  10 31 m (d) 6.63  10 34 m
of particles and as wav e m otion 14. What is the de-Broglie wav elength associated with the
[A IIMS 1983; IIT 1992; UPSEAT 2003] hydrogen electron in its third orbit [A MU (En gg.) 2002]
(a) Diffraction (b)   h / p (a) 9.96  10 10 cm (b) 9.96  10 8 cm
(c) Interference (d) Photoelectric effect
(c) 9.96  10 4 cm (d) 9.96  10 8 cm
5. In which one of the following pairs of experim ental
observ ations and phenom enon does the experim ental 15. If the v elocity of hy drogen m olecule is 5  10 4 cm sec 1 ,
observation correctly account for phenomenon [Ade-Broglie
then its IIMS 1983] wavelength is [MP PMT 2003]
Experimental observa t ion Ph en om en on (a) 2 Å (b) 4 Å
(a) X -ray spectra Charge on the nucleus (c) 8 Å (d) 1 00 Å
(b)  -particle scattering Quantized electron orbit 16. A 2 00g golf ball is m oving with a speed of 5 m per hour.
(c) Em ission spectra The quantization of energy The associated wav e length is (h  6.625  10 34 J - sec)
(d) The photoelectric effect The nuclear atom [MP PET 2003]
6. Which of the following expressions giv es the de-Broglie (a) 10 10 m (b) 10 20 m
relationship[MP PMT 1996, 2004; MP PET /PMT 1998]
 h (c) 10 30 m (d) 10 40 m
(a) h  (b)  
mv mv 17. A cricket ball of 0.5 kg is m ov ing with a v elocity of
m v 100 m / sec . The wavelength associated with its m otion is
(c)   (d)  
hv mh [DCE 2004]
7. de-Broglie equation is (a) 1 / 100cm (b) 6.6  10 34 m
[MP PMT 1999; CET Pune 1998]
(c) 1.32  10 35 m (d) 6.6  10 28 m
(a) n  2d sin (b) E  hv 18. Dual nature of particles was proposed by [DCE 2004]
h (a) Heisenberg (b) Lowry
(c) E  mc 2 (d)  
mv (c) de-Broglie (d) Schrodinger
19. Calculate de-Broglie wavelength of an electron trav elling
8. The de-Broglie wavelength of a particle with m ass 1 gm at 1% of the speed of light [DPMT 2004]
and velocity 100m / sec is[CBSE PMT 1999; EA MCET 1997;
(a) 2.73  10 24 (b) 2.42  10 10
A FMC 1999; AIIMS 2000]
33 (c) 242.2  10 10 (d) None of these
(a) 6.63  10 m (b) 6.63  10 34 m 20. Which is the correct relationship between wavelength and
(c) 6.63  10 35 m (d) 6.65  10 35 m m omentum of particles [Pb. PMT 2000]
9. Minimum de-Broglie wavelength is associated with [RPMT h h
1999] (a)   (b)  
P P
(a) Electron (b) Proton
h P
(c) CO 2 m olecule (d) SO 2 m olecule (c) P  (d) h 
 
10. The de-Broglie wav elength associated with a m aterial 21. The de-Broglie equation applies [MP PMT 2004]
particle is [JIPMER 2000] (a) To electrons only
(a) Directly proportional to its energy (b) To neutrons only
(b) Directly proportional to m om entum (c) To protons only
(c) Inv ersely proportional to its energy (d) All the m aterial object in m otion
(d) Inv ersely proportional t o m om entum
Uncertainty principle and Schrodinger wave
11. An electron has kinetic energy 2.8  10 23 J . de-Broglie equation
wav elength will be nearly
(m e  9.1  10 31 kg ) [MP PET 2000] 1. The uncertainty principle was enunciated by
4 7 [NCERT 1975; Bihar MEE 1997]
(a) 9.28  10 m (b) 9.28  10 m
(a) Einstein (b) Heisenberg
(c) 9.28  10 8 m (d) 9.28  10 10 m (c) Rutherford (d) Pauli
12. What will be de-Broglie wavelength of an electron m ov ing 2. According to heisenberg uncertainty principle
with a velocity of 1.2  10 ms 5 1
[MP PET 2000] [A MU 1990; BCECE 2005]
h
(a) 6.068  10 9 (b) 3.133  10 37 (a) E  mc 2 (b) x  p 
4
(c) 6.626  10 9 (d) 6.018  10 7
 58 Structure of atom
h h (c) Heisenberg, Planck (d) Planck, Heisenberg
(c)   (d) x  p 
p 6 12. The uncertainty in m om entum of an electron is
3. “The position and velocity of a sm all particle like electron 1  10 5 kg  m / s . The uncertainty in its position will be
cannot be simultaneously determined.” This statement is ( h  6.62  10 34 kg  m 2 / s )
[NCERT 1979; BHU 1981, 87]
[A FMC 1998; CBSE PMT 1999; JIPMER 2002]
(a) Heisenber g uncertainty principle
(b) Principle of de Broglie's wav e nature of electron (a) 1.05  10 28 m (b) 1.05  10 26 m
(c) Pauli's exclusion principle (c) 5.27  10 30 m (d) 5.25  10 28 m
(d) Aufbau's principle 13. The uncertainty in the position of a m oving bullet of m ass
h 1 0 gm is 10 5 m . Calculate the uncertainty in its v elocity
4. In Heisenberg's uncertainty equation x  p  ; p
4 [DCE 1999]
stands for 28 28
(a) 5.2  10 m / sec (b) 3.0  10 m / sec
(a) Uncertainty in energy
(b) Uncertainty in v elocity (c) 5.2  10 22 m / sec (d) 3  10 22 m / sec
(c) Uncertainty in m om entum h
14. The equation x .p  shows [MP PET 2000]
(d) Uncertainty in m ass 4
5. Which one is not the correct relation in the following (a) de-Broglie relation
E (b) Heisenberg’s uncertainty principle
(a) h  (b) E  mc 2 (c) Aufbau principle
v
(d) Hund’s rule
h h 15. Which quantum number is not related with Schrodinger
(c) x  p  (d)  
4 mv equation [RPMT 2002]
6. The m aximum probability of finding an electron in the (a) Principal (b) Azim uthal
d xy orbital is [MP PET 1996] (c) Magnetic (d) Spin
(a) Along the x-axis 16. Uncertainty in position of a 0.2 5 g particle is 10 5 .
(b) Along the y-axis Uncertainty of velocity is (h  6.6  10 34 Js) [A IEEE 2002]

(c) At an angle of 45 o from the x and y-axes (a) 1.2  10 34 (b) 2.1  10 29
(d) At an angle of 90 o from the x and y-axes (c) 1.6  10 20
(d) 1.7  10 9
17. The uncertainty in m om entum of an electron is
7. Sim ultaneous determ ination of exact position and
m omentum of an electron is [BHU 1979] 1  10 5 kg m / s . The uncertainity in its position will be
(a) Possible (h  6.63  10 34 Js) [Pb. CET 2000]
(b) Im possible (a) 5.28  10 30
m (b) 5.25  10 28 m
(c) Som etim es possible som etim es im possible (c) 1.05  10 m 26
(d) 2.715  10 30 m
(d) None of the abov e 18. According to Heisenberg’s uncertainty principle, the
8. If uncertainty in the position of an electron is zero, the product of uncertainties in position and v elocities for an
uncertainty in its m om entum would be [CPMT 1988] electron of m ass 9.1  10 31 kg is [
h 3 2 1 5 2 1
(a) Zero (b)  (a) 2.8  10 m s (b) 3.8  10 m s
2 5 2 1
(c) 5.8  10 m s (d) 6.8  10 6 m 2 s 1
h
(c)  (d) Infinite 19. For an electron if the uncertainty in v elocity is  , the
2
uncertainty in its position (x ) is given by [DPMT 2005]
9. The possibility of finding an electron in an orbital was
hm 4
conceiv ed by [MP PMT 1994] (a) (b)
4 hm 
(a) Rutherford (b) Bohr
h 4 m
(c) Heisenberg (d) Schrodinger (c) (d)
4m  h . 
10. Uncertainty principle gav e the concept of
20. Orbital is [DPMT 2005]
(a) Probability
(a) Circular path around the nucleus in which the
(b) An orbital electron rev olv es
(c) Phy sical m eaning of  the  2 (b) Space around the nucleus where the probability of
finding the electron is m axim um
(d) All the abov e
(c) Am plitude of electrons wav e
11. The uncertainty principle and the concept of wave nature (d) None of these
of m atter was proposed by ...... and ...... respectiv ely
[MP PET 1997]
Quantum number, Electronic configuration
(a) Heisenberg, de Broglie (b) de-Broglie, Heisenberg
 Structure of atom 59
and Shape of orbitals 1
(c) 5, 1, 1,  (d) 6, 0, 0, 
1
2 2
1. Be's 4 th electron will hav e four quantum num bers 10. The correct ground state electronic configuration of
[MNR 1985] chromium atom is[IIT 1989, 94; MP PMT 1993; EA MCET 1997;
n l m s ISM Dh anbad 1994; AFMC 1997; Bihar MEE 1996;
(a) 1 0 0 + 1 /2 MP PET 1995, 97; CPMT 1999; Kerala PMT 2003]
(b) 1 1 +1 + 1 /2 (a) [ Ar] 3d 5 4 s 1 (b) [ Ar] 3d 4 4 s 2
(c) 2 0 0 – 1 /2
(d) 2 1 0 + 1 /2 (c) [ AR]3d 6 4 s 0 (d) [ Ar]4 d 5 4 s 1
2. The quantum number which specifies the location of an 11. 2 p orbitals have [NCERT 1981; MP PMT 1993, 97]
electron as w ell as energy is [DPMT 1983] (a) n  1, l  2 (b) n  1, l  0
(a) Principal quantum num ber
(c) n  2, l  1 (d) n  2, l  0
(b) Azim uthal quantum num ber
(c) Spin quantum num ber 12. Electronic configuration of H  is [CPMT 1985]
(d) Magnetic quantum num ber (a) 1s 0 (b) 1s 1
3. The shape of an orbital is given by the quantum number
(c) 1s 2 (d) 1s 1 2 s 1
[NCERT 1984; MP PMT 1996]
13. The quantum numbers for the outerm ost electron of an
(a) n (b) l
(c) m (d) s 1
elem ent are given below as n  2, l  0, m  0, s   . The
4. In a given atom no two electrons can have the same values 2
for all the four quantum num bers. This is called atom s is [EA MCET 1978]
[BHU 1979; A MU 1983; EAMCET 1980, 83; (a) Lithium (b) Bery llium
MA DT Bihar 1980; CPMT 1986, 90, 92; NCERT 1978, 84; (c) Hy drogen (d) Boron
RPMT 1997; CBSE PMT 1991; MP PET 1986, 99] 14. Principal quantum num ber of an atom represents
(a) Hund's rule [EA MCET 1979; IIT 1983; MNR 1990;UPSEAT 2000, 02]
(b) Aufbau's principle (a) Size of the orbital
(c) Uncertainty principle (b) Spin angular m om entum
(d) Pauli's exclusion principle (c) Orbital angular m om entum
5. Nitrogen has the electronic configuration (d) Space orientation of the orbital
1s 2 ,2 s 2 2 p 1x 2 p 1y 2 p 1z and not 1s 2 ,2 s 2 2 p x2 2 p 1y 2 p z0 which is 15. An elem ent has the electronic configuration
determ ined by 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 2 . Its v alency electrons are
[DPMT 1982, 83, 89; MP PMT/PET 1988; EAMCET 1988] [NCERT 1973]
(a) Aufbau's principle (b) Pauli's exclusion (a) 6 (b) 2
principle (c) 3 (d) 4
(c) Hund's rule (d) Uncertainty principle 16. The m agnetic quantum num ber specifies
6. Which one of the following configuration represents a [MNR 1986; BHU 1982; CPMT 1989, 94;
noble gas [CPMT 1983,
MP89, 93;1999;
PET NCERT 1973;
AFMC MPAMU
1999; PMT (Engg.)
1989; 1999]
DPMT 1984]
(a) Size of orbitals (b) Shape of orbitals
2 2 6 2
(a) 1s ,2 s 2 p ,3 s (b) 1s ,2 s 2 p ,3 s1
2 2 6
(c) Orientation of orbitals (d) Nuclear stability
(c) 1s 2 ,2 s 2 2 p 6 (d) 1s 2 ,2 s 2 sp6 ,3 s 2 3 p 6 ,4 s 2 17. Which of the following sets of quantum num bers
represent an impossible arrangement[IIT 1986; MP PET 1995]
7. The electronic configuration of silver atom in ground state
is n l m ms
[CPMT 1984, 93] 1
(a) 3 2 –2 (+ )
(a) [Kr]3d 10
4s 1
(b) [ Xe] 4 f 5 d 6 s
14 10 1
2

(c) [Kr] 4 d 5 s10 1

(d) [Kr] 4 d 9 5 s 2 1
(b) 4 0 0 (–)
2
8. Principal, azimuthal and m agnetic quantum numbers are
respectively related to [CPMT 1988; A IIMS 1999] 1
(c) 3 2 –3 (+ )
(a) Size, shape and orientation 2
(b) Shape, size and orientation 1
(d) 5 3 0 (–)
(c) Size, orientation and shape 2
(d) None of the abov e 18. If n  3 , then the v alue of ' l' which is incorrect
9. Correct set of four quantum numbers for valence electron [CPMT 1994]
of rubidium (Z = 3 7 ) is (a) 0 (b) 1
[IIT 1984; JIPMER 1999; UPSEAT 2003]
(c) 2 (d) 3
1 1 19. Which orbital is dum b-bell shaped
(a) 5, 0, 0,  (b) 5, 1, 0, 
2 2 [MP PMT 1986; MP PET/PMT 1998]
 60 Structure of atom
(a) s -orbital (b) p -orbital [MNR 1988; UPSEAT 1999, 2000; Kerala PMT 2003]
(a) Principal quantum num ber
(c) d -orbital (d) f -orbital
(b) Azim uthal quantum num ber
20. The total number of unpaired electrons in d - orbitals of
atom s of element of atomic number 29 is [CPMT 1983] (c) Magnetic quantum num ber
(d) Spin quantum num ber
(a) 1 0 (b) 1
(c) 0 (d) 5 32. A com pletely filled d -orbital (d 10 ) [MNR 1987]
21. The shape of 2 p orbital is (a) Spherically symmetrical
[CPMT 1983; NCERT 1979] (b) Has octahedral sy m m etry
(a) Spherical (b) Ellipsoidal (c) Has tetrahedral sy m m etry
(c) Dum b-bell (d) Py ram idal (d) Depends on the atom
22. The m agnetic quantum number for a n electron when the 33. If m agnetic quantum number of a given atom represented
v alue of principal quantum num ber is 2 can hav e by –3 , then what will be its principal quantum num ber
[CPMT 1984] [BHU 2005]
(a) 3 v alues (b) 2 v alues (a) 2 (b) 3
(c) 9 v alues (d) 6 v alues (c) 4 (d) 5
23. Which one is the correct outer configuration of chromium 34. The total number of orbitals in an energy level designated
[A IIMS 1980, 91; BHU 1995] by principal quantum number n is equal to
[A IIMS 1997; J&K CET 2005]
(a)     
(a) 2n (b) 2n 2
(b)   
(c) n (d) n 2
(c)       35. The number of orbitals in the fourth principal quantum
(d) num ber will be
     
(a) 4 (b) 8
24. The following has zero valency [DPMT 1991] (c) 1 2 (d) 1 6
(a) Sodium (b) Bery llium 36. Which set of quantum numbers are not possible from the
(c) Alum inium (d) Kry pton following
25. The number of electrons in the valence shell of calcium is
1
[IIT 1975] (a) n  3, l  2, m  0, s  
(a) 6 (b) 8 2
(c) 2 (d) 4 (b) n  3, l  2, m  2, s  
1
26. The valence electron in the carbon atom are [MNR 1982] 2
(a) 0 (b) 2 1
(c) n  3, l  3, m  3, s  
(c) 4 (d) 6 2
27 . For the dum b-bell shaped orbital, the v alue of l is 1
[CPMT 1987, 2003] (d) n  3, l  0, m  0, s  
2
(a) 3 (b) 1 37 . The four quantum number for the valence shell electron
(c) 0 (d) 2 or last electron of sodium (Z = 11) is [MP PMT 1999]
28. Chrom ium has the electronic configuration 4 s1 3d 5 1
(a) n  2, l  1, m  1, s  
rather than 4 s 2 3d 4 because 2
(a) 4 s and 3 d hav e the sam e energy 1
(b) n  3, l  0, m  0, s  
(b) 4 s has a higher energy than 3 d 2
(c) 4 s 1 is m ore stable than 4 s 2 1
(c) n  3, l  2, m  2, s  
(d) 4 s 1 3d 5 half-filled is m or e stable than 4 s 2 3d 4 2
1
29. The electronic configuration of calcium ion (Ca 2  ) is (d) n  3, l  2, m  2, s  
[CMC V ellore 1991] 2
38. The explanation for the presence of three unpaired
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 electrons in the nitrogen atom can be giv en by
(b) 1s 2 ,2 s 2 sp 6 ,3 s 2 3 p 6 ,4 s 1 [NCERT 1979; RPMT 1999; DCE 1999, 2002;
2 2 6
(c) 1s ,2 s 2 p ,3 s 3 p 3d
2 6 2 CPMT 2001; MP PMT 2002; Pb. PMT / CET 2002]
2 2 6 2 6 5 (a) Pauli's exclusion principle
(d) 1s ,2 s sp ,3 s 3 p 3d
2 2 6 2 6 0
(b) Hund's rule
(e) 1s ,2 s 2 p ,3 s 3 p ,4 s (c) Aufbau's principle
30. The structure of external m ost shell of inert gases is (d) Uncertainty principle
[JIPMER 1991]
2 3 2 6 39. The m axim um energy is present in any electron at
(a) s p (b) s p
(a) Nucleus
1 2
(c) s p (d) d 10 s 2 (b) Ground state
31. The two electrons in K sub-shell will differ in (c) First excited state
 Structure of atom 61
(d) Infinite distance from the nucleus 1
(b) n  4 , l  0, m  0, s  
40. The electron density between 1s and 2 s orbital is 2
(a) High (b) Low 1
(c) Zero (d) None of these (c) n  3, l  1, m  1, s  
2
41. For ns orbital, the magnetic quantum number has v alue
1
(a) 2 (b) 4 (d) n  4 , l  2, m  1, s  
2
(c) – 1 (d) 0
53. The angular m omentum of an electron depends on
42. The m axim um num ber of electrons that can be
(a) Principal quantum num ber
accom m odated in the M th shell is (b) Azim uthal quantum num ber
(a) 2 (b) 8 (c) Magnetic quantum num ber
(c) 1 8 (d) 3 2 (d) All of these
43. For a giv en value of quantum num ber l , the num ber of 54. The electronic configuration of copper ( 29 Cu ) is
allowed v alues of m is giv en by
[DPMT 1983; BHU 1980; AFMC 1981;
(a) l  2 (b) 2l  2
CBSE PMT 1991; MP PMT 1995]
(c) 2l  1 (d) l  1
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9 ,4 s 2
44. The num ber of radial nodes of 3 s and 2 p orbitals are
respectively. [IIT -JEE 2005] (b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10 ,4 s1
(a) 2 , 0 (b) 0, 2
(c) 1s 2 .2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 4 p 6
(c) 1 , 2 (d) 2 , 1
45. Which of the sub-shell is circular (d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10
(a) 4 s (b) 4 f 55. The num ber of orbitals in 2 p sub-shell is
(c) 4 p (d) 4 d [NCERT 1973; MP PMT 1996]
(a) 6 (b) 2
46. Which electronic configuration for oxy gen is correct
according to Hund's rule of m ultiplicity (c) 3 (d) 4
56. The number of orbitals in d sub-shell is [MNR 1981]
(a) 1s 2 ,2 s 2 2 p x2 2 p 1y 2 p 1z (b) 1s 2 ,2 s 2 2 p x2 2 p y2 2 p z0
(a) 1 (b) 3
(c) 1s 2 ,2 s 2 2 p x3 2 p 1y 2 p z0 (d) None of these (c) 5 (d) 7
47 . If v alue of azimuthal quantum number l is 2 , then total 57 . A sub-shell l  2 can take how m any electrons
possible v alues of m agnetic quantum num ber will be [NCERT 1973, 78]
(a) 7 (b) 5 (a) 3 (b) 1 0
(c) 3 (d) 2 (c) 5 (d) 6
48. The ty pe of orbitals present in Fe is 58. Pauli's exclusion principle states that
(a) s (b) s and p [MNR 1983; A MU 1984]
(c) s, p and d (d) s, p, d and f (a) Two electrons in the same atom can hav e the sam e
energy
49. The shape of d xy orbital will be
(b) Two electrons in the sam e atom cannot hav e the
(a) Circular (b) Dum b-bell sam e spin
(c) Double dum b-bell (d) Trigonal (c) The electrons tend to occupy different orbitals as far
50. In any atom which sub-shell will have the highest energy as possible
in the following (d) Electrons tend to occupy lower energy orbitals
(a) 3 p (b) 3 d preferentially
(c) 4 s (d) 3 s (e) None of the abov e
51. Which electronic configuration is not observ ing the 59. For d electrons, the azim uthal quantum num ber is
( n  l ) rule
[MNR 1983; CPMT 1984]
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 1 ,4 s 2 (a) 0 (b) 1
2 2 6 2 6 7 2
(b) 1s ,2 s sp ,3 s 3 p 3d ,4 s (c) 2 (d) 3
2 2 6 2 6 5
(c) 1s ,2 s 2 p ,3 s 3 p 3d ,4 s
1
60. For p -orbital, the m agnetic quantum number has v alue
2 2 6 2 6 8
(d) 1s ,2 s 2 p ,3 s 3 p 3d ,4 s
2 (a) 2 (b) 4 , – 4
(c) – 1 , 0, + 1 (d) 0
52. The four quantum numbers of the outermost orbital of K
(atomic no. =19) are [MP PET 1993, 94] 61. For n  3 energy level, the number of possible orbitals (all
1 kinds) are [BHU 1981; CPMT 1985; MP PMT 1995]
(a) n  2, l  0, m  0, s  
2 (a) 1 (b) 3
(c) 4 (d) 9
 62 Structure of atom
62. Which of the following ions is not hav ing the 72. In a potassium atom, electronic energy lev els are in the
configuration of neon following order [EA MCET 1979; DPMT 1991]

(a) F  (b) Mg 2 (a) 4 s  3d (b) 4 s  4 p

(c) 4 s  3d (d) 4 s  3 p
(c) Na  (d) Cl 
63. Elem ents upto atomic number 103 have been synthesized 73. Fe (atom ic num ber = 2 6) atom has the electronic
and studied. If a newly discov ered elem ent is found to arrangement [NCERT 1974; MNR 1980]
have an atomic number 106, its electronic configuration (a) 2 , 8, 8, 8 (b) 2 , 8, 1 6
will be (c) 2 , 8,1 4 , 2 (d) 2 , 8, 1 2 , 4
[A IIMS 1980]
74. Cu 2  will hav e the following electronic configuration
(a) [Rn]5 f ,6 d ,7 s
14 4 2
(b) [Rn]5 f ,6 d 1 ,7 s 2 7 p 3
14
[MP PMT 1985]

(c) [Rn]5 f ,6 d ,7 s
14 6 0
(d) [Rn]5 f ,6 d ,7 s
14 5 1
(a) 1s ,2 s 2 p ,3 s 3 p 3d
2 2 6 2 6 10

64. Ions which have the sam e electronic configuration are (b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9 ,4 s 1
those of
(a) Lithium and sodium (b) Sodium and potassium (c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9
(c) Potassium and calcium (d) Oxy gen and chlorine (d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10 ,4 s 1
65. When the azim uthal quantum num ber has a v alue of 75. Which one is the electronic con figuration of Fe 2
l  0 , the shape of the orbital is [MP PET 1995]
[MA DT Bihar 1982; AIIMS 1989]
(a) Rectangular (b) Spherical
(a) 1s ,2 s 2 p ,3 s 3 p 3d 6
2 2 6 2 6
(c) Dum bbell (d) Unsy m m etrical
66. The m agnetic quantum number for v alency electrons of (b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 4 ,4 s 2
sodium is [CPMT 1988; MH CET 1999]
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 5 ,4 s 1
(a) 3 (b) 2
(d) None of these
(c) 1 (d) 0
76. How m any electrons can be fit into the orbitals that
67 . The electronic configuration of an elem ent with atom ic
number 7 i.e. nitrogen atom is [CPMT 1982, 84, 87] com prise the 3 rd quantum shell n  3
[MP PMT 1986, 87; Orissa JEE 1997]
(a) 1s 2 ,2 s 1 ,2 p x3 (b) 1s 2 ,2 s 2 2 p x2 2 p1y (a) 2 (b) 8
(c) 1s 2 ,2 s 2 2 p 1x 2 p 1y 2 p 1z (d) 1s 2 ,2 s 2 2 p 1x 2 p y2 (c) 1 8 (d) 3 2
77. Which element is represented by the following electronic
68. In a m ulti-electron atom, which of the following orbitals configuration [MP PMT 1987]
described by the three quantum m em bers will hav e the
sam e energy in the absence of m agnetic and electric fields 2p
[A IEEE 2005] 2s   
(1 ) n  1, l  0, m  0 (2 ) n  2, l  0, m  0 1s 
(3 ) n  2, l  1, m  1 (4 ) n  3, l  2, m  0 
(5) n  3, l  2, m  0 (a) Nitrogen (b) Oxy gen
(a) (1 ) and (2 ) (b) (2 ) and (3 ) (c) Fluorine (d) Neon
(c) (3 ) and (4 ) (d) (4 ) and (5) 7 8. If the v alue of azim uthal quantum num ber is 3 , the
possible v alues of m agnetic quantum num ber would be
69. Which of the following represents the electronic
[MP PMT 1987; RPMT 1999; AFMC 2002; KCET 2002]
configuration of an elem ent with atom ic num ber 1 7
[A MU 1982] (a) 0, 1 , 2 , 3 (b) 0, – 1 , – 2 , – 3
(c) 0,  1 ,  2 ,  3 (d)  1 ,  2 ,  3
2 2 6
(a) 1s ,2 s 2 p ,3 s 3 p 1 6
(b) 1s ,2 s 2 p ,3 s 2 3 p 4 ,4 s1
2 2 6
79. Kry pton (36 Kr) has the electronic configuration (18 Ar)
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 5 (d) 1s 2 ,2 s 2 2 p 6 ,3 s 1 3 p 4 ,4 s 2
4 s 2 ,3d 10 ,4 p 6 . The 37 th electron will go into which one of
70. The shape of s -orbital is [NCERT 1978I] the following sub-lev els
(a) Py ram idal (b) Spherical [CBSE PMT 1989; CPMT 1989; EAMCET 1991]
(c) Tetrahedral (d) Dum b-bell shaped (a) 4 f (b) 4 d
7 1. When 3 d orbital is com plete, the new electron will enter (c) 3 p (d) 5 s
the
[EA MCET 1980; MP PMT 1995] 1
80. If an electron has spin quantum num ber of  and a
2
(a) 4 p -orbital (b) 4 f -orbital
m agnetic quantum number of 1 , it cannot be presented
(c) 4 s -orbital (d) 4 d -orbital in an [CBSE PMT 1989; UPSEA T 2001]
 Structure of atom 63
(a) d -orbital (b) f -orbital (c) Cs  (d) K 
(c) p -orbital (d) s -orbital 93. The order of filling of electrons in the orbitals of an atom
will be [
81. The azim uthal quantum num ber is related to
(a) 3d, 4 s, 4 p, 4 d, 5 s (b) 4 s, 3d, 4 p, 5 s, 4 d
[BHU 1987, 95]
(a) Size (b) Shape (c) 5 s, 4 p, 3d, 4 d, 5 s (d) 3d, 4 p, 4 s, 4 d, 5 s
(c) Orientation (d) Spin 94. The quantum number which m ay be designated by s, p, d
82. The total number of electrons that can be accomm odated and f instead of number is BHU 1980]
in all the orbitals having principal quantum number 2 and
azimuthal quantum number 1 is [CPMT 1971, 89, 91] (a) n (b) l
(a) 2 (b) 4 (c) m l (d) m s
(c) 6 (d) 8 95. Which of the following represents the correct sets of the
83. Electronic configuration of C is [CPMT 1975] four quantum num bers of a 4 d electron
[MNR 1992; UPSEAT 2001; J&K CET 2005]
(a) 1s 2 ,2 s 2 2 p 2 (b) 1s 2 ,2 s 2 2 p 3
1
(a) 4 , 3, 2, (b) 4 , 2 , 1 , 0
(c) 1s 2 ,2 s 2 (d) 1s 2 ,2 s 2 2 p 6 2
84. There is no difference between a 2 p and a 3 p orbital 1 1
(c) 4 , 3,  2,  (d) 4 , 2, 1, 
regarding [BHU 1981] 2 2
(a) Shape (b) Size 96. Which of the following statem ents is not correct for an
(c) Energy (d) Value of n electron that has the quantum numbers n  4 and m  2
[MNR 1993]
85. The electronic configuration of chrom ium is
[MP PMT 1993; MP PET 1995; BHU 2001; BCECE 2005]
1
(a) The electron m ay have the quantum num ber s  
2
(a) [ Ne ]3 s 2 3 p 6 3d 4 ,4 s 2 (b) [ Ne ]3 s 2 3 p 6 3d 5 ,4 s 1
(b) The electron m ay have the quantum num ber l  2
(c) [ Ne ]3 s 2 3 p 6 ,4 s 2 4 p 4 (d) [ Ne ]3 s 2 3 p 6 3d 1 ,4 s 2 4 p 3 (c) The electron m ay have the quantum num ber l  3
86. The shape of p -orbital is [MP PMT 1993] (d) The electron m ay hav e the quantum num ber
l  0, 1, 2, 3
(a) Elliptical (b) Spherical
97 . The set of quantum num bers not applicable for an
(c) Dum b-bell (d) Com plex geom etrical electron in an atom is [MNR 1994]
87 . The electronic configuration (outerm ost) of Mn 2 ion (a) n  1, l  1, m l  1, m s  1 / 2
(atom ic num ber of Mn  25 ) in its ground state is (b) n  1, l  0, m l  0, m s  1 / 2
[MP PET 1993]
(c) n  1, l  0, m l  0, m s  1 / 2
(a) 3 d 5 ,4 s 0 (b) 3 d 4 ,4 s 1
(d) n  2, l  0, m l  0, m s  1 / 2
(c) 3 d 3 ,4 s 2 (d) 3d 2 ,4 s 2 4 p 2
98. Correct configuration of Fe 3 [2 6] is
88. The principal quantum number represents [CPMT 1991]
[CPMT 1994; BHU 1995; KCET 1992]
(a) Shape of an orbital
(b) Distance of electron from nucleus (a) 1s ,2 s 2 p ,3 s 3 p 6 3d 5
2 2 6 2

(c) Num ber of electrons in an orbit (b) 1s 2 ,2 s 2 sp 6 ,3 s 2 3 p 6 3d 3 ,4 s 2

(d) Num ber of orbitals in an orbit (c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 6 ,4 s 2
89. When the azim uthal quantum num ber has a v alue of
l  1 , the shape of the orbital is [MP PET 1993] (d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 5 ,4 s 1
(a) Unsy m m etrical (b) Spherically symmetrical 99. Azim uthal quantum number for last electron of Na atom
(c) Dum b-bell (d) Com plicated is
[BHU 1995]
90. How m any electrons can be accommodated in a sub-shell
(a) 1 (b) 2
for which n  3, l  1 [CBSE PMT 1990]
(c) 3 (d) 0
(a) 8 (b) 6 100. A 3 p orbital has [IIT 1995]
(c) 1 8 (d) 3 2 (a) Two spherical nodes
91. For azim uthal quantum num ber l  3 , the m axim um (b) Two non-spherical nodes
num ber of electrons will be [CBSE PMT 1991;
(c) One spherical and one non -spherical nodes
EA MCET 1991; RPMT 2002; CBSE PMT 2002]
(d) One spherical and two non -spherical nodes
(a) 2 (b) 6
101. All electrons on the 4 p sub-shell m ust be characterized
(c) 0 (d) 1 4
by the quantum number(s) [MP PET 1996]
92. An ion has 1 8 electrons in the outerm ost shell, it is
1
[CBSE PMT 1990] (a) n  4 , m  0, s   (b) l  1
2
 4
(a) Cu (b) Th
 64 Structure of atom
1
(c) l  0, s   (d) s  
1 (c) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 4
2 2
(d) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 1 3d 3
102. The electronic configuration of the elem ent of atom ic
num ber 2 7 is 110. Which of the following configuration is correct for iron
[CBSE PMT 1999]
(a) 1s , 2 s 2 p , 3 s 3 p , 4 s () 4 p ()()() 5 s ()
2 2 6 2 6
(a) 1s 2 s 2 p 3 s 3 p 3d
2 2 6 2 6 5

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

(b) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 5
(c) 1s , 2 s 2 p , 3 s 3 p , 3d ()()()(), 4 s ()
2 2 6 2 6
(c) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 7
(d) 1s , 2 s 2 p , 3 s 3 p , 3d ()()()()() 4 s ()
2 2 6 2 6
(d) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 6
103. When the value of the principal quantum number n is 3 , 111. Which of the following set of quantum numbers belong to
the permitted values of the azimuthal quantum num bers highest energy [CPMT 1999]
l and the m agnetic quantum num bers m , are
1
l m (a) n  4 , l  0, m  0, s  
2
0 0
1
(a) 1  1, 0,  1 (b) n  3, l  0, m  0, s  
2
2  2,1, 0,  1,2
1
1 1 (c) n  3, l  1, m  1, s  
2
(b) 2  2, 1,  1 1
(d) n  3, l  2, m  1, s  
3  3,2, 1,  2,3 2
0 0 112. Which quantum number will determine the shape of the
(c) 1 1, 2, 3 subshell [CPMT 1999; Pb. PMT 1998]
2  3,  2, 1,  2,3 (a) Principal quantum num ber
(b) Azim uthal quantum num ber
1 0, 1
(c) Magnetic quantum num ber
(d) 2 0, 1, 2
(d) Spin quantum num ber
3 0, 1, 2, 3
113. For the n  2 energy level, how many orbitals of all kinds
104. The number of possible spatial orientations of an electron are possible [Bih a r CEE 1995]
in an atom is giv en by its
(a) 2 (b) 3
(a) Spin quantum num ber
(c) 4 (d) 5
(b) Spin angular m om entum
(c) Magnetic quantum num ber 114. Which one is in the ground state [DPMT 1996]
(d) Orbital angular m om entum

105. Which of the following sets of orbitals m ay degenerate
(a)   
(a) 2 s, 2 p x , 2 p y (b) 3 s, 3 p x ,3d xy

(c) 1s, 2 s, 3 s (d) 2 p x , 2 p y , 2 p z
106. The set of quantum num bers n  3, l  0, m  0, s  1 / 2 (b)   
belongs to the elem ent 
(a) Mg (b) Na 
(c) Ne (d) F (c)   
107 . An electron has principal quantum num ber 3 . The

num ber of its (i) sub-shells and (ii) orbitals would be
respectiv ely 
[MP PET 1997] (d)   
(a) 3 and 5 (b) 3 and 7 
(c) 3 and 9 (d) 2 and 5
2 115. When the principal quantum number (n  3) , the possible
108. What is the electronic configuration of Cu (Z  29) of
least position [MP PET /PMT 1998; MP PET 2001] v alues of azim uthal quantum num ber ( l ) is
[Bihar MEE 1996; KCET 2000]
(a) [ Ar] 4 s 3d
1 8
(b) [ Ar] 4 s 2 3d 10 4 p 1
(a) 0, 1 , 2 , 3 (b) 0, 1 , 2
(c) [ Ar] 4 s 1 3d 10 (d) [ Ar] 3 d 9 (c) – 2 , – 1 , 0, 1 , 2 (d) 1 , 2 , 3
(e) 0, 1
109. The correct electronic configuration of Ti(Z  22) atom is
116. Which statem ent is not correct for n  5 , m  3
[MP PMT 1999]
2 2 6 2 6 2 2 [CPMT 1996]
(a) 1s 2 s 2 p 3 s 3 p 4 s 3d
1
2 2 6 2 6 4 (a) l  4 (b) l  0, 1, 3; s  
(b) 1s 2 s 2 p 3 s 3 p 3d 2
 Structure of atom 65
(c) l  3 (d) All are correct (c) [ Xe]4 s 3 5 d 5 6 s 2 (d) [ Xe]4 f 6 5 d 2 6 s 2
117 . 2 2
1s 2 s 2 p 3 s 6 1
shows configuration of [CPMT 1996] 125. An e  has m agnetic quantum number as 3 , what is its
(a) Al 3
in ground state (b) Ne in excited state principal quantum number [BHU 1998]

(c) Mg in excited state (d) None of these (a) 1 (b) 2
(c) 3 (d) 4
118. Fiv e v alence electrons of p 15 are labelled as
126. The number of quantum numbers required to describe an
AB X Y Z electron in an atom com pletely is [CET Pu n e 1998]
3s 3p (a) 1 (b) 2
1 (c) 3 (d) 4
If the spin quantum of B and Z is  , the group of
2 127 . The electronic configuration 1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1z
electrons with three of the quantum num ber sam e are
[A FMC 1997; Pb. PMT 1999; CBSE PMT 2001; AIIMS 2001]
[JIPMER 1997]
(a) Oxy gen (b) Nitrogen
(a) AB, XYZ , BY (b) AB
(c) Hy drogen (d) Fluorine
(c) XYZ, AZ (d) AB, XYZ
128. Which one of the following set of quantum numbers is not
119. Electronic configuration of Sc 21 is [BHU 1997] possible for 4 p electron [EA MCET 1998]
(a) 1s 2 s 2 p 3 s 3 p 4 s 3d
2 2 6 2 6 2 1
1
(a) n  4 , l  1, m  1, s  
2 2 6
(b) 1s 2 s 2 p 3 s 3 p 4 s 3d 2 6 1 2 2
1
(c) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 0 3d 3 (b) n  4 , l  1, m  0, s  
2
(d) 1s 2 2 s 2 2 p 6 3 s 2 3 p 2 4 s 2 3d 2
1
120. If n  l  6 , then total possible num ber of subshells (c) n  4 , l  1, m  2, s  
2
would be [RPMT 1997]
1
(a) 3 (b) 4 (d) n  4 , l  1, m  1, s  
2
(c) 2 (d) 5
129. Which of the following orbital is not possible[RPMT 1999]
121. An electron hav ing the quantum num bers (a) 3 f (b) 4 f
1 (c) 5 f (d) 6 f
n  4, l  3, m  0 , s   would be in the orbital
2 130. Which set of quantum numbers for an electron of an atom
[Or issa JEE 1997] is not possible [RPMT ; DCE 1999]
(a) n  1, l  0, m  0, s  1 / 2
(a) 3 s (b) 3 p
(b) n  1, l  1, m  1, s  1 / 2
(c) 4 d (d) 4 f (c) n  1, l  0, m  0, s  1 / 2
122. Which of the following sets of quantum num bers is not (d) n  2, l  1, m  1, s  1 / 2
allowed [Or issa JEE 1997]
131. Electronic configuration of ferric ion is [RPET 2000]
1
(a) n  1, l  0, m  0, s   (a) [ Ar] 3 d 5
(b) [ Ar] 3 d 7
2
(c) [ Ar] 3 d 3
(d) [ Ar] 3 d 8
1
(b) n  1, l  1, m  0, s   132. What is the m aximum number of electrons which can be
2
accom modated in an atom in which the highest principal
1 quantum num ber value is 4 [MP PMT 2000]
(c) n  2, l  1, m  1, s  
2 (a) 1 0 (b) 1 8
(c) 3 2 (d) 54
1 133. Which of the following electronic configurations is not
(d) n  2, l  1, m  0, s  
2 possible
[CPMT 2000]
123. For which of the following sets of four quantum numbers,
2 2 6
an electron will have the highest energy [CBSE PMT 1994] (a) 1s 2 s
2 2
(b) 1s 2 s 2 p
10 2 2
n l m s (c) 3d 4s 4 p (d) 1s 2 2 s 2 2 p 2 3 s 1
(a) 3 2 1 + 1 /2 134. The electronic configuration of an elem ent is
(b) 4 2 1 + 1 /2 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 5 4 s1 . This represents its
(c) 4 1 0 –1 /2 [IIT Screening 2000]
(a) Excited state (b) Ground state
(d) 5 0 0 –1 /2
(c) Cationic form (d) Anionic form
124. The electronic configuration of gadolinium (atom ic no. 135. Which of the following set of quantum num bers is
64) is [CBSE PMT 1997]
possible
8 9 2 7 1 2 [A IIMS 2001]
(a) [ Xe]4 s 5 d 6 s (b) [ Xe]4 s 5 d 6 s
 66 Structure of atom
1 (a) Hund’s rule
(a) n  3; l  2; m  2 and s  
2 (b) Aufbau’s principle
1 (c) Pauli’s exclusion principle
(b) n  3; l  4 ; m  0 and s   (d) Heisenberg’s uncertainty principle
2
145. Which of the following has maximum energy
1
(c) n  4 ; l  0; m  2 and s   [A IIMS 2002]
2 3s 3p 3d
1
(d) n  4 ; l  4 ; m  3 and s   (a)
2
136. Which of the following set of quantum num ber is not 3s 3p 3d
v alid
[A IIMS 2001] (b)
(a) n  1, l  2 (b) 3  2, m  1
(c) m  3, l  0 (d) 3  4, l  2 3s 3p 3d
(c)
137 . Which one pair of atom s or ions will hav e sam e
configuration [JIPMER 2001]
(a) F  and Ne (b) Li  and He  3s 3p 3d
 (d)
(c) Cl and Ar (d) Na and K
138. Which of the following sets of quantum num ber is not 146. The total m agnetic quantum num bers for d-orbital is
possible [MP PET 2001] given by [
1 (a) 2 (b) 0,  1 ,  2
(a) n  3; l  2; m  0; s  
2 (c) 0, 1 , 2 (d) 5
1
(b) n  3; l  0; m  0; s   147 . The outer electronic structure 3 s 2 3 p 5 is possessed by
2 [Pb. PMT 2002; Pb. CET 2001]
1 (a) Cl (b) O
(c) n  3; l  0; m  1; s  
2 (c) Ar (d) Br
1 148. Which of the following set of quantum num ber is not
(d) n  3; l  1; m  0; s  
2 possible [Pb. PMT 2002]
139. Which of the following set of quantum numbers is correct n l m1 m2
for the 19th electron of chromium [DCE 2001]
(a) 3 2 1 + 1 /2
n l m s
(b) 3 2 1 – 1 /2
(a) 3 0 0 1 /2
(c) 3 2 1 0
(b) 3 2 –2 1 /2 (d) 5 2 –1 + 1 /2
(c) 4 0 0 1 /2 149. The configuration 1s 2 , 2 s 2 2 p 5 , 3 s 1 shows[Pb. PMT 2002]
(d) 4 1 –1 1 /2
(a) Excited state of O 2
140. When the v alue of azim uthal quantum num ber is 3 , (b) Excited state of neon
magnetic quantum number can have values[DPMT 2001] (c) Excited state of fluorine
(a) + 1 , 0, – 1
(d) Ground state of fluorine atom
(b) + 2 , + 1 , 0, – 1 , – 2
150. The quantum num ber ‘m’ of a free gaseous atom is
(c) – 3 , – 2 , – 1 , – 0, + 1 , + 2 , + 3 associated with [A IIMS 2003]
(d) + 1 , – 1
(a) The effectiv e v olum e of the orbital
141. The quantum num bers n  2, l  1 represent [A FMC 2002]
(b) The shape of the orbital
(a) 1 s orbital (b) 2 s orbital (c) The spatial orientation of the orbital
(c) 2 p orbital (d) 3 d orbital
(d) The energy of the orbital in the absence of a m agnetic
142. The m agnetic quantum num ber of v alence electron of field
sodium (Na) is [RPMT 2002]
151. Correct statement is [BHU 2003]
(a) 3 (b) 2
(c) 1 (d) 0 (a) K  4 s , Cr  3d 4 s , Cu  3d 4 s
1 4 2 10 2

143. Azimuthal quantum number defines [A IIMS 2002]

(b) K  4 s 2 , Cr  3d 4 4 s 2 , Cu  3d 10 4 s 2
(a) e/m ratio of electron
(b) Spin of electron (c) K  4 s 2 , Cr  3d 5 4 s 1 , Cu  3d 10 4 s 2
(c) Angular m om entu m of electron
(d) K  4 s 1 , Cr  3d 5 4 s 1 , Cu  3d 10 4 s 1
(d) Magnetic m om entum of electron
144. Quantum numbers of an atom can be defined on the basis 152. Number of orbitats in h sub-shell is [BHU 2003]
of [A IIMS 2002] (a) 1 1 (b) 1 5
 Structure of atom 67
(c) 1 7 (d) 1 9 (a) Heisenberg’s principle
153. Electronic configuration (b) Hund’s rule
1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 3d 5 , 4 s 1 represents [CPMT 2003] (c) Aufbau principle
(d) Pauli exclusion principle
(a) Ground state (b) Excited state
162. The electronic configuration of elem ent with atom ic
(c) Anionic state (d) All of these number 24 is [Pb. CET 2004]
154. Which of the following sets is possible for quantum
num bers (a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 4 ,4 s 2
[RPET 2003] (b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10
(a) n  4, l  3, m  2, s  0 (c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 6
1 (d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 5 4 s 1
(b) n  4 , l  4 , m  2, s  
2 163. The m axim um num ber of electrons in p -orbital with
(c) n  4 , l  4 , m  2, s  
1 n  5, m  1 is [Pb. CET 2003]
2 (a) 6 (b) 2
(d) n  4 , l  3, m  2, s  
1 (c) 1 4 (d) 1 0
2 164. Num ber of two electron can have the sam e v alues of ……
155. For principle quantum number n  4 the total number of quantum num bers [UPSEA T 2004]
orbitals having l  3 [A IIMS 2004] (a) One (b) Two
(a) 3 (b) 7 (c) Three (d) Four
(c) 5 (d) 9 165. The number of orbitals present in the shell with n  4 is
156. The num ber of 2 p electrons hav ing spin quantum [UPSEAT 2004]
num ber s  1 / 2 are [KCET 2004] (a) 1 6 (b) 8
(c) 1 8 (d) 3 2
(a) 6 (b) 0
166. Which of the following electronic configuration is not
(c) 2 (d) 3
possible
157 . Which of the following sets of quantum num bers is [MHCET 2003]
correct for an electron in 4 f orbital [A IEEE 2004]
(a) 1s 2 s2 2
(b) 1s ,2 s 2 p 6
2 2
1
(a) n  4 , l  3, m  1, s  
2 (c) [ Ar] 3d 10 ,4 s 2 4 p 2 (d) 1s 2 ,2 s 2 2 p 2 ,3 s 1
1 167 . p x orbital can accom m odate
(b) n  4 , l  4 , m  4 , s  
2 [MNR 1990; IIT 1983; MADT Bihar 1995; BCECE 2005]
1 (a) 4 electrons
(c) n  4 , l  3, m  4 , s  
2 (b) 6 electrons
1 (c) 2 electrons with parallel spins
(d) n  3, l  2, m  2, s  
2 (d) 2 electrons with opposite spins
158. Consider the ground state of (Z  24) . The num bers of 168. The m axim um num ber of electrons that can be
electrons with the azimuthal quantum numbers, l  1 and accom m odated in ' f ' sub shell is
2 are, respectiv ely [A IEEE 2004] [CPMT 1983, 84; MP PET/PMT 1988; BITS 1988]
(a) 1 6 and 4 (b) 1 2 and 5 (a) 2 (b) 8
(c) 1 2 and 4 (d) 1 6 and 5 (c) 3 2 (d) 1 4
159. The four quantum num bers of the v alence electron of 169. The number of electrons which can be accom m odated in
potassium are [DPMT 2004] an orbital is [DPMT 1981; A FMC 1988]
1 1 (a) One (b) Two
(a) 4 , 1 , 0 and (b) 4 , 0, 1 and
2 2 (c) Three (d) Four
1 1 170. The num ber of electrons in the atom which has 2 0
(c) 4 , 0, 0 and  (d) 4 , 1 , 1 and protons in the nucleus[CPMT 1981, 93; CBSE PMT 1989]
2 2
(a) 2 0 (b) 1 0
160. Which of the following electronic configuration is not
possible according to Hund’s rule (c)a l3a0PMT 2004]
[Ker (d) 4 0
(a) 1s 2 2 s 2 (b) 1s 2 2 s 1 171. The m aximum number of electrons accommodated in 5 f
orbitals are [MP PET 1996]
(c) 1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1x (d) 1s 2 2 s 2 2 p x2
(a) 5 (b) 1 0
(e) 1s 2 2 s 2 2 p x2 2 p 1y 2 p 1z (c) 1 4 (d) 1 8
161. The ground state term sy m bol for an electronic state is 172. The m aximum number of electrons in an atom with l  2
governed by [UPSEA T 2004] and n  3 is [MP PET /PMT 1998]
 68 Structure of atom
(a) 2 (b) 6 (a) 6 (b) 4
(c) 1 2 (d) 1 0 (c) 3 (d) 1
173. The configuration 1s 2 2 s 2 2 p 5 3 s 1 shows [A IIMS 1997] 185. 3d 10 4 s 0 electronic configuration exhibits by
(a) Ground state of fluorine atom (a) Zn   (b) Cu  
(b) Excited state of fluorine atom (c) Cd   (d) Hg  
(c) Excited state of neon atom 186. Which of the following m etal ions will hav e m axim um
(d) Excited state of ion O 2 number of unpaired electrons [CPMT 1996]

174. For sodium atom the number of electrons with m  0 will (a) Fe 2 (b) CO 2
be [RPMT 1999] (c) Ni 2 (d) Mn 2
(a) 2 (b) 7 187 . Which of the m etal ion will hav e highest num ber of
(c) 9 (d) 8 unpaired electrons
175. The number of electrons that can be accom m odated in (a) Cu  (b) Fe 2 
dz 2 orbital is [Ku r u ksh et r a CEE (c) Fe 3  (d) Co 2 
2002]
188. The m axim um num ber of unpaired electron can be
(a) 1 0 (b) 1 present in d orbitals are
(c) 4 (d) 2 (a) 1 (b) 3
(c) 5 (d) 7
176. Num ber of unpaired electrons in 1s 2 2 s 2 2 p 3 is
189. The m olecule hav ing one unpaired electron is
[CPMT 1982; MP PMT 1987; BHU 1987; (a) NO (b) CO
CBSE PMT 1990; CET Pune 1998; AIIMS 2000]
(c) CN  (d) O 2
(a) 2 (b) 0
190. A filled or half-filled set of p or d -orbitals is spherically
(c) 3 (d) 1
sy m m etric. Point out the species which has spherical
177. Total number of unpaired electrons in an atom of atom ic
sy m m etry [NCERT 1983]
num ber 2 9 is [CPMT 1984, 93]
(a) Na (b) C
(a) 1 (b) 3
(c) 4 (d) 2 (c) Cl  (d) Fe
191. The atom of the element having atomic number 14 should
17 8. The num ber of unpaired electrons in 1s 2 , 2 s 2 2 p 4 is hav e [A MU 1984]
[NCERT 1984; CPMT 1991; MP PMT 1996, 2002] (a) One unpaired electron (b) Two unpaired electrons
(a) 4 (b) 2 (c) Three unpaired electrons (d)Four unpaired electrons
(c) 0 (d) 1 192. An atom has 2 electrons in K shell, 8 electrons in L shell
179. The m axim um num ber of electrons that can be and 6 electrons in M shell. The num ber of s -electrons
accom m odated in a 3 d subshell is present in that elem ent is [CPMT 1989]
(a) 2 (b) 1 0 (a) 6 (b) 5
(c) 6 (d) 1 4 (c) 7 (d) 1 0
180. The m aximum number of electrons which each sub-shell 193. The num ber of unpaired electrons in carbon atom in
can occupy is [Pb. CET 1989] excited state is [MNR 1987]
(a) 2n 2 (b) 2n (a) One (b) Two
(c) 2(2l  1) (d) (2l  1) (c) Three (d) Four
181. Num ber of unpaired electrons in the ground state of 194. Maxim um num ber of electrons present in ' N ' shell is
bery llium atom is [EA MCET 1984]
(a) 2 (b) 1 (a) 1 8 (b) 3 2
(c) 0 (d) All the abov e (c) 2 (d) 8
182. How m any unpaired electrons are present in Ni 2  cation 195. The number of d electrons in Fe 2 (atom ic num ber of
(atom ic num ber = 2 8) [IIT 1981; MNR 1984; Fe  26 ) is not equal to that of the [MNR 1993]
MP PMT 1995; Kerala PMT 2003] (a) p -electrons in Ne (At. No.= 1 0)
(a) 0 (b) 2
(b) s -electrons in Mg (At. No.= 1 2 )
(c) 4 (d) 6
(c) d -electrons in Fe
183. The number of unpaired electrons in an O 2 m olecule is
[MNR 1983] (d) p -electrons in Cl  (At. No. of Cl = 1 7 )
(a) 0 (b) 1 4
196. A transition m etal X has a configuration [ Ar]3 d in its
(c) 2 (d) 3 3 oxidation state. Its atomic number is[EA MCET 1990]
184. The number of unpaired electrons in a chrom ic ion Cr 3  (a) 2 5 (b) 2 6
(atomic number = 24) is [MNR 1986; CPMT 1992] (c) 2 2 (d) 1 9
 Structure of atom 69
197 . The total num ber of electrons present in all the p - (c) Pauli's exclusion principle
orbitals of brom ine are [MP PET 1994] (d) Uncertainty principle
(a) Fiv e (b) Eighteen 208. According to Aufbau's principle, which of the three
(c) Sev enteen (d) Thirty fiv e 4 d , 5 p and 5 s will be filled with electrons first[MA DT Bi h a r 19
198. Which of the following has the m axim um num ber of (a) 4 d
unpaired electrons [IIT 1996]
(b) 5 p
(a) Mg 2  (b) Ti 3  (c) 5 s
(c) V 3  (d) Fe 2  (d) 4 d and 5 s will be filled sim ultaneously
199. Which of the following has m ore unpaired d -electrons 209. The energy of an electron of 2 p y orbital is [A MU 1984]
[CBSE PMT 1999]
(a) Greater than that of 2 p x orbital
(a) Zn  (b) Fe 2 
(b) Less than that of 2 p x orbital
(c) N 3  (d) Cu 
(c) Equal to that of 2 s orbital
200. Maximum electrons in a d -orbital are [CPMT 1999]
(d) Sam e as that of 2 p z orbital
(a) 2 (b) 1 0
210. Which of the following principles/rules lim its the
(c) 6 (d) 1 4
maximum number of electrons in an orbital to two[CBSE PMT 198
201. The num ber of unpaired electrons in Fe 3  (Z  26) are (a) Aufbau principle
[KCET 2000] (b) Pauli's exclusion principle
(a) 5 (b) 6 (c) Hund's rule of m axim um m ultiplicity
(c) 3 (d) 4 (d) Heisenberg's uncertainty principle
202. How m any unpaired electrons are present in cobalt [ Co] 211. The electrons would go to lower energy lev els first and
m etal [RPMT 2002] then to higher energy lev els according to which of the
(a) 2 (b) 3 following
(c) 4 (d) 7 [BHU 1990; MP PMT 1993]
203. The num ber of unpaired electrons in nitrogen is (a) Aufbau principle
[Pb. CET 2002] (b) Pauli's exclusion principle
(a) 1 (b) 3 (c) Hund's rule of m axim um m ultiplicity
(c) 2 (d) None of these (d) Heisenberg's uncertainty principle
212. Energy of atomic orbitals in a particular shell is in the
204. Which of the following has the least energy
order
(a) 2 p (b) 3 p [A FMC 1990]
(c) 2 s (d) 4 d (a) s  p  d  f (b) s  p  d  f
205. Pauli's exclusion principle states that [CPMT 1983, 84] (c) p  d  f  s (d) f  d  s  p
(a) Nucleus of an atom contains no negativ e charge
213. Aufbau principle is not satisfied by [MP PMT 1997]
(b) Electrons m ove in circular orbits around the nucleus
(a) Cr and Cl (b) Cu and Ag
(c) Electrons occupy orbitals of lowest energy
(d) All the four quantum numbers of two electrons in an (c) Cr and Mg (d) Cu and Na
atom cannot be equal 214. Which of the following explains the sequence of filling the
206. For the energy lev els in an atom , which one of the electrons in different shells [A IIMS 1998; BHU 1999]
following statements is correct [A IIMS 1983] (a) Hund's rule (b) Octet rule
(a) There are sev en principal electron energy lev els (c) Aufbau principle (d) All of these
(b) The second principal energy lev el can hav e four sub- 215. Aufbau principle is obey ed in which of the following
energy levels and contains a maximum of eight electrons electronic configurations [A FMC 1999]
(c) The M energy lev el can hav e m axim um of 3 2 (a) 1s 2 2 s 2 2 p 6 (b) 1s 2 3 p 3 3 s 2
electrons
(c) 1s 2 3 s 2 3 p 6 (d) 1s 2 2 s 2 3 s 2
(d) The 4 s sub-energy level is at a higher energy than
216. Following Hund’s rule which elem ent contains six
the 3 d sub-energy lev el
unpaired electron [RPET 2000]
207 . The statem ents [A IIMS 1982]
(a) Fe (b) Co
(i) In filling a group of orbitals of equal energy , it is
(c) Ni (d) Cr
energetically preferable to assign electrons to em pty
orbitals rather than pair them into a particular 217 . Electron enters the sub-shell for which (n  l) v alue is
orbital. m inim um . This is enunciated as
(ii) When two electron s are placed in two different [RPMT 2000]
orbitals, energy is lower if the spins are parallel. (a) Hund’s rule
are v alid for (b) Aufbau principle
(a) Aufbau principle (c) Heisenberg uncertainty principle
(b) Hund's rule (d) Pauli’s exclusion principle
 70 Structure of atom
218. The atom ic orbitals are progressiv ely filled in order of (a) F  (b) Oxy gen atom
increasing energy . This principle is called as
[MP PET 2001] (c) Mg (d) N 
(a) Hund’s rule (b) Aufbau principle 2. Atom s consists of protons, neutrons and electrons. If the
(c) Exclusion principle (d) de-Broglie rule m ass of neutrons and electrons were m ade half and two
219. The correct order of increasing energy of atom ic orbitals tim es respectively to their actual masses, then the atomic
is m ass of 6 C 12 [NCERT 1982]
[MP PET 2002] (a) Will rem ain approxim ately the sam e
(a) 5 p  4 f  6 s  5 d (b) 5 p  6 s  4 f  5 d (b) Will becom e approxim ately two tim es
(c) 4 f  5 p  5 d  6 s (d) 5 p  5d  4 f  6 s (c) Will rem ain approxim ately half
220. The orbital with maximum energy is [CPMT 2002] (d) Will be reduced by 2 5%
(a) 3 d (b) 5p 3. The increasing order (lowest first) for the v alues of e / m
(c) 4 s (d) 6 d (charge/m ass) for [IIT 1984]
221. p-orbitals of an atom in presence of m agnetic field are (a) e, p, n,  (b) n, p, e, 
[Pb. PMT 2002]
(c) n, p, , e (d) n, , p, e
(a) Two fold degenerate (b) Non degenerate
(c) Three fold degenerate (d) None of these 4. The electronic configuration of a dipositive metal M 2  is
222. Orbital angular momentum for a d-electron is[MP PET 2003] 2 , 8, 14 and its atomic weight is 56 a.m.u. The num ber of
neutrons in its nuclei would be
6h 6h
(a) (b) [MNR 1984, 89; Kerala PMT 1999]
2 2
(a) 3 0 (b) 3 2
12h 12 h
(c) (d) (c) 3 4 (d) 4 2
2 2
223. Number of nodal centres for 2s orbital [RPET 2003] 5. The ratio of the energy of a photon of 2000 Å wavelength
(a) 1 (b) 0 radiation to that of 4000 Å radiation is
(c) 4 (d) 3 [IIT 1986; DCE 2000; JIPMER 2000]
224. The orbital angular m om entum of an electron in 2 s - (a) 1 /4 (b) 4
orbital is [MP
(c)PET
1 /22004] (d) 2
1 h h 6. Discov ery of the nucleus of an atom was due to the
(a) (b)
2 2 2 experiment carried out by [CPMT 1983; MP PET 1983]
h (a) Bohr (b) Mosley
(c) 2 (d) Zero
2 (c) Rutherford (d) Thom son
225. The m aximum num ber of electrons present in an orbit 7. In a Bohr's m odel of atom when an electron jum ps from
l  3 , is [Pb. PMT 2004] n  1 to n  3 , how m uch energy will be em itted or
(a) 6 (b) 8 absorbed [CBSE PMT 1996]
(c) 1 0 (d) 1 4
(a) 2.15  10 11 erg (b) 0.1911  10 10 erg
226. Number of unpaired electrons in Mn 4  is [DPMT 2005]
(a) 3 (b) 5 (c) 2.389  10 12 erg (d) 0.239  10 10 erg
(c) 6 (d) 4 8. The nucleus of an atom can be assum ed to be spherical.
227 . Which of the following sequence is correct as per Aufbau The radius of the nucleus of m ass number A is giv en by
principle [DPMT 2005] 1.25  10 13  A1 / 3 cm Radius of atom is one Å . If the
(a) 3 s  3d  4 s  4 p (b) 1s  2 p  4 s  3d m ass num ber is 64 , then the fraction of the atom ic
(c) 2 s  5 s  4 p  5 d (d) 2 s  2 p  3d  3 p volume that is occupied by the nucleus is [NCERT 1983]
228. Electronic configuration of deuterium atom is (a) 1.0  10 3 (b) 5.0  10 5
[J&K CET 2005]
(c) 2.5  10 2 (d) 1.25  10 13
(a) 1s 1 (b) 2s 2 9. The energy of an electron in the first Bohr orbit of
(c) 2s 1 (d) 1s 2 H atom is 13.6 eV . The possible energy v alue(s) of the
excited state(s) for electrons in Bohr orbits to hy drogen
is(are)
[IIT 1998; Orissa JEE 2005]
(a) 3.4 eV (b) 4.2eV
(c) 6.8 eV (d) 6.8 eV
10. The energy of the electron in the first orbit of He  is
1. Which of the following atom s and ions are isoelectronic  871.6  10 20 J . The energy of the electron in the first
i.e. hav e the sam e num ber of electrons with the neon orbit of hydrogen would be[Roor kee Qu a l ify i n g 1998]
atom
[NCERT 1978] (a)  871.6  10 20 J (b)  435.8  10 20 J
 Structure of atom 71

(c)  217.9  10 20 J (d)  108.9  10 20 J 21. Which of the following electron transition in a hy drogen
atom will require the largest am ount of energy
11. The total number of v alence electrons in 4.2 gm of N 3 [UPSEAT 1999, 2000, 01]
ion is ( N A is the Av ogadro's number) [CBSE PMT 1994] (a) From n  1 to n  2 (b) From n  2 to n  3
(a) 1.6 N A (b) 3.2 N A (c) From n   to n  1 (d) From n  3 to n  5
(c) 2.1 N A (d) 4.2 N A 22. In Bohr series of lines of hy drogen spectrum , the third
line from the red end corresponds to which one of the
12. The Bohr orbit radius for the hy drogen atom (n  1) is following inter-orbit jumps of the electron for Bohr orbits
approximately 0.530 Å . The radius for the first excited in an atom of hy drogen [A IEEE 2003]
state (n  2) orbit is [CBSE PMT 1998; BHU 1999] (a) 3  2 (b) 5  2
(a) 0.13 Å (b) 1.06 Å (c) 4  1 (d) 2  5
(c) 4.77 Å (d) 2.12 Å 23. The v alue of Planck’s constant is 6.63  10 34 Js. The
v elocity of light is 3.0  10 8 ms 1 . Which value is closest to
13. The frequency of a wave of light is 12  10 14 s 1 . The wav e
number associated with this light is [Pb. PMT 1999] the wavelength in nanometres of a quantum of lig ht with
frequency of 8  10 15 s 1 [CBSE PMT 2003]
(a) 5  10 7 m (b) 4  10 8 cm 1
(a) 3  10 7
(b) 2  10 25
(c) 2  10 7 m 1 (d) 4  10 4 cm 1
14. The series lim it for Balm er series of H-spectra is (c) 5  10 18 (d) 4  10 1
[A MU (Engg.) 1999] 24. As electron m oves away from the nucleus, its potential
energy [UPSEA T 2003]
(a) 3 800 (b) 4 2 00
(a) Increases (b) Decreases
(c) 3 6 4 6 (d) 4 000
(c) Rem ains constant (d) None of these
15. The ionization energy of hydrogen atom is 13.6 eV . The
energy required to excite the electron in a hydrogen atom
from the ground state to the first excited state is
(Avogadro’s constant = 6.022 × 10 23) [BHU 1999]

(a) 1.69  10 20 J (b) 1.69  10 23 J

(c) 1.69  10 23 J (d) 1.69  10 25 J
Read the assertion and reason carefully to m ark the correct
16. The energy required to dislodge electron from excited option out of the options giv en below :
isolated H-atom, IE1  13.6 eV is [DCE 2000]
(a) I f both assertion and reason are true and the reason is
(a)  13.6 eV (b)  13.6 eV the correct explanation of the assertion.
(b) I f both assertion and reason are true but reason is not
(c)  13.6 and  3.4 eV (d)  3.4 eV the correct explanation of the assertion.
(c) I f assertion is true but reason is false.
17. The num ber of nodal planes in a p x is (d) I f the assertion and reason both are false.
[IIT Screening 2000] (e) I f assertion is false but reason is true.
(a) One (b) Two 1. Assertion : The position of an electron can be
(c) Three (d) Zero determined exactly with the help of an
18. The third line in Balm er series corresponds to an electron m icroscope.
electronic transition between which Bohr’s orbits in Reason : The product of uncertainty in the
hy drogen m easurement of its m om entum and the
[MP PMT 2001] uncertainty in the m easurem ent of the
position cannot be less than a finite limit.
(a) 5  3 (b) 5  2
[NDA 1999]
(c) 4  3 (d) 4  2 2. Assertion : A spectral line will be seen for a
19. Which of the following has m aximum number of unpaired 2 p x  2 p y transition.
electron (atomic number of Fe 26) [MP PMT 2001]
Reason : Energy is released in the form of wav e of
(a) Fe (b) Fe (II) light when the electron drops from
(c) Fe (III) (d) Fe (IV) 2 p x  2 p y orbital. [A IIMS 1996]
20. The frequency of one of the lines in Paschen series of
3. Assertion : The cation energy of an electron is largely
hy drogen atom is 2.340  10 11 Hz. The quantum num ber determ ined by its principal quantum
n 2 which produces this transition is [DPMT 2001] num ber.
(a) 6 (b) 5 Reason : The principal quantum num ber n is a
m easure of the m ost probable distance of
(c) 4 (d) 3 finding the electron around the nucleus.
[A IIMS 1996]
 72 Structure of atom

4. Assertion : Nuclide 30
Al13 is less stable than 40
Ca 20 Reason : Num ber of orbitals in a shell equals to
2n .
Reason : Nuclides having odd num ber of protons 17. Assertion : Energy of the orbitals increases as
and neutrons are generally unstable
[IIT 1998] 1s  2 s  2 p  3 s  3 p  3d  4 s  4 p
5. Assertion : The atoms of different elem ents hav ing  4 d  4 f  ......
sam e m ass number but different atom ic Reason : Energy of the electron depends
num ber are known as isobars com pletely on principal quantum
Reason : The sum of protons and neutrons, in the num ber.
isobars is always different [A IIMS 2000] 18. Assertion : Splitting of the spectral lines in the
6. Assertion : Two electrons in an atom can hav e the presence of m agnetic field is known as
sam e values of four quantum num bers. stark effect.
Reason : Two electrons in an atom can be present Reason : Line spectrum is sim plest for hy drogen
in the sam e shell, sub-shell and orbital atom .
and have the same spin [A IIMS 2001] 19. Assertion : Thom son’s atom ic m odel is known as
7. Assertion : The v alue of n for a line in Balm er series ‘raisin pudding’ m odel.
of hy drogen spectrum having the highest Reason : The atom is v isualized as a pudding of
wav e length is 4 and 6 . positiv e charge with electrons (raisins)
Reason : For Balm er series of hydrogen spectrum , em bedded in it.
the v alue n1  2 and n2  3 , 4 , 5. 20. Assertion : Atom ic orbital in an atom is designated
by n, l, m l and m s .
[A IIMS 1992]
Reason : These are helpful in designating electron
8. Assertion : Absorption spectrum conists of som e
present in an orbital.
bright lines separated by dark spaces.
21. Assertion : The transition of electrons n3  n 2 in H
Reason : Em ission spectrum consists of dark lines.
[A IIMS 2002] atom will em it greater energy than
n4  n3 .
9. Assertion : A resonance hybrid is always m ore stable
than any of its canonical structures. Reason : n 3 and n 2 are closer to nucleus tan n 4 .
Reason : This stability is due to delocalization of 22. Assertion : Cathode rays are a stream of  -particles.
electrons. [A IIMS 1999]
Reason : They are generated under high pressure
10. Assertion : Cathode ray s do not trav el in straight and high v oltage.
lines.
23. Assertion : In case of isoelectronic ions the ionic size
Reason : Cathode ray s penetrate through thick increases with the increase in atom ic
sheets [A IIMS 1996] num ber.
11. Assertion : Electrons revolving around the nucleus Reason : The greater the attraction of nucleus,
do not fall into the nucleus because of greater is the ionic radius.
centrifugal force.
Reason : Rev olv ing electrons are planetary
electrons.
[A IIMS 1994]
12. Assertion : Threshold frequency is a character istic
for a m etal.
Reason : Threshold frequency is a m axim um
frequency required for the ejection of Discovery and Properties of anode, cathode rays
electron from the m etal surface.
neutron and Nuclear structure
13. Assertion : The radius of the first orbit of hy drogen
atom is 0.52 9 Å.
1 d 2 a 3 c 4 c 5 b
Reason : Radius for each circular orbit
(rn )  0.52 9 Å (n 2 / Z) , where n  1 ,2 ,3 6 a 7 b 8 a 9 d 10 c

and Z  atom ic num ber. 11 b 12 d 13 b 14 a 15 b

14. Assertion : 3 d z 2 orbital is spherically sy m m etrical. 16 b 17 c 18 c 19 c 20 b

Reason : 3 d z 2 orbital is the only d -orbital which 21 a 22 d 23 c 24 b 25 d

is spherical in shape. 26 c 27 b 28 d 29 c 30 a
15. Assertion : Spin quantum number can have the value 31 b 32 d 33 b 34 c 35 c
+ 1 /2 or –1 /2 .
36 a 37 b 38 a 39 d 40 c
Reason : (+ ) sign here signifies the wave function.
16. Assertion : Total number of orbitals associated with 41 c
principal quantum num ber n  3 is 6.
 Structure of atom 73
Atomic number, Mass number, Atomic species 1 b 2 b 3 a 4 c 5 c
6 c 7 b 8 d 9 d 10 a
1 b 2 a 3 b 4 b 5 a
11 a 12 c 13 a 14 b 15 d
6 a 7 c 8 b 9 c 10 b
16 b 17 a 18 c 19 c 20 b
11 b 12 c 13 b 14 c 15 c
16 c 17 c 18 a 19 c 20 a Quantum number, Electronic configuration
21 c 22 b 23 c 24 d 25 b and Shape of orbitals
26 b 27 a 28 a 29 c 30 b
1 c 2 a 3 b 4 d 5 c
31 c 32 d 33 d 34 c 35 c
6 c 7 c 8 a 9 a 10 a
36 c 37 c 38 b 39 d 40 c
11 c 12 c 13 a 14 a 15 d
41 b 42 c 43 a 44 c 45 b
16 c 17 c 18 d 19 b 20 c
46 c 47 d 48 a 49 c 50 c
21 c 22 a 23 c 24 d 25 c
51 a 52 c 53 b 54 a 55 c
26 c 27 b 28 d 29 e 30 b
56 a 57 d 58 c 59 a 60 a
31 d 32 a 33 c 34 d 35 d
61 d 62 b 63 a 64 c 65 b
36 c 37 b 38 b 39 d 40 c
66 a 67 c 68 a 69 d 70 d
41 d 42 c 43 c 44 a 45 a
71 c 72 a 73 b 74 d
46 a 47 b 48 c 49 c 50 b
Atomic models and Planck's quantum theory 51 c 52 b 53 b 54 b 55 c
56 c 57 b 58 e 59 c 60 c
1 c 2 a 3 b 4 b 5 d
6 b 7 c 8 b 9 c 10 a 61 d 62 d 63 d 64 c 65 b

11 b 12 a 13 d 14 b 15 b 66 d 67 c 68 d 69 c 70 b
16 c 17 a 18 c 19 a 20 d 71 a 72 c 73 c 74 c 75 a
21 d 22 c 23 d 24 d 25 c 76 c 77 c 78 c 79 d 80 d
26 a 27 c 28 b 29 c 30 a 81 b 82 c 83 a 84 a 85 b
31 b 32 c 33 d 34 b 35 b
86 c 87 a 88 b 89 c 90 b
36 a 37 c 38 c 39 c 40 a
91 d 92 a 93 b 94 b 95 d
41 c 42 d 43 d 44 a 45 d
96 d 97 a 98 a 99 d 100 c
46 b 47 a 48 c 49 d 50 a
51 a 52 c 53 d 54 c 55 b 101 b 102 d 103 a 104 c 105 d
56 b 57 b 58 a 59 b 60 c 106 a 107 c 108 d 109 a 110 d
61 c 62 b 63 c 64 c 65 b 111 d 112 b 113 c 114 b 115 b
66 b 67 c 68 a 69 b 70 d 116 a 117 c 118 b 119 a 120 a
71 a 72 d 73 a 74 c 75 d
121 d 122 b 123 b 124 b 125 d
76 b 77 a 78 a 79 c 80 a
126 d 127 b 128 c 129 a 130 b
81 a
131 a 132 c 133 d 134 b 135 a
Dual nature of electron 136 a 137 c 138 c 139 c 140 c
141 c 142 d 143 c 144 c 145 b
1 c 2 a 3 a 4 b 5 c
146 d 147 a 148 c 149 b 150 c
6 b 7 d 8 a 9 d 10 d
151 d 152 a 153 a 154 d 155 b
11 c 12 c 13 b 14 b 15 b
156 d 157 a 158 b 159 c 160 d
16 c 17 c 18 c 19 b 20 a
161 c 162 d 163 b 164 c 165 a
21 d
166 d 167 d 168 d 169 b 170 a
171 c 172 d 173 c 174 b 175 d
Uncertainty principle and Schrodinger wave
equation 176 c 177 a 178 b 179 b 180 c
 74 Structure of atom

181 c 182 b 183 c 184 c 185 a 10. (c) This is because chargeless particles do not undergo
any deflection in electric or m agnetic field.
186 d 187 c 188 c 189 a 190 c
11. (b) Neutron and proton found in nucleus.
191 b 192 a 193 d 194 b 195 d 13. (b) Cathode ray s are m ade up of negativ ely charged
196 a 197 c 198 d 199 b 200 b particles (electrons) which are deflected by both the
electric and m agnetic fields.
201 a 202 b 203 b 204 c 205 d 15. (b) Mass of neutron is greater than that of proton, m eson
206 b 207 b 208 c 209 d 210 b and electron.
Mass of neutron = mass of proton + m ass of electron
211 a 212 a 213 b 214 c 215 a
16. (b) Proton is 1 837 (approx 1800) times heav ier than an
216 d 217 b 218 b 219 b 220 d 1
electron. Penetration power 
221 b 222 b 223 a 224 d 225 d mass
226 a 227 b 228 a 18. (c) Nucleus of helium is 2 He 4 m ean 2 neutrons and 2
protons.
Critical Thinking Questions 19. (c) Proton is the nucleus of H  atom ( H  atom dev oid
of its electron).
1 a 2 d 3 d 4 a 5 d 20. (b) Cathode ray s are m ade up of negativ ely charged
6 c 7 b 8 d 9 a 10 c particles (electrons, e  )
11 a 12 d 13 d 14 c 15 b 26. (c) Size of nucleus is m easured in Fermi (1 Ferm i
16 d 17 a 18 b 19 c 20 b  10 15 m) .
21 a 22 a 23 d 24 a 27 . (b) A m olecule of an elem ent is a incorrect statem ent.
The correct statement is “an element of a m olecule”.
Assertion & Reason

1 d 2 d 3 a 4 a 5 c
6 d 7 e 8 d 9 a 10 e
11 b 12 c 13 a 14 d 15 c
16 d 17 c 18 e 19 a 20 e
21 b 22 d 23 d

Discovery and Properties of anode, cathode rays

neutron and Nuclear structure

1. (d) Neutrons and protons in the nucleus and electrons in

the extranuclear region.
2. (a) It consists of proton and neutron and these are also
known as nucleones.
15
3. (c) Radius of nucleus ~ 10 m.
4. (c) Positiv e ions are formed from the neutral atom by the
loss of electrons.
5. (b) The  -ray particle constitute electrons.
6. (a) Jam es Chadwick discov ered neutron (0 n1 ) .
7. (b) Charge/m ass for
2 1 1
n  0,   , p  and e 
4 1 1 / 1837
11
9. (d) The density of neutrons is of the order 10 kg / cc.
 Structure of atom 75
29. (c) Proton is represented by p having charge +1 discovered in 1988 12. (c) Most probable radius = a / Z 0

by Goldstein. where a = 52.9 pm. For helium ion, Z = 2.

0

31. (b) The nature of anode rays depends upon the nature of residual 52 . 9
gas. r =
mp
= 26.45 pm.
2
32. (d) H  (proton) will have very large hydration energy due to its
very small ionic size 13. (b) Four unpaired electron are present in the Fe 2  ion
2
1 Fe 26  [ Ar] 3d 6 ,4 s 0
Hydration energy 
Size 14. (c) Na  has 10 electron and Li  has 2 electron so these are
33. (b) Mass of a proton  1.673  10 24 g different number of electron from each other.
16. (c) P15  2, 8, 5
 Mass of one mole of proton
 9.1  10 24  6.02  10 23  10.07  10 1  1.008 g 17. (c) 8O  1s 2 2 s 2 2 p 4

Mass of a electron  9.1  10 28 g 18. (a) 35 Br

80
 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 10 4 s 2 4 p 5
 Mass of one mole of electron A  80 , Z  35 , N  ?
 9.1  10 28
 6.02  10 23
 54.78  10 5
g  0.55 mg . N  A  Z  80  35  45
35. (c) One mole of electron = 6.023 × 10 electron 23
atomic number (Proton) is 35 and no. of neutron is 45.
Mass of one electron = 9.1 × 10 gm –28

19. (c) 16 O   have more electrons than neutron

8
Mass of one mole of electrons
p  8, e  10, n  8 .
= 6.023 × 10 23  9.1  10 28 gm = 5.48  10 4 gm
20. (a) A 12 and X 13 both are isotopes but have different no. of
= 5.48  10 4  1000 mg = 0.548 gm  0.55 mg .
6 6
neutrons.
36. (a) Charge on proton  1 unit, charge on  particle = + 2 units, 2 : A12 , For A have p  6, e  6 and n  6 and
6
1.
37. (b) m p / me ~ 1837 ~ 1.8  10 3 . 6 X 13 , For B have p  6, e  6 and n  7

38. (a) Splitting of signals is caused by protons attached to adjacent 21. (c) P  20, mass no. (A) = 40
carbon provided these are not equivalent to the absorbing N  A–P = 40  20  20
proton.
39. (d) Nucleus consists of proton and neutron both are called as P  N  20.
nucleon.
22. (b) Electrons in Na   11  1  10
40. (c) Positron (1e 0 ) has the same mass as that of an electron
Electrons in Mg2   12  2  10
(1e ) .
0
40
23. (c) 20 Ca has 20 proton, 20 neutron.
1
41. (c) Electron time lighter than proton so their mass ratio 24. (d) CH 3  6  3  1  8e ,
1837
will be 1 : 1837
H 3 O   3  8  1  10e  ,

Atomic number, Mass number, Atomic species NH 3  7  3  10e  , CH 3  6  3  1  10e 

1. (b) The number of electrons in an atom is equal to its atomic 25. (b)  CONH 2  6  8  7  2  1 (from other atom to form
number i.e. number of protons. covalent bond) = 24.
2. (a) No. of protons = Atomic no. = 25 and no. of neutron = 55 – 26. (b) Complete E.C.  [ Ar]18 3d 10 4 s 2 4 p 6 .
25 = 30.
3. (b) No. of neutrons = mass number – no. of protons. = W – N. Hence no. of e   no. of protons  36  Z .
4. (b) 30 Zn 70 , Zn 2  has No. of Neutrons = 70 – 30 = 40. 28. (a) K   1s 2 2 s 2 2 p 6 3 s 2 3 p 6

5. (a) Na  and Ne are isoelectronic which contain 10 electrons. Cl   1s 2 2 s 2 2 p 6 3 s 2 3 p 6 .

6. (a) One molecule of CO 2 have 22 electrons. 29. (c) Mass no.  At. Wt.
Mass no. = no. of protons + no. of neutrons
7. (c) Cl and Cl  differs in number of electrons. Cl has 17 e 
At. no. = no. of protons.
while Cl  has 18 e  .
30. (b) N 2O  14  8  22
8. (b) CO and CN  are isoelectronic.
CO  6  8  14 and CN   6  7  1  14 . CO 2  6  16  22.
9. (c) Mass of an atom is due to nucleus (neutron + proton). 31. (c) Neutron in 12
 6, , Neutrons in 28
Si  14
6 C 14
10. (b) Atomic number is defined as the number of protons in the
nucleus. Ratio = 6 : 14 = 3 : 7.

11. (b) X 56 A  P  N  Z  N  E  N 33. (d) N 7  1s 2 2 s 2 2 p 3

26

N  A  E  56  26  30 N   1s 2 2 s 2 2 p 2
 76 Structure of atom
C  1s 2 2 s 2 2 p 2 . 62. (b) N 3  , F  and Na  (These three ions have e   10 , hence
they are isoelectronic)
34. (c) O  C  O, linear structure 180 o angle
63. (a) NO 3 and CO 32  consist of same electron and show same
Cl  Hg  Cl, linear structure 180 o angle. isostructural.
35. (c) H   1s 2 and He   1s 2 . 64. (c) Atomic number of chlorine 17 and in Cl  ion total no. of
36. (c) In the nucleus of an atom only proton and neutrons are electron =18.
present. 65. (b) Tritium (H 13 ) has one proton and two neutron.
63
37. (c) Cu 29 Number of neutrons = atomic mass – atomic number = 63
67. (c) X 35  1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 10 4 s 2 4 p 5
– 29 = 34.
38. (b) 21 Protons and 24 Neutrons are present in nucleus and element Total no. of e  is all p–orbitals  6  6  5  17 .
is Sc. 68. (a) Since its nucleus contain 9 proton so its. atomic number is 9
40. (c) 14
, n  14  7  7 and its electronic configuration is 2, 7. So it require one more
7 X
electron to complete its octet. Hence its valency is 1.
42. (c) Cl  have 17 proton, 18 neutron and 18 electron. 69. (d) K 2 S formed by K  and S 2  ion. We know that atomic
43. (a) Number of unpaired electrons in inert gas is zero because they
have full filled orbitals. number of K is 19 and in K  ion its atomic number would be
44. (c) Electrons and Protons are same in neutral atom. 18 similarly atomic number of S is 16 and in form S 2  ion its
48. (d) No. of proton and no. of electron = 18 [ Ar1836 ] and No. of atomic number would be 18 so the K  and S 2  are
neutron = 20 isoelectronic with each other in K 2 S .
Mass number = P + N = 18  20  38. 70. (d) 20 Ca  2, 8, 8, 2
Ca 2   2, 8, 8
231
49. (c) In Xe 89 number of protons and electrons is 89 and No. of
neutrons = A – Z = 231 – 89 = 142. Hence, Ca 2  has 8 electrons each in outermost and
51. (a) NO 2 and O 3 are isostere. The number of atoms in these penultimate shell.
71. (c) Atomic no. of C = 6 so the number of protons in the nucleus =
( 3) and number of electrons (24 ) are same. 6
52. (c) Number of electrons in nitrogen = 7 and number of electron is
72. (a) No. of P  Z  7; No. of electrons in N 3   7  3  10.
oxygen = 8 we know that formula of nitrate ion is NO 3 we
2
also know that number of electron 73. (b) Heavy hydrogen is 1 D .Number of neutrons = 1
= (1 × Number of electrons in nitrogen) 74. (d) Atomic number is always whole number.
+ (3 × number of electrons in oxygen) + 1
= (1 × 7) + (3 × 8) + 1 = 32. Atomic models and Planck's quantum theory
Molecular mass 256
53. (b) Atomicity =   8  S8 . 2. (a) The central part consisting whole of the positive charge and
Atomic mass 32 most of the mass caused by nucleus, is extremely small in size
compared to the size of the atom.
54. (a) In case of N 3  , p  7 and c  10
3. (b) Electrons in an atom occupy the extra nuclear region.
55. (c) Chlorine Cl 17  [Ne ] 4. (b) According to the Bohr model atoms or ions contain one
3s 2
3p 5
electron.
5. (d) The nucleus occupies much smaller volume compared to the
volume of the atom.
7. (c) -particles pass through because most part of the atom is
empty.
Three electron pair
56. (a) Bromine consists of outer most electronic configuration 8. (b) An electron jumps from L to K shell energy is released.
[ Ar] 3d 10 4 s 2 4 p 5 . 9. (c) Neutron is a chargeless particles, so it does not deflected by
electric or magnetic field.
57. (d) Na   1s 2 2 s 2 2 p 6 10. (a) Energy is always absorbed or emitted in whole number or
multiples of quantum.
Mg  1s 2 2 s 2 2 p 6
++

11. (b) Both He and Li  contain 2 electrons each.

18. (c) During the experimental verification of de-Broglie equation,
O 2   1s 2 2 s 2 2 p 6
Davisson and Germer confirmed wave nature of electron.
Cl   1s 2 2 s 2 2 p 6 3 s 2 3 p 6 19. (a) Increases due to absorption of energy and it shows absorption
spectra.
60. (a) Ar1840 = atomic number 18 and no. of Neutron in case of 20. (d) Rutherford -Scattering experiment.
21. (d) It represents Heisenberg’s uncertainty principle.
Ar22
Neutron = Atomic mass – Atomic number E4 22 4 1 E 328
23. (d)  2   ; E4  2   82 kJ / mol.
 40  18  22 E2 4 16 4 4 4
61. (d) Nucleus of tritium contain [H 13 ] c 3  10 8
27. (c) When c     than     1 .5  10 2 m
p 1 , e 1 , n  2  2  10 6
 Structure of atom 77
28. (b) According to quantum theory of radiation, a hot body emits  12.1567  10 8 m
radiant energy not continuously but discontinuously in the
form of small packets of energy called quanta or photons. 3  10 8
c
34
v   24.66  1014 Hz .
h 6 .6  10  12.567  10 8
30. (a) p   3  10 23 kgms 1
 2 .2  10 11 h h
67. (c) We know that   ;  m
n 2h2 mv m
34. (b) Bohr’s radius  . Which is a positive quantity.
4 2 me 2 z The velocity of photon (v) = 3  10 8 m sec 1
40. (a) Gold used by Rutherford in scatting experiment.   1.54  10 8 cm  1.54  10 10 meter
 1 1  6 .626  10 34 Js
41. (c) E  E3  E2  13.6  2  2   1 .9 eV m
 (2) (3)  1 .54  10 10 m  3  10 8 m sec 1
42. (d) R  R 0 ( 1.4  10 13 cm)  A 1 / 3  1.4285  10 32 kg .
q 1q  1 68. (a) The spliting of spectral line by the magnetic field is called
43. (d)        9 .6  10 7  4 .8  10 7 C kg 1 Zeeman effect.
 m  2  m  p 2
69. (b) r  n 2 (excited state n  2 )
44. (a) According to Hydrogen spectrum series.
r  4a 0
45. (d) The electron can move only in these circular orbits where the
h 70. (d) rn  n 2 : An  n 4
angular momentum is a whole number multiple of or it
2 A 2 n 24 2 4 16
is quantised.  4  4   16 : 1
A1 n1 1 1
46. (b) Generally electron moving in orbits according to Bohr‘s
principle. 4 2 mr 2
47. (a) According to the planck’s law that energy of a photon is 71. (a) It will take
nh
directly proportional to its frequency i.e. E  h
49. (d) Bohr’s radius of the hydrogen atom n2
72. (d) rH  0 .529 Å
n 2  0.529Å z
r  ; where z = Atomic number, For hydrogen ; n  1 and z  1 therefore
z
n = Number of orbitals rH  0.529 Å
2 .172  10 18  2 .172  10 18 For Be 3  : Z  4 and n  2 Therefore
51. (a) E 
n2 22
0 .529  2 2
19
 5.42  10 J . rBe 3    0 .529 Å .
4
hc hc
52. (c) E  or   2
13.6 Z eff
 E 73. (a) Eionisation  E  En  eV
n2
6 .64  10 34  3  10 8
  6 .64  10 8 Å  13.6 Z 2 13.6 Z 2 
3  10 8 =   
2
53. (d) rn  r1  n 2  n 2 n12 
r3  0.53  3 2  0.53  9  4.77 Å 13.6  1 2 13.6  1 2
E  h   ; h  13.6  0.85
54. (c) According to Rutherford an atom consists of nucleus which is (1)2 (4 )2
small in size but carries the entire mass (P+ N ).
55. (b) Wavelength of spectral line emitted is inversely proportional to  h  6.625  10 34
1 13.6  0 .85
energy   .   1.6  10 19 = 3.08  10 15 s 1 .
E 6 .625  10 34
1 1 1 1 1 
56. (b) E  ; E1  ; E2  74. (c)
1
 R 2  2 
 8000 16000   n1 n 2 
E1 16000
  2  E1  2 E2 1 1 1 
E2 8000  1 .097  10 7 m 1  2  2 
 1  
c3  10 8 ms 1
58. (a)     5 .0  1014 Hz .    91  10 9 m
 600  10  9 m
13.6 13.6 13.6 We know 10 9  1 nm So   91nm
59. (b) E eV    3.40 eV
n2 22 4 75. (d) r  n 2
r2 (2)2 For I orbit   1
st

65. (b) Bohr radius   2  4.

r1 (1) For III orbit =   3 2  9
rd

1 So it will 9  .
1 1  1 1 
66. (b) v   R  2  2   109678     82258.5 76. (b) Bohr suggest a formulae to calculate the radius and energy of
  n1 n 2  1 4  each orbit and gave the following formulae
  1.21567  10 5 cm or   12.1567  10 6 cm
 78 Structure of atom

n2h2 h 6 .63  10 34

rn     6 .63  10  29 m
4 2 kme 4 Z mv 10 6  10
Where except n 2 , all other unit are constant so rn  n 2 . 15. (b) According to de–Broglie
 E0 h 6 .62  10 20 erg. sec
77. (a) Energy of an electron E   
n2 mv 2
 5  10 4 cm / sec
For energy level (n  2) 6 .023  10 23
13.6 13.6 6 .62  10 27  6 .023  10 23
E   3 .4 eV .  cm  4  10 8 cm  4 Å .
(2) 2
4 2  5  10 4
78. (a) Energy of ground stage (E 0 ) = 13.6 eV and energy level = h 6 .625  10 34
5 16. (c)    10  30 m .
mv 5
13.6 13.6 13.6 0 .2 kg 
E5  eV  =  0 .54 eV . 60  60 ms 1
n2 52 25
17. (c) From de Broglie equation
79. (c) Positive charge of an atom is present in nucleus.
81. (a) For n4  n1 , greater transition, greater the energy difference, h 6 .62  10 34
 = = 1.32  10 35 m .
lesser will be the wavelength. mv 0 .5  100
18. (c) Dual nature of particle was proposed by de-broglie who gave
Dual nature of electron the following equation for the wavelength.
h
h h h 
1. (c) According to de-Broglie equation   or or . mv
mv p mc 19. (b) One percent of the speed of light is
h h h  1 
4. (b)   or or de-Broglie equation. v  1
 (3 .00  10 ms ) = 3.00  10 ms
8 6 1
p mv mc
 100 
5. (c) Emission spectra of different  accounts for quantisation of
energy. Momentum of the electron (p ) = m 
6. (b) According to de-Broglie equation = (9.11  10 31 kg ) (3.00  10 6 ms 1 )
h h h
 , p  mv ,   ,   = 2.73  10 24 kg ms 1
mv p mc
The de-broglie wavelength of this electron is
 h 
7. (d) According to de-Broglie    .
 mv  h 6 .626  10 34
 
p 2 .73  10  24 kgms 1
h 6 .63  10 34
8. (a)    6 .63  10  33 m
mv 10  3  100   2.424  10 10 m .
h 1 20. (a) We know that the correct relationship between wavelength and
9. (d)   . For same velocity   . h
mv m momentum is   . Which is given by de-Broglie.
SO 2 molecule has least wavelength because their molecular p
mass is high. 21. (d) De-broglie equation applies to all the material object in motion.
h
10. (d) de-Broglie equation is   . Uncertainty principle and Schrodinger wave
p
11. (c) Formula for de-Broglie wavelength is
equation
h h 1 2eV 1. (b) The uncertainty principle was enunciated by Heisenberg.
 or    eV  mv 2 or  
p mv 2 m 2. (b) According to uncertainty principle, the product of uncertainties
of the position and momentum, is x  p  h / 4 .
h 6 .62  10 34
  h
2meV 2  9 .1  10  31  2 .8  10  23 5. (c) x  p  is not the correct relation. But correct
4
8
  9.28  10 meter . h
Heisenberg’s uncertainty equation is x  p  .
h 4
12. (c)  , p  mv 7. (b) According to the Heisenberg’s uncertainty principle momentum and
p
exact position of an electron can not be determined simultaneously.
h 6 .62  10 34 h
  8. (d) x. p  , if  x  0 then p   .
mv 9 .1  10  31  1 .2  10 5 4
  6.626  10 9 m . 12. (c) According to x  p 
h
4
13. (b) Mass of the particle (m)  10 6 kg and velocity of the particle
h 6 .62  10 34
(v)  10 ms 1 x    5.27  10 30 m .
p  4 1  10 5  4  3 .14
 Structure of atom 79
13. (a) Uncertainty of moving bullet velocity 10. (a) 3d subshell filled with 5 electrons (half-filled) is more stable
h 6 .625  10 34 than that filled with 4 electrons. 1, 4 s electrons jumps into
v  
4  m  v 4  3 .14  .01  10 5 3 d subshell for more sability.
28 (c) In 2p – orbital, 2 denotes principal quantum number (n) and
 5.2  10 m/sec . 11.
h p denotes azimuthal quantum number (l  1) .
14. (b) x .p  This equation shows Heisenberg’s uncertainty
4 12. (c) Electronic configuration of H  is 1s 2 . It has 2 electrons in
principle. According to this principle the product of uncertainty extra nuclear space.
in position and momentum of particle is greater than equal to
h 13. (a) The electronic configuration must be 1s 2 2 s 1 . Hence, the
. element is lithium (z  3).
4
15. (d) Spin quantum number does not related with Schrodinger 14. (a) Principal quantum no. tells about the size of the orbital.
equation because they always show 1 / 2 , 1 / 2 value. 15. (d) An element has the electronic configuration
h h 1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 2 , (Si) . It’s valency electrons are four.
16. (b) According to x  m  v  ; v 
4 x  m  4 16. (c) The magnetic quantum number specifies orientation of orbitals.
6 .6  10 34 17. (c) If l  2, m  3. (e to e ) .
 5
 2 .1  10  29 m / s
10  0 .25  3 .14  4 18. (d) If n  3 then l  0, 1, 2 but not 3.
h
17. (a) Uncertainity in position x  20. (c) Atomic number of Cu is 29  ( Ar) 4 s 1 3d 10 .
4  p
6 .63  10 34 21. (c) The shape of 2 p orbital is dumb-bell.
  5.28  10 30 m .
4  3 .14  (1  10 5 ) 22. (a) When the value of n  2 , then l  1 and the value of
18. (c) Given that mass of electron  9.1  10 31
kg m  1, 0,  1 i.e. 3 values.
34 2 1
Planck’s constant  6.63  10 kg m s 23. (c) Cr24  ( Ar) 3d 5 4 s 1 electronic configuration because half filled
h orbital are more stable than other orbitals.
h
By using x  p  ; x  v  m  24. (d) Kr has zero valency because it contains 8 electrons in
4 4
outermost shell.
where : x = uncertainity in position
25. (c) 2 electron in the valence shell of calcium Ca 20  (2, 8, 8, 2) .
v = uncertainity in velocity
h 27. (b) Value of l  1 means the orbital is p (dumb-bell shape).
x  v 
4  m 28. (d) Cr has Ar4 s 1 3d 5 electronic configuration because half
34
6 .63  10 filled orbital are more stable than other orbitals.
  5.8  10 5 m 2 s 1 .
4  3 .14  9 .1  10 31 31. (d) The two electrons will have opposite spins.
33. (c) If m = –3, then l = 3, for this value n must be 4.
Quantum number, Electronic configuration
and Shape of orbitals 2n 2
34. (d) No. of electrons  2n 2 hence no. of orbital   n2 .
2
3. (b) The shape of an orbital is given by azimuthal quantum number 2n 2
'l' . 35. (d) No. of electrons  2n 2 hence no. of orbital   n2 .
2
5. (c) Hund’s rule states that pairing of electrons in the orbitals of a
36. (c) If n  3 then l  0 to n  1 & m  l to l
subshell (orbitals of equal energy) starts when each of them is
singly filled. 37. (b) Na 11  2, 8, 1  1s 2 , 2 s 2 2 p 6 , 3 s 1
2 2 6
6. (c) 1s , 2 s , 2 p represents a noble gas electronic configuration. n  3, l  0, m  0, s  1 / 2
7. (c) The electronic configuration of Ag in ground state is 38. (b) Hund’s rule states that pairing of electrons in the orbitals of a
[Kr] 4 d 10 5 s 1 . subshell (orbitals of equal energy) starts when each of them is
singly filled.
8. (a) n, l and m are related to size, shape and orientation
respectively. 39. (d) As a result of attraction, some energy is released. So at infinite
distance from the nucleus energy of any electron will be
9. (a) Electronic configuration of Rb(37 ) is maximum. For bringing electrons from  to the orbital of
any atom some work has to be done be electrons hence it bill
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 10 4 s 2 4 p 6 5 s 1 loose its energy for doing that work.
So for the valence shell electron (5 s 1 ) 40. (c) This space is called nodal space where there is no possibility of
oressene of electrons.
1 (d) For s orbital l  0 m  0 .
n  5, l  0, m  0 , s   41.
2
42. (c) For M th shell, n  3; so maximum no. of electrons in M th
shell  2n 2  2  3 2  18 .
 80 Structure of atom
43. (c) m  l to l including zero. 67. (c) N  1s 2 ,2 s 2 2 p1x , 2 p1y , 2 p1z . Hund’s rule states that pairing
7
44. (a) Number of radial nodes = (n  l  1) of electrons in the orbitals of a subshell (orbitals of equal
For 3s: n = 3, l = 0 energy) starts when each of them is singly filled.
(Number of radial node = 2) 68. (d) (4) and (5) belong to d -orbital which are of same energy.
For 2p: n =2, l = 1 69. (c) Atomic no. 17 is of chlorine.
(Number of radial node = 0) 70. (b) The s-orbital has spherical shape due to its non- directional
nature.
45. (a) It consists only s orbital which is circular.
71. (a) According to the Aufbau’s principle the new electron will enter
46. (a) Hund’s rule states that pairing of electrons in the orbitals of a
subshell (orbitals of equal energy) starts when each of them is in those orbital which have least energy. So here 4 p -orbital
singly filled. has least energy then the others.
47. (b) If value of l is 2 then m  2,  1, 0,  1,  2 . m  l to l 72. (c) According to Aufbau’s principle.
including zero. 73. (c) 1s 2 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 3d 6  2, 8, 14, 2 .
(5 values of magnetic quantum number)
74. (c) Ground state of Cu 29  1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 10 4 s1
48. (c) s, p, d orbitals present in Fe
Cu 2   1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9 .
Fe26  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 , 4 s 2 3d 6
50. (b) According to Aufbau rule. 76. (c) No. of electrons in 3 rd shell = 2n 2 = 2 (3) 2  18
51. (c) 3d subshell filled with 5 electrons (half-filled) is more stable 77. (c) F9  1s 2 2 s 2 2 p 5
than that filled with 4 electrons. 1, 4 s electrons jumps into
3 d subshell for more sability. 78. (c) When l  3 then
m  3,  2,  1, 0,  1,  2,  3 . m  l to l including
52. (b) K19  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 , 4 s 1
zero.
for 4 s1 electrons. 80. (d) m  1 is not possible for s orbital (l  0).
1 84. (a) Both 2p and 3p-orbitals have dumb-bell shape.
n  4, l  0, m  0 and s   .
2 85. (b) 3d subshell filled with 5 electrons (half-filled) is more stable
54. (b) 3d subshell filled with 5 electrons (half-filled) is more stable than that filled with 4 electrons. 1, 4 s electrons jumps into
than that filled with 4 electrons. 1, 4 s electrons jumps into 3 d subshell for more sability.
3 d subshell for more sability. 86. (c) The shape of 2 p orbital is dumb-bell.
55. (c) It has 3 orbitals p x , p y , p z .
87. (a) 25 Mn  [ Ar] 3d 5 4 s 2 -2 electrons Mn 2   [ Ar] 3d 5 4 s 0
57. (b) If l  2 then it must be d orbital which can have 10
electrons. 89. (c) For p-orbital, l  1 means dumb-bell shape.
59. (c) for d orbital l  2 . 91. (d) l  3 means f subshell maximum number of e in f subshell = –

14.
60. (c) m  l to l including zero.
93. (b) As per Aufbau principle.
61. (d) When n  3 shell, the orbitals are n 2  3 2  9 .
94. (b) l  0 is s, l = 1is p and l  2 is d and so on hence s p d may
No. of electrons  2n 2 be used in state of no..
2n 2 1
Hence no. of orbital   n2 . 95. (d) For 4 d, n  4, l  2, m  2,1,0,1,2, s   .
2 2
62. (d) Configuration of Ne = 1s 2 2 s 2 2 p 6 96. (d) m cannot be greater than l ( 0, 1).

(a) For n  1, l  0.
2 2 6
F = 1s 2 s 2 p 97.

Na  = 1s 2 2 s 2 2 p 6 99. (d) Na 11  1s 2 2 s 2 p 6 3 s 2
Mg   = 1s 2 2 s 2 2 p 6 1
n  3, l  0, m  0 and s   .
 2 2 6 2 6 2
Cl = 1s 2 s 2 p 3 s 3 p .
102. (d) According to Aufbau’s rule.
63. (d) Unh106  [Rn] 5 f 14 , 6 d 5 , 7 s 1
105. (d) 2 p x , 2 p y , 2 p z sets of orbital is degenerate.
 
64. (c) K and Ca have the same electronic configuration
106. (a) Mg12 have 1s 2 2 s 2 2 p 6 3 s 2 electronic configuration
(1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 )
1
65. (b) For s-orbital, l  0. n  3, l  0, m  0, s   .
2
66. (d) 3s 1 is valency electrons of Na for this
1
n  3, l  0, m  0, s 
2
 Structure of atom 81
107. (c) The principle quantum number n  3 . Then azimuthal 146. (d) m  (2l  1) for d orbital l  2 m  (2  2  1)  5 .
quantum number l  3 and number of orbitals 147. (a) The atomic number of chlorine is 17 its configuration is
 n 2  3 2  9 . 3 and 9 1s 2 2 s 2 2 p 6 3 s 2 3 p 5
108. (d) 29 Cu  [ Ar] 3d 10 4 s 1 , Cu 2  [ Ar] 3d 9 .4 s 0 . 148. (c) n l m1 m2
Ground state of Cu 29
 1s 2 s 2 p 3 s 3 p 3 d 4 s
2 2 6 2 6 10 1
3 2 1 0
2
Cu  1s , 2 s 2 p ,3 s 3 p 3 d .
2 2 6 2 6 9 This set (c) is not possible because spin quantum number
1
110. (d) 1s 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 6
2
it shows electronic values   .
2
configuration of Iron.
149. (b) The ground state of neon is 1s 2 2 s 2 2 p 6 on excitation an
111. (d) Orbitals are 4 s, 3 s, 3 p and 3 d . Out of these 3d has highest
energy. electron from 2 p jumps to 3 s orbital. The excited neon

113. (c) For the n  2 energy level orbitals of all kinds are possible configuration is 1s 2 2 s 2 2 p 5 3 s 1 .
2n , 2 2  4 . 152. (a) s p d f g h
l=0 1 2 3 4 5
114. (b) n  2 than no. of orbitals  n 2 , 2 2  4 Number of orbitals  5  2  1  11
118. (b) For both A & B electrons s  1 / 2 & 1 / 2 respectively, 153. (a) It is the ground state configuration of chromium.
n  3, l  0, m  0 155. (b) n  4  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 3d 10 , 4 s 2 , 4 p 6 , 4 d 10 , 4 f 14
119. (a) According to Aufbau’s rule. So l  (n  1)  4  1  3 which is f orbit contain 7 orbital.
120. (a) Possible number of subshells would be (6s, 5p, 4d). 156. (d) 2p have contain maximum 6 electron out of which there are 3
121. (d) For f orbital l  3 . are of + 1/2 spin and 3 are of – 1/2 spin
123. (b) 4d-orbital have highest energy in given data.  
   +1/2 –1/2
125. (d) If m  3, l  3 and n  4.
157. (a) For 4f orbital electron, n  4
127. (b) N 714  1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1z . l  3 (Because 0, 1, 2, 3)
128. (c) m can't be greater than l. s, p, d, f
130. (b) n  1 and m  1 not possible for s-orbitals. m = + 3, + 2, +1, 0, – 1, – 2, – 3
s = +1/2
131. (a) Fe 26  [ Ar] 3d 6 4 s 2
158. (b) Cr  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 3 d 5 , 4 s1
Fe 3   [ Ar] 3d 5 4 s 0 .
24
l 1 l 1 l  2
132. (c) Maximum number of electron (We know that for p the value of l  1 and for d , l  2)
 2n 2 (where n  4 )  2  4 2  32 .
For l  1 total number of electron = 12
133. (d) When 2p orbital is completely filled then electron enter in the
For l  2 total number of electron = 5.
3s. The capacity of 2p orbital containing e  is 6. So
159. (c) Atomic number of potassium is 19 and hence electronic
1s 2 , 2 s 2 2 p 2 3 s 1 is a wrong electronic configuration the
configuration will be 1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 4 s1
write is 1s 2 2 s 2 2 p 3 .
Hence for 4 s1 electron value of Quantum number are
134. (b) This electronic configuration is Cr (chromium element) in the
ground state Principal quantum number n  4
 1s 2 s 2 p 3 s 3 p 3 d 4 s
2 2 6 2 6 5 1 Azimuthal quantum number l  0
Magnetic quantum number m  0
137. (c) No. of electron are same (18) in Cl  and Ar .
Spin quantum number s  1 / 2
138. (c) For s-subshell l  0 then should be m  0.
160. (d) According to Hund's rule electron first fill in unpaired form in
139. (c) 19 electron of chromium is 4 s1
th
vacant orbital then fill in paired form to stabilized the molecule
n  4 , l  0, m  0, s  
1 by which 1s 2 ,2 s 2 ,2 p x2 is not possible. According to Hund's
2 rule. Because 2 p x , py , p z have the same energy level so
140. (c) The value of m is – l to l including zero so for l = 3, m would electron first fill in unpaired form not in paired form so it
be –3, –2, –1, 0, +1, +2, +3.
should be 1s 2 , 2 s 2 , 2 p1x ,2 p1y .
141. (c) l  1 is for p orbital.
161. (c) It is governed by Aufbau principle.
142. (d) Magnetic quantum number of sodium ( 3 s1 ) final electron is m 162. (d) The electronic configuration of atomic number
= 0.
24  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 ,3 p 6 ,3d 5 , 4 s1
143. (c) Generally azimuthal quantum number defines angular
momentum. 163. (b) The maximum number of electron in any orbital is 2.
 82 Structure of atom
164. (c) According to pauli principle 2 electron does not have the same
value of all four quantum number. They have maximum same
value are 3.
192. (a) Shell = K, L, M = 1s 2 2 s 2 2 p 6 3 s 2 3 p 4
165. (a) Number of orbitals  n 2  4 2  16 .
Hence the number of s electron is 6 in that element.
166. (d) We know from the Aufbau principle, that 2p orbital will be
filled before 3s orbital. Therefore, the electronic configuration 193. (d) C 6  1s 2 , 2 s 2 2 p 2 (Ground state)
1s 2 , 2 s 2 , 2 p 2 , 3 s1 is not possible.
 1s 2 2 s1 2 Px1 2 p1y 2 p1z (Excited state)
167. (d) Each orbital may have two electrons with opposite spin.
In excited state no. of unpaired electron is 4.
168. (d) Maximum no. of electrons in a subshell  2(2l  1) for f-
194. (b) Max. no. of electrons in N-shell n  4 
subshell, l = 3 so 14 electrons accommodated in f -subshell.
169. (b) Each orbital has atleast two electron.  2n2  2  4 2  32 .
170. (a) Nucleus of 20 protons atom having 20 electrons. 195. (d) 26 Fe  [ Ar] 3d 6 , 4 s 2
174. (b) For m  0 , electron must be in s-orbital. Fe2   [ Ar] 3d 6 , 4 s0
176. (c) In this type of electronic configuration the number of unpaired
electrons are 3. Number of d-electrons = 6
=3 17 Cl  [ Ne ] 3 s 2 , 3 p 5
1s 2s 2p Cl   [ Ne ] 3 s 2 , 3 p 6
177. (a) Atomic number of Cu is 29 so number of unpaired electrons is
1 Number of p-electrons = 6.
4 s1 3d 10 196. (a) Electrons in the atom  18  4  3  25 i.e. Z  25 .
Cu  (Ar) 197. (c) The atomic number of bromine is 35 and the electronic
configuration of Br is
1s 2 2s 2 2p4 Br35  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 3d 10 , 4 s 2 , 4 p 5
178. (b) O8 
total electron present in p-orbitals of Br is –
Unpaired electron 2 p 6  3 p 6  4 p 5  17.
181. (c) Be 4  1s , 2 s = (Ground state)
2 2

198. (d) Fe 2  has 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 6 configuration with 4

Number of unpaired electrons in the ground state of Beryllium unpaired electron.
atom is zero.
182. (b) Two unpaired electrons are present in 199. (b) Fe 2  [ Ar] 3d 6 4 s 0

Ni   (z  28) cation =4

Ni28  [ Ar]
Fe 2  consist of maximum 4 unpaired electrons.
3d 8 4s 0

201. (a) Fe 3  (z  26)

183. (c) O 2  1s 2 2 s 2 2 p 6 3 s 2 3 p 4
Fe 3   [ Ar] 3d 5 4 s 0
3s 2 3p4
=5
3d 4s
2 Unpaired electrons
3
184. (c) Cr24  ( Ar) 3d 5
4 s 1 but Cr24  ( Ar) 3d 4 s 3 0
Total no. of unpaired electron=5

185. (a) Zn 30  [ Ar] 3d 10 4 s 2 202. (b) Co 27  [ Ar] 3d 7 4 s 2

Zn    [ Ar] 3d 10 4 s 0 3d 7

186. (d) Mn 2 ion will have five (maximum) unpaired electrons

3 unpaired electron are present in cobalt metal.
[Ar] 203. (b) According to Hund's rule, the pairing of electrons will not
occur in any orbital of a subshell unit and unless, all the
3d 5 4s available of it have one electron each.
3 Electronic configuration of
187. (c) Fe ion will have five (maximum) unpaired electrons.
N 14  1 s 2 , 2 s 2 , 2 p x 2 p y 2 p z
1 1 1
190. (c) Due to full filled d-orbital Cl  has spherical symmetry. 7

191. (b) Atomic number 14 leaving 2 unpaired electron Hence it has 3 unpaired electron in 2p-orbital.

14 Si  1s 2 s 2 p 3 s 3 p
2 2 6 2 2 204. (c) 2 s orbital have minimum energy and generally electron filling
increases order of energy according to the Aufbau’s principle.

1s 2s 2p 3s 3p
 Structure of atom 83
205. (d) According to Pauli’s exclusion principle no two electrons in the 1
same atom can have all the set of four quantum numbers (iii) Proton  1
identical. 1
206. (b) The second principal shell contains four orbitals viz 1
(iv) electron   1837 .
2 s, 2 p x , 2 p y and 2 p z . 1 / 1837
207. (b) Follow Hund’s multiplicity rules. 4. (a) Metal is 56 M 2  (2, 8, 14) than n  A  Z
208. (c) According to the Aufbau’s principle, electron will be first enters
in those orbital which have least energy. So decreasing order of  56  26  30 .
energy is 5 p  4 d  5 s. 5. (d) E  hv  h
c
i.e. E 
1
210. (b) No two electrons in an atom can have identical set of all the  
four quantum numbers. E1  4000
 2   2.
212. (a) In particular shell, the energy of atomic orbital increases with E2 1 2000
the value of l .
6. (c) Rutherford discovered nucleus.
214. (c) Aufbau principle explains the sequence of filling of orbitals in
increasing order of energy. 7. (b) According to Bohr’s model E  E1  E3
215. (a) According to Aufbau principle electron are filling increasing 2 .179  1011
order of energy. Therefore the electronic configuration  2 .179  10 11 
9
1s 2 2 s 2 2 p 6 obeys Aufbau principle.
8
  2 .179  10 11  1.91  10 11  0.191  10 10 erg
216. (d) Electronic configuration of the Cr24 is [ Ar] 4 s1 3d 5 or 9
=6 Since electron is going from n  1 to n  3 hence energy is
absorbed.
3d 4s
8. (d) Radius of nucleus  1.25  10 13  A1 / 3 cm
217. (b) According to the Aufbau principle electron filling minimum to
higher energy level.
 1.25  10 13  64 1 / 3  5  10 13 cm
219. (b) According to Aufbau principle electron are filled in various
atomic orbital in the increasing order of energy Radius of atom = 1Å  10 8 cm.
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f <
5 d < 6p < 7 s . Volume of nucleus (4 / 3) (5  10 13 )3

220. (d) According to Aufbau’s rule. Volume of atom (4 / 3) (10  8 )3
222. (b) We know that for d-electron l  2.  1.25  10 13 .
h h
  l(l  1) ;   2 (2  1) 13.6
2 2 9. (a) Values of energy in the excited state  eV
n2
h h 13.6
  2 (2  1) ;  6 .   3 .4 eV in which n  2, 3, 4 etc.
2 2 4
223. (a) Number of nodal centre for 2 s orbitals ( n  1)  2  1  1 .
10. (c) E1 He   E1 H  z 2
224. (d) Since s-orbital have l  0
 871.6  10 20  E1 H  4
h h
Angular momentum = l (l  1)  = 0 0 .
2 2 E1 H  217.9  10 20 J
225. (d) Azimuthal quantum number (l) = 3 shows the presence of f
orbit, which contain seven orbitals and each orbital have 2 11. (a) 42g of N 3 ions have 16 N A valence electrons 4.2g of N 3
electrons. Hence 7  2  14 electrons. 16 N A
ion have   4 .2  1 .6 N A .
227. (b) According to Aufbau principle. 42
228. (a) Atomic number of deuterium = 1; 1 D2  1s1 12. (d) Ist excited state means n  2

Critical Thinking Questions r  r0  2 2  0.53  4  2.12 Å

13. (d) Frequency   12  1014 s 1 and velocity of light

1. (a) F  have the same number of electrons with the neon atom.
c  3  10 cm s 1 . We know that the wave number
10
2. (d) No change by doubling mass of electrons however by reducing
mass of neutron to half total atomic mass becomes 6  3 v 12  1014
   4  10 4 cm 1
instead of 6  6 . Thus reduced by 25%. c 3  1010
e 0 14. (c) The last line in any series is called series limit. Series limit for
3. (d) for (i) neutron  0 Balmer series is 3646 Å.
m 1
13.6 13.6
(ii)   particle 
2
 0.5 15. (b) E   3.4 eV
4 n2 4
 84 Structure of atom
We know that energy required for excitation E  E 2  E1 6. (d) We know from the Pauli exclusion principle, that two electrons
in the same atom can not have same value of all four quantum
 3.4  (13.6)  10.2 eV numbers. This means each electron in an atom has only one set
Therefore energy required for excitation of electron per atom of values for n, l, m and s . Therefore both the Assertion and
10.2 Reason are false.
  1.69  10  23 J 7. (e) We know that the line in Balmer series of hydrogen spectrum the
6.02  10 23
highest wavelenght or lowest energy is between n1  2 and
17. (a) The number of nodal plane are present in a p x is one or no.
n 2  3 . And for Balmer series of hydrogen spectrum, the value of
of nodal place = l
n1  2 and n 2  3,4,5 . Therefore the Assertion is false but the
for p x orbital l = 1
Reason is true.
8. (d) We know that Absorption spectrum is produced when white
x light is passed through a substance and transmitted light is
analysed by a spectrograph. The dark spaces corresponds to the
Nodal plane light radiation absorbed by the substance. And emission
18. (b) In Balmer series of hydrogen atomic spectrum which electronic spectrum is produced by analysing the radiant energy emitted
by an excited substance by a spectrograph. Thus discontinuous
transition causes third line O  L , n2  5  n1  2
spectra consisting of a series of sharp lines and separated by
1 1 1  dark bands are obtained. Therefore both the Assertion and
20. (b)    RH  2  2  Reason are false.
  n1 n2  9. (a) We know that a resonance hybrid or the actual molecule is
always more stable than any of its canonical structures which is
1 1 1 
  R H  2  2   n2  3 for Paschen series. also called hypothetical or imaginary structures. This stability is
  3 n2  due to delocalization of electrons and is measured in terms of
resonance energy or delocalization energy, it is defined as the
1 1  difference in internal energy of the resonance hybrid and the
21. (a) E 2  2 most stable canonical structure. Therefore both the Assertion
 n2 n1  and Reason are true and the Reason is a correct explantion of
c 3  10 8 the Assertion.
23. (d)     3 .75  10 8 10. (e) We know that cathode rays cast shadows of solid objects placed
v 8  1015
in their path. During experiment performed on these rays,
 3.75  10 8  10 9 nm  4  101 nm . fluorescene (flash of light) is observed in the region, outside
the shadow. This shows that cathode rays travel in straight
Assertion & Reason lines. We also known that cathode rays penetrate through a
thin sheet of metals but are stopped by thick sheets. Therefore
both Assertion and Reason are false.
1. (d) The assertion is false but the reason is true exact position and
exact momentum of an electron can never be determined as 11. (b) We know that electrons are revolving around the nucleus at
according to Hesenberg’s uncertainity principle even with the high speed in circular paths. The centrifugal force (which arises
due to rotation of electrons) acting outwards, balances the
help of electron microscope because when e  beam of electrostatic force of attraction (which arises due to attraction
electron microscope strikes the target e  of atom, the impact between electrons and nucleus). This prevent the electron from
falling into the nucleus. We also know that Rutherford’s model
causes the change in velocity of e  thus attempt to locate the of atom is comparable to the “solar system”. The nucleeus
e  changes ultimately, the momentum & position of e  . represent the sun whereas revolving electrons represent the
h planets revolving around the sun. Thus revolving electron are
x .p   0.57 ergs sec/ gm. also called planetary electrons. Therefore both Assertion and
4 Reason are true but Reason is not a correct explanation of
2. (d) Both assertion and reason are false. 2 p x and 2 p y orbitals Assertion.
are degenerate orbitals, i.e., they are of equal energy and hence 12. (c) Assertion is true but Reason is false. Threshold frequency is a
no possibility of transition of electron. minimum frequency required for the emission of electrons
from the metal surface.
3. (a) We know that principal quantum number represent the main
energy level or energy shell. Since each energy level is 13. (a) Both assertion and reason are true and reason is the correct
associated with a definite amount of energy, this quantum explanation of assertion.
number determines to a large extent te energy of an electron. It n2h2 n2
also determines the average distance of an electron around the Radius, r   0 .529 Å. rn also increases
nucleus. Therefore both Assertion and Reason are true and the 4e mZ
2
Z
Reason is a correct explanation of the Assertion. indicating a greater separation between the orbit and the
nucleus.
4. (a) It is observed that a nucleus which is made up of even number
of nucleons (No. of n & p ) is more stable than nuclie which 14. (d) Both assertion and Reason are false. Only s -orbital is
consist of odd number of nucleons. If number of neutron or spherically symmertrical. Shape of different d orbitals is as
proton is equal to some numbers i.e., 2,8, 20, 50, 82 or 126 below.
(which are called magic numbers), then these passes extra 15. (c) Assertion is true but reason is false. Spin angular momentum
stability. of the electron, a vector quantity, can have two orientations
5. (c) The assertion that the isobars are the atoms of different (represented by + and – sign) relative to a chosen axis. These
elements having same mass number but different atomic two orientation are distinguished by the spin quantum number
number, is correct but reason is false because atomic mass is 1 1
m s equals to  or  . These are called the two spin
sum of number of neutron and protons which should be same 2 2
for isobars.
 Structure of atom 85
states of the electron and are normaly represented by the two
arrows  (spin up) and  (spin down) respectively.
16. (d) Both assertion and reason are false. Total number of orbitals
associated with Principal quantum number n  3 is 9. One
3 s orbital + three 3 p orbital + five 3 d orbitals. 
Therefore there are a total number of nine orbitals. Number of
orbitals in a shell equals to n 2 .
17. (c) Assertion is true but reason is false. The order
1s  2s  2 p  3 s  3 p  3d  .... is true for the energy of
an electron in a hydrogen atom and is solely determined by
Principal quantum number. For multielectron system energy
also depends on azimuthal quantum number. The stability of
an electron in a multi electron atom is the net result of the
attraction between the electron and the uncleus and the
repulsion between the electron and the rest of the electron
present. Energies of different subshell (azimuthal quantum
number) present within the same principal shell are found to
be in order of s  p  d  f .
18. (e) Assertion is false but reason is true. Splitting of the spectral
lines in the presence of a magnetic field is known as Zeeman
effect or in electric field it is known as stark effect. The
splitting of spectral lines is due to different orientations which
the orbitals can have in the presence of magnetic field.
19. (a) Both assertion and reason are true and reason is the correct
explanation of assertion.
20. (e) Assertion is false but reason is true. Atomic orbital is
designated by n, l and m l while state of an electron in an
atom is specified by four quantum numbrs n, l, m l and m s .
21. (b) Both assertion and reason are true but reason is not the
correct explanation of assertion. The difference between the
energies of adjacent energy levels decreases as we move away
from the nucleus. Thus in H atom
E2  E1  E3  E2  E4  E3 ......
22. (d) Both assertion and reason are false. Cathode rays are stream of
electrons. They are generated through gases at low pressure
and high voltage.
23. (d) Both assertion and reason are false. In case of isoelectronic, i.e.,
ions, having the same number of electrons and different
nuclear charge, the size decreases with increase in atomic
number.
Ion At. No. No. of electrons Ionic radii
Na+ 11 10 0.95Å
Mg2+ 12 10 0.65Å
Al3+ 13 10 0.50Å
 86 Structure of atom

1. The correct set of quantum numbers for the unpaired electron of (c) n  2; l  1; m  0l s  1 / 2
chlorine atom is [IIT 1989; MP PET 2004]
n l m (d) n  4 ; l  3; m  2; s  1 / 2

(a) 2 1 0 8. The uncertainty in the position of an electron ( mass =

(b) 2 1 1 9.1  10 28 g) moving with a velocity of 3.0  10 4 cm s 1
(c) 3 1 1 accurate upto 0.001% will be
(d) 3 0 0 h
(Use in the uncertainty expression, where
2. The orbital diagram in which the Aufbau's principle is violated is 4 [IIT 1988; AMU 1999]

2s 2px 2 py 2 pz h  6.626  10 27 erg  s ) [CBSE PMT 1995]

(a) 1.92cm (b) 7.68 cm

(a)   
(c) 5.76cm (d) 3.84 cm
(b)    
9. The orbital angular momentum of an electron in s orbital is
(c)     [IIT 1996; AIEEE 2003; MP PET 2004]
(d)     1 h
(a)  . (b) Zero
3. The mass of neutron is nearly 2 2
[MNR 1988; UPSEAT 1999, 2000, 02] h h
(c) (d) 2.
(a) 10 23
kg (b) 10 24 kg 2 2
10. Values of the four quantum numbers for the last electron in the
(c) 10 26 kg (d) 10 27 kg
atom are n  4, l  1, m  1 and s  1 / 2 . Atomic number of
4. Which electronic level would allow the hydrogen atom to absorb a the atom will be
photon but not to emit a photon
(a) 22 (b) 32
[IIT 1984; CPMT 1997]
(c) 33 (d) 36
(a) 3s (b) 2p
11. The atomic weight of an element is 39. The number of neutrons in
(c) 2s (d) 1s 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
5. Which of the following is not correct for electron distribution in the
ground state [AIIMS 1982] (a) 19, 20, 19 (b) 19, 19, 20

4s 3d (c) 20, 19, 19 (d) 20, 19, 20

12. The electrons identified by quantum numbers n and l (i)
(a) Co (Ar)      
n  4, l  1 (ii) n  4, l  0 (iii) n  3, l  2 (iv) n  3, l  1
(b) Ni(Ar)       can be placed in order of increasing energy from the lowest to
highest, as [IIT 1999]
(c) Cu (Ar)      
(a) (iv) < (ii) < (iii) <(i)
(d) Zn(Ar)       (b) (ii) < (iv) < (i) < (iii)
6. If electron, hydrogen, helium and neon nuclei are all moving with (c) (i) < (iii) < (ii) < (iv)
the velocity of light, then the wavelengths associated with these (d) (iii) < (i) < (iv) < (ii)
particles are in the order [MP PET 1993]
13. Ground state electronic configuration of nitrogen atom can be
(a) Electron > hydrogen > helium > neon represented by [IIT 1999]
(b) Electron > helium > hydrogen > neon
(a)     
(c) Electron < hydrogen < helium < neon
(d) Neon < hydrogen < helium < electron (b)     
7. From the given sets of quantum numbers the one that is inconsistent
with the theory is [IIT Screening 1994]
(c)     
(a) n  3; l  2; m  3; s  1 / 2

(b) n  4 ; l  3; m  3; s  1 / 2
 Structure of atom 87

(d)      (c) 10 20 (d) 6.02  10 20

18. Which of the following have the same number of unpaired electrons
in ‘d’ orbitals [Roorkee 2000]
14. Which of the following statements (s) is (are) correct
(a) Cr (b) Mn
[IIT 1998]
(c) Fe 3+
(d) Co 3+

5 1
(a) The electronic configuration of Cr is [ Ar] 3d 4 s (Atomic 19. The quantum numbers + 1/2 and – 1/2 for the electron spin
represent [IIT Screening 2001]
no. of Cr  24 )
(a) Rotation of the electron in clockwise and anticlockwise
(b) The magnetic quantum number may have a negative value direction respectively
(c) In silver atom, 23 electrons have a spin of one type and 24 of (b) Rotation of the electron in anticlockwise and clockwise
the opposite type (Atomic no. of Ag  47 ) direction respectively
(c) Magnetic moment of the electron pointing up and down
(d) The oxidation state of nitrogen in HN 3 is 3 respectively
15. The position of both an electron and a helium atom is known within (d) Two quantum mechanical spin states which have no classical
1.0nm and the momentum of the electron is known within analogue
50  10 26 kg ms 1 . The minimum uncertainty in the 20. The de-Broglie wavelength of a tennis ball of mass 60 g moving with
a velocity of 10 metres per second is approximately
measurement of the momentum of the helium atom is
[CBSE PMT 1998; AIIMS 2001] (a) 10 33 metres (b) 10 31 metres

(a) 50kg ms 1 (b) 60 kg ms 1 (c) 10 16 metres (d) 10 25 metres

21. Which of the following are isoelectronic and isostructural
(c) 80  10 26 kg ms 1 (d) 50  10 26 kg ms 1
NO 3 , CO 32  , ClO3 , SO 3 [IIT Screening 2003]
16. Which of the following pair of orbitals posses two nodal planes [RPMT 2000]
(a)  2
NO 3 , CO 3 (b) SO 3 , NO 3
(a) p xy , d x 2 y 2 (b) d xy , d zx
(c) ClO3 , CO 32  (d) CO 32  , SO 3
(c) p xy , d zx (d) d z 2 , d x 2 y 2 22. 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
17. The number of atoms in 0.004 g of magnesium are
(a) 8, 26, 10 (b) 10, 24, 20
[AFMC 2000]
(c) 8, 22, 24 (d) 12, 20, 22
(a) 4  10 20
(b) 8  10 20

(SET -2)

1. (c) Electronic configuration of Cl is

3s 2 3p So for the unpaired electron (3 p1z ) :

1
[ Ne ] 3 s 2 3 p 5 or [Ne ] n  3, l  1, m  1, S  
2
3 p 2 x 3p 2 y 3p 1 z
 88 Structure of atom
2. (b) According to Aufbau principle the orbitals of lower energy (2s) 1
should be fully filled before the filling of orbital of higher 14. (a,b,c) The oxidation state of nitrogen in HN 3 is  .
energy starts. 3

3. (d) Mass of neutron  1.675  10 27 kg . 1

HN 3 : 1  3 x  0  3 x  1 or x 
4. (d) 1s-orbital is of lowest energy. Absorption of photon can raise 3
the electron in higher energy state but emission is not possible.
15. (d) The product of uncertainties in the position and the
5. (c) The correct electronic configuration
momentum of a sub atomic particle  h/4 . Since x is
Cu 29  [ Ar], 4 s1 3d 10
same for electron and helium so p must be same for both
the particle i.e. 50  10 26 kg ms 1 (given).
1
6. (a)   , m e  m H  m He  m Ne . 16. (b) d xy and d zx has two modal planes.
m
7. (a) When l  2, m  3 . 17. (c) No. of atoms in magnesium =
0 .004
 6 .023  10 23 =10 20

24
8. (a) p  m  v
18. (a,b,c) Cr , Mn and Fe3  have 5 unpaired electron in d-
0 .001 orbitals.
p  9.1  10  28  3.0  10 4 
100 24 Cr  3d 5 4 s1  5

P  2.73  10 24 25 Mn  3d 5 4 s 2  5

h 6 .626  10 27 26 Fe3   3d 5 4 s 0  5

Hence x  
p  4 2 .73  10  28  4  3 .14 19. (a,d) Both statement are correct.
h 6 .63  10 34
x  1.92 cm. 20. (a)    10  33 m
mv 60  10  3  10
9. (b) For 2s orbital, l = 0; azimuthal quantum number is not show
21. (a) NO 3 and CO 32  consist of same electron and show same
angular momentum for the 2 s orbitals.
isostructural.
h 22. (b) (Cs 35 )  1s 2 , 2 s 2 , 2 p 6 ,3 s 2 ,3 p 6 ,3d 10 ,4 s 2
Angular momentum  l(l  1) 0.
2
4 p 6 ,4 d 10 ,5 s 2 ,5 p 6 ,6 s1
10. (d) Atomic number is 36 and element is Kr .
Cs   1s 2 , 2 s 2 ,2 p 6 , 3 s 2 ,3 p 6 ,3d 10 , 4 s 2 ,
11. (a) 39
K 19 , P  19 , E  19 , N  20
4 p 6 ,4 d 10 , 5 s 2 ,5 p 6
12. (a) (i) 4 p (ii) 4 s (iii) 3d (iv) 3 p order of increasing energy is
Total no. of e  in s-orbitals  10
3 p  4 s  3d  4 p.
Total no. of e  in p-ortbitals  24
13. (a,d) According to Hund’s principle.
Total no. of e  in d-ortbitals  20 .

***