02. Atomic Structure [1-35]

7/28/2019 02. Atomic Structure [1-35]

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... 1Chemistry / Atomic Structure

Chapter Objectives

Atomic Structure

1. Sub-Atomic particles

2. Atomic models3. Plancks quantum theory

4. Photo electric effect

5. Bohrs model

6. Hydrogen spectrum

7. Limitation of Bohrs theory

8. Dual nature of electron9. Heisenbergs uncertaintiy principle

10. Quantum mechanical model of H-atom

11. Quantum numbers

12. Electronic configurations


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Concept Notes

Sub-Atomic Particles

Experiments of Cathode Rays (Discovery of Electron)

When electron discharges from a high potential source and passed through a discharge tube evacuated toa pressure of 0.01 mm or less, rays are emitted from the cathode. These are commonly known as cathoderay. Cathode rays comprise of negatively-charged particles called electrons.

High voltage

Gas atlow pressure

Cathoderays To vacu um

pump

Cathode +

Anode

Discharge tube experiment-Production of cathode rays

+

Properties of cathode rays(i) They travel in straight lines at right angle to the cathode surface.(ii) They are material particles as they produce mechanical motion in a small paddle wheel placed in

their path. The name electron to these material particles was given by Stony.(iii) They are deflected from their path by electric and magnetic fields as they are electrically charged.

Direction of deflection indicates that they are negatively charged.

Explanation of discharge phenomenon:

At low pressure the electrons can move through a considerably large distance before attaching to anothermolecule forming a negative ion which is not possible at high pressure. These free electrons and thepositive ions are responsible for conduction of electricity.

At very low pressure around 0.01 mm of Hg the walls of tube begin to glow this is called florescence. Thecolour of glow depends on nature of glass. It is yellowish green for soda glass.If the pressure is still decreased the current through the gas gradually decreases and finally the tube stopsconducting.In vacuum there are no charges and hence no conduction occurs.

Note:Charge on electron was given by R. A. Mullikans by an oil drop experiment. Its value was found to be1.6 1019 coulombs which is taken as one unit negative charge.


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e/m for electrons (cathode rays) is a fundamental property, i.e. it does not depend upon:(1) Metal used as cathode.(2) Gases used in discharge tube.Thomson determined e/m value as 1.76 108 coulombs/g

By knowing charge over an electron and e/m ratio, we can get the value of mass.

Mass of electron m =19

8

28 31

e 1.6 10 coulombs / electron

e /m 1.76 10 coulombs/ g

9.1 10 g or 9.1 10 kg

=

=

This is rest mass of electron which is1

1837times the mass of a hydrogen atom.

Mass of electron moving with a velocity v is given by

o2

2

mmv

1c

= (c = Velocity of light, v = Velocity of elelctron)

It is clear that with increase in velocity, mass of electron would increase but the velocity cannot equal orexceed the velocity of light.

Anode Rays (Discovery of proton)

It is a well-known fact that the atom is electrically neutral. The presence of negatively-charged electrons inthe atom emphasized the presence of positively-charged particles.

To detect the presence of positively-charged particles, the discharge tube experiment was carried out, inwhich a perforated cathode was used. Gas at low pressure was kept inside the tube. On passing highvoltage between the electrodes it was observed that some rays were emitted from the side of the anode.These rays passed through the holes in the cathode and produced green fluorescence on the oppositeglass wall coated with ZnS. These rays consist of positively-charged particles.

+

ZnSCoating

Perforatedcathode

H gas ins ide

at low pressure2

For anode rays, e/m is not fundamental property as different gases used have different mass on C-12scale.Highest e/m is for hydrogen gas.Mass of proton is found to be 1.672 x 1024 g or 1.0075 amu (1 amu= 1.66 x 1024 g)


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Some of the well known fundamental particles present in an atom are-protons, electrons and neutrons.Many others were discovered later viz positron, neutrinos etc.

cimotabuSselcitrap

lobmyS egrahctinU ssamtinU niegrahCbmoluoC

umanissaM

notorP1p1 1+ 1 0106.1+ 91 528700.1

nortueN0n1 0 1 0 566800.1

nortcelE1e0 1 elbigilgeN 01206.1 91 01984.5 4

Chadwick in 1932 discovered the neutron by bombarding elements like beryllium withfast moving -particles. He observed that some new particles were emitted whichcarried no charge and had mass equal to that of proton.

Atomic Models

J.J. Thomsons Model of an AtomAn atom consisted of a uniform sphere of positive charge in which the electrons were embedded, so as tomake the atom electrically neutral. This is known as plum-pudding model.

Positivesphere

Electron

Thomson's Atomic Mo del

Do You Know?


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Advantages of Thomsons ModelThis model was quite attractive as it could explain several observations available at that time.1. It could explain why only negatively charged particles are emitted when metal is heated and never

positive charged particles.

2. It could also explain the formation of ions and ionic compounds of chemistry.

Disadvantages1. Almost free passage of cathode rays through an atom was not consistent with Thomsons model for

that the atom should have a lot of empty space.2. The phenomenon of - rays scattering could not be explained by this model.

Rutherfords ExperimentThe experiment involved scattering of massive -particles by very thin foils of metals such as gold, silveretc., of thickness 0.0004 mm. The position of the particles after passing through the foil was ascertainedby the flash produced on a ZnS screen.

Radium

Block of lead

=-rays

Thin gold plate

Circular ZnSscreen

Mo st of particlesstrike here

=

Slit

Observationsa. Most of the -particles passed through the metal foil without any deflection in their path.b. Some -particles were deflected through small angles.

c. Very few -particles (1 in 20,000) were deflected by an angle greater than 90or even were deflectedback completely.

Conclusionsa. Most of the space in the atom is empty.b. The nucleus is surrounded by negatively-charged electrons, which revolve around the nucleus.c. An atom consists of an extremely small dense positively-charged nucleus which is present at the

centre of the atom. Size of nucleus is 105 to 106 times less thanthe size of atom.This model is called Rutherfords model


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Drawbacks of Rutherfords Model

According to Maxwells equation of electro magnetism when a charged particle moves with acceleration, itcontinuously loses energy in the form of electromagnetic radiation. So should be the case with electrons,

and as a result of loss of energy, its motion should slow down and it should eventually fall into the nucleusas it will no longer be able to withstand the attractive forces of the nucleus. As a result the atom shouldcollapse. But we all know that an atom is quite stable. Rutherford was unable to account for the stability ofthe atom.

Note: Atomic number and mass number Atom is electrically neutral hence it is necessary for thenumber of protons to be equal to the number of electrons in order to balance the charges.Atomic number (Z) = Number of protons present in the nucleus = Number of electrons present in the atom.

Mass number of an element (A) = Number of protons + Number of neutrons.

Representation of an element. An element is represented as YAZwhere,Y is the symbol of elementA is the mass numberZ is the atomic number

IIIIIllustrative Example

Example 1: Sodium is represented as Na2311 . The mass number (A) of sodium is 23 and atomic

number (Z) is 11.Atomic weight of Na is 23 g, i.e., 1 mole or 6.023 1023 atoms of Na weigh 23 g.

Then, weight of 1 atom of Na = 2323 23 amu

6.023 10=

Where, 1 amu = 231

g6.023 10

Electromagnetic Waves

Electromagnetic waves are waves consisting of oscillating electric and magnetic fields. The basic sourceof electromagnetic waves is an accelerated charge, this produces changing electric magnetic field whichconstitute the wave. Both the components have same wavelength and frequency and travel in planesperpendicular to each other and also perpendicular to the direction of propagation of electromagnetic wave.

x

y


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Characteristics of a wave(i) Wavelength: It is represented by . Units are m, cm or .

1 = 108 cm = 1010 m

(ii) Frequency: The number of times a wave passes through a given point in 1 s.(iii) Velocity: The linear distance travelled by a crest or a trough in 1 s. Its unit is cms1.

(iv) Wave number: The number of waves present in 1 cm length. It is represented by1

.

Electromagnetic spectrumThe arrangement of various electromagnetic radiations in order of their increasing (or decreasing)

wavelengths (or frequencies) is known as electromagnetic spectrum.

Electromagnetic Wavelength (A ) Origin

radiation

1. Cosmic rays Up to 0.01 Originate from outer space and penetrate the

earths atmosphere to reach its surface.2. Gamma( ) rays 0.0-0.1 Produced from nuclei of radioactive substances

during disintegration of nuclei.3. X-rays 0.1-150 Produced by placing a metal plate in the path

of fast moving cathode rays.4. Ultraviolet (UV) 150-3800 Present in suns rays. Lab sources include H-

or Xe lamps etc.5. Visible radiations 3800-7600 Produced from stars, arc lamps and any other

source whose light is visible.6. Infrared (IR) 7600 6 106 Heat waves. Produced by incandescent bodies.

7. Microwaves 3 106

3 109

Produced by special generators.8. Radiowaves 3 107 3 1014 Generate from alternating electric current of

high frequencies.

Plancks quantum theory

The three key points of Plancks quantum theory are as follows.a. Radiant energy is absorbed or emitted discontinuously in the form of small packets of energy

called quanta. In case of light energy these packets are called photons.b. Energy of each photon is directly proportional to the frequency of radiation. This is mathematically

expressed asE

E = h

= c

h is called Plancks constanth = 6.626 1034 Js.

c. Total amount of energy emitted or absorbed is given asnhc

E nhv= =

n = 1, 2, 3, ... (no of photons)


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Example 1: It has been found that gaseous iodine molecules dissociate into separate atoms afterabsorption of light at wavelengths less than 4,995 . If each quantum is absorbed by onemolecule of I

2, what is the minimum energy in k cal/mole, needed to dissociate I

2by this

photochemical process?Solution: E(per mole) = N

A.h

E Nhc mole Js m s

mA= =

( . ) ( . ) ( . )6 023 10 6 626 10 3 0 10

4995 10

23 1 34 8 1

10

239 61

4184. /

.kJ mole

kcal

kJb g

.HG

IKJ = 57.3 k cal/mole

Photoelectric Effect

Sir J.J. Thomson observed that electrons are emitted instantaneously from a clean metal plate in vacuum

when a beam of light falls on it. This is called the photoelectric effect. Usually such an effect is produced

by a radiation in the UV region and also in some cases in the visible region. Photoelectric effect is an

illustration of the particle nature of light. Photoelectric emissions are associated with the following facts.

(a) Electrons are emitted instantaneously from a clean metal plate when irradiated with a radiation of

frequency equal to or greater than some minimum frequency, called the threshold frequency.

The energy corresponding to this frequency is known as the work function.

(b) Kinetic energy of the emitted electrons depends upon the frequency of the incident radiation and

not on its intensity. The kinetic energy increases linearly with the increase in the frequency of

radiation.

(c) The number of electrons emitted is proportional to the intensity of the incident radiation.

Suppose, the threshold frequency of light required to eject electrons from a metal is 0 and hence

the minimum energy required is 0h . If the frequency of light is higher than 0 (Let ), then theexcess energy will be converted to the kinetic energy of the electron.

E.Khh 0 +=

20 mv

2

1hh +=

02 hhmv

2

1=


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M ax

KE

fre qu en cy ( )

M ax

KE

Intensity (I)


Example 1: Calculate the kinetic energy of a photoelectron emitted by Na surface when light of

wavelength 400 nm strikes on it. (Given work function of Na = 2.28 eV)

Solution: Energy of photon = h =hc

=

6 626 10 3 10

400 10

34 8 1

9

.

J s m s

m= 4.969 10-19 J

Work function = 2.28 eV= 2.28 1.602 1019 J (1 eV = 1.602 10-19J)= 3.653 1019 JNow, KE = h Work function = 4.969 10-19 J 3.653 10-19 J= 1.316 1019 J

Types of Spectrum

Spectra is classified into two types depending upon the source of radiation.(a) Emission spectra(b) Absorption spectra

(a) Emission spectraEmission spectrum is obtained when the radiations from a source (electric discharge through a gasat low pressure) are passed directly through a prism and received on a photographic plate.There are two types of emission spectra:

(i) Continuous spectra


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Beam

Prism

Photographic plate

W hite light

When white light is passed through a prism, it splits into seven different colours from violet to red.The colour band is continuous and merge into other colour bands before and after it. Thus, we get acontinuous spectrum.(ii) Line spectra:

Prism

Photograph ic plate

When an electric discharge is passed through a gas at low pressure, light is emitted. If this light ispassed through a prism and on to a photographic plate, it is observed that lines separated from eachother are obtained on the photographic plate. Thus, we get a line spectra.

(b) Absorption spectraWhen white light from the sun or from a bulb is passed through the vapours or solution of a chemicalsubstance, i.e. sodium vapours and then passed through a prism and on to a photographic film,some dark lines are observed in the continuous spectrum. The appearance of dark lines in thecontinuous spectrum is due to the absorption of certain wavelength by the sodium vapours.


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Hydrogen Spectrum

On supplying energy to the hydrogen atom, the electron in its ground state gets excited to higher energylevels. It then tends to return to its ground state by emitting the energy absorbed in the form of radiations

which produce lines which collectively form the hydrogen spectrum.

Spectral lines of hydrogen

Energy of an electron in Bohrs orbit =2 2 4 2

2 2

2 Z e mk

n h

hcE h = =

f iE E E = E

f

= Energy in final orbitE

i= Energy in initial orbit

2 4 2

3 2 2f i

1 2 Z e mk 1 1

ch n n

=

For hydrogen Z = 1

4 2 4

3 2 2f i

1 2 e mk e 1 1

ch n n

=

4 4 2

H3

2 e me kRch

=

RH

is Rydbergs constantR

H= 1,09,678 cm1 = 1.09 107m1

H 2 2f i

1 1 1R

n n

=

1=

= Wave number

Spectral series Value of n1

Value of n2

Spectral region

n2 = (n1+ 1), (n1+ 2) + ...

LYMAN 1 2, 3, 4, 5 ... UV

BALMER 2 3, 4, 5, 6 ... VISIBLE

PASCHEN 3 4, 5, 6, 7 ... IR

BRACKETT 4 5, 6, 7, 8 ... IR

PFUND 5 6, 7, 8, 9 ... IR


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Maximum number of lines produced when the electron jumps from the nth level to the ground state

=n(n 1)

2

Bohrs model of atom

Neil Bohr on the basis of Plancks quantum theory overcame the drawbacks of Rutherfords model andsuggested a modified model of an atom which retained the key features of Rutherfords model and introducedthe concept of stationary orbits.

Bohrs postulates:i. An atom comprises of a small positively-charged nucleus in the centre in which the entire mass is

concentrated.ii. Electrons revolve around the nucleus in circular orbits, these orbits have a fixed value of energy.

They are popularly known as stationary orbits. When an electron revolves in a circular orbit, energyis neither emitted nor absorbed. This is known as quantization of energy.

iii. Electron can revolve only in those orbits whose angular momentum is an integral multiple of the

factorh

2.

i.e. =

nhmur

2where n=1,2,3.........

What is angular momentum of electron in Bohrs second orbit?

Answer: Angular momentum of electron in Bohrs second orbit h /=

iv. Energy is emitted or absorbed when an electron jumps from one orbit to another.

2 1E E E =

Bohrs theory is used fora. Calculation of the radius of the orbit in which the electron revolves.b. Velocity of electron in the orbit.

c. Energy of the electron

Do You Know?


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Calculation of radius of an electron in the orbit

+

nucleus m u

r

electron

2

r

According to Coulombs law, the electrostatic force of attraction (acting inwards)

2

2

a r

kZe

F = (k is a constant with value 9 109

Nm2

/C2

)

Centrifugal force Fc =2mu

r(acting outwards)

In order to maintain the motion of the electron in the orbitFa = Fc

2 2

2

mu kZe

r r=

u2 =2kZe

mrAccording to Bohrs postulates

mur =nh

2

u =nh

2 mr

2 22

2 2 2

n hu

4 m r=

2 2 2

2 2 2

n h kZe

mr4 m r=

=

2 2

2 2

n hr

4 mkZe

For hydrogen Z = 1For n = 1, Z = 1r = 0.529


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Radius of an electron in the nth orbit of an atom with atomic number Z can be calculated as

2

nn

r 0.529 Z

=

Calculation of velocity of electron in an orbit

2 2

2

mu kZe

r r= ... (i)

mur =nh

2... (ii)

Dividing (i) and (ii)

u =2kZ2 e

nh

nh

e2u

2= u is in cm/s, for H atom

Z = 1, k in cgs unit,22 e

unh

=

Velocity of an electron in nth orbit of an atom with atomic number Z is given by

8n

Zu 2.18 10 cm/ s

n

=

Number of revolutions per second =Velocity of electron in an orbit

Circumference of orbit

Calculation of energy of electron in an orbit:Total energy = Potential energy + Kinetic energyTE = PE + KE

PE =2kZe

r

KE =21mu

2=

2kZe

2r(from (i))

TE =2

21 kZemu2 r

TE =2 2kZe kZe

2r r


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TE =2kZe

2r

Putting value of r in the above expression

TE =

2 2 4 2

2 2

2 Z e mk

n h

The expression for energy is given by

E =2.18 10112

2

Z

nerg/atom

= 2

18

2

Z2.18 10 J/atom

n

=13.62

2

Z

neV/atom

1 eV is defined as the amount of energy gained by an electron when it accelerates through a field of 1 V.


Example 1: Calculate the wavelength in Angstrom of the photon that is emitted when an electron inBohr orbit n = 2 returns to the orbit n = 1 in the hydrogen atom. The ionization potential ofhydrogen atom in ground state is 2.17 10-11 erg per atom.

Solution: The ionization potential of the ground state of hydrogen atom is 2.171011 erg.

E1 =2.17 1011 erg.

Since2

11

221

n2

1017.2E;

n

EE

== erg

=0.5425 1011 ergE

2 E

1=0.5425 1011erg (2.17 1011) erg

E = 16275 10 11. erg

E hhc

= =

E = 1.6275 1011 erg

16275 106 626 10 3 1011

27 10 1

..

=

erg

erg s cm s

=

6 626 10 3 10

16275 10

27 10 1

11

.

.

erg s cm s

erg

= 12.2138 106 cm = 1221.4 108 = 1,221.4


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Reduced mass

According to Bohr, in an atom electron is revolving in a circular orbit around the stationary nucleus. But

nucleus will be stationary only if, the mass of the nucleus is infinite and we knowmass of electron 1

mass of proton 1836

=

so for lighter elements we cant assume nucleus as stationary. Actually the nucleus oscillates slightlyabout the center of gravity (nucleus and electron) so in Rydberg constant R

4

2 30

e mR

8 h C=

mass of electron (m) is replaced by reduced mass

mM

m M =

+wherem = mass of electronM = mass of nucleus.

Facts in favour of Bohrs theory

(i) The frequencies of the spectral line of the hydrogen spectrum, as calculated using Bohrs theoryare in close agreement with their experimental values.

(ii) The value of Rydbergs constant as calculated from Bohrs theory is in full agreement with thevalue obtained from spectroscopic studies.

(iii) The radii and energy of permissible orbits are consistent with their experimental values.(iv) The absorption and emission spectra of H and H-like atoms is quite well explained.

Drawbacks of Bohrs theory

1. The theory explains the spectrum of only hydrogen and hydrogen-like species.2. The splitting of spectral lines in magnetic field is called Zeeman effect, and the splitting of lines in

electric field is called Stark effect. These phenomenon could not be explained by Bohr.3. It provides no proof for postulate which states quantization of angular momentum

nhmur

2=

4. It is not in accordance with dual nature of electron, as according to Bohr, electron behaves as a

particle and has no wave nature associated with it.5. It was unable to account for Heisenbergs uncertainty principle.

Dual nature of electrons

de Broglie pointed out that matter has both particle and wave nature.The phenomenon of electron diffraction established the wave nature of electrons and the phenomenon ofphotoelectric effect and black body radiation established the particle nature of electrons.de Broglie derived an expression for calculating the wavelength of the wave associated with an electron.

= h/p (p =momentum)

Wavelengthp Momentum


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The above equation is called de Broglie equation and the wavelength is called de Broglies wavelength.Let a photon of mass m moves with a velocity c around the nucleus and be associated with a wave of

wavelength . From Plancks equation, we have

E = hv

and Einsteins equation givesE = mc2

Combining both equations, we havemc2 = hv

mc2 = hc/or mc = h/ or = h/mc

= h/mass Velocityor = h/p

IIIIIllustrative Examples

Example 1: Calculate the momentum of a particle which has a de Broglies wavelength of 3 .10 34 2 1(1 10 m, h 6.6 10 kg m s ) = =

Solution:34 2 1

10

24 1

h 6.6 10 kgm smv

3 10 m

Momentum 2.2 10 kgms

= =

=

Heisenbergs Uncertainty Principle

According to the principle, It is impossible to determine exactly both the position and the momentum(or velocity) of an electron or any other microscopic moving particle at the same time. Anyattempt to locate an electron changes its momentum.

If x is the uncertainty in position and p is the uncertainty in momentum, then according to Heisenbergsprinciple, these two quantities are related as follows:

h

x . p

4

( a constant)

hx . m v

4

hx . v

4 m

x Uncertainty in positionv Uncertainty in velocity


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Explanation: Let us try to measure both the position and momentum of an electron. To find the position of

the electron we have to use light so that photon of light strikes the electron. As a result of lighting the

position as well as the velocity of electron are disturbed but the accuracy with which the position of a

particle can be measured depends upon wavelength of light used. The uncertainity in measurement of

position using wavelength is . Therefore shorter the wavelength the greater is the accuracy. But shorterwavelength means higher frequency and hence higher energy. When the high energy photon strikes the

electron it changes its speed as well as direction. But this is not true for macroscopic moving particles. So

Heisenbergs uncertainity principle has no significance in everyday life.


Example 1: Calculate the uncertainty in position of an electron if the uncertainty in its velocity is 5.7 x105ms1.(h = 6.6 1034 Kg m2s1 and m = 9.1 1031 kg)

Solution:

34 2 1

5 1 31

10

hx p

4

hx

4 . v . m

6.6 10 kgm s

4 3.14 57 10 ms 9.1 10 kg

1 10 m

=

=

=

=

Concept of Orbital

Following Heisenbergs uncertainty principle, we replace the concept of orbit (introduced by Bohr) by theconcept of orbital. Orbital refers to three-dimensional space around the nucleus within which the probabilityof finding an electron of given energy is maximum.

Quantum or Wave Mechanical Model of Atom

This new model of atom was put forward by Schrodinger taking into account the de Broglie concept of dual

nature and Heisenbergs uncertainity principle. He described the motion of the electron in three-dimensionalspace in terms of a mathematical equation called Schrodinger wave equation.

( )2 2 2 2

2 2 2 2

8 mE V 0

x y z h

+ + + =

Where (psi) is the amplitude of the electron wave at a point with coordinates x, y and z. E is the totalenergy and V is the potential energy of the electron.

is also called wave function. 2 gives the probability of finding the electron at (x, y, z). The acceptablesolutions of the above equation for the energy E are called eigen values and the corresponding wave

function are called eigen functions. These functions have to be single valued, continuous and finite.


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Schrodinger wave equation can be written as

( )2 2 2 2

2 2 2 2

8 mE V 0

x y z h

+ + + =

( )2

2

2

8 mE V 0

h

+ =

Where,2 2 2

2

2 2 2x y z

= + +

is called laplacian operator.

This equation can be rewritten as

( )2

2

2

8 mE V

h

=

2 2

2

hV E

8 m

+ =

H E =

Where,

22

2

hH V

8 m

= +

is called Hamiltonian operator.

In this operator, first term represents kinetic energy operator ( )T and second term represents potential

energy operator ( )V i.e. H T V= + . Hence Schrodinger wave equation can also be written as

( ) T V E+ =

Plot of radial wave function R

The plot of radial wave function as a function of distance r from the nucleus gives the information about how

the radial wave function changes with distance r and about the presence of nodes where the change of sign

of R occurs.

The square of radial wave function R2 for an orbital gives the probability density of finding the electron at a

point along a particular radial line. Since the atoms have spherical symmetry, it is more useful to discuss

the probability of finding the electron in a spherical shell between the spheres of radius (r + dr) and r. The

volume of the shell is equal to 4r2

dr.This probability which is independent of direction is called radial probability and is equal to 2 24 r drR .

Radial probability function (= 4r2 R2) gives the probability of finding the elecron at a distance r from thenucleus regardless of direction.

The plots of radial wave function R, radial probability density R2 and radial probability function 4r2 R2 for 1s,2s and 2p atomic orbitals as a function of distance r from the nucleus are shown in the figure.


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R

r

1s

R

r

2s

R

r

2p

Node

R2

r

1s

r

2s

r

2p

R2

R2

Node

4

rR

2

2

r

1s

r

2s

r

2p

4

rR

2

2

4

rR

2

2

At node, the value of radial function changes from positive to negative.

It has been found that ns-orbitals have (n 1) nodes and np orbitals have n 2 nodes.

The radial probability function for the 1s orbital initially increases with increase in distance from the nucleus.

It reaches a maximum at a distance very close to the nucleus and then decreases. The maxima in the

curve corresponds to the distance at which the probability of finding the electron is maximum. This distance

is called the radius of maximum probability.

The radial probability function curve for 2s orbital shows two maxima, a smaller one near the nucleus and

a bigger one at a larger distance. In between these two maxima it passes through a zero value indicating

that there is zero probability of finding the electron at that distance. The point at which the probability of

finding the electron is zero is called a nodal point.


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The distance of maximum probability for a 2p electron is slightly less than that for 2s electron. However, in

contrast to 2p curve, there is a small additional maxima in the 2s curve. This indicates that the electron in

2s orbital spends some of its time near the nucleus. In other words, the 2s electron penetrates a little

closer to the nucleus and is therefore held more tightly than the 2p electron. So electron in 2s orbital is

more stable and has lower energy than an electron in 2p orbital.

Plots of angular wave function

The plots of the angular wave function , and angular probability |, |2 for s and pz orbitals are shownbelow (fig a, b)

s p z p z

(a) (b)

For an s-orbital, the angular part is independent of angle and is therefore of constant value. Hence the

graph is circular or to be precise spherical. For the pz orbital, we get two tangent spheres The px and py are

identical in shape but are oriented along x and y axes respectively.

In the angular wave function plots, the distance from the centre is proportional to the numerical values of , in that direction and is not the distance from the centre of nucleus.

The angular probability density plots can be obtained by squaring the angular function plots. For s-orbital,

the squaring causes no change in shape. For both p and d orbitals, however, on squaring the plot tends to

become more elongated (fig b).

For a hydrogen atom, wave function of principal quantum number, n, there is a total of (n 1) nodes thatoccur at finite values of the radial distance, r. The number of angular nodes is just equal to the angularmomentum quantum number, l.

Thus, we haveangular nodes = l,radial nodes = n l 1.total nodes = n 1,

We should note that the total number of nodes in an atomic wave function is sometimes stated to be nrather than n 1. In this case, the node that always occur at r = is being included in the count.


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Shapes of Orbitals

s-orbital:- s-orbitals do not have directional character. They are spherically symmetrical.The s-orbital of higher energy levels are also spherically symmetrical. They are more diffused and have

spherical shells within them where probability of finding the electron is zero.In the s-orbital, number of nodes is (n 1)

y Node

z

x

2s orbital

p-orbital:- p orbital has a dumb-bell shape and it has a directional character.The two lobes of a p-orbital are separated by a plane that contains the nucleus and is perpendicular to thecorresponding axis. Such plane is called a nodal plane because there is no probability of finding theelectron.

y

z

x+

px

y

z

x

+

py

y

z

x

+

pz

In the absence of an external electric or magnetic field, the three p-orbitals of a particular energy level havesame energy and are degenerate. In the presence of an external magnetic field or electric field this degeneracyis removed.


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d-orbitals:- For d-orbitals five orientations are possible viz., dxy, dyz, dxz, 2 2x yd

,2zd . All these five

orbitals in the absence of magnetic field are equivalent in energy and are degenerate.

The shapes of the orbitals are as follows:

y

x

dxy

z

x

dxz

z

x

dyz

These three d orbitals are similar. The maximum probability of finding the electron is in lobes which aredirected in between the axes. Nodal region is along the axes.

z

y

x

These two d-orbitals are similar. Probability of finding the electron is maximum along the axes and thenodal region is in between the axes.

Quantum Numbers

These are used to determine the region of probability of finding a particular electron in an atom.

(a) Principal quantum number (n):This denotes the energy level or the principal or main shell to which an electron belongs. It canhave only integral values 1, 2, 3 etc. The letters K, L, M ... are also used to designate the value of n.Thus, an electron in the K shell has n = 1, that is L shell has n = 2 and so on.

Example 1: The principal quantum number of 2s-electron is ___.Solution: n = 2


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(b) Azimuthal quantum numbers (l):This denotes the orbital (Sub-level) to which an electron belongs. It gives an idea about the shapeof the orbital .l can have any value from 0 to (n 1), for a given value of n,

i.e. l = 0, 1, 2, ... (n 1)

foeulaV l 0 1 2 3

llehsbuS s p d f

The value of orbital angular momentum of the electron for a given value of l ish

( 12

+

l l .

Question: What will be the value of angular momentum of d orbitals?

Answer: Angular momentum of d-orbitalh

2(2 1)2

= +

(For d-orbital value of l= 2)h

62

=

(c) Magnetic quantum number (m):It gives us the idea about the orientations an orbital can have in space in the presence of magneticfield. The values of m depend on l orbital quantum number.Total number of alue of m=(2l+1) and it varries froml to +l.

For example, for l = 0 the value of magnetic quantum number m is also equal to zero, i.e. s-orbital

can have only one orientation in space in presence of magnetic field.

(d) Spin quantum number (s):The electron while moving round the nucleus in an orbit also rotates or spins about its own axis

either in a clockwise direction or in an anticlockwise direction. Its value is +1

2or

1

2corresponding

to clockwise or anticlockwise spin.

(1) The value of spin angular momentum for a given value of s is s s h( )12

(2) The spin magnetic moment of electron (excluding orbital magnetic momentum) is given

by n(n 2)BM (Where n Numberof unpairedelectrons)effective = + =

Do You Know?

Do You Know?


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Distribution of Electrons in an Atom:- Electronic configuration

The filling up of orbitals with electrons takes place according to certain rules which are given below:

(i) The maximum number of electrons in a main shell is equal to 2n2, where n is the principal quantumnumber.

(ii) The maximum number of electrons in a sub-shell like s, p, d, f is equal to 2(2l + 1), where l is theazimuthal quantum number for the respective orbitals. Thus s, p, d, f can have a maximum of2, 6, 10 and 14 electrons respectively.

(a) Aufbau Principle

According to this principle, Electrons are added progressively to the various orbitals in the order ofincreasing energy.

What does the word Aufbau mean?Aufbau is a German term which means building up or construction.

The energy of various orbitals increase in the order given below:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s ...

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

4f

Do You Know?


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(i) A new electron enters the orbitals for which (n + l) is minimum, e.g. if we consider 3d and 4sorbitals, the electron will first enter 4s-orbitals in preference to 3d.This is because the value of (n + l) for 4s-orbitals is less (4 + 0 = 4) than that for 3d-orbital(3 + 2 = 5)

(ii) In case where (n + l ) values are the same, the new electron enters the orbital for which n isminimum, e.g. in a choice between 3d and 4p for which (n + l) values are same(3 + 2 = 5, 4 + 1 = 5), the electron will prefer to go to the 3d-orbital, since n is lower for thisorbital.

(b) Paulis Exclusion Principle

It states that it is impossible for two electrons in a given atom to have same set of all four quantumnumbers.ORAn orbital can have a maximum of two electrons and that too with opposite spin.

Example:

(a) n = 2, l = 0, m = 0, s = +1/2n = 2, l = 0, m = 0, s =1/2

(b) n = 2, l = 1, m = 0, s = +1/2n = 2, l = 1, m = 0, s = 1/2n = 2, l = 1, m = +1, s = +1/2n = 2, l = 1, m = +1,s =1/2n = 2, l = 1, m =1, s = +1/2n = 2, l = 1, m =1, s =1/2

(c) Hunds Rule of Maximum Multiplicity

According to this rule, electrons enter the orbitals (e.g. s, px, py, pz ...) in the same sub-level in sucha way as to give maximum number of unpaired electrons. In other words it means that pairing beginswith the introduction of the fourth electron sixth electron in d subsheel and so on in pd subshell.

What is the electronic configuration of Cu (Z = 29)?1s2 2s2 2p6 3s2 3p6 3d10 4s1

Exceptional Electronic Configuration

Some atoms such as copper and chromium exhibit exceptional electronic configuration.For example:Cr(z = 24) has an electronic configuration

1s2 2s2 2p6 3s2 3p6 3d5 4s1

It is because of the extra stability associated with the half-filled and completely filled orbitals.

Ask Yourself ?


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1. Account for the following. Limits your answer to two sentences. Atomic weights of most of the

elements are fractional. (1979)

2. Nitrogen atom has atomic number 7 and oxygen has atomic number 8, then the total number ofelectrons in azide ion is(a) 20 (b) 21(c) 22 (d) 19

3. The increasing order (lowest first) for the value of e/m (charge/mass) for electron (e), proton(p),neutron (n) and alpha particle ( ) is: (1984)(a) e, p, n, (b) n, p, e, (c) n, p, , e (d) n, , p, e

4. Rutherfords scattering experiment is related to the size of the : (1983)(a) nucleus (b) atom (c) electron (d) neutron

5. With what velocity should an -particle travel towards the nucleus of a copper atom so as to arriveat a distance 1013 meter from the nucleus of the copper atom? (1997)

6. The number of photons emitted in 10 hours by a 60 W sodium lamp is(photon = 5893 A )(a) 6.40 1023 (b) 6.40 1024(c) 6.40 1025 (d) 5.40 1024

*7. Calculate the accelerating potential that must be imparted to a proton beam to give it an effectivewavelength of 0.005 nm.

*8. A photon of 300 nm is absorbed by a gas and thus re-emits two photons. One re-emitted photon haswavelength 496 nm. Calculate the energy of other photon re-emitted out.

*9. What accelerating potential is needed to produce an electron beam with an effective wavelengthof 0.09A ?

(a) 100 kV (b) 18.6 kV(c) 1 kV (d) 12.2 kV

10. The minimum energy required to overcome the attractive forces between electron and the surface ofAg metal is 7.52 1019 J. What will be the maximum kinetic energy of electron ejected out from Agwhich is being exposed to UV light of = 360 A?

11. The photo electric emission requires a threshold frequency v0. For a certain metal 1 2200 = and

2 1900 = produce electrons with a maximum kinetic energy KE1 and KE2. If KE2 = 2KE1,

calculate v0 and corresponding 0 .

Take-Off


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12. An electron beam can undergo diffraction by crystals. Through what potential should a beam of

electrons be accelerated so that its wavelength becomes equal to 1.54 A . (1997)

13. A photon of wavelength 4000 strikes a metal surface, the work function of metal being 2.13eV.Calculate, the energy of Photon in eV, (ii) the kinetic energy of emitted photoelectron, and (iii) thevelocity of photoelectron.

14. Following data were collected for the photoelectric emission of an electron from an element X,

(in nm) KE(in eV)

254 1.93

313 0.9

365 0.5

405 0.2

Calculate the wave function of an element X and also the value of Plancks constant.

15. Iodine molecule dissociates into atoms after absorbing light to 4500 A . If one quantum of radiationis absorbed by each molecule, calculate the kinetic energy of iodine atoms.(Bond energy of I

2= 240 kJ mol1) (1995)

16. In hydrogen atom, electrons are excited to 4th energy level. The number of lines that mayappear in the spectrum will be(a) 4 (b) 6 (c) 10 (d) 12

17. The velocity of an electron in the first Bohr orbit of a hydrogen atom is 6.32 104 m/s. Its velocity inthe second orbit would be(a) 2.16 104 (b) 1.5 104 (c) 3.16 104 (d) 3.16 108

18. The electron energy in hydrogen atom is given by

12

2

21.7 10E ergs

n

= . Calculate the energy

required to remove an electron completely from the n = 2 orbit. What is the longest wavelength(in cm) of light that can be used to cause this transition? (1984 )

19. Estimate the difference in energy between 1st and 2nd Bohrs orbit for a hydrogen atom. At whatminimum atomic number, a transition from n = 2 to n = 1 energy level would result in the emission

of X-rays with 83.0 10 m = ? Which hydrogen atom-like species does this atomic number corre-spond to? (1993)

20. Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohrsorbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionisation potential of the groundstate hydrogen atom is 2.17 1011 erg per atom. (1982)


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34. What is the wavelength emitted during the transition of electron in between two levels of He+ ionwhose sum is 5 and difference is 3?(a) 200 A (b) 250 A

(c) 244 A (d) 240 A35. Consider the hydrogen atom to be a proton embedded in a cavity of radius a

0(Bohrs radius) whose

charge is neutralized by the addition of an electron to the cavity in vacuum, infinitely slowly. Esti-mate the average total energy of an electron in its ground state in a hydrogen atom as the work donein the above neutralisation process. Also, if the magnitude of the average kinetic energy is half themagnitude of the average potential energy, find the average potential energy. (1996)

*36. 1.8 g of hydrogen atoms are excited by the radiations. The study of spectra indicates that 27% ofthe atoms are in third energy level and 15% of atoms in second energy level and rest in ground state.Ionization energy of H atom is 13.6 eV. Calculate

(i) the number of atoms present in third and second energy levels,(ii) the total energy evolved when all the atoms returned to ground state.

*37. The ionization energy of H is 13.6 eV, it is exposed to electromagnetic waves of wavelength 1028and gives out induced radiations. Find the wavelength of these induced radiations.

38. The angular momentum of an electron in a Bohrs orbit of H-atom is 4.2178 1034 kg m2/s.Calculate the spectral line emitted when electron falls from this level to next lower level.

39. Which electronic level allows the hydrogen atom to absorb a photon but not emit photon?(a) 3s (b) 1s(c) 2p (d) 3d

*40. The temperature at which the de Broglie wavelength of helium atom is 0.62 Awill be(Atomic mass of helium = 4.04)(a) 710.5 K (b) 750.4 K(c) 410.75 K (d) 570.4 K

41. A dust particle has mass equal to 1011 g, diameter of 104 cm and velocity 104 cm s1. The errorin measurement of velocity is 0.11. Calculate the uncertainty in its position. Commenton the result.

42. What is the maximum precision with which the momentum of an electron may be known if the

position is determined within 0.0001.

43. (a) The Schrodinger wave equation for hydrogen atom is:

( )

//

/

3 2r ao

2s 1 2o o

r1 12 e

a a4 2

=

Where ao

is Bohrs radius. Let the radial node in 2s be at ro. Then find r

oin terms of a

o.

(b) A base ball having mass 100 g moves with velocity 100 m/sec. Find out the value ofwavelength of base ball.


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44. The principal quantum number of an atom is related to the : (1983)(a) size of the orbital (b) spin angular momentum(c) orientation of the orbital in space (d) orbital angular momentum

45. The electrons, identified by quantum numbers n and l(i) n = 5, l = 1 (ii) n = 5, l = 0 (iii) n = 4, l = 1 (iv) n = 4, l = 2, can be placed in order of increasingenergy, from the lowest to highest, as(a) (ii) < (iii) < (iv) < (i) (b) (iii) < (ii) < (iv) < (i)(c) (i) > (iv) > (ii) > (iii) (d) (i) > (iv) > (iii) > (ii)

46. Which of the following sets of quantum numbers is not allowed?(a) n = 3, l = 1, m = +2 (b) n = 3, l = 1, m = +1(c) n = 3, l = 0, m = 0 (d) n = 3, l = 2, m 2

47. Which of the following sets of quantum number is/are not allowed.

(I) n = 4, l = 3, m = - 1, s = +1

2(II) n = 2, l = 3, m = + 1, s = -

1

2

(III) n = 3, l = 0, m = + 1, s = +1

2(IV) n = 2, l = 2, m = + 1, s = +

1

2

(a) only I (b) II, III and IV (c) only II and III (d) only IV

48. Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z = 37) is :(1984)

(a) 5, 0, 0,1

2+ (b) 5, 1, 0, 1

2+ (c) 5, 1, 1, 1

2+ (d) 6, 0, 0, 1

2+

49. Which of the following atoms would be expected to be most paramagnetic?(a) 3Li (b) 4Be

(c) 5B (d) 6C

50. The ratio of magnetic moments of Cr, Cu, Fe is

(a) 4 :1: 8 (b) 6 : 1 : 4

(c) 2 :1: 8 (d) 8 :1: 8

51. A 3d orbital has

(a) zero radial and two angular nodes(b) two radial and two angular nodes.(c) three radial and three angular nodes(d) two radial and zero angular nodes


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52. Ground state electronic configuration of oxygen atom can be represented by

(a) (b)

(c) (d)

53. Give reason why the ground state outermost electronic configuration of silicon is : (1985)

and not

3p3s 3s 3p

54. Which of the following statement(s) is(are) correct?(a) Oxygen molecule is diamagnetic(b) The electronic configuration of Au is 5d106s1

(c) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type

(d) The oxidation state of nitrogen is N3H is3

55. What is the maximum number of electrons that may be present in all the atomic orbitals withprincipal quantum number 3 and the azimuthal quantum number 2? (1985)

56. A compound of vanadium has a magnetic moment of 1.73 BM. Work out the electronicconfiguration of the vanadium ion in the compound. (1997)

57. Write electonic configuration of the following: C(z = 6), Ar(z = 18), Br(z = 35), Cs (z = 55)

58. Write electronic configuration of Mn2+ ion and calculate (i) number of unpaired electrons, (ii) magnetic

moment and find whether the ion is paramagnetic or diamagnetic.

(a) Statement I is True, StatementII is True;Statement II is a correct explanation for Statement I(b) Statement I is True, Statement II is True;StatementII is not a correct explanation for Statement I(c) StatementI is True, StatementII is False(d) Statement I is False, StatementII is True.

59. Statement 1: - An orbital cannot have more than two electronsStatement 2:- The two electrons in an orbital create opposing magnetic field.

60. Statement 1:- On increasing the intensity of radiation, the number of photoelectrons ejected andtheir KE increases.Statement 2:- Greater intensity means, greater energy which in turn means greater frequency ofradiation

61. Statement 1:- The configuration of B atom cannot be 1s2 2s2

Statement 2:- Hunds rule demands that configuration should display maximum multiplicity.


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62. Statement 1:- Angular momentum of e in 1s, 2s, 3s etc. is zeroStatement 2:- 1s, 2s 3s ... all have spherical shape

63. Statement 1:- In Rutherfords gold foil experiment, very few -particles are deflected back.Statement 2:- Nucleus present inside the atom is heavy.

64. StatementI : Band gap in germanium is small.BecauseStatementII: The energy spread of each germanium atomic energy level is infinitesimally small.

(2007)

65. Match the followingColumn I Column II(i) Aufbau principle P. Line spectrum invisible region

(ii) de Broglie Q. Orientation of e in an orbital(iii) Hunds rule R. h m = = (iv) Balmer series S. Electronic configuration

66. Column I Column II(i) Thomson P. Exclusion principle(ii) Pauli Q. Atomic model(iii) Bohr R. Cathode rays(iv) Chadwick S. Neutron

67. Column I Column II(i) Cathode rays P. Helium nuclei(ii) Alpha particles Q. Dumbell(iii) X-rays R. electrons(iv) P-orbitals S. Electro magnetic radiation

68. According to Bohrs theoryE

n= total energy

Kn

= Kinetic energy

Vn= Potential energy

rn

= Radius of nth orbit (2006)

Match the following:Column I Column II(A) V

n/ K

n= ? (P) 0

(B) If radius of nth oribit , ?xnE x = (Q)1

(C) Angular momentum in lowest orbital (R)2

(D) , ?y

n

1z y

r = (S) 1


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Passage

Based upon the information given below, answer the questions that follow:

According to Bohrs theory for single electron species. The radius of orbit of an e

2

n

n

r 0.52 9 Z= ,

the velocity of e in an orbit,6

nz

2.18 10 m/ sn

= and, the Energy of e in an orbit

2

n 2

ZE 13.6 eV / atom

n= .

Where n is the orbit in which e revolves and Z is the atomic number of species

69. The radius of second orbit of He+ ion is:(a) 0.529 (b) 0.529 2

(c) 0.529 4 (d) 0.529 1

2

70. The ratio of velocities of electron in third orbit of H-atom to that of electron in second orbit of Li2+ ionis:

(a)1

2(b)

3

2(c)

2

9(d)

1

3

71. The energy required to remove an electron from 1st excited state of H-atom is:(a) + 13.6 eV (b) + 3.4 eV (c) + 1,9 eV (d) zero


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Answer Key

Atomic Structure

2. c 3. d 4. a

5. 6.34104 ms1 6. b 7. 32.85 V8. 2.63 1019 J 9. b 10. 47.68 1019 J

11. 1.14831015 ms1 , 2.612 107 m 12. 63.57 V

13. 55.87 10 m/ s 14. 346.38 10 Js 15. 2.161020 J

16. b 17. c 18. 3.67105 cm

19. 16.351019 J, 2, He+ 20. 1220 A 21. 104.7 10 m22. 102.38 10 m. 23. 61.09 10 m/ s 24. 127419.5 cm

25. 494.5 kJ/mole 26. 6.6 107 m 27. 2.92 1015 s1

28. 3 29. a 30. c

31. b 32. d 33. a

34. c 35. ,2 2

o o o o

e e

4 a 6 a

36. (i) 1.62 1023 atoms, (ii) 830.50 J

37. 1.028 107 m, 1.216 107 m, 6.5688 107 m38. 1.8 104 cm 39. b 40. c

41. 5.27 106 42. 22 15.28 10 Kg ms

43. (a) 2ao

(b) 6.6261025A 44. a 45. b

46. a 47. b 48. a49. d 50. a 51. a52. d 54. b, c 55. 1056. 1s2 2s2 2p6 3s2 3p6 3d1 4s0 59. b 60. d61. b 62. b 63. b

64. c 65. (i) S, (ii) R, (iii) Q, (iv) P

66. (i) R, (ii) P, (iii) Q, (iv) S 67. (i) R, (ii) P, (iii) S, (iv) Q

68. A R, B Q, C P, D S 69. b 70. c

71. b

Documents

02. Atomic Structure [1-35]