STRUCTURE OF ATOM
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Atom
is the smallest indivisible particle of the matter. Atom is made of electron,
proton and neutrons.
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PARTICLE
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ELECTRON
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PROTON
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NEUTRON
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Discovery
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Sir. J. J.
Thomson
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Goldstein
(1886)
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Chadwick
(1932)
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(1869)
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Nature of charge
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Negative
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Positive
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Neutral
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Amount of
charge
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1.6 x 10-19Coloumb
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1.6 x 10-19Coloumb
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0
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Mass
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9.11 x 10-31kg
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1.672614 x 10-27kg
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1.67492 x10-27kg
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Electrons were discovered using cathode ray
discharge tube experiment. 
Nucleus was discovered by Rutherford in 1911.
Nucleus was discovered by Rutherford in 1911.
Cathode ray discharge tube experiment: A
cathode ray discharge tube madeof glass is taken with two electrodes. At very
low pressure and high voltage,current starts flowing through a stream of
particles moving in the tube fromcathode to anode. These rays were called
cathode rays. When a perforatedanode was taken, the cathode rays struck the
other end of the glass tube atthe fluorescent coating and a bright spot on the
coating was developed
Results:
a.
Cathode rays consist of negatively
charged electrons.
b.
Cathode rays themselves are not visible
but their behavior can be observed with
help of fluorescent or phosphorescent
materials.
c. In
absence of electrical or magnetic field cathode rays travel in straight lines
d. In
presence of electrical or magnetic field, behaviour of cathode rays is similar
to that shown by electrons
e. The
characteristics of the cathode rays do not depend upon the material of the
electrodes and the nature of the gas present in the cathode ray tube.
Charge to mass ratio of an electron was
determined by Thomson. The chargeto mass ratio of an electron as 1.758820 x 1011
C kg-1
Charge on an electron was determined by R A
Millikan by using an oil dropexperiment. The value of the charge on an electron
is -1.6 x 10-19C. The mass on an electron was determined by
combining the results ofThomson’s experiment and Millikan’s oil drop experiment.
The mass of anelectron was determined to be 9.1094 x 10-31kg. Discovery of protons and canal rays: Modified
cathode ray tube experimentwas carried out which led to the discovery of
protons. Characteristics of positively charged
particles:
a. Charge to mass ratio
of particles depends on gas from which these originate
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b. The
positively charged particles depend upon the nature of gas present in the
cathode ray discharge tube
c.
Some of the positively charged particles
carry a multiple of fundamental of electrical charge.
d.
Behaviour of positively charged
particles in electrical or magnetic field is opposite to that observed for
cathode rays
Neutrons were discovered by James Chadwick
by bombarding a thin sheet of beryllium by α- particles. They are electrically
neutral particles having a mass slightly greater than that of the protons. Atomic number (Z) : the number of protons
present in the nucleus (Moseley1913). Mass Number (A) :Sum
of the number of protons and neutrons present in thenucleus. Thomson model of an atom: This model
proposed that atom is considered asa uniform positively charged
sphere and electrons are embedded in it.An important feature of Thomson model
of an atom was that mass of atom isconsidered to be evenly spread over the
atom.Thomson model of atom is also called as Plum pudding, raisin pudding
orwatermelon modelThomson model of atom was discarded because it could not
explain certainexperimental results like the scattering of α- particles by thin metal foils.
Observations from α- particles
scattering experiment by Rutherford:
a.
Most
of the α- particles passed through gold foil un deflected
b.
A
small fraction of α- particles got
deflected through small angles
c.
Very
few α- particles did
not pass through foil but suffered large deflection nearly180o
Conclusions Rutherford drew from α- particles
scattering experiment:
a.
Since
most of the α-particles
passed through foil undeflected, it means most of the space in atom is empty
b.
Since
some of the α-particles are
deflected to certain angles, it means that there is positively mass present in
atom
c.
Since
only some of the α-particles
suffered large deflections, the positively charged mass must be occupying very
small space
d.
Strong
deflections or even bouncing back of α-particles
from metal foil were due to direct collision with positively charged mass in
atom
Rutherford’s model of atom:
This
model explained that atom consists ofnucleus which is
concentrated in a very small volume. The nucleus comprisesof protons and
neutrons. The electrons revolve around the nucleus in fixedorbits. Electrons
and nucleus are held together by electrostatic forces ofattraction. Drawbacks of Rutherford’s model of atom:
a. According
to Rutherford’s model of atom, electrons which are negatively charged particles
revolve around the nucleus in fixed orbits. Thus,
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b. theelectrons
undergo acceleration. According to electromagnetic theory of Maxwell, a charged
particle undergoing acceleration should emitelectromagnetic radiation. Thus, an
electron in an orbit should emitradiation. Thus, the orbit should shrink. But
this does not happen.
c. The
model does not give any information about how electrons aredistributed around
nucleus and what are energies of these electrons
Isotopes: These are the
atoms of the same element having the same atomicnumber but different
mass number.e g 1H1,1H2,1H3 Isobars: Isobars are the
atoms of different elements having the same massnumber but different
atomic number.e g 18Ar40
, 20Ca40 Isoelectronic species:
These are those species which have the same numberof electrons. Electromagnetic radiations:
The radiations which are associated withelectrical and magnetic fields
are called electromagnetic radiations. When anelectrically charged particle
moves under acceleration, alternating electricaland magnetic fields are
produced and transmitted. These fields aretransmitted in the form of waves.
These waves are called electromagneticwaves or electromagnetic radiations. Properties of electromagnetic radiations:
a. Oscillating
electric and magnetic field are produced by oscillating charged particles.
These fields are perpendicular to each other and both areperpendicular to the
direction of propagation of the wave.
b.
They do not need a medium to travel.
That means they can even travel in vacuum.
Characteristics of electromagnetic radiations:
a. Wavelength:
It
may be defined as the distance between two neighbouring crests or
troughs of wave as shown. It is denoted by λ.
b. Frequency
(ν):
It
may be defined as the number of waves which passthrough a particular
point in one second.
c.
Velocity (v): It is defined as the distance
travelled by a wave in
onesecond. In vacuum all types of
electromagnetic radiations travel with thesame velocity. Its value is 3 X108m
sec-1.
It is denoted by v
d.
Wave number: Wave number 
is defined as the number of wavelengths per
unit length.
is defined as the number of wavelengths per
unit length. Velocity = frequency x wavelength c = νλ Planck's Quantum Theory-
o
The radiant energy is emitted or
absorbed not continuously but discontinuously in the form of small discrete
packets of energy called
‘quantum’. In case of light , the
quantum of energy is called a ‘photon’
o The energy of
each quantum is directly proportional to the frequency of
the radiation, i.e. E α
υ or E= hυ where h= Planck’s constant =
6.626 x 10-27
Js
o
Energy is always emitted or absorbed as
integral multiple of this quantum. E=nhυ Where n=1,2,3,4,.....
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Black body: An ideal body,
which emits and absorbs all frequencies, is calleda black body. The
radiation emitted by such a body is called black bodyradiation. Photoelectric effect:
The phenomenon of ejection of electrons from thesurface of metal when
light of suitable frequency strikes it is calledphotoelectric effect. The
ejected electrons are called photoelectrons. Experimental results observed for the
experiment of Photoelectric effect-
o
When beam of light falls on a metal
surface electrons are ejectedimmediately.
o
Number of electrons ejected is
proportional to intensity or brightness of light
o
Threshold frequency (vo): For each metal
there is a characteristicminimum frequency below which photoelectric effect is
not observed. Thisis called threshold
frequency.
o
If frequency of light is less than the
threshold frequency there is noejection of electrons no matter how long it
falls on surface or how high isits intensity.
Photoelectric work function (Wo): The minimum
energy required to ejectelectrons is called photoelectric work function.Wo= hvo
Energy of the ejected electrons :
Dual behavior of electromagnetic radiation-
The light possesses both particle and wave like properties, i.e., light has
dual behavior . whenever radiation interacts with matter, it displays particle
like properties.(Black body radiation and photoelectric effect) Wave like
properties are exhibited when it propagates(interference an diffraction) When a white light is passed through a prism,
it splits into a series ofcoloured bands known as spectrum. Spectrum is of two types: continuous and line
spectrum
a. The
spectrum which consists of all the wavelengths is called continuous
spectrum.
b.
A spectrum in which only specific
wavelengths are present is known as a line
spectrum. It has bright lines with dark
spaces between them.
Electromagnetic spectrum is a continuous
spectrum. It consists of a range ofelectromagnetic radiations arranged in the
order of increasing wavelengths ordecreasing frequencies. It extends from radio
waves to gamma rays. Spectrum is also classified as emission and
line spectrum.
o Emission spectrum: The spectrum of
radiationemitted by a substance that has absorbed energy
is called an emissionspectrum.
o
Absorption spectrum is the spectrum
obtained when radiation is passedthrough a sample of material. The sample
absorbs radiation of
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The
study of emission or absorption spectra is referred as spectroscopy. Spectral
Lines for atomic hydrogen:
Rydberg equation
R = Rydberg’s constant = 109677
cm-1
Bohr’s model for hydrogen atom:
a. An
electron in the hydrogen atom can move around the nucleus in a circular path of
fixed radius and energy. These paths are called orbits orenergy levels. These
orbits are arranged concentrically around thenucleus.
b.
As long as an electron remains in a particular
orbit, it does not lose or gain energy and its energy remains constant.
c. When
transition occurs between two stationary states that differ inenergy, the
frequency of the radiation absorbed or emitted can becalculated
d.
An electron can move only in those
orbits for which its angularmomentum is an integral multiple of h/2π
The radius of the nth orbit is given
byrn =52.9 pm x n2 Z energy of electron in nth orbit is :
Limitations of Bohr’s model of atom:
a. Bohr’s
model failed to account for the finer details of the hydrogen spectrum.
b. Bohr’s
model was also unable to explain spectrum of atoms containing more than one
electron.
Dual behavior of matter: de
Broglie proposed that matter exhibits dualbehavior i.e. matter shows
both particle and wave nature. de Broglie’s relation is
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Heisenberg’s uncertainty principle: It
states that it is impossible to determine simultaneously, the exact
position and exact momentum (or velocity) of an electron.The product of their
uncertainties is always equal to or greater than h/4π.
Heisenberg’s uncertainty principle rules out
the existence of definite pathsor trajectories of electrons and other similar
particles Failure of Bohr’s model:
a. It ignores the
dual behavior of matter.
b.
It contradicts Heisenberg’s uncertainty
principle.
Classical mechanics is based on Newton’s laws
of motion. It successfullydescribes the motion of macroscopic particles but
fails in the case ofmicroscopic particles.
Reason: Classical mechanics ignores the concept of
dual behaviour of matter especially for sub-atomic particles and the
Heisenberg’s uncertainty principle.
Quantum mechanics is
a theoretical science that deals with the study of themotions of the
microscopic objects that have both observable wave like andparticle like
properties. Quantum mechanics is based on a fundamental
equation which is calledSchrodinger equation. Schrodinger’s equation: For a system (such as
an atom or a molecule whoseenergy does not change with time) the Schrödinger
equation is written as:
When Schrödinger equation is solved for
hydrogen atom, the solution givesthe possible energy levels the electron can
occupy and the correspondingwave function(s) of the electron associated with
each energy level.Out of the possible values, only certain solutions are
permitted. Eachpermitted solution is highly significant as it corresponds to a
definite energystate. Thus, we can say that energy is quantized. ψ gives us the
amplitude of wave. The value of ψhas no
physicalsignificance. Ψ2gives
us the region in which the probability of finding an electron ismaximum.
It is called probability density.
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Orbital: The region of space around the
nucleus where the probability offinding an electron is maximum is called an
orbital.
Quantum numbers: There are a set of four
quantum numbers which specifythe energy, size, shape and orientation of an
orbital. To specify an orbital only three quantum numbers are required while to
specify an electron all four quantum numbers are required.
Principal quantum number (n):It identifies
shell, determines sizes and
energy of orbitals
Azimuthal quantum number (l): Azimuthal
quantum number. ‘l’ is also known as orbital angular momentum or
subsidiary quantum number. l. It identifies sub-shell, determines the shape of
orbitals, energy of orbitals in multi-electron atoms along with principal
quantum number and orbital angular
momentum, i.e., The number of orbitals in a subshell = 2l
+ 1. For a given value of n, it can have n values ranging from 0
to n-1. Total number of subshells in a particular shell is equal to the value
of n.
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Subshell
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s
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p
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d
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f
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g
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notation
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Value of ‘l’
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0
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1
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2
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3
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4
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Number of
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1
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3
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5
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7
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9
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orbitals
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Magnetic quantum number or Magnetic orbital
quantum number (ml):
Itgives
information about the spatial orientation of the orbital with respect
tostandard set of co-ordinate axis.For any sub-shell (defined by ‘l’ value)
2l+1 values of ml are possible.For each value of l, ml
= – l, – (l –1), – (l–2)... 0,1...
(l – 2), (l–1), l
Electron spin quantum number (ms): It
refers to orientation of the spin of theelectron. It can have two values
+1/2 and -1/2. +1/2 identifies the clockwisespin and -1/2 identifies the anti-
clockwise spin.
The region where this probability density
function reduces to zero is callednodal surfaces or simply nodes.
Radial nodes: Radial nodes occur when the
probability density of wave functionfor the electron is zero on a spherical
surface of a particular radius. Numberof radial nodes = n – l – 1
Angular nodes: Angular nodes occur when the
probability density wavefunction for the electron is zero along the directions
specified by a particularangle. Number of angular nodes = l
Total number of nodes = n – 1
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Shape of p and d-orbitals
Shielding effect or screening effect:
Due to the presence of electrons in theinner shells, the electron in the
outer shell will not experience the full positivecharge on the nucleus.
So,
due to the screening effect, the net positive charge experienced by theelectron
from the nucleus is lowered and is known as effective nuclear charge.
Effective nuclear charge experienced by the orbital decreases with increase of
azimuthal quantum number (l).
Aufbau Principle:
In the ground state of the atoms, the orbitals are filled inorder of
their increasing energies
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n+l rule-Orbitals with lower value of (n+l)
have lower energy. If two orbitals have the same value of (n+l) then orbital
with lower value of nwill have lower energy.
The order in which the orbitals are filled
isas follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s...
Pauli Exclusion Principle:
No two electrons in an atom can have the same setof four quantum
numbers. Only two electrons may exist in the same orbitaland these electrons
must have opposite spin.
Hund’s rule of maximum multiplicity:
Pairing of electrons in the orbitalsbelonging to the same subshell (p, d
or f) does not take place until eachorbital belonging to that subshell has got
one electron each i.e., it is singlyoccupied.
Electronic configuration of atoms:Arrangement
of electrons in different orbitals of an atom. The electronic
configuration of differentatoms can be represented in two ways.
a.
sapbdc......
notation.
b.
Orbital diagram:, each orbital of the
subshell is represented by a box and the
electron
is represented by an arrow (↑) a positive spin or an arrow (↓)
a negative spin.
Stability of completely filled and half filled
subshells:
a. Symmetrical
distribution of electrons- the completely filled or half filled sub-shells have
symmetrical distribution of electrons in
them and are more stable.
b. Exchange
energy-The two or more electrons with the same spin present in the degenerate
orbitals of a sub-shell can exchange their position and the energy released due
to this exchange is called exchange energy. The number of exchanges is maximum
when the subshell is either half filled or completely filled. As a result the
exchange energy is maximum and so is the stability.
ONE MARK QUESTIONS
1. 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.
2.
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 106 m-1
3. What
are the values of n and l for 2p orbital? Ans. n=2 and l= 1
4. Which
of the following orbitals are not possible? 1p, 2s, 3f and 4d Ans. 1p and 3f
are not possible.
5. Write
the electronic configuration of the element having atomic number 24. Ans. 1s2
2s2
2p6
3s2
3p6
3d5
4s1
6.
What atoms are indicated by the
following electronic configurations?
25
Ans.
a. Boron b. Scandium
7. 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.
8.
State Pauli Exclusion Principle.
Ans. No two electrons in an atom can
have the same set of four quantum numbers or an orbital can have maximum two
electrons and these must have opposite spin. 9. When α- rays hit a thin foil of
gold, very few α- particles is deflected back.
What does it prove?
Ans.
There is a very small heavy body present within the atom. 10. 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.
TWO MARKS QUESTIONS
1. 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) 3517Cl (b) 23392U
2. Using
s,p,d and f notation, describe the orbital with the following quantum numbers-
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(a) n=1,l=0
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(b) n=3, l=1
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(c) n=4, l=2 (d) n=4, l=3
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Ans.
(a) 1s
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(b) 3p (c)4d
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(d) 4f
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3. How many electrons in an atom have the
following quantum numbers?
a. n=4, ms=
-1/2 b.
n =3 , l=o
Ans. (a) 16 electrons (b)
2 electrons.
4.
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 8135Br
5. (i)
The energy associated with the first orbit in the hydrogen atom is -2.18 x 10-18J/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/ 52 = -8.72 x 10-20
J
(ii) For H atom, rn=
0.529 x n2 r5
= 0.529 x 52 = 13.225 A0=
1.3225 nm
6. Explain , giving reasons, which of the
following sets of quantum numbers are
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not possible.
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(a) n=0, l=0;
ml = 0, ms= + ½
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(c)n=1, l=0; ml = 0, ms= - ½
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(b) n=1, l=1;
ml =- 0, ms= + ½
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(d)
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n=2, l=1; ml = 0, ms= + ½
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Ans.
(a) Not possible because n≠ 0 (c)
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Not possible because when n=1, l≠1
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7.
(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 mlfor
this electron.
Ans.(a) minimum value of n= 5 (b)n=3, l=2, ml
= -2, -1, 0, +1, +2
8. Calculate the total
number of angular nodes and radial nodes present in 30 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
9.
Mention the draw backs of Rutherford’s
atomic model.
Ans. 1. It could not explain the
stability of an atom.
2.
It could not explain the line spectrum
of H- atom.
10. State
de-Broglie concept of dual nature of matter. How do dual nature of electron
verified?
Ans. Just as light has
dual nature, every material particle in motion has dual nature (particle nature
and wave nature). The wave nature has been verified by Davisson
and Germer’sexperiment
whereas particle nature by scintillation experiment.
THREE MARKS QUESTIONS
1. State
(a)Hund’s Rule of maximum Multiplicity (b) Aufbau Principle (c) n+l rule
Ans.(a)
Pairing of electrons in the orbitals belonging to the same subshell (p, d or f)
does not takeplace until each orbital belonging to that subshell has got one
electron each i.e., it is singly occupied.
(b)In the ground state of the atoms, the orbitals
are filled inorder of their increasing energies
(c) Orbitals with lower value of (n+l) have lower
energy.If two orbitals have the same value of (n+l) then orbital with lower
value of n will have lower energy.
2.
Write down the quantum numbers n and l for the following orbitals
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a. 2p
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b. 3d
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c.
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5f
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Ans.
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a. n=2, l= 1
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b. n= 3, l=2
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c. n= 5, l=3
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3. Write the 3 points of difference between
orbit and orbital.
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Ans.
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Orbit
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orbital
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1. An orbit is a well defined
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1. An orbital is the three dimensional
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circular path
around the
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space around
the nucleus within
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nucleus in which the
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which the probability of
finding an
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electrons
revolve
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electron is
maximum(upto 90 %)
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2.
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It represents
the planar
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2. It represents the three dimensional
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motion of an
electron around
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motion of an
electron around the
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the nucleus
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nucleus
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3.
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All orbits are
circular and
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3.
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Different
orbitals have different
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disc like
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shapes, i.e., s-orbitals are
spherically
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symmetrical,
p-orbitals are dumb-bell
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shaped and so on.
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4. State
Heisenberg’s uncertainty principle.calculate the uncertainty in the position of
an electron if the uncertainty in its velocity is 5.7 x 105
m/s.
Ans.
It states that it is impossible to determine simultaneously, the exact position
and exact momentum (or velocity) of an electron. The product of their
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uncertainties
is always equal to or greater than h/4π.
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Δx x (m x Δv) = h/4ᴨ
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Δx = h/4ᴨ x m
x Δv =
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6.6 x 10-34
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= 1.0 x 10-10 m
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4
x 3.14 x 9.1 x 10-31
x 5.7 x 105
x 3.14 x 9.1 x 10-31
x 5.7 x 105
5.
Write 3 points of differences between
electromagnetic waves and matterwaves.
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Electromagnetic
waves
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Matter waves
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1.
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These are
associated with
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1.
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These are not
associated with
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electric and
magnetic
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electric and
magnetic field.
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fields
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2.
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They do not
require any
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2.
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They require
medium for
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medium for
propagation.
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propagation
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3.
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They travel
with the same
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3.
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They travel
with lower speeds
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speed a that
of light
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not constanr
for all matter
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waves
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6.
(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 × 10–3kg)
=
0.1098 × 10–3
+ 31
= 0.1098 × 1028
= 1.098 × 1027
(ii) Mass of one electron = 9.10939 × 10–31 kg
Mass
of one mole of electron = (6.022 × 1023)
× (9.10939 ×10–31 kg) = 5.48 × 10–7
kg
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Charge on one mole of electron =
(1.6022 × 10–19C) (6.022 × 1023)
=
9.65 × 104 C
7.
Find
energy of each of the photons which
(i) correspond to light
of frequency 3× 1015Hz. (ii) have
wavelength of 0.50 Å.
Ans.(i)
Energy (E) of a photon is given by the expression,
E
=
Where,
h = Planck’s constant =
6.626 × 10–34Js ν = frequency
of light = 3 × 1015Hz Substituting
the values in the given expression ofE:
E = (6.626 × 10–34) (3 × 1015) E = 1.988 × 10–18J
(ii) Energy (E) of a
photon having wavelength (λ)is given by the expression,
h = Planck’s constant =
6.626 × 10–34Js
c
= velocity of light in vacuum = 3 × 108m/s
Substituting the values in the given expression of E:
8. 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.Theni=
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,
29
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:
9. 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 1s22s2
2p6
3s2
3p64s2
3d10
(iii)
Copper
10. Give the number of electrons in the
species,
H2 and
Ans. Number of electrons
present in hydrogen molecule (H2)
= 1 + 1 = 2
∴
Number of electrons in = 2 – 1 = 1 Number of electrons in H2
= 1 + 1 = 2
Number of electrons present in oxygen
molecule (O2) = 8 + 8 = 16
∴ Number
of electrons in = 16 – 1 = 15
FIVE MARKS QUESTIONS
WITH ANSWERS
30
1. 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 i
4. Bohr’s
model could not explain the ability of atoms to form molecules by chemical
bonds
Since a 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.
2. 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):
31
6.53 × 10–7 m
Hence, the threshold wavelengthis
653 nm.
(b) From the expression,,
we get:
Where, ν0=
threshold frequency h = Planck’s constant
Substituting the values
in the given expression ofν0:
(1
eV = 1.602 × 10–19J) ν0=
4.593 × 1014s–1
Hence, the threshold
frequency of radiation (ν0)
is 4.593 × 1014s–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 × 10–20J
Since K.E
v = 4.52 × 105ms–1
Hence, the velocity of
the ejected photoelectron (v) is 4.52 × 105ms–1.
3. (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:
32
2.
n= 3, l = 2,
ml= 1 , ms=
+1/2
3.
n= 4, l =
1, ml=
0 , ms=
+1/2
4.
n = 3, l= 2,
ml= –2 , ms=
–1/2
5.
n = 3, l= 1,
ml= –1 ,ms=
+1/2
6.
n = 4, l= 1,
ml= 0 , ms=
+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)Forn = 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.
4. (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
2s2 2p63s2
3p3 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
2s2 2p6
3s23p2
The orbital picture of Si can be represented as:
From the orbital
picture, silicon has two unpaired electrons.
(c) Chromium (Cr):
Atomic number = 24
33
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
2s2
2p6 3s23p6
4s2 3d6
The orbital picture of chromium is:
From the orbital picture, iron has four
unpaired electrons.
HOTS QUESTIONS WITH ANSWERS
1. 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
2p6 3s23p63d104s24p6
Total no. of s-electrons = 8, total no. of p-electrons = 18. Difference = 10
No. of d- electrons = 10
2.
What is the minimum product of
uncertainty in position and momentum of an electron?
Ans.h/4π
3. Which
orbital is non-directional ? Ans. s- orbital
4. What
is the difference between the notations l and L ? Ans. l represents the
sub-shell and L represent shell.
5. How
many electrons in an atom can have n + l = 6 ? Ans. 18
6.
An
anion A3+ has 18 electrons. Write the atomic number of
A.
Ans.15
7. 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
34