Chapter 1 Structure of Atom by TEACHING CARE online tuition and coaching classes
John Dalton 1808, believed that matter is made up of extremely minute indivisible particles, called atom which can takes part in chemical reactions. These can neither be created nor be destroyed. However, modern researches have conclusively proved that atom is no longer an indivisible particle. Modern structure of atom is based on Rutherford’s scattering experiment on atoms and on the concepts of quantization of energy.
The works of J.J. Thomson and Ernst Rutherford actually laid the foundation of the modern picture of the atom. It is now believed that the atom consists of several subatomic particles like electron, proton, neutron, positron, neutrino, meson etc. Out of these particles, the electron, proton and the neutron are called fundamental subatomic particles and others are nonfundamental particles.
Electron
 It was discovered by J. Thomson (1897) and is negatively charged particle. Electron is a component particle of cathode rays.
 Cathode rays were discovered by William Crooke’s & J. Thomson (1880) using a cylindrical hard glass tube fitted with two metallic electrodes. The tube has a side tube with a stop cock. This tube was known as
discharge tube. They passed electricity (10,000V) through a discharge tube at very low pressure ( 10 ^{2} to
10 ^{3} mm Hg). Blue rays were emerged from the cathode. These rays were termed as Cathode rays.
(3) Properties of Cathode rays
 Cathode rays travel in straight
 Cathode rays produce mechanical effect, as they can rotate the wheel placed in their
 Cathode rays consist of negatively charged particles known as electron.
 Cathode rays travel with high speed approaching that of light (ranging between 10 9
 Cathode rays can cause
to 10 11 cm/sec)
 Cathode rays heat the object on which they fall due to transfer of kinetic energy to the
 When cathode rays fall on solids such as Cu, X – rays are
 Cathode rays possess ionizing power e., they ionize the gas through which they pass.
 The cathode rays produce scintillation the photographic plates.
 They can penetrate through thin metallic
 The nature of these rays does not depend upon the nature of gas or the cathode material used in discharge
 The e/m (charge to mass ratio) for cathode rays was found to be the same as that for an e ^{–}
(1.76 ´ 108 coloumb per gm). Thus, the cathode rays are a stream of electrons.
Note : ® When the gas pressure in the discharge tube is 1 atmosphere no electric current flows through the tube. This is because the gases are poor conductor of electricity.
 The television picture tube is a cathode ray tube in which a picture is produced due to fluorescence on the television screen coated with suitable material. Similarly, fluorescent light tubes are also cathode rays tubes coated inside with suitable materials which produce visible light on being hit with cathode
 S. Mullikan measured the charge on an electron by oil drop experiment. The charge on each electron is – 1.602 ´ 10 ^{19} C.
 Name of electron was suggested by S. Stoney. The specific charge (e/m) on electron was first determined by J.J. Thomson.
 Rest mass of electron is 1 ´ 1028 gm = 0.000549amu = 1 / 1837 of the mass of hydrogen atom.
 According to Einstein’s theory of relativity, mass of electron in motion is, m¢=
Where u = velocity of electron, c= velocity of light. When u=c than mass of moving electron =¥.
 Molar mass of electron = Mass of electron × Avogadro number = 483 ´ 10 ^{4}.
 1 ´ 10^{27} electrons =1gram.
 1 mole electron = 5483 mili gram.
 Energy of free electron is≈ The minus sign on the electron in an orbit, represents attraction between the positively charged nucleus and negatively charged electron.
 Electron is universal component of matter and takes part in chemical
 The physical and chemical properties of an element depend upon the distribution of electrons in outer
 The radius of electron is 28 ´ 10 ^{12} cm.
 The density of the electron is = 17 ´ 10 17 g / mL .
Example : 1 The momentum of electron moving with 1/3^{rd} velocity of light is (in g cm sec^{–1})
(a)
9.69 ´ 10^{–}^{8}
(b)
8.01´ 10^{10}
(c)
9.652 ´ 10 ^{–}^{18}
(d) None
Solution: (c) Momentum of electron, ‘p’ = m¢ ´ u
Where m¢ is mass of electron in motion = m ; Also u = c / 3
\ Momentum = 9.108 ´ 10 ^{–}^{28} ´ 3 ´ 10^{10}
3
= 9.108 ´ 10 ^{–}^{28} ´ 3 ´ 10^{10}
0.94 ´ 3
= 9.652 ´ 10 ^{–}^{18} g cm sec ^{–}^{1}
Example: 2 An electron has a total energy of 2 MeV. Calculate the effective mass of the electron in kg and its speed.
Assume rest mass of electron 0.511 MeV.
(a)
2.9 ´ 108
(b)
8.01´ 108
(c)
9.652 ´ 108
(d) None
Solution: (a) Mass of electron in motion = 2 amu
931
= 2 ´ 1.66 ´ 10^{–}^{27} kg = 3.56 ´ 10 ^{–}^{30} kg
931
Let the speed of the electron be u.
(1 amu = 931 MeV)
(1 amu = 1.66 ´ 10^{–}^{27} kg )

m¢ = m
0.511 ´ 1.66 ´ 10^{–}^{27}
or 3.56 ´ 10^{–}^{30} = 931 =
0.911´ 10^{–}^{30}
1 – (u / c)2
æ u ö^{2}
1 – ç 3 ´ 108 ÷
æ u ö^{2}
è ø
2 16 8
or 1 – ç 3 ´ 108 ÷
= 0.06548
or u
= 9 ´ 10 ´ 0.93452 or u = 2.9 ´ 10 m
è ø
Example: 3 A electron of rest mass 1.67 ´ 10^{–}^{27}
and momentum.
kg is moving with a velocity of 0.9c (c = velocity of light). Find its mass
(a)
10.34 ´ 1019
(b)
8.01´ 10^{10}
(c)
9.652 ´ 10 ^{–}^{18}
(d) None
Solution: (a) Mass of a moving object can be calculated using Einsten’s theory of relativity :
m¢ = m = rest mass (given), u = velocity (given), c = velocity of light
m¢ = = 3.83 ´ 10 ^{–}^{27} kg
Momentum ‘ p‘ = m¢ ´ u
p = 3.83 ´ 10^{–}^{27} ´ 0.9c = 10.34 ´ 10^{–}^{19} kg ms^{–}^{1}
 Proton was discovered by Goldstein and is positively charged It is a component particle of anode rays.
 Goldstein (1886) used perforated cathode in the discharge tube and repeated Thomson’s experiment and observed the formation of anode These rays also termed as positive or canal rays.
(3) Properties of anode rays
 Anode rays travel in straight
 Anode rays are material particles.
 Anode rays are positively
 Anode rays may get deflected by external magnetic
 Anode rays also affect the photographic
 The e/m ratio of these rays is smaller than that of
 Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the It is maximum when gas present in the tube is hydrogen.
 These rays produce flashes of light on ZnS
(4) Charge on proton = 1.602 ´ 10 ^{19} coulombs = 4.80 ´ 10 ^{10} e.s.u.
 Massof proton = Mass of hydrogen atom= 00728amu = 1.673 ´ 10 ^{24} gram = 1837 of the mass of electron.
 Molar mass of proton = mass of proton ´ Avogadro number = 008 (approx).
 Proton is ionized hydrogen atom (H ^{+} ) e., hydrogen atom minus electron is proton.
 Proton is present in the nucleus of the atom and it’s number is equal to the number of
 Mass of 1 mole of protons is » 007 gram.
 Charge on 1 mole of protons is » 96500 coulombs.
 The volume of a proton (volume =
4 pr ^{3} ) is » 1.5´ 10 ^{38} cm^{3} .
3
 Specific charge of a proton is 58 ´ 10^{4} Coulomb/gram.
 Neutron was discovered by James Chadwick (1932) according to the following nuclear reaction,
 The reason for the late discovery of neutron was its neutral
 Neutron is slightly heavier (0.18%) than
(4) Mass of neutron = 1.675 ´ 10 ^{24} gram = 1.675 ´ 10 ^{27} kg = 1.00899 amu »
 Specific charge of a neutron is
 Density = 5 ´ 10 ^{14} gram / c.c.
 1 mole of neutrons is » 008 gram.
 Neutron is heaviest among all the fundamental particles present in an
 Neutron is an unstable It decays as follows :
mass of hydrogen atom.
0 n1
¾¾® 1 H^{1} + 1 e^{0} +
0n 0
neutron
proton electron
anti nutrino
 Neutron is fundamental particle of all the atomic nucleus, except hydrogen or
Comparison of mass, charge and specific charge of electron, proton and neutron
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^{–24} 
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 
 The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C^{12} , e. 1.660 ´ 10 ^{–}^{27} kg .
Other non fundamental particles
Particle  Symbol  Nature  Charge esu
´10–10 
Mass
(amu) 
Discovered by  
Positron  e ^{+} ,1e ^{0} , b ^{+}  + 0
–
+
– 
+ 4.8029  0.000548
6 
Anderson (1932)  
Neutrino  n  0  <  Pauli (1933) and Fermi (1934)  
0.00002  
Antiproton  p ^{–}  – 4.8029  1.00787  Chamberlain Sugri (1956) and
Weighland (1955) 

Positive  mu  m +  + 4.8029  0.1152  Yukawa (1935)  
meson  
Negative  mu  m –  – 4.8029  0.1152  Anderson (1937)  
meson 
(1) Atomic number or Nuclear charge
 The number of protons present in the nucleus of the atom is called atomic number (Z).
 It was determined by Moseley as,
where, n = X – rays frequency Z= atomic number of the metal a & b are constant.
 Atomic number = Number of positive charge on nucleus = Number of protons in nucleus = Number of electrons in nutral
 Two different elements can never have identical atomic number.
(2) Mass number
 The sum of proton and neutrons present in the nucleus is called mass
Mass number (A) = Number of protons + Number of neutrons or Atomic number (Z) or Number of neutrons = A – Z .
 Since mass of a proton or a neutron is not a whole number (on atomic weight scale), weight is not necessarily a whole
 The atom of an element X having mass number (A) and atomic number (Z) may be represented by a symbol,
Note : ® A part of an atom up to penultimate shell is a kernel or atomic core.
 Negative ion is formed by gaining electrons and positive ion by the loss of
 Number of lost or gained electrons in positive or negative ion =Number of protons ± charge on
(3) Different Types of Atomic Species
Atomic species Similarities Differences Examples
Isotopes
(i) Atomic No. (Z)
(i) Mass No. (A)
 ^{1} H, ^{2} H, ^{3} H
(Soddy)
 of protons
(iii) No. of electrons
 of neutrons
(iii) Physical properties
1 1 1
(ii)
 Electronic configuration
(v) Chemical properties
 Position in the periodic table
Isobars (i) Mass No. (A)
(ii) No. of nucleons
(i) Atomic No. (Z)
 of protons, electrons and neutrons
(iii) Electronic configuration
 Chemical properties
(v) Position in the perodic table.
(iii)
(i)
(ii)
Isotones No. of neutrons (i) Atomic No.
(i)
30 Si, 31 P, 32 S
(ii) Mass No., protons and electrons.
14
(ii)
15 16
(iii) Electronic configuration
(iii)
 ^{13} C, ^{14} N
6 7
(iv) Physical and chemical properties
 Position in the periodic
Isodiaphers Isotopic No.
(N – Z) or (A – 2Z)
(i) At No., mass No., electrons, protons,
(i) 92 U 235 , 90 Th231
 K ^{39} , F ^{19}
neutrons.
(ii) Physical and chemical properties.
19 9

(iii)
Isoelectronic species
(i) No. of electrons
 Electronic configuration
At. No., mass No. (i) N 2 O, CO2 , CNO^{–} (22e ^{–} )
 CO, CN ^{–} , N 2 (14e ^{–} )
 H ^{–} , He, Li ^{+} , Be ^{2}^{+} (2e ^{–} )
Isosters (i) No. of atoms
 of electrons
(iii) Same physical and chemical properties.
P^{3}^{–}, S^{2}^{–}, Cl ^{–}, Ar, K^{+}and Ca^{2}^{+}(18e^{–})
 N _{2} and CO
 CO_{2} and N _{2}O
 HCl and F_{2}
 CaO and MgS
 C_{6} H_{6} and
Note : ® In all the elements, tin has maximum number of stable isotopes (ten).
 Average atomic weight/ The average isotopic weight
= % of 1st isotope ´ relative mass of 1st isotope + % of 2nd isotope ´ relative mass of 2nd isotope 100
Example : 4 The characteristics X– ray wavelength for the lines of the ka series in elements X and Y are 9.87Å and 2.29Å
respectively. If Moseley’s equation = 4.9 ´ 10^{7} (Z – 0.75) is followed, the atomic numbers of X and Y are
Solution : (a)
(a) 12, 24 (b) 10, 12 (c) 6, 12 (d) 8, 10
n = c
l
= = 5.5132 ´ 10^{8}
= = 11.4457 ´ 10^{8}
using Moseley’s equation we get
\ 5.5132 ´ 10^{8} = 4.9 ´ 10^{7} (Zx – 0.75)
…..(i)
and 11.4457 ´ 10^{8} = 4.90 ´ 10^{7} (Zy – 0.75)
On solving equation (i) and (ii) Zx = 12, Zy = 24.
….. (ii)
Example : 5 If the straight line is at an angle 45° with intercept, 1 on number Z is 50.
 axis,
calculate frequency n when atomic
(a) 2000 s ^{–}^{1}
(b) 2010 s ^{–}^{1}
(c) 2401 s ^{–}^{1}
(d) None
Solution : (c)
= tan 45° = 1 = a
ab=1
\ = 50 – 1 = 49 Z
n = 2401 s ^{–}^{1}.
Example : 6 What is atomic number Z when n = 2500 s ^{–}^{1} ?
(a) 50 (b) 40 (c) 51 (d) 53
Solution : (c) =
= Z – 1,
Z = 51.
Example : 7 Atomic weight of Ne is 20.2. Ne is a mixutre of
Ne 20
and Ne ^{22} . Relative abundance of heavier isotope is
(a) 90 (b) 20 (c) 40 (d) 10
Solution:(d) Average atomic weight/ The average isotopic weight
= % of 1st isotope ´ relative mass of 1st isotope + % of 2nd
100
isotope ´ relative mass of 2nd
isotope
\ 20.2 = a ´ 20 + (100 – a) ´ 22 ;
100
\ a = 90 ; per cent of heavier isotope = 100 – 90 = 10
Example : 8 The relative abundance of two isotopes of atomic weight 85 and 87 is 75% and 25% respectively. The average atomic weight of element is
(a) 75.5 (b) 85.5 (c) 87.5 (d) 86.0
Solution:(b) Average atomic weight/ The average isotopic weight
= % of 1st isotope ´ relative mass of 1st isotope + % of 2nd
100
isotope ´ relative mass of 2nd
isotope
= 85 ´ 75 + 87 ´ 25 = 85.5
100
Example : 9 Nitrogen atom has an atomic number of 7 and oxygen has an atomic number of 8. The total number of electrons in a nitrate ion is
(a) 30 (b) 35 (c) 32 (d) None
Solution : (c) Number of electrons in an element = Its atomic number
So number of electrons in N=7 and number of electrons in O=8.

Formula of nitrate ion is NO ^{–}
So, in it number of electrons
= 1 ´ number of electrons of nitrogen +3 ´ number of electrons of oxygen +1 (due to negative charge)
= 1 ´ 7 + 3 ´ 8 + 1 = 32
Example :10 An atom of an element contains 11 electrons. Its nucleus has 13 neutrons. Find out the atomic number and approximate atomic weight.
(a) 11, 25 (b) 12, 34 (c) 10, 25 (d) 11, 24
Solution : (d) Number of electrons =11
\ Number of protons = Number of electron =11 Number of neutrons = 13
Atomic number of element = Number of proton = Number of electrons =11 Further, Atomic weight = Number of protons + Number of neutrons =11 + 13=24
Example : 11 How many protons, neutrons and electrons are present in (a) ^{31} P (b) ^{40} Ar (c) ^{108} Ag ?
15 18 47
Solution : The atomic number subscript gives the number of positive nuclear charges or protons. The neutral atom contains an equal number of negative electrons. The remainder of the mass is supplied by neutrons.
Atom  Protons  Electrons  Neutrons 
31 P  15  15  31 – 15=16 
40 Ar
18 
18  18  40 – 18=22 
108 Ag
47 
47  47  108 – 47=61 
Example :12 State the number of protons, neutrons and electrons in C^{12} and C^{14} .
Solution : The atomic number of C^{12} is 6. So in it number of electrons = 6 Number of protons =6; Number of neutrons =12 – 6=6
The atomic number of C^{14} is 6. So in it number of electrons = 6
Number of protons = 6; Number of neutrons =14 – 6=8
Example :13 Predict the number of electrons, protons and neutrons in the two isotopes of magnesium with atomic number 12 and atomic weights 24 and 26.
Solution : Isotope of the atomic weight 24, i.e. 12 Mg ^{24} . We know that Number of protons = Number of electrons =12
Further, Number of neutrons = Atomic weight – Atomic number =24 – 12 =12
Similarly, In isotope of the atomic weight 26, i.e. 12 Mg ^{26} Number of protons = Number of electrons =12 Number of neutrons = 26 – 12 = 14
 Light and other forms of radiant energy propagate without any medium in the space in the form of waves are known as electromagnetic radiations. These waves can be produced by a charged body moving in a magnetic field or a magnet in a electric e.g. a – rays, g – rays, cosmic rays, ordinary light rays etc.
 Characteristics : (i) All electromagnetic radiations travel with the velocity of light. (ii) These consist of electric and magnetic fields components that oscillate in directions perpendicular to each other and perpendicular to the direction in which the wave is
 A wave is always characterized by the following five characteristics:
 Wavelength : The distance between two nearest crests or nearest troughs is called the It is denoted by l (lambda) and is measured is terms of centimeter(cm), angstrom(Å), micron( m ) or nanometre (nm).
1Å = 10 ^{8} cm = 10 ^{10} m
1m = 10 ^{4} cm = 10 ^{6} m
1nm = 10 ^{7} cm = 10 ^{9} m
1cm = 10^{8} Å = 10^{4} m = 10^{7} nm
 Frequency : It is defined as the number of waves which pass through a point in one It is denoted by the symbol n (nu) and is expressed in terms of cycles (or waves) per second (cps) or hertz (Hz).
ln = distance travelled in one second = velocity =c
 Velocity : It is defined as the distance covered in one second by the wave. It is denoted by the letter ‘c’. All electromagnetic waves travel with the same velocity, e., 3 ´ 10^{10} cm / sec .
Thus, a wave of higher frequency has a shorter wavelength while a wave of lower frequency has a longer wavelength.
 Wave number : This is the reciprocal of wavelength, i.e., the number of wavelengths per centimetre. It is denoted by the symbol n (nu bar). It is expressed in cm^{1} or m^{1} .
 Amplitude : It is defined as the height of the crest or depth of the trough of a wave. It is denoted by the letter ‘A’. It determines the intensity of the
The arrangement of various types of electromagnetic radiations in the order of their increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum.
Name  Wavelength (Å)  Frequency (Hz)  Source 
Radio wave  3 ´ 10^{14} – 3 ´ 10^{7}  1 ´ 10^{5} – 1 ´ 10^{9}  Alternating current of high 
frequency  
Microwave  3 ´ 10^{7} – 6 ´ 10^{6}  1 ´ 10^{9} – 5 ´ 10^{11}  Klystron tube 
Infrared (IR)  6 ´ 10^{6} – 7600  5 ´ 10^{11} – 3.95 ´ 10^{16}  Incandescent objects 
Visible  7600 – 3800  3.95 ´ 10^{16} – 7.9 ´ 10^{14}  Electric bulbs, sun rays 
Ultraviolet (UV)  3800 – 150  7.9 ´ 10^{14} – 2 ´ 10^{16}  Sun rays, arc lamps with 
mercury vapours  
XRays  150 – 0.1  2 ´ 10^{16} – 3 ´ 10^{19}  Cathode rays striking metal 
plate  
g – Rays  0.1 – 0.01  3 ´ 10^{19} – 3 ´ 10 ^{20}  Secondary effect of 
radioactive decay  
Cosmic Rays  0.01 zero  3 ´ 10 ^{20} – infinity  Outer space 
Atomic spectrum
 Spectrum is the impression produced on a photographic film when the radiation (s) of particular wavelength (s) is (are) analysed through a prism or diffraction It is of two types, emission and absorption.
 Emission spectrum : A substance gets excited on heating at a very high temperature or by giving energy and radiations are emitted. These radiations when analysed with the help of spectroscope, spectral lines are A substance may be excited, by heating at a higher temperature, by passing electric current at a very low pressure in a discharge tube filled with gas and passing electric current into metallic filament. Emission spectra is of two types,
 Continuous spectrum : When sunlight is passed through a prism, it gets dispersed into continuous bands of different colours. If the light of an incandescent object resolved through prism or spectroscope, it also gives continuous spectrum of
 Line spectrum : If the radiations obtained by the excitation of a substance are analysed with help of a spectroscope a series of thin bright lines of specific colours are obtained. There is dark space in between two consecutive This type of spectrum is called line spectrum or atomic spectrum..
 Absorption spectrum : When the white light of an incandescent substance is passed through any substance, this substance absorbs the radiations of certain wavelength from the white light. On analysing the transmitted light we obtain a spectrum in which dark lines of specific wavelengths are observed. These lines constitute the absorption The wavelength of the dark lines correspond to the wavelength of light absorbed.
 Hydrogen spectrum is an example of line emission spectrum or atomic emission
 When an electric discharge is passed through hydrogen gas at low pressure, a bluish light is
 This light shows discontinuous line spectrum of several isolated sharp lines through
 All these lines of Hspectrum have Lyman, Balmer, Paschen, Barckett, Pfund and Humphrey These spectral series were named by the name of scientist discovered them.
 To evaluate wavelength of various Hlines Ritz introduced the following expression,
Where R is universal constant known as Rydberg’s constant its value is 109, 678 cm^{1} .
 Thomson regarded atom to be composed of positively charged protons and negatively charged
The two types of particles are equal in number thereby making atom electrically neutral.
 He regarded the atom as a positively charged sphere in which negative electrons are uniformly distributed like the seeds in a water
 This model failed to explain the line spectrum of an element and the
scattering experiment of Rutherford.
 Rutherford carried out experiment on the bombardment of thin (10^{–4} mm) Au foil with high speed positively charged a – particles emitted from Ra and gave the following observations, based on this experiment :
 Most of the a – particles passed without any
 Some of them were deflected away from their
 Only a few (one in about 10,000) were returned back to their original direction of
 The scattering of a –
particles µ
1 .

sin ^{4} æ q ö
2
è ø
 From the above observations he concluded that, an atom consists of
 Nucleus which is small in size but carries the entire mass e. contains all the neutrons and protons.
 Extra nuclear part which contains This model was similar to the solar system.
(3) Properties of the Nucleus
 Nucleus is a small, heavy, positively charged portion of the atom and located at the centre of the
 All the positive charge of atom (i.e. protons) are present in
 Nucleus contains neutrons and protons, and hence these particles collectively are also referred to as nucleons.
 The size of nucleus is measured in Fermi (1 Fermi = 10^{–13} cm).
 The radius of nucleus is of the order of
1.5 ´ 10 ^{13} cm. to
6.5 ´ 10 ^{13} cm.
i.e.
1.5
to 6.5
Fermi.
Generally the radius of the nucleus ( rn ) is given by the following relation,
This exhibited that nucleus is 105 times small in size as compared to the total size of atom.
 The Volume of the nucleus is about
10 39 cm3
and that of atom is
10 ^{24} cm^{3} ,
i.e., volume of the
nucleus is 10 ^{15}
times that of an atom.
 The density of the nucleus is of the order of nucleus is spherical than,
1015 g cm3
or 10^{8}
tonnes
cm3
or 1012 kg / cc . If
(4) Drawbacks of Rutherford’s model
 It does not obey the Maxwell theory of electrodynamics, according to it “A small charged particle moving around an oppositely charged centre continuously loses its energy”. If an electron does so, it should also continuously lose its energy and should set up spiral motion ultimately failing into the
 It could not explain the line spectra of H – atom and discontinuous spectrum
Example:14 Assuming a spherical shape for fluorine nucleus, calculate the radius and the nuclear density of fluorine nucleus of mass number 19.
Solution : We know that,
r = (1.4 ´ 10 ^{–}^{13} )A^{1} ^{/} ^{3}
= 1.4 ´ 10 ^{–}^{13} ´ 19^{1} ^{/} ^{3}
= 3.73 ´ 10^{–}^{13} cm
(A for F=19)
Volume of a fluorine atom = 4 pr ^{3}
3
= 4 ´ 3.14 ´ (3.73 ´ 10^{–}^{13} )^{3}
3
= 2.18 ´ 10^{–}^{37} cm^{3}
Mass of single nucleus = Mass of one mol of nucleus = 19 g
Avogadro’ s number 6.023 ´ 10 ^{23}
Thus Density of nucleus = Mass of single nucleus = 10 ´ 1
= 7.616 = 10^{13} g cm^{–}^{1}
Volume of single nucleus 6.023 ´ 10 ^{23} 2.18 ´ 10^{–}^{37}
Example: 15 Atomic radius is the order of 10^{–}^{8} cm, and nuclear radius is the order of 10 ^{–}^{13} cm. Calculate what fraction of atom is occupied by nucleus.
Solution : Volume of nucleus = (4 / 3)pr ^{3} = (4 / 3)p ´ (10^{–}^{13} )^{3} cm^{3}
Volume of atom = (4 / 3)pr ^{3} = (4 / 3)p ´ (10 ^{–}^{8} )^{3} cm^{3}
\ Vnucleus = 1039
= 10^{–}^{15} or V
= 10 ^{–}^{15} ´ V
Vatom
1024
nucleus
atom
Planck’s Quantum theory
 Max Planck (1900) to explain the phenomena of ‘Black body radiation’ and ‘Photoelectric effect’ gave quantum This theory extended by Einstein (1905).
 If the substance being heated is a black body (which is a perfect absorber and perfect radiator of energy) the radiation emitted is called black body radiation.
 Main points
 The radiant energy which is emitted or absorbed by the black body is not continuous but discontinuous in the form of small discrete packets of energy, each such packet of energy is called a ‘quantum‘. In case of light, the quantum of energy is called a ‘photon‘.
 The energy of each quantum is directly proportional to the frequency (n ) of the radiation, e.
E µ n
or E = hn = hc
l
where, h = Planck’s constant = 6.62×10^{–27} erg. sec. or 6.62 ´ 10 ^{34} Joules sec .
 The total amount of energy emitted or absorbed by a body will be some whole number
Hence E = nhn , where n is an integer.
 The greater the frequency (i.e. shorter the wavelength) the greater is the energy of the
thus,
 Also
E1 E2
E = E
= n _{1}
n _{2}
 E
= l_{2}
l_{1}
, hence,
hc = hc + hc
or 1 = 1 + 1 .
1 2 l
l_{1} l_{2}
l l_{1} l_{2}
Example: 16 Suppose 10 ^{–}^{17} J
of energy is needed by the interior of human eye to see an object. How many photons of
green light (l = 550 nm) are needed to generate this minimum amount of energy
(a) 14  (b) 28  (c) 39  (d) 42  
Solution : (b)  Let the number of photons  required =n 
n hc = 10^{–}^{17} ;
l
n = 10^{–}^{17} ´ l =
hc
10^{–}^{17} ´ 550 ´ 10^{–}^{9}
6.626 ´ 10^{–}^{34} ´ 3 ´ 10^{8}
= 27.6 = 28 photons
Example: 17 Assuming that a 25 watt bulb emits monochromatic yellow light of wave length 0.57m. The rate of emission of quanta per sec. will be
(a) 5.89 ´ 10^{13} sec ^{–}^{1}
(b) 7.28 ´ 10^{17} sec ^{–}^{1}
(c) 5 ´ 10^{10} sec ^{–}^{1}
(d) 7.18 ´ 10^{19} sec ^{–}^{1}
Solution: (d) Let n quanta are evolved per sec.
né hc ù = 25J sec 1 ; n 6.626 ´ 10 ^{–}^{34} ´ 3 ´ 10^{8}
= 25 ;
n = 7.18 ´ 10^{19} sec ^{–}^{1}
êë l úû
0.57 ´ 10 ^{–}^{6}
 When radiations with certain minimum frequency
(n 0 )
strike the surface of a metal, the electrons are
ejected from the surface of the metal. This phenomenon is called photoelectric effect and the electrons emitted are called photoelectrons. The current constituted by photoelectrons is known as photoelectric current.
 The electrons are ejected only if the radiation striking the surface of the metal has at least a minimum
frequency
(n 0 )
called Threshold frequency. The minimum potential at which the plate photoelectric current
becomes zero is called stopping potential.
 The velocity or kinetic energy of the electron ejected depend upon the frequency of the incident radiation and is independent of its
 The number of photoelectrons ejected is proportional to the intensity of incident
 Einstein’s photoelectric effect equation : According to Einstein,
Maximum kinetic energy of the ejected electron = absorbed energy – threshold energy
where, n 0
and l0
are threshold frequency and threshold wavelength.
Note : ® Nearly all metals emit photoelectrons when exposed to u.v. light. But alkali metals like lithium, sodium, potassium, rubidium and caesium emit photoelectrons even when exposed to visible light.
 Caesium (Cs) with lowest ionisation energy among alkali metals is used in photoelectric
 This model was based on the quantum theory of radiation and the classical law of It gave new idea of atomic structure in order to explain the stability of the atom and emission of sharp spectral lines.
 Postulates of this theory are :
 The atom has a central massive core nucleus where all the protons and neutrons are The size of the nucleus is very small.
 The electron in an atom revolve around the nucleus in certain discrete Such orbits are known as
stable orbits or non – radiating or stationary orbits.
 The force of attraction between the nucleus and the electron is equal to centrifugal force of the moving
Force of attraction towards nucleus = centrifugal force
 An electron can move only in those permissive orbits in which the angular momentum (mvr) of the
electron is an integral multiple of h / 2p . Thus, mvr = n h
2p
Where, m = mass of the electron, r = radius of the electronic orbit, v = velocity of the electron in its orbit.
 The angular momentum can be
h , 2h , 3h ,…. nh .
This principal is known as quantization of
2p 2p 2p 2p
angular momentum. In the above equation ‘n’ is any integer which has been called as principal quantum number. It can have the values n=1,2,3, (from the nucleus). Various energy levels are
designed as K(n=1), L(n=2), M(n=3)—— etc. Since the electron present in these orbits is associated
with some energy, these orbits are called energy levels.
 The emission or absorption of radiation by the atom takes place when an electron jumps from one stationary orbit to
 The radiation is emitted or absorbed as a single quantum (photon) whose energy difference in energy DE of the electron in the two orbits Thus, hn = DE
hn is equal to the
Where ‘h’ =Planck’s constant, n = frequency of the radiant energy. Hence the spectrum of the atom will have certain fixed frequency.
 The lowest energy state (n=1) is called the ground state. When an electron absorbs energy, it gets excited and jumps to an outer It has to fall back to a lower orbit with the release of energy.
(3) Advantages of Bohr’s theory
 Bohr’s theory satisfactorily explains the spectra of species having one electron, viz. hydrogen atom,
He ^{+} , Li^{2}^{+} etc.
 Calculation of radius of Bohr’s orbit : According to Bohr, radius of orbit in which electron moves is
where, n =Orbit number, m =Mass number [9.1 ´ 10^{31}kg], e =Charge on the electron [1.6 ´ 10 ^{19} ]
Z =Atomic number of element, k = Coulombic constant [9 ´ 10^{9} Nm ^{2}c ^{2} ]
After putting the values of m,e,k,h, we get.
 For a particular system [e.g., H, He^{+} or Li^{+2}]
r µ n^{2} [Z = constant]
Thus we have
r n^{2}
1 = 1
i.e., r : r
: r ……….. :: 1 : 4 : 9……. r < r < r


2 1 2 3
2 2
1 2 3
 For particular orbit of different species
r µ 1 [Z =constant] Considering A and B species, we have rA
Z r_{B}
= ZB
ZA
Thus, radius of the first orbit H,
 Calculation of velocity of electron
He ^{+} , Li ^{+2} and
Be ^{+3} follows the order:
H > He ^{+} > Li^{+2} > Be ^{+3}
For H atom, Vn
= 2.188 ´ 10^{8} cm. sec 1
n
 For a particular system [H, He^{+} or Li^{+2}]
V µ 1 [Z = constant] Thus, we have,
n
V1 = n2 V2 n1
The order of velocity is V1
 V2
 V3
……… or
1
V1 : V2 : V3……………. :: 1 : 2 :
1 ……..
3
 For a particular orbit of different species
V µ Z [n =constant] Thus, we have
 For H or He^{+} or Li^{+2}, we have
H < He ^{+} < Li ^{+2}
V1 : V2 = 2 : 1; V1 : V3 = 3 : 1;
V1 : V4
= 4 : 1
 Calculation of energy of electron in Bohr’s orbit
Total energy of electron = K.E. + P.E. of electron = kZe 2 – kZe 2
= – kZe ^{2}
2r r 2r
Substituting of r, gives us
E = – 2p ^{2} mZ ^{2}e ^{4} k ^{2}
n2 h2
Where, n=1, 2, 3…….. ¥
Putting the value of m, e, k, h, p we get
E = 21.8 ´ 10 ^{12} ´ Z 2 erg per atom = 21.8 ´ 10 ^{19} ´ Z 2 J per atom(1J = 10^{7} erg)
n2 n2
E = 13.6 ´ Z 2 eV per atom(1eV = 1.6 ´ 10–19 J) = 313.6 ´ Z 2 kcal./ mole
(1 cal = 4.18J)
n2 n2
or – 1312 Z 2kJmol 1
n2
 For a particular system[H, He^{+} or Li^{+2}]




E µ – 1 [Z =constant] Thus, we have E1 2

n2 E2 2
The energy increase as the value of n increases
 For a particular orbit of different species




E µ –Z ^{2} [n =constant] Thus, we have EA 2

EB 2
For the system H, He^{+} , Li^{+2}, Be^{+3} (nsame) the energy order is H > He ^{+} > Li^{+2} > Be ^{+3}
The energy decreases as the value of atomic number Z increases.
When an electron jumps from an outer orbit (higher energy) energy) n1 , then the energy emitted in form of radiation is given by
n_{2} to an inner orbit (lower
2p 2 k 2 me 4 Z 2 æ 1 1 ö 2 æ 1 1 ö

DE = En – En =
ç – ÷Þ DE = 13.6Z
h2 n2 n2
ç – ÷ eV / atom
n2 n2

è 1 2 ø è 1 2 ø
1 DE
2p ^{2}k ^{2}me ^{4} Z ^{2} æ 1 1 ö
As we know that
E = hn , c = nl and n = l
= hc , =
ch^{3}
ç
ç n2
÷
n2 ÷
è 1 2 ø
1 2 æ 1 1 ö 2p 2 k 2me 4
This can be represented as l = n
= RZ
ç – ÷ where, R =
n2 n2
ch3
R is known as Rydberg
è 1 2 ø
constant. Its value to be used is 109678cm1 .
(4) Quantisation of energy of electron
 In ground state : No energy In ground state energy of atom is minimum and for 1^{st} orbit of Hatom, n=1.
\ E_{1} = 13.6eV.
 In excited state : Energy levels greater than n_{1} are excited i.e. for H atom n_{2} , n_{3} , n_{4}
state. For H– atom first excitation state is = n_{2}
are excited
 Excitation potential : Energy required to excite electron from ground state to any excited
Ground state ¾¾® Excited state
I^{st} excitation potential =
E2 – E1 = 3.4 + 13.6 = 10.2 eV.

II^{nd} excitation potential = E – E = 1.5 + 13.6 = 12.1 eV.
 Ionisation energy : The minimum energy required to relieve the electron from the binding of
 Ionisation potential : V
ionisation
= Eionisation
e
 Separation energy : Energy required to excite an electron from excited state to
S.E. =
E¥ – Eexcited .
 Binding energy : Energy released in bringing the electron from infinite to any orbit is called its binding energy (B.E.).
Note : ® Principal Quantum Number ‘n‘ = .
 Spectral evidence for quantisation (Explanation for hydrogen spectrum on the basisof bohr atomic model)
 The light absorbed or emitted as a result of an electron changing orbits produces characteristic absorption or emission spectra which can be recorded on the photographic plates as a series of lines, the optical spectrum of hydrogen consists of several series of lines called Lyman, Balmar, Paschen, Brackett, Pfund and These spectral series were named by the name of scientist who discovered them.
 To evaluate wavelength of various Hlines Ritz introduced the following expression,
n 1 n é 1 1 ù
= l = c = Rên2 – n2 ú
ë 1 2 û
where, R is = 2p 2 me 4
ch^{3}
= Rydberg’s constant
It’s theoritical value = 109,737 cm^{–1} and It’s experimental value = 109,677.581cm^{1}
This remarkable agreement between the theoretical and experimental value was great achievment of the Bohr model.
 Although H atom consists only one electron yet it’s spectra consist of many spectral lines as shown in
Humphrey ser  
Pfund
series 

Brackett series  
Paschen
series 

Balmer series  
 Comparative study of important spectral series of Hydrogen
S.No. Spectral
series 
Lies in the Transition
region n2 > n1 
l = n2n2
1 2 ^{max} (n^{2} – n^{2})R 2 1 
l = n2
1 min R 
lmax = n2
2 l_{min} n^{2} – n^{2} 2 1 
(1) Lymen  Ultraviolet n1 = 1
region n = 2,3,4.. ¥ 2
Visible n1 = 2 region n_{2} = 3,4,5.. ¥
Infra red n_{1} = 3 region n2 = 4,5,6………. ¥ 
n_{1} = 1 and n_{2} = 2
l = 4 max 3R
n_{1} = 2 and n_{2} = 3
l = 36 max 5R n_{1} = 3 and n_{2} = 4
l = 144 max 7R 
n_{1} = 1 and n_{2} = ¥
l = 1 min R
n_{1} = 2 and n_{2} = ¥
l = 4 min R n_{1} = 3 and n_{2} = ¥
l = 9 min R 
4 
series  3  
(2) Balmer 

series  9  
5  
(3) Paschen  16  
series  7 
 If an electron from n^{th} excited state comes to various energy states, the maximum spectral lines
obtained will be = n= principal quantum number.
as n=6 than total number of spectral lines =
6(6 – 1) = 30 = 15.
2 2
 Thus, at least for the hydrogen atom, the Bohr theory accurately describes the origin of atomic spectral
(6) Failure of Bohr Model
 Bohr theory was very successful in predicting and accounting the energies of line spectra of hydrogen
i.e. one electron system. It could not explain the line spectra of atoms containing more than one electron.
 This theory could not explain the presence of multiple spectral
 This theory could not explain the splitting of spectral lines in magnetic field (Zeeman effect) and in electric field (Stark effect). The intensity of these spectral lines was also not explained by the Bohr atomic
 This theory was unable to explain of dual nature of matter as explained on the basis of De broglies
 This theory could not explain uncertainty
 No conclusion was given for the concept of quantisation of
Example: 18 If the radius of 2^{nd} Bohr orbit of hydrogen atom is r_{2}. The radius of third Bohr orbit will be
 4 r (b) 4r (c)
9 r (d) 9r
Solution : (c)
9 2
r = n2 h2
\ r_{2} = 2^{2}
2 4 3 2
\ r = 9 r
4p ^{2}mZe ^{2}
r_{3} 3^{2}
3 4 2
Example: 19 Number of waves made by a Bohr electron in one complete revolution in 3^{rd} orbit is (a) 2 (b) 3 (c) 4 (d) 1
Solution : (b) Circumference of 3^{rd} orbit =
2pr3
According to Bohr angular momentum of electron in 3^{rd} orbit is
mvr
= 3 h or h = 2pr_{3}
^{3} 2p
mv 3
by DeBroglie equation,
l = h mv
\l = 2pr3
3
\2pr3
= 3l
i.e. circumference of 3^{rd} orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in 3^{rd} orbit is three.
Example: 20 The degeneracy of the level of hydrogen atom that has energy – R11 is
16
(a) 16 (b) 4 (c) 2 (d) 1
Solution : (a)
E = – RH
n n2
\ – RH
n2
= – RH
16
i.e. for 4 ^{th} subshell

n=4 1
1=0
m=0 +1
one s
0 +1
three p
five d
seven f
i.e. 1+3+5+7=16, \ degeneracy is 16
Example: 21 The velocity of electron in the ground state hydrogen atom is 2.18 ´ 10^{8} ms ^{–}^{1}. Its velocity in the second orbit would be
(a) 1.09 ´ 10^{8} ms ^{–}^{1}
(b) 4.38 ´ 10^{8} ms ^{–}^{1}
(c) 5.5 ´ 10^{5} ms ^{–}^{1}
(d) 8.76 ´ 10^{8} ms ^{–}^{1}
Solution : (a) We know that velocity of electron in n^{th} Bohr’s orbit is given by
v = 2.18 ´ 10^{6} Z m / s
n
for H, Z = 1
2.18 ´ 10^{6}
v1 =
v2 =
1
2.18 ´ 10^{6}
2
m / s
m / s = 1.09 ´ 10^{6}
m / s
Example: 22 The ionization energy of the ground state hydrogen atom is second orbit would be
2.18 ´ 10^{–}^{18} J.
The energy of an electron in its
(a) 1.09 ´ 10^{–}^{18} J
(b) – 2.18 ´ 10^{–}^{18} J
(c) – 4.36 ´ 10^{–}^{18} J
(d) – 5.45 ´ 10 ^{–}^{19} J
Solution : (d) Energy of electron in first Bohr’s orbit of Hatom
E = – 2.18 ´ 10^{–}^{18} J
n2
( ionization energy of H = 2.18 ´ 10^{–}^{18} J )
E = – 2.18 ´ 1018 J = 5.45 ´ 10^{–}^{19} J
2 22
Example: 23 The wave number of first line of Balmer series of hydrogen atom is 15200 cm^{–}^{1} . What is the wave number of
first line of Balmer series of
Li 3+
ion.
(a) 15200cm^{–}^{1}
(b) 6080 cm^{–}^{1}
(c) 76000cm^{–}^{1}
(d) 1,36800 cm^{–}^{1}
Solution : (d) For
Li ^{3}^{+} v = v
for H ´ z ^{2} =15200 ×9= 1,36800 cm^{–}^{1}
Example: 24 The Bohr orbit radius for the hydrogen atom (n = 1) is approximately 0.530Å. The radius for the first excited state (n = 2) orbit is (in Å)
(a) 0.13 (b) 1.06 (c) 4.77 (d) 2.12
Solution : (d) The Bohr radius for hydrogen atom (n = 1) = 0.530Å
The radius of first excited state (n = 2) will be = 0.530 ´ n2 = 0.530 ´ (2)2 = 2.120 Å
Z
Example: 25 How many chlorine atoms can you ionize in the process following process :
Cl + e ^{–} ® Cl ^{–} for 6 ´ 10 ^{23} atoms
1
Cl ® Cl^{+} + e^{–},
by the energy liberated from the
Given electron affinity of Cl = 3.61eV, and IP of Cl = 17.422 eV
(a) 1.24 ´ 10 ^{23}
atoms (b) 9.82 ´ 10 ^{20}
atoms (c)
2.02 ´ 10^{15}
atoms (d) None of these
Solution : (a) Energy released in conversion of 6 ´ 10 ^{23} atoms of Cl ^{–} ions = 6 ´ 10 ^{23} × electron affinity
= 6×10 ^{23} ´ 3.61 = 2.166 ´ 10 ^{24} eV.
Let x Cl atoms are converted to Cl ^{+} ion Energy absorbed = x ´ ionization energy
x ´ 17.422 = 2.166 ´ 10 ^{24} ;
x = 1.243 ´ 10 ^{23}
atoms
Example: 26 The binding energy of an electron in the ground state of the He atom is equal to 24eV. The energy required to remove both the electrons from the atom will be
(a) 59eV (b) 81eV (c) 79eV (d) None of these
Solution : (c) Ionization energy of He
= Z 2 ´ 13.6
n2
= 22 ´ 13.6
12
= 54.4eV
Energy required to remove both the electrons
= binding energy + ionization energy
= 24.6 + 54.4 = 79eV
Example: 27 The wave number of the shortest wavelength transition in Balmer series of atomic hydrogen will be (a) 4215 Å (b) 1437Å (c) 3942Å (d) 3647Å
1 _{2} æ 1 1 ö
_{2} æ 1 1 ö
Solution : (d) l
= RZ
ç – ÷
n2 n2
=109678 ´ 1
´ ç 22 – ¥ 2 ÷
shortest
è 1 2 ø è ø
l = 3.647 ´ 10 ^{–}^{5} cm = 3647 Å
Example: 28 If the speed of electron in the Bohr’s first orbit of hydrogen atom is x, the speed of the electron in the third Bohr’s orbit is
(a) x/9 (b) x/3 (c) 3x (d) 9x
Solution : (b) According to Bohr’s model for hydrogen and hydrogen like atoms the velocity of an electron in an atom is
quantised and is given by v µ 2pZe 2
nh
so v µ 1
n
in this cass n = 3
Example: 29 Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is
(a) n=1 to n=2 (b) n = 3 to n = 8
 n = 2 to n = 1
 n = 8
to n = 3
Solution : (b) Absorption line in the spectra arise when energy is absorbed i.e., electron shifts from lower to higher orbit, out of a & b, b will have the lowest frequency as this falls in the Paschen series.
Example: 30 The frequency of the line in the emission spectrum of hydrogen when the atoms of the gas contain electrons in the third energy level are
(a) 1.268 ´ 10^{14} Hz
and 2.864 ´ 10^{16} Hz
(b) 3.214 ´ 10^{10} Hz
and 1.124 ´ 10^{12} Hz
(c) 1.806 ´ 10^{12} Hz and 6.204 ´ 10^{15} Hz
(d) 4.568 ´ 10^{14} Hz
and 2.924 ´ 10^{15} Hz
Solution : (d) If an electron is in 3^{rd} orbit, two spectral lines are possible
 When it falls from 3^{rd} orbit to 2^{nd}
In equation
n = 3.289 ´ 10^{15} é 1 – 1 ù
ê n2 n2 ú
ë 1 2 û

n = 3.289 ´ 10^{15} é 1
^{1} ë 2^{2}
– 1 ù = 3.289 ´ 10^{15} ´ 5 =

3^{2} û 36
4.568 ´ 14^{14} Hz
 When it falls from 3^{rd} orbit to 1^{st} orbit :
n = 3.289 ´ 10^{15} ´ é1 – 1 ù = 3.289 ´ 10^{15} ´ 8 = 2.924 ´ 10^{15} Hz
2 êë1
32 úû 9
Example: 31 If the first ionisation energy of hydrogen is per atom is
2.179 ´ 10 ^{–}^{18} J
per atom, the second ionisation energy of helium
(a)
8.716 ´ 10 ^{–}^{18} J
(b) 5.5250 kJ
(c) 7.616 ´ 10^{–}^{18} J
(d) 8.016 ´ 10 ^{–}^{13} J
Solution : (a) For Bohrs systems : energy of the electron µ Z 2
n2
Ionisation energy is the difference of energies of an electron
(E¥ ),
when taken to infinite distance and Er
when present in any Bohr orbit and Ea
electron in any Bohr orbit.
is taken as zero so ionisation energy becomes equal to the energy of
Z 2

EH µ H
H
Z 2

; EHe µ He
He
or EH =
EHe
1
2 ´ 2
[as ZH = 1, Z He = 2, nH = 1, nHe = 1]
or EHe = EH ´ 4 = 2.179 ´ 10 ^{–}^{18} ´ 4 = 8.716 ´ 10 ^{–}^{18} Joule per atom.
Example: 32 The ionization energy of hydrogen atom is 13.6eV. What will be the ionization energy of He ^{+}
(a) 13.6eV (b) 54.4eV (c) 122.4eV (d) Zero
Solution : (b) I.E. of He ^{+} = 13.6eV ´ Z ^{2}
13.6eV ´ 4 = 54.4eV
Example: 33 The ionization energy of He ^{+}
(a) 19.6 ´ 10^{–}^{18} J atom^{1}
(c) 19.6 ´ 10 ^{–}^{19} J atom^{1}
is 19.6 ´ 10^{–}^{18} J atom^{–1}. Calculate the energy of the first stationary state of Li ^{+}^{2}
(b) 4.41 ´ 10^{–}^{18} J atom^{–1} (d) 4.41 ´ 10 ^{–}^{17} J atom^{–}^{1}
Solution : (d) I.E. of He ^{+} = E ´ 2^{2} (Z
for
He = 2)
I.E. of Li ^{2}^{+} = E ´ 3^{3} (Z for Li=3)
\ I.E.(He ^{+} ) = 4
or I.E. (Li ^{2}^{+} ) = 9 ´ I.E.(He ^{+} )
= 9 ´ 19.6 ´ 10 ^{–}^{18}
= 4.41 ´ 10^{–}^{17}
J atom^{–1}
I.E.(Li ^{2}^{+} ) 9 4 4
 In 1915, Sommerfield introduced a new atomic model to explain the fine spectrum of hydrogen
 He gave concept that electron revolve round the nucleus in elliptical orbit. Circular orbits are formed in special conditions only when major axis and minor axis of orbit are
 For circular orbit, the angular momentum = component e. only angle changes.
nh where n= principal quantum number only one
2p
 For elliptical orbit, angular momentum = vector sum of 2 In elliptical orbit two components are,
 Radial component (along the radius) = n h
^{r} 2p
Where, n _{r} = radial quantum number
 Azimuthal component = n _{f} h
2p
Where, n _{f} = azimuthal quantum number
So angular momentum of elliptical orbit = n
h + n h
Angular momentum = (n + n ) h
^{r} 2p ^{f} 2p
r f 2p
 Shape of elliptical orbit depends on,
Length of major axis = n
Length of minor axis nf
= nr + nf nf
 n fcan take all integral values from l to ‘n’ values of n r depend on the value of n f . For n = 3,
n f can have values 1,2,3 and n r can have (n –1) to zero i.e. 2,1 and zero respectively.
Thus for n = 3, we have 3 paths
n  n f  n r  Nature of path 
3  1  3  elliptical 
2  1  elliptical  
3  0  circular 
The possible orbits for n = 3 are shown in figure.
Thus Sommerfield showed that Bohr’s each major level was composed of several sublevels. therefore it provides the basis for existance of subshells in Bohr’s shells (orbits).
(7) Limitation of Bohr sommerfield model :
 This model could not account for, why electrons does not absorb or emit energy when they are moving in stationary
 When electron jumps from inner orbit to outer orbit or vice –versa, then electron run entire distance but absorption or emission of energy is
 It could not explain the attainment of expression of explain Zeeman effect and Stark
nh for angular momentum. This model could not
2p
 In 1924, the french physicist, Louis de Broglie suggested that if light has both particle and wave like nature, the similar duality must be true for Thus an electron, behaves both as a material particle and as a wave.
 This presented a new wave mechanical theory of matter. According to this theory, small particles like electrons when in motion possess wave
 According to debroglie, the wavelength associated with a particle of mass m,
given by the relation
moving with velocity v is
where h = Planck’s constant.
 This can be derived as follows according to Planck’s equation,
E = hn = h.c
l
æ c ö


ç
è ø
energy of photon (on the basis of Einstein’s mass energy relationship), E = mc ^{2}
equating both
hc = mc ^{2}
l
or l = h
mc
which is same as deBroglie relation. (
mc = p)
 This was experimentally verified by Davisson and Germer by observing diffraction effects with an electron Let the electron is accelerated with a potential of V than the Kinetic energy is
1 mv2 = eV ;
2
m2 v 2 = 2eVm
mv =
= P ;
l = h
 If Bohr’s theory is associated with deBroglie’s equation then wave length of an electron can be determined in bohr’s orbit and relate it with circumference and multiply with a whole number
2pr = nl or l = 2pr
n
From deBroglie equation, l = h . Thus
h = 2pr
or mvr = nh
mv mv n 2p
Note : ® For a proton, electron and an a particle moving with the same velocity have debroglie wavelength in the following order : Electron > Proton > a – particle.
 The deBroglie equation is applicable to all material objects but it has significance only in case of microscopic Since, we come across macroscopic objects in our everyday life, debroglie relationship has no significance in everyday life.
Solution : (b) KE = 1 mv ^{2} = 4.55 ´ 10^{–}^{25} J
2

2´ 4.55 ´ 10 ^{–}^{25}
v 9.1 ´ 10 ^{–}^{31}
=1´10^{6} ;
v = 10^{3}
m / s
DeBroglie wavelength l = h =
mv
6.626 ´ 10 ^{–}^{34}
9.1´10 ^{–}^{31} ´10^{3}
= 7.28 ´ 10 ^{–}^{7} m
Example: 35 The speed of the proton is one hundredth of the speed of light in vacuum. What is the de Broglie wavelength?
Assume that one mole of protons has a mass equal to one gram, h = 6.626´10 ^{–}^{27} erg sec
Solution : (b)
(a) 3.31 × 10^{–}^{3} Å
m = 1 g
6.023 ´ 10 ^{23}
(b) 1.33 ×
10^{–}^{3} Å
(c) 3.13 × 10 ^{–}^{2} Å
(d) 1.31 ×
10^{–}^{2} Å
l = h =
mv
6.626 ´ 10^{–}^{27}
1 ´ 3 ´ 10^{8} cm sec ^{–}^{1}
´ 6.023
´ 10 ^{23}
= 1.33
´ 10
11 cm
 One of the important consequences of the dual nature of an electron is the uncertainty principle, developed by Warner Heisenberg.
 According to uncertainty principle “It is impossible to specify at any given moment both the position and momentum (velocity) of an electron”.
Mathematically it is represented as ,
Where Dx = uncertainty is position of the particle, Dp = uncertainty in the momentum of the particle
Now since Dp = m Dv
So equation becomes,
Dx. mDv ³ h or
4p
Dx ´Dv ³
h
4pm
The sign ³ means that the product of
Dx and
Dp (or of
Dx and
Dv ) can be greater than, or equal to but
never smaller than
h . If Dx is made small, Dp increases and vice versa.
4p
 In terms of uncertainty in energy, DE and uncertainty in time Dt, this principle is written as,
DE . Dt ³ h
4p
Note :® Heisenberg’s uncertainty principle cannot we apply to a stationary electron because its velocity is 0 and position can be measured accurately.
Example: 36 What is the maximum precision with which the momentum of an electron can be known if the uncertainty in the position of electron is ± 0.001Å ? Will there be any problem in describing the momentum if it has a value
of h
2pa0
, where a_{0} is Bohr’s radius of first orbit, i.e., 0.529Å?
Solution :
Dx . Dp = h
4p
Dx = 0.001Å = 10^{–}^{13} m
\ Dp =
6.625 ´ 10 ^{–}^{34}
4 ´ 3.14 ´ 10 ^{–}^{13}
= 5.27 ´ 10 ^{–}^{22}
Example: 37 Calculate the uncertainty in velocity of an electron if the uncertainty in its position is of the order of a 1Å.
Solution : According to Heisenberg’s uncertainty principle
Dv. Dx »
h
4pm
Dv » h
4pm.Dx
= 6.625 ´ 10^{–}^{34}
4 ´ 22 ´ 9.108 ´ 10^{–}^{31} ´ 10^{–}^{10}
7
= 5.8 ´ 10^{5} m
sec ^{–}^{1}
Example: 38 A dust particle having mass equal to
10^{–}^{11} g, diameter of
10^{–}^{4} cm
and velocity
10^{–}^{4} cm sec ^{–}^{1} . The error in
measurement of velocity is 0.1%. Calculate uncertainty in its positions. Comment on the result .
Solution :
Dv = 0.1´ 10^{–}^{4}
100
= 1´ 10^{–}^{7} cm sec^{–}^{1}
Dv. Dx =
h
4pm
\ Dx =
6.625 ´ 10^{–}^{27}
4 ´ 3.14 ´ 10^{–}^{11} ´ 1 ´ 10^{–}^{7}
= 5.27 ´ 10^{–}^{10} cm
The uncertainty in position as compared to particle size.
= Dx = 5.27 ´ 10 ^{–}^{10}
= 5.27 ´ 10 ^{–}^{6} cm
diameter
10 4
The factor being small and almost being negligible for microscope particles.
 Schrodinger wave equation is given by Erwin Schrödinger in 1926 and based on dual nature of
 In it electron is described as a three dimensional wave in the electric field of a positively charged
 The probability of finding an electron at any point around the nucleus can be determined by the help of Schrodinger wave equation which is,
¶ 2 Y  + ¶ 2 Y  + ¶ 2 Y  +  8p 2m  (E  –  V)Y  =  0 
¶x 2  ¶y 2  ¶z 2  h2 
Where
x, y
and z are the 3 space coordinates, m = mass of electron, h = Planck’s constant,
E = Total energy, V = potential energy of electron, Y = amplitude of wave also called as wave function.
¶ = stands for an infinitesimal change.
 The Schrodinger wave equation can also be written as :
Where Ñ = laplacian operator.
(5) Physical Significance of Y and Y ^{2}
 The wave function Y represents the amplitude of the electron The amplitude Y is thus a function of space coordinates and time i.e. Y = Y(x, y, z………….. times)
 For a single particle, the square of the wave function probability of finding the particle at that
(Y ^{2} )
at any point is proportional to the
 If Y ^{2}
is maximum than probability of finding e –
is maximum around nucleus. And the place where
probability of finding
e ^{–} is maximum is called electron density, electron cloud or an atomic orbital. It
is different from the Bohr’s orbit.
 The solution of this equation provides a set of number called quantum numbers which describe specific or definite energy state of the electron in atom and information about the shapes and orientations of the most probable distribution of electrons around the
Quantum numbers
 Each orbital in an atom is specified by a set of three quantum numbers (n, l, m) and each electron is designated by a set of four quantum numbers (n, l, m and s).
(2) Principle quantum number (n)
 It was proposed by Bohr’s and denoted by ‘n’.
 It determines the average distance between electron and nucleus, means it is denoted the size of
 It determine the energy of the electron in an orbit where electron is
 The maximum number of an electron in an orbit represented by this quantum number as energy shell in atoms of known elements possess more than 32
 It gives the information of orbit K, L, M, N——– –.
 The value of energy increases with the increasing value of
 It represents the major energy shell or orbit to which the electron
 Angular momentum can also be calculated using principle quantum number
2n ^{2} . No
(3) Azimuthal quantum number (l)
 Azimuthal quantum number is also known as angular quantum Proposed by Sommerfield and denoted by ‘l’.
 It determines the number of sub shells or sublevels to which the electron
 It tells about the shape of
 It also expresses the energies of subshells
s < p < d < f
(increasing energy).
 The value of l = (n – 1) always where ‘n’ is the number of principle
(vi) Value of l  =  0  1  2  3……… (n1) 
Name of subshell  =  s  p  d  f 
Shape of subshell  =  Spherical  Dumbbell  Double dumbbell  Complex 
 It represent the orbital angular Which is equal to h
2p
 The maximum number of electrons in subshell = 2(2l + 1)
s – subshell ® 2 electrons
p – subshell ® 6 electrons
d – subshell ® 10 electrons
f – subshell ® 14 electrons.
 For a given value of ‘n’ the total value of ‘l’ is always equal to the value of ‘n’.
 The energy of any electron is depend on the value of n & l because total energy = (n + l). The electron enters in that sub orbit whose (n + l) value or the value of energy is
(4) Magnetic quantum number (m)
 It was proposed by Zeeman and denoted by ‘m’.
 It gives the number of permitted orientation of
 The value of m varies from –l to +l through
 It tells about the splitting of spectral lines in the magnetic field e. this quantum number proved the Zeeman effect.
 For a given value of ‘n’ the total value of ’m’ is equal to n ^{2} .
 For a given value of ‘l’ the total value of ‘m’ is equal to(2l + 1).
 Degenerate orbitals : Orbitals having the same energy are known as degenerate e.g. for p
subshell px py pz
 The number of degenerate orbitals of s subshell =0.
(5) Spin quantum numbers (s)
 It was proposed by Goldshmidt & Ulen Back and denoted by the symbol of ‘s’.
 The value of
‘ s‘
is + 1/2 and – 1/2, which is signifies the spin or rotation or direction of electron on it’s
axis during movement.
 The spin may be clockwise or
 It represents the value of spin angular momentum is equal to h
2p
s(s + 1).
 Maximum spin of an atom = 1 / 2 ´ number of unpaired
Magnetic field
–1/2
S N
 This quantum number is not the result of solution of schrodinger equation as solved for Hatom.
Distribution of electrons among the quantum levels
n  l  m  s  Designation of
orbitals 
Electrons
present 
Total no. of
electrons 
1 (K shell)  0  0  +1/2, –1/2  1s  2  2 
2 (L shell)  0  0  +1 / 2, – 1 / 2  2s  2ù  
1 
+1
0 –1 
+ 1 / 2, – 1 / 2
+ 1 / 2, – 1 / 2 + 1 / 2, – 1 / 2 
2p 
ú
ú 6úû 
8 
3 (M shell)  0  0  +1 / 2,1 / 2
+ 1 / 2,1 / 2 + 1 / 2,1 / 2 +1 / 2,1 / 2
+ 1 / 2,1 / 2ù + 1 / 2,1 / 2ú ú + 1 / 2,1 / 2ú + 1 / 2,1 / ú 2ú + 1 / 2,1 / 2úû 
3s  2 ù
ú ú 6 ú ú ú ú ú ú ú ú ú ú ú ú 10ú û 

+1  
1  0  3p  
–1  
18  
+2 

+1  3d  
2  0  
–1  
–2  
0  0  +1 / 2,1 / 2
+ 1 / 2,1 / 2ù + 1 / 2,1 / 2ú ú + 1 / 2,1 / 2úû
+ 1 / 2,1 / 2ù + 1 / 2,1 / 2ú ú + 1 / 2,1 / 2ú + 1 / 2,1 / ú 2ú + 1 / 2,1 / 2úû
+ 1 / 2,1 / 2ù + 1 / 2,1 / 2ú ú + 1 / 2,1 / 2ú + 1 / 2,1 / ú 2ú + 1 / 2,1 / 2ú + 1 / 2,1 / ú 2ú + 1 / 2,1 / 2ú û 
4s  2 ù
ú ú 6 ú ú ú ú ú ú ú ú ú ú ú 10ú ú ú ú ú ú ú ú ú ú ú 14úû 

+1  
1  0  4p  
–1  
+2 

+1  
2  0  4d  32  
–1  
–2  
4(N shell)  
+3  
+2  
3 
+1
+0 
4f  
–1  
–2  
–3 
Shape of orbitals
(1) Shape of ‘s’ orbital
 For ‘s’ orbital l=0 & m=0 so ‘s’ orbital have only one unidirectional orientation i.e. the probability of finding the electrons is same in all
 The size and energy of ‘s’ orbital with increasing ‘n’ will be
1s < 2s < 3s < 4s.
 It does not possess any directional s orbital has spherical shape.
(2) Shape of ‘p’ orbitals
 For ‘p’ orbital l=1, & m=+1,0,–1 means there are three ‘p’ orbitals, which is symbolised as
px , py , pz .
 Shape of ‘p’ orbital is dumb bell in which the two lobes on opposite side separated by the nodal
 porbital has directional
Z 
Y Nodal Plane X
Px orbital 
Nodal Plane 
Nodal Plane 
Z 
Y
X
Py orbital 
Nodal Plane
Nodal Plane 
Z 
Y
X
Pz orbital 
(3) Shape of ‘d’ orbital
 For the ‘d’ orbital l =2 then the values of ‘m’ are –2,–1,0,+1,+2. It shows that the ‘d’ orbitals has five orbitals as dxy , dyz , dzx , dx2 –y2 , dz 2 .
 Each ‘d’ orbital identical in shape, size and
 The shape of d orbital is double dumb bell .

 It has directional
(4) Shape of ‘f’ orbital
 For the ‘f’ orbital l=3 then the values of ‘m’ are –3, –2, –1,0,+1,+2,+3. It shows that the ‘f’ orbitals
have seven orientation as
fx(x ^{2} – y ^{2} ), fy(x ^{2} – y ^{2} ), fz(x ^{2} – y ^{2} ), fxyz , fz ^{3} , fyz ^{3} and fxz ^{2} .
 The ‘f’ orbital is complicated in shape.
The distribution of electrons in different orbitals of atom is known as electronic configuration of the atoms. Filling up of orbitals in the ground state of atom is governed by the following rules:
(1) Aufbau principle
 Auf bau is a German word, meaning ‘building up’.
 According to this principle, “In the ground state, the atomic orbitals are filled in order of increasing energies e. in the ground state the electrons first occupy the lowest energy orbitals available”.
 In fact the energy of an orbital is determined by the quantum number n and l with the help of (n+l) rule or Bohr Bury
 According to this rule
 Lower the value of n + l, lower is the energy of the orbital and such an orbital will be filled up
 When two orbitals have same value of (n+l) the orbital having lower value of “n” has lower energy and such an orbital will be filled up first .
Thus, order of filling up of orbitals is as follows:
1s < 2s < 2p < 3s < 3p < 4s < 4 p < 5s < 4d < 5 p < 6s < 6 f < 5d
(2) Pauli’s exclusion principle
 According to this principle, “No two electrons in an atom can have same set of all the four quantum numbers n, l, m and s .
 In an atom any two electrons may have three quantum numbers identical but fourth quantum number must be
 Since this principle excludes certain possible combinations of quantum numbers for any two electrons in an atom, it was given the name exclusion Its results are as follows :
 The maximum capacity of a main energy shell is equal to 2n^{2}
 The maximum capacity of a subshell is equal to 2(2l+1)
 Number of subshells in a main energy shell is equal to the value of n.
 Number of orbitals in a main energy shell is equal to n ^{2} .
 One orbital cannot have more than two
 According to this principle an orbital can accomodate at the most two electrons with spins opposite to each It means that an orbital can have 0, 1, or 2 electron.
 If an orbital has two electrons they must be of opposite
Correct Incorrect
(3) Hund’s Rule of maximum multiplicity
 This rule provides the basis for filling up of degenerate orbitals of the same subshell.
 According to this rule “Electron filling will not take place in orbitals of same energy until all the available orbitals of a given subshell contain one electron each with parallel spin”.
 This implies that electron pairing begins with fourth, sixth and eighth electron in p, d and f orbitals of the same subshell
 The reason behind this rule is related to repulsion between identical charged electron present in the same
 They can minimise the repulsive force between them serves by occupying different
 Moreover, according to this principle, the electron entering the different orbitals of subshell have parallel spins. This keep them farther apart and lowers the energy through electron exchange or
 The term maximum multiplicity means that the total spin of unpaired correct filling of orbitals as per this
Energy level diagram
e ^{–} is maximum in case of
The representation of relative energy levels of various atomic orbital is made in the terms of energy level diagrams.
One electron system : In this system 1s ^{2} level and all orbital of same principal quantum number have same energy, which is independent of (l). In this system l only determines the shape of the orbital.
Multiple electron system : The energy levels of such system not only depend upon the nuclear charge but also upon the another electron present in them.

5 4s 4p 4d 4f

5 d
4 f
6 s
5 p
3s 3p 3d 4 d
5 s
3 4 p
 d
 s
2s 2p 3 p

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



d^{4} (1) (2) (3)
3 exchanges by 1st e– 2 exchanges by 2nd e^{–} Only 1 exchange by 3rd e^{–}
To number of possible exchanges = 3 + 2 + 1 =6
d^{5} (1)
 (4)
4 exchanges by 1st e^{–}

1 exchange by 4th e^{–}
(2)


3 exchanges by 2nd e^{–}
(3)
2 exchange by 3rd e^{–}
To number of possible exchanges = 4 + 3 + 2 +1 = 10
 On the basis of the elecronic configuration priciples the electronic configuration of various elements are given in the following table :
Electronic Configuration (E.C.) of Elements Z=1 to 36
Element  Atomic Number  1s  2s  2p  3s  3p  3d  4s  4p  4d  4f 
H  1  1  
He  2  2  
Li  3  2  1  
Be  4  2  2  
B  5  2  2  1  
C  6  2  2  2  
N  7  2  2  3  
O  8  2  2  4  
F  9  2  2  5  
Ne  10  2  2  6  
Na  11  2  2  6  1  
Mg  12  2  
Al  13  2  1  
Si  14  10  2  2  
P  15  electrons  2  3  
S  16  2  4  
Cl  17  2  5  
Ar  18  2  2  6  2  6  
K  19  2  2  6  2  6  1  
Ca  20  2  
Sc  21  1  2  
Ti  22  2  2  
V  23  3  2  
Cr  24  5  1  
Mn  25  5  2  
Fe  26  6  2  
Co  27  18  7  2  
Ni  28  electrons  8  2  
Cu  29  10  1  
Zn  30  10  2  
Ga  31  10  2  1  
Ge  32  10  2  2  
As  33  10  2  3  
Se  34  10  2  4  
Br  35  10  2  5  
Kr  36  2  2  6  2  6  10  2  6 
 The above method of writing the electronic configurations is quite Hence, usually the electronic configuration of the atom of any element is simply represented by the notation.
 (i) Elements with atomic number 24(Cr), 42(Mo) and 74(W) have ns^{1}(n – 1)d ^{5} configuration and not
ns ^{2} (n – 1)d ^{4} due to extra stability of these atoms.
(ii) Elements with atomic number 29(Cu), 47(Ag) and 79(Au) have ns^{1}(n – 1)d^{10} configuration instead of

ns ^{2} (n – 1)d ^{9} due to extra stability of these atoms.
Cr (24) [Ar]
4s^{1}
Cu (29) [Ar]
4s^{1}
 In the formation of ion, electrons of the outer most orbit are lost. Hence, whenever you are required to write electronic configuration of the ion, first write electronic configuration of its atom and take electron from
outermost orbit. If we write electronic configuration of Fe ^{2}^{+} (Z = 26, 24 e ^{–} ), it will not be similar to Cr (with 24 e ^{–} ) but quite different.
Fe[Ar] 4s ^{2} 3d ^{6} üï

Fe ^{2}^{+} [Ar] 4s ^{∘} 3d ^{6} ýï
outer most orbit is 4^{th} shell hence, electrons from 4s have been removed to make
Fe ^{2}^{+} .
 Ion/atom will be paramagnetic if there are unpaired Magnetic moment (spin only) is
m = BM (Bohr Magneton). (1BM = 9.27 ´ 10 ^{24} J / T) where n is the number of unpaired electrons.
 Ion with unpaired electron in d or f orbital will be Thus, Cu ^{+}
with electronic configuration
[Ar]3d^{10}
(blue).
is colourless and Cu ^{2}^{+}
with electronic configuration [Ar]3d ^{9} (one unpaired electron in 3d) is coloured
 Position of the element in periodic table on the basis of electronic configuration can be determined as,
 If last electron enters into ssubshell, psubshell, penultimate dsubshell and anti penultimate fsubshell then the element belongs to s, p, d and f – block
 Principle quantum number (n) of outermost shell gives the number of period of the
 If the last shell contains 1 or 2 electrons (i.e. for sblock elements having the configuration ns^{1}^{2} ), the group number is 1 in the first case and 2 in the second
 If the last shell contains 3 or more than 3 electrons (i.e. for pblock elements having the configuration
ns ^{2} np^{1}^{6} ), the group number is the total number of electrons in the last shell plus 10.
 If the electrons are present in the (n –1)d orbital in addition to those in the ns orbital (i.e. for dblock elements having the configuration (n –1) d^{1}^{10}ns^{1}^{2} ), the group number is equal to the total number of electrons present in the (n –1)d orbital and ns