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STRUCTURE OF ATOM II

An Ideal Learner’s school of chemistry 1

STRUCTURE OF ATOM Discovery of Electron:

A cathode ray tube is made of glass containing two electrodes, and a gas kept at very low pressures and at very high

voltages. When sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of

particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were

called cathode rays or cathode ray particles. The flow of current from cathode to anode was further checked by

making a hole in the anode and coating the tube behind anode with phosphorescent material zinc sulphide. When

these rays, after passing through anode, strike the zinc sulphide coating, a bright spot on the coating is developed.

In the absence of electrical or magnetic field,

these rays travel in straight lines . While in the

presence of electrical or magnetic field, it deflected

towards the positive plate of electric field, suggesting

that the cathode rays consist of negatively charged

particles, called electrons. The characteristics of

cathode rays (electrons) do not depend upon the

material of electrodes and the nature of the gas

present in the cathode ray tube. From this it is clear

that electrons are basic constituent of all the atoms.

Charge to Mass Ratio of Electron

J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (me) by using cathode ray tube and

applying electrical and magnetic field perpendicular to each other as well as to the path of electrons. Thomson

argued that the amount of deviation of the particles from their path in the presence of electrical or magnetic field

depends upon:

(i) The magnitude of the negative charge on the particle-- greater the magnitude of the charge on the particle, greater

is the interaction with the electric or magnetic field and thus greater is the deflection.

(ii) The mass of the particle — lighter the particle, greater the deflection.

(iii) The strength of the electrical or magnetic field — the deflection of electrons from its original path increases

with the increase in the voltage across the electrodes, or the strength of the magnetic field.

By carrying out accurate measurements on the amount of deflections observed by the electrons on the electric field

strength or magnetic field strength, Thomson was able to determine the value of e/me as:

Where me is the mass of the electron in kg and e is the magnitude of the charge on the electron in coulomb (C).

Since electrons are negatively charged, the charge on electron is –e.

Charge on the Electron

Based on oil drop experiment R.A. Millikan determined the charge on the electrons as – 1.6 × 10–19

C. The present

accepted value of electrical charge is – 1.6022 × 10–19

C. Thus the mass of the electron is 9.1094×10–31

kg

Discovery of Protons and Neutrons

Goldstein, working on a discharge tube with a perforated cathode and at a low pressure, observed a new type of rays

streaming behind the cathode. These rays were named anode rays or canal rays. Further investigations of these rays

showed that they consist of positively charged material particles and they depend upon the nature of gas present in

the cathode ray tube. Also the charge to mass ratio of the particles is found to depends upon the nature of the gas

taken in the discharge tube.

It was observed that e/me ratio was maximum when hydrogen gas was taken in the discharge tube. This shows

that positive ion formed from hydrogen is the lightest. These positively charged particles are called protons. Their

charge is 1.6022x10-19

C and mass is 1.67x10-27

Kg

Neutrons were discovered by Chadwick (1932) by bombarding a thin sheet of beryllium by α -particles.

When electrically neutral particles having a mass slightly greater than that of the protons was emitted. He named

these particles as neutrons.



ATOMIC MODELS

Thomson Model of Atom (plum pudding model)

J. J. Thomson proposed that an atom possesses a spherical shape in which the positive charge is uniformly

distributed. The electrons are embedded into it in such a manner that the atom as a whole is electrically neutral. An

important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom.

Although this model was able to explain the overall neutrality of the atom, but was not consistent with the results of

later experiments.

Rutherford’s Nuclear Model of Atom

In Rutherford’s α–particle scattering experiment,a stream of high energy α–particles from a radioactive source was

directed at a thin foil of gold metal. The thin gold foil had a circular fluorescent zinc sulphide scree n around it.

Whenever α–particles struck the screen, a tiny flash of light was produced at that point.

The results of scattering experiment were,

(i) most of the α– particles passed through the gold foil undeflected.

(ii) a small fraction of the α–particles was deflected by small angles.

(iii) a very few α– particles bounced back, that is, were deflected by nearly 180°.

On the basis of these observations, Rutherford concluded that,

(i) Most of the space in the atom is empty as most of the α–particles passed through the foil undeflected.

(ii) A few positively charged α– particles were deflected. Thus positive charge has to be concentrated in a very small

volume that repelled and deflected the positively charged α– particles.

(iii) Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to

the total volume of the atom.

On the basis of above observations and conclusions, Ruther ford proposed the nuclear model of atom. According to

this model :

(i) The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This

very small portion of the atom was called nucleus by Rutherford.

(ii)The nucleus is surrounded by electrons that move around the nucleus with a very high speed in circular paths

called orbits.

(iii) Electrons and the nucleus are held together by electrostatic forces of attraction.

Thus, Rutherford’s model of atom resembles the solar system, with the nucleus in the place of the sun and the

electrons in the place of the planets.

Atomic Number and Mass Number

The number of protons present in the nucleus is equal to atomic number (Z). In order to keep the electrical

neutrality, the number of electrons in an atom is equal to the number of protons . Thus,

Atomic number (Z) = number of protons in the nucleus of an atom

= number of electrons in a nuetral atom

The mass of the nucleus is due to protons and neutrons. Protons and neutrons present in the nucleus are

collectively known as nucleons. The total number of nucleons is termed as mass number (A) of the atom. Thus,

Mass number (A) = number of protons (Z) + number of neutrons (n)

Isobars and Isotopes

Isobars are the atoms with same mass number but different atomic number. Example,

and .

Atoms with identical atomic number but different mass number are known as Isotopes. Example,

.

Here number of neutrons found to be different. Since chemical properties of atoms are controlled by the number of

electrons, all the isotopes of a given element show same chemical properties.

Drawbacks of Rutherford Model

According to the electromagnetic theory of Maxwell, charged particles when accelerated should emit

electromagnetic radiation. Therefore, an electron in an orbit will emit radiation, and move closer and closer to the

nucleus. The orbit will thus continue to shrink and electron would finally fall in the nucleus. But this does not

happen. Thus, the Rutherford model cannot explain the stability of an atom.



Also this model does not say nothing about the electronic structure of atoms i.e., how the electrons are distributed

around the nucleus and what are the energies of these electrons.

Electromagnetic Radiation

James Maxwell suggested that when electrically charged particle moves under acceleration, alternating electrical

and magnetic fields are produced and transmitted. These fields are transmitted in the forms of waves called

electromagnetic waves or electromagnetic radiation. This electric and magnetic fields are perpendicular to each

other and both are perpendicular to the direction of propagation of the wave.

Electromagnetic waves do not require medium and

can move in vacuum. The arrangement of visible as

well as invisible electromagnetic waves according to

the wave length or frequency is known as the

electromagnetic spectrum.

These radiations are characterised by the properties,

namely, frequency (ν ) and wavelength (λ).

Frequency (ν ) is defined as the number of waves that pass a given point in one second. Its unit in SI system is

Hertz (Hz)

Wavelength is the distance between two neighbouring crests or troughs of a wave. Unit is Angstrom or nanometer.

The other commonly used quantity is the wave number ( ̅). It is defined as the number of wavelengths per unit

length or it is the reciprocal of wavelength.

Planck’s Quantum Theory

According to this theory, radiant energy is emitted or absorbed not continuously but discountinuously in the form of

small packets of energy called quanta. Each quanta is associated with a definite amount of energy . In case of light,

the quantum of energy is often called photon. Max planck showed that, the amount of energy associated with a

quantum of radiation is proportional to the frequency of radiation. i.e. E α ν or E = hν

Where h is Planck’s constant ( 6.625x 10-34

Js). But ν=c/Λ

Therefore , E=

A body can emit or absorb energy only in the integral multiples of quantum, i.e. E= nhν Where n = 1,2,3,4……

Photoelectric Effect

H. Hertz in some of his famous experiments, observed that when light of a certain frequency strikes the surface of a

metal, electrons are ejected from the metal. The phenomenon of ejection of electrons from the surface of metal when

light of a suitable frequency strikes on it, is known as photoelectric effect. The ejected electrons are called

photoelectrons.

Characteristics of photoelectric effect:

1. For each metal, a certain minimum frequency of incident light is needed to eject electrons. This is known as

threshold frequency (ν0)

2. The kinetic energy of ejected electrons is independent of the intensity of the incident light but varies linearly with

its frequency.

3.The number of ejected electrons from the metal surface depends upon the intensity of the incident radiation. The

greater the intensity, the larger is the number of ejected electrons.

Einstein was explained the photoelectric effect using Planck’s quantum theory . According to this theory, when a

photon of light of frequency ν0 strikes an electron in a metal, it imparts its entire energy to the electron. This energy

enables the electrons to break away from the atom by overcoming the attractive influence of the nucleus. Thus each

photon can eject one electron.

Now suppose a photon of the light falling on a metal surface having frequency higher than the threshold frequency,

ν; some of its energy is consumed to separate the electron from the metal and the remaining energy will be imparted

to the ejected electrons to give it certain kinetic energy.



Then by quantum theory, hν= + ½ mv2 where is the binding energy of the electron known as work function or

threshold energy. Expressing in terms of threshold frequency, =h ν0

Thus , hν= h ν0+ ½ mv2 or ½ mv

2 = h(ν-ν0)

Dual Behaviour of Electromagnetic Radiation:

The particle nature of light could explain the black body radiation and photoelectric effect. But it was not consistent

with the known wave behaviour of light which could account for the phenomena of interference and diffraction. The

only way to resolve this problem was to accept the idea that light possesses both particle and wave-like properties,

i.e., light has dual behaviour. Whenever radiation interacts with matter, it displays particle like properties and it

exhibits wave like properties when it propagates.

Atomic spectra:

The speed of light depends upon the nature of the medium through which it passes. As a result, the beam of light is

deviated or refracted from its original path as it passes from one medium to another. Thus when a ray of white light

is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength.

Hence white light is spread out into a series of coloured bands called spectrum. The light of red colour which has longest wavelength is deviated the least while the violet light, which has shortest wavelength is deviated the most.

The spectrum of white light see from violet at 7.50 ×1014

Hz to red at 4×1014

Hz. Such a spectrum is called

continuous spectrum. When electromagnetic radiation interacts with matter, atoms and molecules may absorb

energy and reach to a higher energy state. With higher energy, these are in an unstable state. For returning to their

normal energy state, the atoms and molecules emit radiations in various regions of the electromagnetic spectrum.

Emission and Absorption Spectra:

The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. To

produce an emission spectrum, energy is supplied to a sample by heating it.

When a continuous radiation is passed through a gas or a solution of some salt and the transmitted light is analysed,

we obtain a spectrum in which dark lines are observed in an otherwise continuous spectrum. These dark lines

indicate that the radiations of corresponding wavelengths have been absorbed by the substance from the white light.

Such a spectrum containing a few dark lines due to absorption of light is known as absorption spectrum. The study

of emission or absorption spectra is referred to as spectroscopy.

Line Spectrum of Hydrogen

When an electric discharge is passed through gaseous hydrogen, the H2 molecules dissociate and the energetically

excited hydrogen atoms produced emit electromagnetic radiation of discrete frequencies. The hydrogen spectrum

consists of several series of lines named after their discoverers. These are Lyman series, Balmer series, Paschen

series, Bracket series, and Pfund series. Lyman series appear in the ultra violet region, Balmer series appear in the

visible region while the other three series lie in the infra red region.

Balmer showed that if spectral lines are expressed in terms of wavenumber ( ̅), then the visible lines of the

hydrogen spectrum obey the following formula :

Where n is an integer equal to or greater than 3

Later on Rydberg gave a more general formula which is applicable to all series in the hydrogen spectrum. That is;

Where n1=1,2........ n2= n1+ 1, n1+ 2...... The value 109,677 cm–1

is called the Rydberg constant for hydrogen.

For Lyman series : n1= 1 n2 = 2,3,4,…..

For Balmer series : n1= 2 n2 = 3,4,5,…….

For Paschen series : n1= 3 n2 = 4,5,6,…..

For Bracket series : n1= 4 n2 = 5,6,….

For Pfund series : n1= 5 n2 = 6,7,8,…



BOHR’S MODEL FOR HYDROGEN ATOM

Bohr’s model for hydrogen atom is based on the following postulates:

i) The electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy.

These paths are called orbits, stationary states or allowed energy states.

ii) The energy of an electron in the orbit does not change with time. However, the electron will move from a lower

stationary state to a higher stationary state when required amount of energy is absorbed by the electron or energy is

emitted when electron moves from higher stationary state to lower stationary state.

iii) The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ

in energy by ΔE, is given by :

Where E1and E2 are the energies of the lower and higher allowed energy states respectively.

This expression is commonly known as Bohr’s frequency rule.

iv) The angular momentum of an electron in a given stationary state is quantized.

That is ,angular momentum

According to Bohr’s theory for hydrogen atom:

a) The stationary states for electron are numbered n= 1,2,3... These integral numbers are known as Principal

quantum numbers.

b) The radii of the stationary states are expressed as: rn = n2a0 where a0 = 52.9 pm. Thus the radius of the first

stationary state, called the Bohr orbit, is 52.9 pm.

c) The energy of the stationary state is given by the expression,

When the electron is free from the influence of nucleus, the energy is taken as zero. The electron in this situation is

associated with the stationary state of Principal Quantum number n= ∞ When the electron is attracted by the

nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for negative

electronic energy.

d) Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen

atom.

Here ; and

e) It is also possible to calculate the velocities of electrons moving in these orbits and the magnitude of velocity of

electron increases with increase of positive charge on the nucleus and decreases with increase of principal quantum

number.

Explanation of Line Spectrum of Hydrogen:

Line spectrum of hydrogen atom can be explained by using Bohr’s model. Energy is absorbed if the electron moves

from the orbit of smaller Principal quantum number to the orbit of higher Principal quantum number, whereas the

energy is emitted if the electron moves from higher orbit to lower orbit. The energy gap between the two orbits is

given by equation,

ΔE= Ef– Ei Thus;

= Where ni and nf stand for initial orbit and final orbits .

But frequency,

In case of absorption spectrum, nf> ni and the term in the parenthesis is positive and energy is absorbed. On the

other hand in case of emission spectrum ni> nf, Δ E is negative and energy is released.

In case of large number of hydrogen atoms, different possible transitions can be observed and thus leading to large

number of spectral lines. The brightness or intensity of spectral lines depends upon the number of photons of same

wavelength or frequency absorbed or emitted.



Limitations of Bohr’s Model:

Bohr’s model account for the stability and line spectra of hydrogen atom and hydrogen like ions. However, Bohr’s

model was too simple to account for the following points.

i) It fails to account for the finer spectrum. This model unable to explain the spectrum of atoms other than hydrogen.

Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field

(Zeeman effect) or an electric field (Stark effect).

ii) It could not explain the ability of atoms to form molecules by chemical bonds.

QUANTUM MECHANICAL MODEL OF THE ATOM

Dual Behaviour of Matter:

de Broglie proposed that matter, like radiation, should also exhibit dual behaviour i.e., both particle and wave like

properties. This means that just as the photon has momentum as well as wavelength, electrons should also have

momentum as well as wavelength, de Broglie, from this analogy, gave the following relation between wavelength

(λ) and momentum (p) of a material particle

Where m is the mass of the particle, v its velocity and p its momentum.

de Broglie’s prediction was confirmed experimentally when it was found that an electron beam undergoes dif

fraction, a phenomenon characteristic of waves. This fact has been put to use in making an electron microscope.

According to de Broglie, every object in motion has a wave character. The wavelengths associated with ordinary

objects are so short that their wave properties cannot be detected. The wavelengths associated with electrons and

other subatomic particles can however be detected experimentally.

Heisenberg’s Uncertainty Principle:

Heisenberg’s a uncertainty principle states that it is impossible to determine simultaneously, the exact position and

exact momentum (or velocity) of a microscopic moving particles like electron. Mathematically, it can be given a;

or ( )

or

where Δx is the uncertainty in position and Δp is the uncertainty in momentum , Δv is the uncertainty in velocity of

the particle.

Quantum mechanics:

Quantum mechanics is a theoretical science that deals with the study of the motions of the microscopic objects that

have both observable wave like and particle like properties. Quantum mechanics was developed independently by

Werner Heisenberg and Erwin Schrödinger.

Quantum Mechanical Model of Atom:

Based on de-Broglie’s idea of matter wave and uncertainty principle a new model of atom was developed. Called

quantum mechanical model. Here behavior of an electron in an atom is given by an equation known as Schrodinger

wave equation.

i.e;

+

+

+

( E-V ) = 0

where x,y and z are Cartesian co-ordinates, h- Plank’s constant, m,V – mass and potential energy of electron, E-

total energy and ψ is the amplitude of the electron wave called wave function.

On solving Schrodinger equation we get several solutions and these solutions are called wave functions. Among

these wave functions some of them are physically acceptable and they are known as ‘eigen functions’. Each eigen

function represents a possible state of electron in the atom and is called an orbital wave function or simply orbital.

And an orbital may be defined as the ‘ region of space around the nucleus where the probability of finding the

electron is maximum.

Each orbital has a definite amount of energy, hence it is quantized. And the energy of an electron is given by;

En = -2п2me

4/ n

2h

2 where m and e – mass and energy of electron; n- principal quantum number.



Probability Density ( ψ2) :

Orbital wave function ψ itself has no physical

significance, it simply represents the amplitude of

electron wave. But its square ψ2 is a measure of the

probability of finding an electron in unit volume of

space around the nucleus of an atom. Hence ψ2 is

called probability density.

A graph of ψ2 as a function of distance from the

nucleus is called probability density curve.

Probability density curve for ‘1s’ and ‘2s’ orbitals are

given below. From the graph it is clear that ψ2 is

maximum at the nucleus and decreases with increase

in distance from nucleus and the probability of

finding the electron does not become zero even at

very large distance from the nucleus.

Concept of orbitals:

When the schrodinger equation for the electron in a hydrogen atom is solved, we get several solutions. These

solutions are called wave functions. Only some of these wave functions are physically acceptable. These acceptable

wave functions are known as eigen functions. Each eigen function represents a possible state of electron in the atom

and is called an orbital wave function or simply orbital. And an orbital may be defined as the region of space around

the nucleus where the probability of finding the electron is maximum.

Quantum Numbers:

Quantum numbers are identification numbers for an individual electron in an atom and describe the position and

energy of an electron in an atom. There are four quantum numbers. They are:

1. Principal Quantum number (n): It determines the size and energy of the shell and is denoted by ‘n’ having

the values 1,2,3,4,.. corresponds to K,L,M,N,… energy and size increases with increasing the value of ‘n’.

it also determines the average distance of an electron from the nucleus.

2. Azimuthal Quantum Number (l) : It determines the shape of orbitals and is denoted by ‘l’ having the

values from 0 to (n-1). Each value of ‘l’ represents a particular sub shell. These levels are represented as

s,p,d,f,.. according to the value of ‘l’ is 0,1,2, 3 etc.

3. Magnetic Quantum number (m): the number of orbitals in a given energy sublevel within a principal

energy level is given by the magnetic quantum number and is denoted by ‘m’, having the values from –l to

+l including 0. Thus the total of (2l+1) values. Each value of ‘m’ represents a particular orbital. For

example if l=0, m=0 i.e. s- subshell can have only one orbital.

4. Spin Quantum Number(s): It describes the spin of the electron around its own axis and is denoted by ‘s’,

having two values, i.e. + ½ or – ½ . Two electrons in the same orbital have opposite spin. i.e.

Shape of ‘S’-orbital:

‘s’ orbitals are non directional and spherically symmetrical. Ie.,the probability of finding the electron is the same in

all directions. The density of charge cloud is maximum at the nucleus and becomes small at large distance. In the

case of 2s orbitals there is a region where the probability of finding the electron is zero. This is called a ‘node’ or

nodal surface. It has been found that ns-orbital has (n– 1) nodes.

Shape of ‘P’-orbitals:

There are three ‘P’ orbitals in P-subshell namely ,Px,Py and Pz. Each P- orbital consists of two lobes about a

particular axis. Hence they posses dump-bell shape.The probability density functions for p-orbital also pass through

value zero. The number of nodes are given by (n –l-1). Besides the radial nodes, the probability density functions

for p-orbitals are zero at the plane (s), passing through the nucleus (origin). These are called angular nodes or ‘nodal

plane’. Number of angular nodes is given by ‘l’. Thus the total number of nodes are given by (n–1), i.e., sum of l

angular nodes and (n –l – 1) radial nodes.



Shape of ‘d’-orbitals:

There are five ‘d’-orbitals in d- sub shell namely, dxy, dxz,dyz,dx2-y

2 and dz

2 .The dxy,dxz dyz and dx

2-y

2 are

double dumb bell shaped. The dz2 orbital has a dumb bell shaped curve about the z-axis havng a circular collar in

xy plane. They have l angular nodes and (n-l-1) radial nodes.

Aufbau Principle:

According to this principle, electrons are filled in various orbitals in the increasing order of the orbital energies. The

orbital with a lower energy is filled first before the filling of the orbitals of higher energy occurs. The energy

sequence of orbitals is decided by the following two rules.

1. Orbitals are filled in the order of increasing (n+l) values. This means that between 3d and 4s orbitals, the 4s

orbital is filled before 3d orbitals.

2. If two orbitals have the same (n+l), the one with lower n will be filled first. Thus between 2p and 3s , 2p

will be filled before 3s.

Based on these, the orbitals can be arranged in the following order.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s...

Pauli’s Exclusion Principle:

This principle states that, ‘No two electrons in an atom can have the same values for all the four quantum numbers’.

Thus in any atom two electrons may have the same values for three quantum numbers but the fourth must be

different.

Thus an orbital can have a maximum of two electrons and those posses opposite spin. Hence this rule can also be

defined as ‘no orbital can accommodate more than two electrons and these two must have opposite spin’.

Hunds Rule of Maximum Multiplicity:

According to this rule electron pairing will not takes place in orbitals of same energy until each orbital is first singly

filled. This suggests that it is difficult for an electron to enter an orbital which already has an electron than to enter

an unoccupied orbital of the same energy.

Stability of half filled and completely filled orbitals:

The extra stability of half filled and completely filled configurations is due to the symmetry of orbitals and the large

exchange energy of electrons. Exchange energy is the energy decrease in the change of position of electrons from

one orbital to another of the same sub-level. In the case of d5 and d

10; more electrons can exchange their positions

than in d4 and d

9 configurations. The greater exchange of half filled and completely filled configurations gives them

more stability.



Q :1(i) Calculate the total number of electrons present in one mole of methane.

(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14

C.

(Assume that mass of a neutron = 1.675 × 10–27

kg).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP. Will the answer change

if the temperature and pressure are changed?

(i) Number of electrons present in 1 molecule of methane (CH4)

{1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 × 1023

molecules of methane

= 6.022 × 1023

× 10 = 6.022 × 1024

(ii) (a) Number of atoms of 14

C in 1 mole= 6.023 × 1023

Since 1 atom of 14

C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14

C is (6.023 × 1023

) ×8.

Or, 14 g of 14

C contains (6.022 × 1023

× 8) neutrons.

Number of neutrons in 7 mg

= 2.4092 × 10

21

(b) Mass of one neutron = 1.67493 × 10–27

kg

Mass of total neutrons in 7 g of 14

C = (2.4092 × 1021

) (1.67493 × 10–27

kg) = 4.0352 × 10–6

kg

(iii) (a) 1 mole of NH3 = {1(14) + 3(1)} g of NH3 = 17 g of NH3 = 6.022× 1023

molecules of NH3

Total number of protons present in 1 molecule of NH3 = {1(7) + 3(1)} = 10

Number of protons in 6.023 × 1023

molecules of NH3 = (6.023 × 1023

) (10) = 6.023 × 1024

⇒ 17 g of NH3 contains (6.023 × 1024

) protons.

Number of protons in 34 mg of NH3

= 1.2046 × 1022

(b) Mass of one proton = 1.67493 × 10–27

kg

Total mass of protons in 34 mg of NH3 = (1.67493 × 10–27

kg) (1.2046 × 1022

) = 2.0176 × 10–5

kg

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions.

Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Q:2: Find energy of each of the photons which (i) correspond to light of frequency 3× 1015

Hz.

(ii) have wavelength of 0.50 Å.

(i) Energy (E) of a photon is given by the expression, E = hʋ

Where, h = Planck’s constant = 6.626 × 10–34

Js ν = frequency of light = 3 × 1015

Hz

Substituting the values in the given expression of E:

E = (6.626 × 10–34

) (3 × 1015

)

E = 1.988 × 10–18

J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

h = Planck’s constant = 6.626 × 10

–34 Js c = velocity of light in vacuum = 3 × 10

8 m/s


Q.3:What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Energy (E) of a photon = hν

Energy (En) of ‘n’ photons = nhν



Where, λ = wavelength of light = 4000 pm = 4000 ×10–12

m c = velocity of light in vacuum = 3 × 108

m/s

h = Planck’s constant = 6.626 × 10–34

Js

Substituting the values in the given expression of n:

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 10

16.

Q.4:What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an

energy level with n = 4 to an energy level with n = 2?

The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the

transition is given by the relation,


E = – (4.0875 × 10–19

J)

The negative sign indicates the energy of emission.

Wavelength of light emitted

Substituting the values in the given expression of λ:

Q.5:What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the

ground state?

A total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the nth

level drops down to the ground state is given by

.

Given, n = 6

Number of spectral lines = 15

Q.6:(i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18

J atom–1

. What is the

energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

E5 = –8.72 × 10

–20 J

(ii) Radius of Bohr’s nth

orbit for hydrogen atom is given by, rn = (0.0529 nm) n2

For, n = 5 r5 = (0.0529 nm) (5)2

Thus , r5 = 1.3225 nm



Q.7:Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms

–1.

According to de Broglie’s equation,

Where, λ = wavelength of moving particle m = mass of particle v = velocity of particle h = Planck’s constant

Substituting the values in the expression of λ:

Hence, the wavelength of the electron moving with a velocity of 2.05 × 10

7 ms

–1 is 3.548 × 10

–11 m.

Q.8:Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na+, K

+, Mg

2+, Ca

2+, S

2–, Ar

Isoelectronic species have the same number of electrons.

Number of electrons in sodium (Na) = 11

Number of electrons in (Na+) = 10

A positive charge denotes the loss of an electron.

Similarly,

Number of electrons in K+ = 18

Number of electrons in Mg2+

= 10

Number of electrons in Ca2+

= 18

A negative charge denotes the gain of an electron by a species.

Number of electrons in sulphur (S) = 16

∴ Number of electrons in S2-

= 18

Number of electrons in argon (Ar) = 18

Hence, the following are isoelectronic species:

1) Na+ and Mg

2+ (10 electrons each)

2) K+, Ca

2+, S

2– and Ar (18 electrons each)

Q.9:(i) Write the electronic configurations of the following ions: (a) H–

(b) Na+ (c) O

2–(d) F

–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3 s1

(b) 2p3 and

(c) 3p5?

(iii) Which atoms are indicated by the following configurations?

(a) [He] 2s1 (b) [Ne] 3s

2 3p

3 (c) [Ar] 4s

2 3d

1.

(i) (a) H– ion

The electronic configuration of H atom is 1s1.

A negative charge on the species indicates the gain of an electron by it.

∴ Electronic configuration of H– = 1s

2

(b) Na+

ion

The electronic configuration of Na atom is 1s2 2s

2 2p

6 3s

1.

A positive charge on the species indicates the loss of an electron by it.

∴ Electronic configuration of Na+ = 1s

2 2s

2 2p

6 3s

0 or 1s

2 2s

2 2p

6

(c) O2–

ion

The electronic configuration of 0 atom is 1s2 2s

2 2p

4.

A dinegative charge on the species indicates that two electrons are gained by it.

∴ Electronic configuration of O2–

ion = 1s2 2s

2 p

6

(d) F– ion

The electronic configuration of F atom is 1s2 2s

2 2p

5.

A negative charge on the species indicates the gain of an electron by it.

∴ Electron configuration of F– ion = 1s

2 2s

2 2p

6

(ii) (a) 3s1

Completing the electron configuration of the element as 1s2 2s

2 2p

6 3s

1.

∴ Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11

(b) 2p3


2 2p

3.



∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7


(c) 3p5


2 2p

5.

∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

∴ Atomic number of the element = 9 (iii) (a) [He] 2s

1

The electronic configuration of the element is [He] 2s1 = 1s

2 2s

1.


Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).

(b) [Ne] 3s2 3p

3

The electronic configuration of the element is [Ne] 3s2 3p

3= 1s

2 2s

2 2p

6 3s

2 3p

3.


Hence, the element with the electronic configuration [Ne] 3s2 3p

3 is phosphorus (P).

(c) [Ar] 4s2 3d

1

The electronic configuration of the element is [Ar] 4s2 3d

1= 1s

2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

1.


Hence, the element with the electronic configuration [Ar] 4s2 3d

1 is scandium (Sc).

Q.10:What is the lowest value of n that allows g orbitals to exist?

For g-orbitals, l = 4.

As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to

(n – 1).

∴ For l = 4, minimum value of n = 5

Q.11: An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

For the 3d orbital:

Principal quantum number (n) = 3 Azimuthal quantum number (l) = 2

Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2 Q.12:An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the

electronic configuration of the element.

(i) For an atom to be neutral, the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is 1s2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10.

Q.13: How many electrons in an atom may have the following quantum numbers?(a) n = 4, ms= -1/2

(b) n = 3, l = 0

(a) Total number of electrons in an atom for a value of n = 2n2

∴ For n = 4,

Total number of electrons = 2 (4)2

= 32

The given element has a fully filled orbital as 1s2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10.

Hence, all the electrons are paired.

∴ Number of electrons (having n = 4 and ms = -1/2) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital.

Therefore, the number of electrons having n = 3 and l = 0 is 2.

1 . Which of the following conclusions could not be

derived from Rutherford’s α -particle scattering

experiement?

(i) Most of the space in the atom is empty.

(ii) The radius of the atom is about 10 –10

m while

that of nucleus is 10–15

m.

(iii) Electrons move in a circular path of fixed energy

called orbits.

(iv) Electrons and the nucleus are held together by

electrostatic forces of attraction.

2 . Which of the following options does not represent

ground state electronic configuration of an atom?

(i) 1s2 2s

2 2p

6 3s

2 3p

6 3d

8 4s

2

(ii) 1s2 2s

2 2p

6 3s

2 3p

6 3d

9 4s

2

(iii) 1s2 2s

2 2p

6 3s

2 3p

6 3d

10 4s

1

(iv) 1s2 2s

2 2p

6 3s

2 3p

6 3d

5 4s

1

3 . Which of the following statement is not correct

about the characteristics of cathode rays?

(i) They start from the cathode and move towards the

anode.

(ii) They travel in straight line in the absence of an

external electrical or magnetic field.

(iii) Characteristics of cathode rays do not depend

upon the material of electrodes in cathode ray tube.

(iv) Characteristics of cathode rays depend upon the

nature of gas present in the cathode ray tube.



4. Which of the following statements about the

electron is incorrect?

(i) It is a negatively charged particle.

(ii) The mass of electron is equal to the mass of

neutron.

(iii) It is a basic constituent of all atoms.

(iv) It is a constituent of cathode rays.

5. Which of the following properties of atom could

be explained correctly by Thomson Model of atom?

(i) Overall neutrality of atom.

(ii) Spectra of hydrogen atom.

(iii) Position of electrons, protons and neutrons in

atom.

(iv) Stability of atom.

6. Two atoms are said to be isobars if.

(i) they have same atomic number but different mass

number.

(ii) they have same number of electrons but different

number of neutrons.

(iii) they have same number of neutrons but different

number of electrons.

(iv) sum of the number of protons and neutrons is

same but the number of protons is different.

7. The number of radial nodes for 3 p orbital is ___.

(i) 3 (ii) 4 (iii) 2 (iv) 1

8. Number of angular nodes for 4d orbital is _____.

(i) 4 (ii) 3 (iii) 2 (iv) 1

9. Which of the following is responsible to rule out

the existence of definite paths or trajectories of

electrons?

(i) Pauli’s exclusion principle.

(ii) Heisenberg’s uncertainty principle.

(iii) Hund’s rule of maximum multiplicity.

(iv) Aufbau principle.

10. Total number of orbitals associated with third

shell will be __________.

(i) 2 (ii) 4 (iii) 9 (iv) 3

11. Orbital angular momentum depends on _______.

(i) l (ii) n and l (iii) n and m (iv) m and s

12. Chlorine exists in two isotopic forms, Cl-37 and

Cl-35 but its atomic mass is 35.5. This indicates the

ratio of Cl-37 and Cl-35 is approximately

(i) 1:2 (ii) 1:1 (iii) 1:3 (iv) 3:1

13. The pair of ions having same electronic

configuration is __________.

(i) Cr 3+

, Fe 3+

(ii) Fe 3+

, Mn2+

(iii) Fe 3+

, Co 3+

(iv) Sc 3+

, Cr 3+

14. For the electrons of oxygen atom, which of the

following statements is correct?

(i) Z eff for an electron in a 2s orbital is the same as

Z eff for an electron in a 2p orbital.

(ii) An electron in the 2s orbital has the same energy

as an electron in the 2p orbital.

(iii) Z eff for an electron in 1s orbital is the same as Z

eff for an electron in a 2s orbital.

(iv) The two electrons present in the 2 s orbital have

spin quantum numbers ms but of opposite sign.

15. If travelling at same speeds, which of the

following matter waves have the shortest

wavelength?

(i) Electron (ii) Alpha particle (He 2+ )

(iii) Neutron (iv) Proton

16. Which of the following sets of quantum numbers

are correct?

n l m

(i) 1 1 +2

(ii) 2 1 +1

(iii) 3 2 –2

(iv) 3 4 –2

17. In which of the following pairs, the ions are iso-

electronic?

(i) Na + , Mg

2+ (ii) Al

3+ , O

–

(iii) Na+ , O

2– (iv) N

3– , Cl

–

14. Neutrons can be found in all atomic nuclei except in one case. Which is this atomic nucleus and what does it

consists of?

Ans. Hydrogen atom. It consists of only one proton.

15. Calculate wave number of yellow radiations having wavelength of 5800 A0 .

Ans. Wave number = 1/ wavelength

Wavelength = 5800 A0 = 5800 x 10

-10 m Wave number = 1/5800 x 10

-10 m = 1.72 x 10

6 m

-1

16. What are the values of n and l for 2p orbital?

Ans. n=2 and l= 1

17. Which of the following orbitals are not possible? 1p, 2s, 3f and 4d

Ans. 1p and 3f are not possible.

18. What atoms are indicated by the following electronic configurations?

a. 1s2 2s2 2p

1 b. [Ar]4s

2 3d

1

Ans. a. Boron b. Scandium

19. What is the relationship between frequency and wavelength of light?

Ans. velocity of light = frequency x wavelength. Frequency and wavelength are inversely proportional to each other.



20. What is the difference between a quantum and a photon?

Ans. The smallest packet of energy of any radiation is called a quantum whereas that of light is called photon.

21. Write the complete symbol for the atom with the given atomic number (Z) and mass number (A).

(a) Z = 17, A = 35 (b) Z = 92 , A = 233

Ans. (a) 35

17 Cl (b) 233

92U

22. Using s,p,d and f notation, describe the orbital with the following quantum numbers-

(a) n=1,l=0 (b) n=3, l=1 (c) n=4, l=2 (d) n=4, l=3

Ans. (a) 1s (b) 3p (c)4d (d) 4f

23. How many electrons in an atom have the following quantum numbers?

a. n=4, s = -1/2 b. n =3 , l=o

Ans. (a) 16 electrons (b) 2 electrons.

24. An element with mass number 81 contains 31.7 % more neutrons as compared to protons. Assign the atomic

symbol.

Ans. Mass number = 81, i.e., p + n = 81

If protons = x, then neutrons = x + 31.7 X x = 1.317 x

100

x+1.317x = 81 or 2.317x = 81

x=35

Thus proton = 35, i.e., atomic no. = 35

Hence symbol is 81

35 Br

25. (i) The energy associated with the first orbit in the hydrogen atom is -2.18 x10 -18

J/atom.

What is the energy associated with the fifth orbit

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Ans. (i) En = -2.18 x 10 -18

/ n2 E5 = -2.18 x 10

-18 / 5

2 = -8.72 x 10

-20 J

(ii) For H atom, r n = 0.529 x n 2 r5 = 52.9 x 5

2 = 1322.5 pm = 1.3225 nm

26. Explain , giving reasons, which of the following sets of quantum numbers are not possible.

(a) n=0, l=0; m = 0, s = + ½ (c)n=1, l=0; l = 0, s = - ½

(b) n=1, l=1; m =- 0, s = + ½ (d) n=2, l=1; m = 0, s = + ½

Ans. (a) Not possible because n≠ 0 (c) Not possible because when n=1, l≠1

(b) Possible (d) Possible

27. (a)What is the lowest value of n that allows g orbitals to exist?

(b)An electron is in one of the 3d orbitals, Give the possible values of n , l and m for this electron.

Ans.(a) minimum value of n= 5

(b)n=3, l=2, m l = -2, -1, 0, +1, +2

28. Calculate the total number of angular nodes and radial nodes present in 3p orbitals.

Ans. For 3p orbitals, n=3, l= 1

Number of angular nodes = l= 1

Number of radial nodes = n-l-1 = 3-1-1= 1

29. Write down the quantum numbers n and l for the following orbitals

a. 2p b. 3d c. 5f

Ans. a. n=2, l= 1 b. n= 3, l=2 c. n= 5, l=3

30. Write the 3 points of difference between orbit and orbital.

Ans.

Orbit Orbital

1. An orbit is a well defined circular path around the

nucleus in which the electrons revolve

1. An orbital is the three dimensional space around

the nucleus within which the probability of finding an

electron is maximum(upto 90 %)



2. It represents the planar motion of an electron

around the nucleus

2. It represents the three dimensional motion of an

electron around the nucleus

3. All orbits are circular and disc like

3. Different orbitals have different shapes,

i.e.-orbitals are spherically symmetrical,

p-orbitals are dumb-bell shaped and so on.

31. Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is 5.7 x 10 5 m/s.

Ans. Δx x (m x Δv) = h/4ᴨ

Δx = h/4ᴨ x m x Δv = 6.6 x 10 -34

= 1.0 x 10 -10

m

4 x 3.14 x 9.1 x 10 -31

x 5.7 x 10 5

32. Write 3 points of differences between electromagnetic waves and matter waves.

Electromagnetic waves Matter waves

1. These are associated with electric and magnetic

fields

1. These are not associated with electric and magnetic

field.

2. They do not require any medium for propagation. 2.They require medium for propagation

3. They travel with the same speed as that of light 3. They travel with lower speeds

not constant for all matter waves

33. (i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons

Ans. (i) Mass of one electron = 9.10939 × 10 –31

kg

Number of electrons that weigh 9.10939 × 10 –31

kg = 1

Number of electrons that will weigh 1 g = (1 × 10 –3

kg)

= 1x10-3

Kg = 1.098 × 10 27

9.10939x10-31

Kg

(ii) Mass of one electron = 9.10939 × 10 –31

kg

Mass of one mole of electron = (6.022 × 10 23

) × (9.10939 ×10 –31

kg) = 5.48 × 10 –7

kg

Charge on one electron = 1.6022 × 10 –19

coulomb

Charge on one mole of electron = (1.6022 × 10 –19

C) (6.022 × 1023

)

= 9.65 × 104 C

34. Find energy of each of the photons which (i) correspond to light of frequency 3× 10 15 Hz.

(ii) have wavelength of 0.50 Å.

Ans.(i) Energy (E) of a photon is given by the expression, E = hʋ

Where,

h = Planck’s constant = 6.626 × 10 –34

Js ν = frequency of light = 3 × 10 15

Hz

Substituting the values in the given expres sion of E:

E = (6.626 × 10 –34

) (3 × 10 15

) E = 1.988 × 10 –18

J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

E= hc

λ

h = Planck’s constant = 6.626 × 10 –34

Js

c = velocity of light in vacuum = 3 × 10 8 m/s


35. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an

energy level with n = 4 to an energy level with n= 2?



Ans.Then ni = 4 to nf = 2 transition will give rise to a

spectral line of the Balmer series. The energy

involved in the transition is given by the relation,


E = (4.0875 × 10

–19 J)

The negative sign indicates the energy of emission.

Wavelength of light emitted

Substituting the values in the given expression of λ:

36. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the

electronic configuration of the element (iii)Identify the element .

Ans.(i)For an atom to be neutral, the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is 1s 2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

10

(iii). Copper

37. Give the number of electrons in the species , H2+ ,H 2 and O2

+

Ans. Number of electrons present in hydrogen molecule (H 2 ) = 1 + 1 = 2

∴ Number of electrons in H2+

= 2 – 1 = 1

Number of electrons in H 2 = 1 + 1 = 2

Number of electrons present in oxygen molecule (O2 ) = 8 + 8 = 16

Number of electrons in O2+= 16-1 = 15

38. What are the draw backs of Bohr’s atomic model? Show that the circumference of the Bohr orbit for the

hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around

the orbit.

Ans.1.Bohr’s model failed to account for the finer details of the hydrogen spectrum.

2. Bohr’s model was also unable to explain spectrum of atoms containing more than one electron.

3. Bohr’s model was unable to explain Zeeman effect and Stark effect.

4. Bohr’s model could not explain the ability of atoms to form molecules by chemical bonds.

Since hydrogen atom has only one electron, according to Bohr’s

Postulate, the angular momentum of that electron is

given by:

Where, n = 1, 2, 3, …

According to de Broglie’s equation:

Substituting the value of ‘mv’ from expression (2) in

expression (1):

Since‘2πr’ represents the circumference of the Bohr

orbit (r), it is proved by equation (3) that the

circumference of the Bohr orbit of the hydrogen atom

is an integral multiple of de Broglie’s wavelength

associated with the electron revolving around the

orbit.



39. State photo electric effect. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold

wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with

awavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Ans. Photoelectric effect: The phenomenon of ejection of electrons from the surface of metal when light of suitable

frequency strikes it is called photoelectric effect. The ejected electrons are called photoelectrons.

It is given that the work function (W0 ) for caesium atom is 1.9 eV.

(a) From the expression, , we get:

Where,

λ0 = threshold wavelength

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of

(λ0 ):

Hence, the threshold wavelength λ0 is 653 nm.

(b) From the expression, , we get:

Where,

ν 0 = threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν0:

(1 eV = 1.602 × 10

–19 J)

ν0 = 4.593 × 1014

s –1

Hence, the threshold frequency of radiation (ν0 ) is

4.593 × 10 14

s –1

.

(c) According to the question:

Wavelength used in irradiation (λ) = 500 nm

Kinetic energy = h (ν – ν0 )

= 9.3149 × 10

–20 J

Kinetic energy of the ejected photoelectron

= 9.3149 × 10 –20

J

Since K.E

v = 4.52 × 10

5 ms

–1

Hence, the velocity of the ejected photoelectron (v) is

4.52 × 10 5 ms

–1 .

40. (a)The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If

any

of these combination(s) has/have the same energy lists:

1. n= 4, l = 2, m = –2 , s = –1/2

2. n= 3, l = 2, m = 1 , s = +1/2

3. n= 4, l = 1, l = 0 , s = +1/2

4. n = 3, l= 2, l = –2 , s = –1/2

5. n = 3, l= 1, l = –1 , s = +1/2

6. n = 4, l= 1, l = 0 , s = +1/2

(b)Among the following pairs of orbitals which orbital will experience the larger effective nuclearcharge?

(i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p

Ans.(a)For = 4 and l = 2, the orbital occupied is 4d. For n = 3 and l = 2, the orbital occupied is 3d.

For n = 4 and l = 1, the orbital occupied is 4p. Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the

4d, 3d, 4p,3d, 3p, and 4p orbitals respectively.

Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) =6(4p) < 1 (4d).



(b)Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron

atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.

(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than

the electron(s) in the 3s orbital.

(ii) 4d will experience greater nuclear charge than 4fsince 4d is closer to the nucleus.

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

41. (i) The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective

nuclear charge from the nucleus? (ii) Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe

Ans. (i) the electrons in the 3p orbital of silicon will

experience a more effective nuclear charge than

aluminium.

(ii) (a) Phosphorus (P): Atomic number = 15

The electronic configuration of P is:

1s2 2s

2 2p

6 3s

2 3p

3

The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three

unpaired electrons.

(b) Silicon (Si): Atomic number = 14

The electronic configuration of Si is:

1s2 2s

2 2p

6 3s

2 3p

2

The orbital picture of Si can be represented as:

From the orbital picture, silicon has two unpaired

electrons.

(c) Chromium (Cr): Atomic number = 24

The electronic configuration of Cr is:

1s2 2s

2 2p

6 3s

2 3p

6 4s

1 3d

5

The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired

electrons.

(d) Iron (Fe): Atomic number = 26

The electronic configuration is:

1s2 2s

2 2p

6 3s

2 3p

6 4s

2 3d

6

The orbital picture of chromium is:

From the orbital picture, iron has four unpaired

electrons.

42. Give the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to

the difference between the numbers of total p and total s electrons.

Ans. electronic configuration of Kr ( atomic no.=36) =1s2 2s

2 2p

6 3s

2 3p

6 3d

10 4s

2 4p

6

Total no. of s-electrons = 8, total no. of p-electrons = 18. Difference = 10 No. of d- electrons = 10

43. What is the minimum product of uncertainty in position and momentum of an electron?

Ans.h/4π

44. Which orbital is non-directional ?

Ans. s- orbital

45. What is the difference between the notations l and L ?

Ans. l represents the sub-shell and L represent shell.

46. How many electrons in an atom can have n + l = 6 ?

Ans. 18

47. An anion A 3+

has 18 electrons. Write the atomic number of A.

Ans.15

48. Arrange the electron (e), protons (p) and alpha particle (α) in the increasing order for the values of e/m

(charge/mass).

Ans. α<p < e

1. (iii) 2. (ii) 3. (iv) 4. (ii) 5. (i) 6. (iv) 7. (iv) 8. (iii) 9. (ii) 10. (iii) 11. (i) 12. (iii) 13. (ii) 14. (iv) 15. (ii) 16. (ii), (iii)

17. (i), (iii)

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