# CHAPTER # 3 ATOMIC STRUCTURE

CHAPTER # 3 ATOMIC STRUCTURE
CHAPTER # 3

ATOMIC STRUCTURE

Introduction

The theory explains that matter was made from small indivisible particles called atoms. According to ancient era, this theory was put on a sound scientific basis by John Dalton in 1808.

Dalton’s Concept

In early 1800s, John Dalton proposed his theory that matter was composed of minute discrete particles called atoms (Greek = indivisible).

Recent Concept

Today, it is well established that atoms are complex organization of matter and energy. Many particles have been discovered within the atom. These sub atomic particles include electron, proton, neutron, several types of mesons, hyprons, etc.

Evidence for the presence of electrons, protons and neutrons in the atom is derived through the experiment.

Passage of Electricity through Solutions
In this experiment, Faraday’s passed electricity through an electrolytic solution. He observed that when two  metal plates (electrodes) are placed in an electrolytic solution and electricity is passed through this solution,  ions (present in this solution) move towards their respective electrodes. In other words, these ions move  towards the oppositely charged electrodes to give up their charge and liberate as a neutral particles.
Faraday also determined the charge of different ions and the amount of elements from the electrolytic  solution. Due to this experiment, presence of charged particle in the structure of atom is discovered. This basic unit of electric charge was later named as electron by Stoney in 1891.

Conclusion

There is some elementary unit of electric charge associated with these ions which can be calculated. The ions were observed to carry some integral multiple of this charge and the basic charge was later named by as an electron.

Example

Ionization of Electrolytic Solution

NaCl Current Na+ + Cl-

A+ anode (oxidation) (lose of particles)

2Cl- Cl2 + 2e-

At cathode (Reduction) (gain of particle)

2 Na+ + 2e- 2 Na

CROOK’S TUBE OR DISCHARGE TUBE EXPERIMENT
(Passage of electricity through gases under low pressure)

Introduction

The conduction of electricity through gas was first studied by GEISSLER followed by W-CROOKES in 1879. Electron was the first sub-atomic particles of an atom which was discovered in an electrical discharge tube called Crook’s Tube by J.J.Thomson in 1896.

Construction of the discharge tube

The discharge tube consists of long glass tube with circular electrodes sealed into each end. A small opening is made on its one side which is connected to a vacuum pump. The two electrodes are connected to the positive and negative terminal of a very high voltage battery.

Working of Discharge tube

At normal pressure (760mm Hg), a very high voltage about 4000 volt is connected to the tube. A sharp noise is heard by reducing pressure with the help of vacuum pump. The noise disappears, the gas attains a violet colour, then pink and finally whole tube begins to glow.

When pressure is lowered to about 1m.m. Hg, a dark space appears which is separated from cathode by a bluish glow at about 0.1mm. The rest of the glow in the tube is broken into several alternate bright and dark patches called “striations”. At further low pressure, the blue glow separates from cathode and a dark space appears in between called Crook’s Dark Space. At further low pressure, the Crook’s and Faraday’s dark space appear growing in size & the positive column shrinks, till at 0.001 mm Hg. The whole tube is filled with crook’s dark space and radiation called “cathode rays” emit from cathode electrode.

Nature of Cathode Rays
1. They travel in a straight line. This is proved by the fact that they produce a sharp shadow of opaque  substances placed in their path.
2. They emerge normally from cathode.
3. They can penetrate in small thickness of metals.
4. They can exert mechanical pressure showing they possess K.E.
5. They produce x-ray when they strike on thick metal plates.
6. They are deflected by electrostatic field.
7. Their deflection in electrostatic and magnetic field show that they carry negative charge. 8. They consist of stream of particles. This is proved by the fact that they rotate a light metallic  pinwheel placed in their path.
9. Their nature is independent of the nature of metal of electrode and the gas used in the tube.

Conclusion
Different discharge tubes with different electrodes and gases were tried by a number of workers besides  Thomsan. All the experiment gave the same value for the charge of the mass ratio (e/m). This shows that  electrons could be produced from any kind of matter and hence were a constituent’s part of all matter.

Discovery of Protons
(passage of electricity through at low pressure also resulted in the discovery of proton)
Logic for proton The atom is electrically neutral so when electrons were found to be generated from cathode in a discharge  tube, it was natural to know the positively charge particles were emitting from the anode

Goldstein’s experiment
Goldstein’s discovered that in the discharge tube not only the cathode rays are present but some other rays  are also traveling in a direction opposite to that of cathode rays.
He performed the experiment in 1886 by using a new modified tube. The cathode tube is used in this  experiment. Goldstein discovered that when two electrodes are connected to the two terminal of high voltage  (battery) and the tube is evacuated in order to reduce the pressure, cathode rays and positive rays are  produced. Positive rays travel towards the cathode and pass through the holes or canals to the other side of  the cathode.  Since these rays pass through the canals of the cathodes; hence, they are also called canal rays.

PROPERTIES OF CANAL RAYS
i) They travel in the straight path opposite to the cathode rays.
ii) They can produce sharp shadow of an opaque body that is placed behind the cathode in their  path.
iii) They cause mechanical motion in a small pinwheel is placed in their path.
iv) They can produce fluorescence when they strikes on the wall of the discharged tube. This  property is also observed when Zinc Sulphide plate is placed in their path.
v) These rays are deflected by the electric as well as magnetic field. They bend towards negative  pole in the electromagnetic field.
vi) The charge to mass ratio (e/m) of these rays depend upon the nature of the gas used in the  discharge tube. It is highest for hydrogen.
vii) The magnitude of charge of the canal rays was found to be 1.602 x 10-19 coulombs.  viii) The mass of these rays is virtually the same as that of atom.

Conclusion
These are the positive charged particles which he called PROTON and proton are the fundamental  particle of an atom.

(Confirmation of electron and proton)
Definition: The spontaneous and continuous emission of rays at constant rate by same elements is called Radioactivity. And, the elements are called Radioactive elements.

History The phenomena was first observed by Henry Becqueral while working on a mineral pitch blinde. Most of rays emitted by this mineral consisted of electron.

Later, Pierrie Curie and Marie curie isolated Radium, which was highly radioactive.

Rutherford used simple method for the detection and separation of this radiation. He placed a piece of radioactive substance (e.g. Ra) in a lead pot. The rays emitted by the radioactive substance were allowed to pass through two oppositely charged plates. The rays bending towards the negative plate carry +ve charge and are called Alpha (α). Those deflected towards the positive plate carry a - ve charge and are named as Beta (β) rays. Those rays which remained undeflected carry no charge and are called Gamma ray ( γ )

Result of Radioactivity: An element after giving out radiation breaks down to a more stable element, e.g. U on emission of α - particles would be converted to Th.

238U 234 Th + He4

The emission of radiation would continue until the formation of lead is produced which is more stable.

Conclusion
The evidence of radioactivity shows that the atom is not an indivisible particle. If it can emit electrons and Helium nuclei, it must have a substructure of its own.

PROPERTRIES OF ALPHA ( α) RAYS

i) Charge: They carry positive charge.
ii) Nature: They consist of fast moving helium nuclei in which α - particles consists of 2p and 2n. Hence, each has a charge of +2 units and mass of 4-a.m.u. Symbol for α - particles is 2He4.
iii) Velocity: Their velocity is in the range of C/10 to C/100, where C is speed of light.
iv) Range: They have very small range about 1 to 2cm in air 0.1mm. Aluminum absorbs them.
v) Ionization power: They have strong ionization power as they knock out electrons from the atoms  of gas through which they pass.
vi) Effect on photoplates: They block photoplates and produce discontinuous flouresence on striking  Zn-S screen.
PROPERTIES OF BETA PARTICLES
i) Charge: They carry negative charge.
ii) Nature: Beta rays consist of fast moving electrons. The charge is –1 unit, i.e. 1.6 x 10-19Coulomb.  Their mass is 1/1836 that of H atom. Symbol is β0.
iii) Velocity: Their velocity is tremendous. It reaches up to 98% of light.
vii) Range: They have larger range then alpha-ray. They penetrate thick Aluminum foil.  viii) Ionization power: Ionization power of β ray is 1000 times less then x-ray. This is due to their  small mass.
ix) Effect on photo plate: They affect protoplast more then x-ray and also produce florescence on  Zns screen.
PROPERTIES OF GAMMA RAYS
i) Charge: They carry no charge.
ii) Nature: Gamma rays particles have no range. They are electromagnetic radiation. They are also called very penetrating x-ray their is b/w 0.0001 Å and 0.1 Å.
iii) Velocity: Their velocity is equal to the velocity of light.
iv) Range: They are most penetrating. They can penetrate 15 to 20cm of head block.
v) Ionizing power: Their ionizing power is very small and 100 times less then the β-rays.  vi) Effect on photo plate: They blacken photo plate.

CHAD WICK EXPERIMENT (Discovery of Neutrons)
When a light element is bombarded by x-particles, these x-particles leave the nucleus in an unstable  disturbed state which on settling down to stable condition sends out radioactive rays. The phenomenon is  know as Artificial Radioactivity.
In 1933, Chadwick identified a new particle obtained from the bombardment of Beryllium by x-particles. It  had a unit mass and carried no charge. Its was name “Neutron”. The process for this reaction is given blow.

4Be9+2He4 6C12+0n1

With the help of artificial radioactivity, the particle “Neutron” was discovered.

PLANK’S QUANTUM THEORY
Introduction
This theory was proposed in 1901 by German physicist Max-plank to account for the radiation emitted by  hot bodies.
Postulate
According to this theory:
1. A hot body absorbs or emits energy discontinuously.
2. Energy is absorbed or emitted in specific amount called gamma or photons.
3. The amount of energy depends upon the frequency of radiation, i.e.
E α υ
E = h υ
h = Plank’s constant

Conclusion
Energy change occurs in small packets of energy h υ or multiple of those packet, i.e.  2h υ, 3h υ ………………..nh υ
It is called Quantization of Energy.

Spectroscopy
The branch of science that deals with the study of absorption or emission of electromagnetic radiations is  called spectroscopy.  The instrument used is called spectrometer. When radiation emitted or absorbed are passed through a prism  in a spectroscope, they are separated into component wave lengths to form an image on a photographic plate  which is called spectrum.
Types of spectra
Spectra are of two types depending upon light is emitted or absorbed by substances:
a) Emission Spectrum
b) Absorption Spectrum
Emission Spectrum
When an element absorbs sufficient energy from a flame or an electric arc, it emits radiation and spectra  obtained are called Emission spectrum.
Emission Spectra are of two types:
a) Continuous Spectrum
b) Line Spectrum

Continuous Spectrum

White light radiant energy coming from the sun or from any incandescent lamp, it is composed of light  waves in the range of 4000-8000 A0. Each of it has a characteristic colour. When a beam of white light is  passed through a prism, different wavelengths are bend through different angle. When received on a screen,  these form a continuous series of colour bend in the order of “VIBGYOR”. There is no line of demarcation  among different colours. This series of bands that form a continuous rainbow of colours is called continuous  spectrum.

Line Spectrum
When an element in the vapour or gaseous state is heated with a flame or in a discharge tube, the atoms are  excited and emit light radiations of a characteristic colour. If we pass these light radiations through prism and receive on a photographic plate, the Spectrum obtained is  found to be consisting of bright lines.  Such a spectrum in which each line represents a specific wave length of radiation emitted by the atom is  referred to as line Spectrum or atomic Spectrum.

Rutherford’s atomic model
(Evidence for nucleus and arrangement of particles.)
After discovery of electrons, protons and neutrons, there was question about their arrangements in the atom.  J.J Thomson in 1898 put forward a model of atom. According to him:
a) The atom is a spherical mass of positive charge and the electrons are embed in it at random.  b) The electrons are stationary.
c) There is no empty space in the atom.
d) The positive charge due to protons is unformal and spreaded throughout the atom.

Ruther ford’s Experiment
Rutherford performed an experiment in 1911 to test the validity of Thomson model of atom.  In actual experiment, he bombarded alpha particles of He on Al, Au or Ag. (0.0004cm thickness) and  observed the following points:
1. Most of Alpha particles passed through without deflection.
2. A few (1 in 8000) particles suffered a deflection greater than 90.
3. Rarely a particle came back or returned on the same path.

Assumption Drawn from this model
1) Atom has a tiny dense central core or the nucleus which contains practically the entire mass of the  atom leaving the rest of the atom almost empty.
2) The entire positive charge of the atom is located on the nucleus, while electrons were distributed in  vacant space around it.
3) The electrons move in orbits or closed circular paths around the nucleus like planets around the sun.  4) The greater part of the atomic volume comprises of empty space in which electrons revolve and spin.

Conclusion
1) Since most of the atom is empty, so most of alpha particles pass but some are deflected.  2) Small angle deflections are due to attraction between alpha particles and electrons.  3) Large angle deflections are due to repulsion between alpha particles and nucleus.

Defects of Rutherford Model (Weakness of Rutherford’s Theory)
There are two main defects in Rutherford’s atomic model as pointed out by  Neil Bohr in 1912:
1) Classical Mechanics
According to classical theory, any charged particle, if accelerated, must emit  energy. If the revolving electron emits energy, its energy would decrease and the radius of the orbit would become smaller. This would go on till the electron falls into nucleus. In actual practice, this does not happen. Rutherford’s theory could not explain this.
2) Spectroscopic Theory
An electron moving continuously around nucleus must radiate energy whose spectra should be continuous spectrum. But actually the spectrum obtained from the atom is discontinuous or line spectrum. Rutherford’s theory could not explain it as well.

BOHR’S ATOMIC THEORY
Introduction
This theory was put forward by Danish Scientist NEIL’S BOHR in 1913. The purpose was:
1. To remove the defects in Rutherford’s Atomic theory.
2. To explain the line spectrum of hydrogen.
Assumptions of Bohr’s theory
Bohr’s based his assumptions on plank’s Quantum theory. Followings are the assumptions of Bohr’s  Theory:
1. There exists the possibility that electron in certain orbits may not give out or absorb energy or radiation.
2. Electron revolving in any of such orbits is completely stable and its energy remains stable or constant. Such orbits are called stationary orbits. Since in a certain orbit, electron has fixed energy;  hence, Bohr’s orbits are also called the energy levels.
3. Electrons can jump from one energy state to another. If it jumps to higher energy state, it absorbs energy to equal to the energy difference (∆E) between the two orbits. When it falls back to lower  level, it emits energy difference (∆E).

If E2 = Energy of higher state.

E1= Energy of lower state.

Then ∆E = E2- E1

∆E = H υ

Where “h” is called plank’s constant. Each transition of electron leaves a spectrum.  4. Stationary states are only those orbits for which the product of momentum ( mν) and circumefrence ( 2πr ) is equal to plank’s constant (h) or integral multiple of (h) this product is also called “ THE  ACTION” Mathematically.

m v x 2 π r = nh

m v r = nh / 2π

(mvr is called Angular Momentum )

According to equation,

Only those stationary states or orbits are possible for which the angular momentum ( mvr ) is an integral multiple of h/2π .

“ n “ is called principal Quantum Number . Its values are n = 1,2,3,4,----

It is equal to number of orbit .

Expression for the Radius of BOHR’S ATOM

( Radius of nth orbit of an atom)

If we know the atomic number of an atom and orbit number (n), we can find the radius of nth orbit as  follows:

Derivation:

Let Z = atomic No

N = orbit No

E- charge of electron in nth orbit.

E+ = charge of proton in nucleus

Ze+ = total charge on nucleus

According to columb’s Law.

The electrostatic force between e- and nucleus will be

Fe = k ze+ . e- (Centripetal force which is denoted by Fe)   r2

The values of e+ and e’ are same = 1.6 x 10-19 columb.

Fe = k. Ze2

-------------

r2

There is another force in the atom which keeps the electron in their orbits, this is called centrifugal force  Which is denoted by Fc

Fc =2

For equation: Fe = Fc

KZe ² = ²

r² r

Kze² = mv²r2

r

Kze2 = mv2

r = k Ze2

mv2

r = K. Ze2 . 1

m v2

Now, we will find the velocity v. According to second postulate of Bohr:

mvr = n h

2π

Squaring both the sides

v2 = n2 h2

4π2 m2 r2

1 = 4π2 m2 v2

V2 n2 h2

Putting the value of 1 _ in equation (1), we get

v2

r =k Ze2 . 4 π2 m r2

m n2 h2

1/r = KZe2 . 4 π2mr2

m n2h2

1/r = KZe2. 4 π2m

n2h2

r = n2h2 (PROVED)

4 π2mKZe2

Radius of Bohr’s atom for nth orbit

Radius of 1st Orbit of Hydrogen atom

(r1 or x =1 )

For Hydrogen Atomic Number = xo = 1

For 1st Orbit n = 1

Putting in equation 2, we get

r1 = ao = h2 / 4 π m e2

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

ao = 0.529 A0

This a0 is the value of radius of 1st orbit of hydrogen atom, and is called Bohr’s radius.  Relation between r and a0

r = n2 h2 .

4 π2 m Ze2

But for hydrogen Z =1 and h2 / 4π2 m Ze2 = a0

r = n2 a0

This expression can be used to find radius of any orbit of hydrogen atom.  For 1st Orbit = n =1

2nd Orbit = n = 2

3rd Orbit = n = 3

“n” is number of orbit which is called Principal Quantum Number.

Calculate the radius of second orbit of hydrogen atom in Angstrom unit. Q) Bohr’s radius is 0.529 A0Calculate radius of 2nd and 3rd Orbit of hydrogen atom.

DATA

Bohr’s radius = a0 = 0.529 A0

For 2nd Orbit = n =2

3rd Orbit = h = 3

Solution

Now

R = n2 ao

For 2nd Orbit

r = (2)2 x 0.529

r = 4 x 0.529

r = 2.11 A0

For 3rd Orbit

r = (3)2 x 0.529

r = 9 x 0.529

r = 4.76 A0

Derivation of energy of electron in an orbit
Energy of an electron in an orbit is sum of KE and PE.

KE = ½ mv2

PE = - Ze2 / r

PE of electron is work done in bringing electrons from α to a distance r from nucleus. Now,

En = KE + PE

= ½ mv2 + ( -Ze2 / r)

= ½ mv2 - rZe2 --------------- eq 1

we know that

2 2

mv =

Ze

r

r

2

mv2 = rZe2

now subs the value of mv2 in eq (1)

E = 21 mv2 = rZe2

E = 21rZe2 = rZe2

Substituting the value of r = [ n2 h2 / 4 π2m Ze2]  In equation ( 1 ) we get,

E = - Ze2 = -2 π2m Z2e4

2[n2h2/4π2mZe2] n2 h2

Expression for frequency and Wave number

According to Bohr’s quantum theory when electron jumps from higher energy. Sate (E2) to lower energy state (E1) the energy difference is equal to the energy of photon emitted i.e.,

h ν = E2 – E1

h ν = -2 π2m Z2e4 – (-2 π2m Z2e4)

n22 h2 n12h2

h ν = -2 π2m Z2e4 + 2 π2m Z2e4

n22 h2 n12h2

h ν = 2 π2m Z2e4 + 2 π2m Z2e4)

n12 h2 n22h2

h ν = 2 π2m Z2e4 - 2 π2m Z2e4

n12 h2 n22h2

h ν = 2 π2m Z2e4 ( 1/n12 –1/n22)

h2

h ν = 2 π2m e4 Z4 ( 1/n12 –1/n22)

h2

ν = 2 π2m e4 Z2 ( 1/n12 –1/n22)

h3

This is the expression of frequency. Now ν = v . c

Where ν is wave number. It is no of wave/cm.

v . c = 2 π2m e4 Z2 ( 1/n12 –1/n22)

h3

v = 2 π2m e4 Z2( 1/n12 –1/n22)

ch3

Now for hydrogen z =1. on substituting the vale of constant is

2 π2m e4 , it was found to be equal to 109678cm-1. This value is denoted by RH is called Rydberg constant  ch3

now.

v = RH ( 1/n12 –1/n22) (PROVED)

Hydrogen Atom Specturm (proved)

Balmer Series

The simplest element is hydrogen which contains only one electron in its valence shell. Balmer studied the spectrum of hydrogen for this purpose in 1885. He used hydrogen gas in the discharge tube. Balmer observed that hydrogen atom spectrum consist of a series of lines called Balmer Series. Balmer determined the wave number of each of the line in the series and found that the series could be derived from a simple formula.

ν = RH ( 1/n12 –1/n22

where n1 = 2 and n2 = 3rd, 4th, 5th, etc.

and Rn is a constant called Rydberg Constant for hydrogen

RH = 109678 cm-1

Lyman Series

Lyman series is obtained when the electron returns to the ground state, i.e. from higher energy level to lower  energy level.

Where n2 = 2,3,4,5,6, etc.

This series of lines belongs to ultraviolet region of spectrum. It is represented by the following equation:  ν = RH ( 1/n12 –1/n22

Pasahen Series

Paschen series is obtained when the electron returns to the 3rd shell.

n = 3 from the higher energy levels.

n2 = 4, 5, 6…… etc.

This series belongs to the infrared region. The equation for Pasahen series may be written as:  ν = RH ( 1/32 –1/n22

n2 = 4,5,6, etc.

Bracket Series

This series is obtained when an electron jumps from a higher energy level to the 4th energy level. The  equation for this series:

V = RH ( 1/42 –1/n22

n2 = 5,6,7… etc.

P Fund Series

P-Fund series is obtained when an electron jumps from an energy level to 5th energy level.  V = RH ( 1/52 –1/n22

n = 6,7,8… etc.

Heisenberg’s Uncertainty Principle

According to Bohr’s theory, an electron was considered to be particle. However, electron also behaves like a wave according to de-Borglie. Due to this dual nature of electron, Heisenberg gave a principle known as Heisenberg uncertainty principle. It is stated as: It is impossible to calculate the position and momentum of a moving electron simultaneously. If one was known exactly, it would be impossible to know the other exactly. Therefore, if the uncertainty in the determination of momentum is px and the uncertainty is x, then according to this principle, the product of these two uncertainties may be written as:

Δ P x ΔX = h

So, if one of these quantities is known exactly, then the uncertainty in its determination is zero and the other uncertainty will become infinite, which is according to the principle.

Energy Levels and Sub Energy Levels

According to Bohr’s atomic theory, electrons revolve around the nucleus in circular orbits which are present at definite distance from the nucleus. These orbits are associated with definite energy of the electron increasing outwards from the nucleus. Hence, these orbits are referred to as “Energy Levels” and are designed as 1,2,3,4… etc., or L, M, N, etc.

The spectral line which corresponds to the transition of an electron from one energy level to another consists of several separate close lines as double triplets and so on. It indicates that some of the electrons of the given energy level have different energies or the electron belonging to some energy level may differ in their energy. So energy levels are accordingly divided into sub energy levels and are denoted by letters s, p, d, f (sharp, principal, diffuse and fundamental).

The numbers of sublevels in a given energy level or shell is equal to the value of n, e.g. in third shell where n = 3, three sublevels (s, p and) are possible.

The maximum number of electrons in each shell can be calculated by using formula: Total number of electrons in shell n = 2n2

Total numbers of electron in shell K = (n = 1) are 2 (1)2 = 2

Total numbers of electron in shell L = (n = 2) are 2 (2)2 = 8

Total numbers of electron in shell M = (n = 3) are 2 (3)2 = 18

Total numbers of electron in shell N = (n = 4) are 2 (4)2 = 32

The number of sub energy levels present in an energy level (orbit) are equal to its values of n.  K Shell ( n = 1) has only one orbit “s”.

L Shell ( n = 2) has only one orbit “s, p”.

M Shell ( n = 3) has only one orbit “s, p, d”.

N Shell ( n = 4) has only one orbit “s, p, d, f”.

Orbitals and Quantum Numbers

By using mathematical methods known as wave mechanics, Schrodingre in 1926, calculated the probability of locating the electrons in a region of space about the nucleus. Thus, on the basis of wave mechanics, the path of electrons cannot be probably determined. However, we can say that there are the regions in which probability of finding an electron is maximum. Such regions are called orbitals. Each orbital in an atom is completely described by four quantums.

Quantum numbers

• Principal quantum number

• Azimuthal quantum number or subsidiary quantum number

• Spin quantum number

• Magnetic quantum number

Principal Quantum Number (n)

Principal quantum number specifies the size of the orbital. It is denoted by symbol n. Its value may be 1, 2,  3, 4… etc., for respective orbitals. Therefore, the size (radius) of the orbital = 0.529 A.n2 (n = 1, 2, 3 …) and

energy of orbital n = En = 2 π2m e2 Z2

n2 h2

The energy level K, L, M, N, O, etc.

Corresponds to n = 1, 2, 3, 4, 5, … etc.

If n = 1 the electron is in K = shell

n = 2 the electrons is in L = shell

n = 3 the electrons is in M = shell

Azimuthal Quantum Number

It cannot measure the angular momentum of electron, as value of “l” increases, the angular momentum also increases. It depends upon the energy of electron. It also indicates the shape of the orbital. If n is known, then the value “ι”of can be calculated using formula (1 = n – 1)

If n = 1 l = 0 n - 1

l = 0 1 - 1

l = 0 0 = (s) circular (2, electrons)

If n = 2 l = 0 n - 1

l = 0 2 – 1

l = 0 1 – (dumbbell)

l = 0 0 = (s, p) circular (8 electron)

If n = 3 l = 0 n - 1

ι = 0 3 – 1

ι = 0 2

ι = 0,1,2 (s, p, d) circular (18 electron)  style="color: black; font-family: "Times New Roman","serif"; mso-fareast-font-family: "Times New Roman";">The energy of sub shell are in the order of s>p>d>f.

Magnetic Quantum Number (m)

Due to angular momentum of an electron, magnetic field is developed. This magnetism can be measured by magnetic quantum number. It depends upon the value of ‘l’ and gives different orientations of an orbital in space in applied magnetic field. It can be calculated by the formula

For –p –orbital we can calculate the Orientation for p – orbital = 1, where l = 1 then  m = -1 0 + 1

m = -1, 0, + 1

(px), (py), (pz) (Orientation)

d-orbital = l=2 then

m = -2 -1 0 + 1+2

m = -2, -1, 0, 1, 2

pz2 , px2y2 , pxy, pyz, pxz (Orientation)

Spin Quantum Number (s)

It specifies the spin of electron in an orbital. Clockwise spin is denoted by (+1/2) and anticlockwise is denoted by (-1/2).

Example

Spin of two electrons of He atom in s – orbital can be shown as above:

S = 1/2

Pauli’s Exclusion Principle

Introduction

When rules of Quantum mechanics are applied to an atom having maximum than one electron. This rule  puts a limit to the values of quantum numbers of an electron.

Definition

This rule was given by Wolf Gang Pauli in 1925. According to this rule, “In an atom, no two electrons can  have the same set of all four quantum number.”

Explanation

According to this rule, in an element, two atoms are not exactly alike. They must differ in some respect.  Thus, all maximum two electrons can have only three quantum numbers are same. The fourth one is always  different.

Example

Consider He (atomic number = 2)

Its configuration is 1s2

Because both electrons are in the first orbit, so for both n = 1

Because both electrons are in the s orbit, so for both l= 0

Because both electrons are in l = 0, so for both m = 0

Now, according to Plank’s principle, maximum three quantum can be same; and hence, the fourth quantum  must be different.

So for one electron = S = +1/2

For other electron = S = -1/2

They should have opposite spin.

Shapes of Orbitals

The shape of s orbital is spherical so the probability of finding the electrons is maximum because the electrons are uniformly distributed around the nucleus. The p – orbitals are dum-bel shaped
These are oriented in space along the three mutually perpendicular axis (x, y, z) and are called px, py, pz. These are degenerated orbitals of equal energy. Each p-orbital has two lobes, one of which is labeled (+) and the other (-).

Electronic Configuration

The distribution of electron in the available orbitals is proceeded according to the following rules: • Pauli’s Exclusion Principle

• Aufbau Principle

(n + 1) rule

The details of these rules and principles are given below:

Aufbau Principle

It states that, “the orbitals are filled up with electrons in the increasing order of their energy” starting with the 1st orbital. OR

Hypothetically, the electronic configuration of the atoms can be constructed by placing the electrons in the lowest available orbitals until the total number of electrons added is equal to the atomic number Z.

1s2

2s2 2p6

3s2 3p6 3d10

4s2 4p6 4d10 4f14

5s2 5p6 5d10

6s2 6p6

7s2

In order is :

1s2 , 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6

5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6s2, 7s2

n + 1 rule

According to this rule,

The orbital with the lowest value of (n + 1) is filled first. However, when the two orbitals have the same value of (n + 1), the orbital with the lower value of n is filled first.

The electronic configuration according to (n + 1) rule is stated as:

1s2 , 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6

5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6s2, 7s2

Example:

1s 2s 2p 3s 3p

n =1 n = 2 n = 2 n = 3 n = 3

l = 0 l = 0 l = 1 l = 0 l = 1

n+1=1 (n+1)=2 (n+1)=3 (n+1) = 3 (n+1) =4  Hund’s rule

When there are available orbitals of equal energy of electrons, they would go in separate particles having  same spin; rather then go in the same orbitals and have paired spin. In other words, the arrangement (12) is  more stable than (11) provided that these two orbitals are of equal energy.

Px Py Pz Px Py Pz

A B

According to Hund’s rule structure (A) is possible.

or

In other words, we can say that electrons are distributed in the orbitals of a sub shell in such a way as to give the same direction of spin.

Example:

16O8

Z = 8 = 1s2, 2s2, 2px, 2py, 2pz

14 N7  19 F9

Z = 7 = 1s2, 2s2, 2px, 2py, 2pz  Z = 9 = 1s2, 2s2, 2px, 2py, 2pz

Definition

It is defined as half of the distance between two homo nuclear atoms in diatomic molecules; such as, H2, O2,  N2, etc.

Range and units

angstrom (A)

Pi meter (pm)

Nanometer (n.m.)

The range of atomic radii is 0.7A0 to 2.9A0

1Ao = 10-10m = 100m = 0.1n.m

Explanation
In case of hetronuclear molecules; like AB, the bond length is calculated which is (rA + rB) and if radii of any one is known, the radii of the other can be calculated For the elements present in the periodic table, the atomic radius decreases from left to right due to more  attraction on the valence shell electron, but it increases down the group with increase of number of shell.

Definition

Ionic radius is defined as the distance between the nucleus of an ion and the point up to which the nucleus has influence on its electron cloud.

Explanation

When an electron is removed from a neutral atom, the atom is left with an excess of positive charge called a cation.

Example

Na Na+ + e-

Mg Mg+2 + 2e

But when an electron is added in a neutral atom, a negative ion or anion is formed.  Cl’ + e′ ⮴ Cl

O + 2e′ ⮴ O-2

Cations have smaller ionic radii than parent neutral atoms or K+1 is smaller than K0

This is because due to loss of electrons.

• The effective nuclear charge increases, pulling the remaining electrons more strongly.
• Generally, valence shell is lost, i.e. number of orbits decreases.

Example

Li -e Li+

Li > Li+

3A0 1.2A0

Anions have larger ionic radii than neutral atom
This is due to gain of electrons. The repulsion between the valence electrons increases which results in the increase in the size of –ve ion as compared to parent atom. Thus,

F -e+ F-

1.3A0 2.7A0 F < F -

Ionization Potential

The energy required to completely remove the most loosely bounded electron from an atom in gaseous state is the energy of that atom.

or

The energy required to remove the most loosely held electrons from an isolated gaseous atom in its ground state is called ionization potential.

or
It is defined as the minimum energy required to remove the least strongly bounded electron from a neutral gaseous atom, or ion or molecule. It is also called ionization energy.

Unit

Its units are:

Kj/mole

Electron volt (ev)

Example:

To remove an electron of H atom, we require 1312K.J mole-1 energy, so it is the I.P of H-atom.   H 1. P H+ + e-

1312Kj/mole

Factors affecting I.P:

It is effected by,

Atomic Radius : Greater the atomic radius, the less is the I.P values

Nuclear Charge: Greater the nuclear charge, greater the I.P values

I. P Nuclear Charge

Shielding effect: Greater the shielding effect, less the value of I.P

Types of Ionization potential

I.P is of following three types:

1st Ionization potential

The energy requires to remove the first electron from the neutral gaseous atom is called 1st ionization potential.

M (g) M+ (g) + e/

Electron Affinity

Definition

The amount of energy liberated by an atom when an electron is added in it is called electron affinity. or

The energy released when an electron is added to a gaseous atom to produce gaseous negative ionis called electron affinity.

It shows that this process is an exothermic change which is represented as:

Cl- + e′ ⮴ Cl- ΔH = -348kj/mole

Unit

The unit of electron affinity is kilo-joule per mole.

Explanation

when a neutral gaseous atom gains an electron, it looses some energy. Therefore, the electron is generally exothermic.

O-1 + e′ ⮴ O-2 Δ E = -141kj/mole

But when a second electron is added in the uni negative ion, then the incoming electron is repelled by the negative ion and the energy is absorbed in this process.

O + e′ ⮴ O Δ E = +844kj/mole

Hence, the first electron affinity is exothermic, but the successive electron affinities are endothermic.

Factors Affecting Electron Affinity

The magnitude of electron affinity depends on the following factors:

Atomic Size: With the increase in atomic size electron affinity decreases.

Nuclear Charge: Electron affinity value increases with the increase in nuclear charge. Shielding effect: With the increase in shielding effect, the electron affinity value increase.

Electron affinity and Periodic table

In the periodic table the electron affinity value increases from left to right in a period and decreases from top to bottom in a group.

Electron Affinity period Increasing

Group

Decreasing

Electronegativity

The electronegativity of an element is a measure of the attraction that an atom has for the shared electron in combined state.

The ability of an atom to attract shared electrons to itself is called electronegativity. or

The tendency of the bonded atom in a molecule to attract shared pair of electron is called electronegativity.
Explanation

When an atom forms a covalent bond with the other atom, then a pair of electron is shared between the atoms. Now each atom tends to acquire this shared pair of electrons. Therefore, both the atoms exert a force to attract the electron pair. This is the electronegativity of the atoms.

Factors Affecting Electronegativity

The magnitude of electronegativity depends on the following factors:

Atomic Size: With the increase in atomic size Electronegativity decrease.

Nuclear Charge: Increases the nuclear charge increases the Electronegativity.

Shielding Effect: Increases the shielding effect, the Electronegativity value increases from left to right in a period and decreases from top to bottom in a group.

Electronegativity and Periodic Table

In the periodic table, the electronegativity value increases from left to right in a period and decreases from top to bottom in a group.

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