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 sub-atomic particles like electron, proton, neutron, positron, neutrino, meson etc. Out of these particles, the electron, proton and the neutron are called fundamental subatomic particles and others are non-fundamental particles.

Electron 

• 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 ´ 10-28 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 ´ 10²⁷ 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´ 10¹⁰

(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 ´ 10¹⁰

3

= 9.108 ´ 10 ^–28 ´ 3 ´ 10¹⁰

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      ö²

 

1 – ç 3 ´ 108 ÷

 

 

æ      u       ö²

 

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 ´ 10-19

(b)

8.01´ 10¹⁰

(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 cm³ .

3

 

• Specific charge of a proton is 58 ´ 10⁴ 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 H¹ + -1 e⁰ +

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 × 108	9.58 × 104	Zero

• The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C¹² , 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
Anti-proton		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

 

29 Cu65 , 24 Cr 55

(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 Ca²⁺(18e^–)

• N ₂ and CO
• CO₂ and N ₂O
• HCl and F₂
• CaO and  MgS
• C₆ H₆ 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 ´ 10⁷ (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 ´ 10⁸

 

=                               = 11.4457 ´ 10⁸

 

using Moseley’s equation we get

 

\ 5.5132 ´ 10⁸ = 4.9 ´ 10⁷ (Zx – 0.75)

…..(i)

 

and 11.4457 ´ 10⁸ = 4.90 ´ 10⁷ (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.

3

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 C¹² and C¹⁴ .

Solution :            The atomic number of C¹² is 6. So in it number of electrons = 6 Number of protons =6; Number of neutrons =12 – 6=6

The atomic number of C¹⁴ 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 = 10⁸ Å = 10⁴ m = 10⁷ 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 ´ 10¹⁰ 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 ´ 1014 – 3 ´ 107	1 ´ 105 – 1 ´ 109	Alternating  current   of   high
			frequency
Microwave	3 ´ 107 – 6 ´ 106	1 ´ 109 – 5 ´ 1011	Klystron tube
Infrared (IR)	6 ´ 106 – 7600	5 ´ 1011 – 3.95 ´ 1016	Incandescent objects
Visible	7600 – 3800	3.95 ´ 1016 – 7.9 ´ 1014	Electric bulbs, sun rays
Ultraviolet (UV)	3800 – 150	7.9 ´ 1014 – 2 ´ 1016	Sun  rays,   arc   lamps   with
			mercury vapours
X-Rays	150 – 0.1	2 ´ 1016 – 3 ´ 1019	Cathode rays  striking  metal
			plate
g – Rays	0.1 – 0.01	3 ´ 1019 – 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 H-spectrum 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 H-lines 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     .

ç   ÷

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 10-5 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 cm³ ,

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 cm-3

or 10⁸

tonnes

cm-3

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 cm³

 

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 = 10¹³ 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 cm³

Volume of atom = (4 / 3)pr ³ = (4 / 3)p ´ (10 ^–8 )³ cm³

 

\ Vnucleus    = 10-39

= 10^–15 or V

= 10 ^–15 ´ V

 

Vatom

10-24

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

= l₂

l₁

, hence,

hc = hc + hc

or 1 = 1 + 1 .

 

1         2                      l

l₁    l₂

l     l₁    l₂

 

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 ´ 10⁸

= 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 ´ 10¹³ sec ^–1

(b) 7.28 ´ 10¹⁷ sec ^–1

(c) 5 ´ 10¹⁰ sec ^–1

(d) 7.18 ´ 10¹⁹ sec ^–1

 

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

né hc ù = 25J sec -1 ; n 6.626 ´ 10 ^–34 ´ 3 ´ 10⁸

= 25 ;

n = 7.18 ´ 10¹⁹ 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 photo-electrons. 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 ⁺ , Li²⁺ 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^-31kg], e =Charge on the electron [1.6 ´ 10 ^-19 ]

Z =Atomic number of element, k = Coulombic constant [9 ´ 10⁹ Nm ²c ^-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 n² [Z = constant]

 

 

Thus we have

r        n²

1 =   1

i.e., r : r

: r ……….. :: 1 : 4 : 9……. r < r < r

 

n

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 ´ 10⁸ 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 ²e 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 = 10⁷ 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]

n

=

n

2

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

1

n2                                                                  E2          2

The energy increase as the value of n increases

• For a particular orbit of different species

Z

=

Z

A

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

B

EB                  2

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

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

 

 

 

When an electron jumps from an outer orbit (higher energy) energy) n1 , then the energy emitted in form of radiation is given by

n₂ to an inner orbit (lower

 

2p 2 k 2 me 4 Z 2 æ 1     1 ö                       2 æ 1     1 ö

 

2                  1

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 ²k ²me ⁴ Z ² æ 1     1 ö

 

As we know that

E = hn , c = nl and n = l

= hc , =

ch³

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 109678cm-1 .

(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 H-atom, n=1.

\ E₁ = -13.6eV.

 

• In excited state : Energy levels greater than n₁ are excited i.e. for H- atom n₂ , n₃ , n₄

state. For H– atom first excitation state is = n₂

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.

 

3         1

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 H-lines 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

ch³

= 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     (n2 – n2)R 2         1	l      = n2 1   min       R	lmax =     n2 2      lmin     n2 – n2 2              1
(1)       Lymen	Ultraviolet      n1 = 1 region              n   = 2,3,4.. ¥ 2     Visible             n1 = 2 region  n2 = 3,4,5.. ¥   Infra      red  n1  =  3 region                      n2 = 4,5,6………. ¥	n1 = 1 and n2 = 2 l       = 4 max      3R    n1 = 2 and n2 = 3   l       = 36 max      5R  n1 = 3 and n2 = 4   l       = 144 max      7R	n1 = 1 and n2 = ¥ l      = 1 min     R    n1 = 2 and n2 = ¥   l      = 4 min     R  n1 = 3 and n2 = ¥   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 r₂. The radius of third Bohr orbit will be

 

• 4 r (b) 4r                                (c)

9 r                                            (d) 9r

 

 

Solution : (c)

9 2

r =      n2 h2

\ r₂ = 2²

2                                       4 3                                                      2

\ r = 9 r

 

4p ²mZe ²

r₃     3²

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   = 2pr₃

 

 

³        2p

mv          3

 

by De-Broglie 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 sub-shell

 

2         3       +2     –     1      0       +1     +2     –3     –     2     –       1     0     +1  +2     +3

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 ´ 10⁸ ms ^–1. Its velocity in the second orbit would be

 

(a) 1.09 ´ 10⁸ ms ^–1

(b) 4.38 ´ 10⁸ ms ^–1

(c) 5.5 ´ 10⁵ ms ^–1

(d) 8.76 ´ 10⁸ ms ^–1

 

Solution : (a) We know that velocity of electron in n^th Bohr’s orbit is given by

v = 2.18 ´ 10⁶ Z m / s

n

for H, Z = 1

2.18 ´ 10⁶

 

 v1  =

 

 v2  =

1

2.18 ´ 10⁶

2

m / s

 

m / s = 1.09 ´ 10⁶

 

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 H-atom

 

E = – 2.18 ´ 10^–18 J

n2

( ionization energy of H = 2.18 ´ 10^–18 J )

 

E   = – 2.18 ´ 10-18 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 ´ 10¹⁵

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 ´ 10¹⁴ Hz

and 2.864 ´ 10¹⁶ Hz

(b) 3.214 ´ 10¹⁰ Hz

and 1.124 ´ 10¹² Hz

 

(c) 1.806 ´ 10¹² Hz and 6.204 ´ 10¹⁵ Hz

(d) 4.568 ´ 10¹⁴ Hz

and 2.924 ´ 10¹⁵ 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 ´ 10¹⁵ é 1 – 1 ù

 

ê n2       n2 ú

ë 1         2 û

 

ê

n = 3.289 ´ 10¹⁵ é 1

¹                       ë 2²

– 1 ù = 3.289 ´ 10¹⁵ ´ 5 =

ú

3² û                                  36

4.568 ´ 14¹⁴ Hz

 

• When it falls from 3^rd orbit to 1^st orbit :

n = 3.289 ´ 10¹⁵ ´ é1 – 1 ù = 3.289 ´ 10¹⁵ ´ 8 = 2.924 ´ 10¹⁵ 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

n

EH   µ    H  

H

Z 2

n

; 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 ´ 2² (Z

for

He = 2)

 

I.E. of Li ²⁺ = E ´ 3³ (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 sub-levels. 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 de-broglie, 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

l ÷

n =

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 de-Broglie 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 de-Broglie’s equation then wave length of an electron can be determined in bohr’s orbit and relate it with circumference and multiply with a whole number

2pr = nl or l = 2pr

n

 

From de-Broglie 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 de-broglie wavelength in the following order : Electron > Proton > a – particle.

• The de-Broglie 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, de-broglie relationship has no significance in everyday life.

 

Solution : (b)   KE = 1 mv ² = 4.55 ´ 10^–25 J

2

 

2 =

2´ 4.55 ´ 10 ^–25

v           9.1 ´ 10 ^–31

=1´10⁶ ;

v = 10³

m / s

 

De-Broglie wavelength l = h   =

mv

6.626 ´ 10 ^–34

9.1´10 ^–31 ´10³

= 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 ´ 10⁸ 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 a₀ 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 ´ 10⁵ 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 co-ordinates, 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 co-ordinates 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……… (n-1)
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 H-atom.

Distribution of electrons among the quantum levels

 

n	l	m	s	Designation of orbitals	Electrons present	Total no. of electrons
1 (K shell)	0	0	+1/2, –1/2	1s	2	2
2 (L shell)	0	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
• p-orbital 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 .

Z

• 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 2n²
  • The maximum capacity of a subshell is equal to 2(2l+1)
  • Number of sub-shells 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 sub-shell.
• 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.

 

6 p

5    4s                    4p                                                 4d                                                           4f

4

5 d

4 f

6 s

5 p

3s                    3p                                                 3d                                                                                                                4 d

5 s

3                                                                                                                                      4 p

• d
• s

2s                    2p                                                                                                                                                                     3 p

2

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 multi-electron 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 sub-shell 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

Half-filled and completely filled sub-shell 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

 

d⁴ (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

 

 

d⁵ (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 ns¹(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 ns¹(n – 1)d¹⁰ configuration instead of

3d5

ns ² (n – 1)d ⁹ due to extra stability of these atoms.

 

 

Cr (24)      [Ar]

 

4s¹

Cu (29)      [Ar]

4s¹

 

• 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

þ

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]3d¹⁰

(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 s-subshell, p-subshell, penultimate d-subshell and anti penultimate f-subshell 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 s-block 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 p-block elements having the configuration

ns ² np^1-6 ), the group number is the total number of electrons in the last shell plus 10.

• If the electrons are present in the (n –1)d orbital in addition to those in the ns orbital (i.e. for d-block elements having the configuration (n –1) d^1-10ns^1-2 ), the group number is equal to the total number of electrons present in the (n –1)d orbital and ns