484 74 479MB
English Pages 848 Year 2005
Revised Sixth Edition
• '
.•
(NEW REVISED SIXTH EDITION)
s.o. Pillai
B.Sc. (Hons.), M.Sc., Ph.D.
Professor of Physics (RetJL) Anna University, Chennai India
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•l'"E88
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NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835724, Ansari Road, Datyaganj, New Delhi - 11 the classical theory. It can revolve only in a few widely separated permitted orbits While moving along these
oibih round the nucleus, an electron docs mH radiau- energy. These non-radiating orbits arc called stationary orbits.
fiii)
The permissible orbits of an electron revolving round J nucleus are those for which rhe angular momentum of the electron is an integral multiple ot ftfln, where k is Pian-ck*s constant. Thus for
any permitted orbit*
j
ft
V*
':
or
Ffr)dr
f Ze*dr PE~ 1 4lt£0r;
Thai is, Zr2
P-E«
4neur„
Similarly K.E of the electron -r-iwvj Substituting the value of Imv* from equation (2>4h we get
Thus the total energy of the electron in the n* orbit is
E^PE^KE Ze*
Ze1
*
Ze*
4ke0fh Substituting the value of ra from equation (2.7)
(240)
The negative sign of the energy expression shows that the electron is bound to the nucleus and some work must be done to pull jt away. It further follows from equation (2.10) that as n increases, £„ becomes
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H
Solid State Phyxii r
le^-% negative and hence its algebraical value increases. I'he electron. therefore has minimum energy when it is in its innermost orbit, fl = L The stale of the atom with the electron revolving in the innermost orbit is called the ground or normal state and u obviously the most stable state of the atom.
VIL
CALCULATION OF rn AND En FOR HYDROGEN ATOM " ■■'
■
nV r - —------ ; it fliZe
with 2 - i
..
I
. {Refer equation (2.7)]
n 6.626 x l0~w)?(8.854x IO'11) *“ n(9.109 x 1G*)(1.602 x KT'-f I I
r, - («’) (0.529 x 10' '*) metre
(2.11)
t Thus radius of the first orbit of hydrogen, r - 0.529 x 10rf° m.
This is called the Bohr radius (r#). f; =
Thus
= 0.0529 nm I
rj = 2
and
>.
r, = rtV,
The energy of the electron in the
orbit is
tnZ^e* Efl *------- 2~"j t 8 £* ft n
with
= 1
[Refer equation (2.10)|
_ 8 me* E* A; where 8 *—=—? * n1 8ejh2
Thus
-
e *
(2.12)
tory
;
n!x8(8.354x 1 fl-11)’(6.626 X 10M)2 k
„
2.179XI0'1*. . =------------ j------ joule
2.179x IO-11 '^1.602 x IO-1*)”
।
4—(2.13) fl*
when
A - 1 E;*-13.6 eV
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Jfevfrw ofAtomic Structure
9
This corresponds to the ground state energy of the aiom, and is called the ionirarion potential of hydrogen atom. Similarly, the energy of the atom with the electron in the second otbit is
and that with ihe electron in the third orbit 13.6 a
*
-L511 eV and so on.
BOHR'S INTERPRETATION OF HYDROGEN SPECTRUM
VIII.
If an electron jumps from an outer initial orbit n, of higher energy to an inner f inal orbit
of lower energy,
the frequency of the radiation emitted is given by
U 8€jfcJL"? mZ1**
Hence
mA‘[
1 «e}*4«i
i
X
I 5 «s.
i
(2.14)
l
c
/nZ:e4
>-
8ejh!L«:J
mzv ur
8eJcAJtni
"1-
The reciprocal ol the wavelength is called the wave number.
Thus
(2.151
mZ’e4 86} cA’
w'ithZ = I is called Rydberg Constant for hydri'gen IHu; Thus (9.109 x 10~>;) (1,602 x 10'7____
S.8.854x 10 b;)‘(3x 10'} (6.626 x 10 '7
Rh - 1.0961 x 10’ tif' An atom is said to be excited, if the electron ts raised to an orbit of higher energy. The electron can be
completely freed from the influence of the nucleus by supphinp sufficient energy and the minimum energy
needed for this is called ionisation energy. For hydrogen
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I0
Solid Suu tr Physics
me*Z2
I
1
SejA’Ll Thus
with Z «I
(2.16)
E = 13 6 eV
As long as the electron remains in its orbit, no energy is radiated, but whenever an electron jumps from an outer orbit to an inner orbit, energy is emitted in the form of radiation. When the hydrogen atom is
subjected to an external source of energy, the electron jumps from lower energy state to higher energy state. The atom is said to be excited, The excited state is not stable and hence the electron returns to its ground stare in about IQ ’ second. The excess of energy is now emitted in the form of radiations of different
wavelengths. In a hydrogen discharge lube there are very large number of hydrogen atoms which arc excited, and radiate energy. In some atoms the electrons may jump from the second orbit to the first orbit in some others from the third to the second or first and so on. The different wavelengths due to different transitions of the elections constitute spectral series which are characteristics of the atom emitting them.
X
SPECTRAL SERIES OF HYDROGEN
Lyman series When electrons jump from second, third, .« etc. orbits io the first orbit* the spectral lines are in the ultra
violet region. Here, n] - i and n3 = J, 3,4... Thus
-8 Vi = g
f15 and u, - /tJ —
and soon.
This is identified as Lyman series.
Belmer Series When electrons jump from outer orbits to the second orbit (n, - 2, n2 - 3.4, we get lines of Balmer series
i.e.
with n. - 3,4,5,".
This series is called Balmer series and Lies in the visible region of the spectrum, The first line in the series = 3} is called Ha line, the second line (n2 = 4) is called and so an.
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Re i'iw of A tomic Structure
Fig. 2.3
11
Quantum jynips giving rise to diflercm spectral series of hydrogen
Paschen series Paschen series lines in the infrared region are due to the transition of electrons from outer orbits to the third orbit Thus
with n2 - 4,5,....
Brackett series If «| ■ 4 and
« 5,5.7.,., we get Brackett series lines.
with «3 = 5,6,7, ft,, ii .etc.
Pfund series If «. =s5 and «2 ~ 6,7,8...,we get the Pfund series lines. Fur this series,
o ~RH
r J. s2
with
7, ft,
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12
Solid Slate Physics
Brackett and Pfund series he in the far infrared region of the hydrogen spectrum. By putting «, = (fl| + 1) in each one of the series, we get the longest wavelength of the series. Similarly by putting
nt = «• in each one of the scries, we get rhe shortest wavelength of the series (or series limit).
n;a>(E+l3 6)
6 (13 2) 5 (13-1) 4 (12-8)
3
(12-1)
2 (10 2)
1 Fig. 2.4
X.
(0)
Energy level diagram of hydiogert.
SHORTCOMINGS OF BOHR’S THEORY
Bohr's theory, based on circular electron orbits, was able to explain successfully a number of experimentally observed facts and has correctly predicted the spectral tines of neutral hydrogen atom and singly ionized
helium atom in terms of the principal quantum number n However, the theory fails to explain the following facts:
(i >
It Codd not account the spectra of atoms more complex than hydrogen.
(ii)
H fails to give any information regarding the distribution and unangetnem of electrons in
(Hl I
atoms. It does not explain the ex pen mentally observed variations er intensity of the spectral lines of an element.
(iv)
It fails to explain the transitions of electrons from one level to archer, the rate al which they
occur or the ie^et7ww rules which apply to them. (v)
it fails to account the/me ffw/wre of spectral lines Actually, it was found that when spectral
fines emitted by an atom are examined, each line is composed of several lines closely packed
together Bohr's theory does ixn Throw any hghi on it.
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Review of Atomic Structure
13
(vi)
The theory cannot be used for the quantitative study of chemical bonding.
(vii)
It was found that when electric or magnetic field is applied to the atom, each spectral hoe splits into several lines. The former one is called Stark effect whiki the later Zee^w« effect.
Bohr's theory fails to explain these effects.
XI. SOMMERFELD’S RELATIVISTIC ATOM MODEL According to Bohr the lines of hydrogen-like spectrum should each have a well-defined wavelength. Care ful spectroscopic examination showed that the
and
lines in the hydrogen spectrum are not single
Each spectra! line actually consisted of several very close lines packed together. Michelson found (hat under high resolution. H.f line can be resolved into two close components, with a wavelength separation 0.D13 nm This is called the fine structure of the spectral lines. Bohr's theory could not explain this fine structure.
Sommerfeld succeeded partially in explaining the observed fine structure of spectral lines by intro ducing the following main modifications in Bohr s theory:
(i)
Sommerfeld suggested that the path of an electron around the nucleus in general, is an ellipse
with (he nucleus at one of the foci The circular orbits of Bohr are a special case of this. (ii)
The velocity of the electron moving in an elliptical orbit varies considerably ai different parts
of the orbit This causes relativistic variation in the mass of the moving electron. Therefore Sommerfeld took into account the relativistic variation of the mass of rhe electron with velocity. Hence this model of the atom is called the rehmwtic aia/n mode/.
XII. ELLIPTICAL ORBITS FOR HYDROGEN An electron moving in the field of ihe nucleus describes elliptical orbits* with ihe nucleus at one focus Cir cular orbits arc only special cases of ellipses. When the electron moves along a circular orbit, the angular coordinate is sufficient to describe its motion In □□ elliptical orbit, the position of the electron at my time
instant is fixed by two coordinates namely the angular coordinate (0) and the radial coordinate (ri Here r is the radius vector and o is the angle which the radius vector makes with the major axis of the ellipse. Consider an electron of mass m and linear tangential
velocity v revolving in the elliptical orbit. TTiis tangential
t
velocity of the electron can he resolved into two components: one along ihc radius vector called radial velocity and the other
|
>***"^*^
perpendicular to the radius vector called the transverse vefoctty. Corresponding to these velocities, the electron has two
r
momenta: one along the radius vector called radial momentum
\
b 0
fI gf X
J
and the other perpendicular to the radius vector known as azimuthal momentum or angular momentum. So in the case of elliptic motion, both the angle © and the radius vector r vary periodically, as shown in Fig.2.5,
Fig. 2.$
Sommcrfdd’i model of the atom.
Thus U« momenta associated with both these coordinates (0 and r) may be quantized in accordartce
with Bohr’s quantum condition. The two quantisation cordilions. are
(2.17) ptdr = nfi
(2.18)
14
Solid State Physics
and
where
system,
arc the two quantum numbers introduced by Sommerfeld. Since they stand for one periodic
-4. Here n, is known as radial quantum number and rq as angular or azimuthal quantum
number, n is known as principal quantum number.
We have now to quantize the momenta associated with both the radial and angular coordinates. The two equations are
J P/T0 ■ V and
The integrals arc taken over one complete cycle of variation of the respective coordinates* Both n*
and nf are integers- Furthermore, the sum of nt and a, is equal io the principal quantum number n. The force on the electron is due to the electrostatic a I traction to the nucleus and acts along the radius
vector There is no force at right angles to this radius vector, so the transverse component of the acceleration & zero ihroughoui the atom. Hence
r^L Therefore r I dr
i
dt „
iszrw.
must be a constant.
Equation (2 J 7) can be easily evaluated* Thus,
0 .(1 +fcos$)J a(l -e3)
11 dr | rl^J
esin0 (1 + ecos0)
15
(2.22)
Referring to equation (X13)
id* 1 dr Substituting for mr — (= pj and -—-from equation (2,22), wc get al raQ ecos0.
d^ = n,A
Referring to equation (2.19) Esini> J + ECOS$.
cty - nrh
21W Li i-ECOS^J put w -tsin^ arj^ v
ffr
(2.23)
1 1 -FECOS0
The above equation may now be written as
«0- ^JJo
1+ecos*
I r^ECOSfrfo
2/tJJo
I +ecos$
I J + E COS0
-1 rf0*=n,
(214)
Sofutrofl of the inu^raL*
P.nns/rinhitirl m ntt^ri a
16
.Wirf Sfui e Pfry
1+ECOS$
Jo Hint: Proper^ of def inite integral
f */(*>&-2 C/UM* if/f2^x)=/Cx)
Jo
Jo
In the integral
l+ecos^
Jo
I+
ecos^
1 +ECOS(2rt-Q) = 1 + £COS$ c*
Jo
f"
tf®
d$
Jo l+ecoa^
(1 +ecos)
I he MsJutjon of this integral is obtained as follows: Let
p,;
fflZ? _ 2 2neop; 0*m, =0.mr = ±^
J?
1?
1? Fig. 2.11
XX.
Quantum representation of electrons in A’-shelL
APPLICATION OF PAULI’S EXCLUSION PRINCIPLE
U) Find the maximum number of electrons that n - 2 orbit can accomodate according to Pauli's exclusion principle.
ForL• shell
n=2 i -0J
- 0,+L-1
Hence, the possible different combinations of the four quantum numbers are:
n
2222222
I
0
0
M*
0
0-1-1
m’
1 1 1 222
2
111111
| “2
1
1
0
0
1 1 1 2222
1: can now be seen that in no two vertical columns, all the four values of quantum numbers are identical. This means that the maximum number of electrons that can exist tn L-shell is 8.
(ii) Maximum number of electrons in orbitals or sub shells Consider an orbit with principal quantum numbers and orbital quantum number I For finding the number of electrons in any orbital* it is clear that the values of a and / are fixed. Only the values of
and ni, can
change. Now rnf — -1 to + / including zero.
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28
Solid State Physics
Thus fi?r & given value of /, there are {2i + 1) values of mr Now for each value of mr
. i 1 values, namely: +
1
can have only two
•
Hence. there can be only 2(21 + I) different combinations of (n< /, rty,
Thus the maximum number
of electrons in any sub-shell with orbital quantum number / = 2(2/ + 1). For example: j-sub shell (/ = 0) can have a maximum of 2(2/ +1) =2 electrons, p*sub shell (/ - 1) can have a maximum of
2(2/ +1) - 6 electrons and so on.
(ill) Maximum number of electrons in main shells The main shell consists of a number of sub-shells. called &rbitols< According to Pauli* s exclusion principle the maximum number of electrons in an orbit with principal quantum number n is equal to W. This can be proved as under" Consider an orbit with principal quantum number n and orbital quantum number L It has already been
shown that the maximum number of electrons in an orbital is equal to 2(2/ +1) with I = 0< L 2t,„ (w-1), Thus the total number of electrons with principal quantum number n is given by
S 2(2/+1) fMfl
M
2n(ji * 1) i
1-0
(2.42a)
=c J
2
Table 2» A 8
/ a/, 2
The distnbuiton of electrons in the various stales Number of electrons in sub-sMH
"',±1
«/ H ta +o
1 I
0 0
0 0
+1/2 -1/2
2
1?
0 0
+1/2 -1/2
2
2?
2
0 0
2 2 2 2 2 2
)
1 1 1 1 1
-1 —1 0 a t 1
+1/2 -1/2 + 1/2 -1/2 + 1/2 -J/2
3 3
0 0
0 0
+ 1/2 -j/2
r 2
Number of eletirrms in main shells
2(K)
8(L)
6
2
3?
(Co/ud,)
Copyriatiied material
Rivriwo* AikmuMuMI _______________
-
- •-------------------------
-•
- -
M
-*-**--+ —
Table 2.1 The distribution ol electrons in the various stales
: I 04
/ = 0,l,2, m, = -1 to + I
s’
n
Number of electrons In subshells
Number of electrons
In main shells (in2)
...» (n-1) -
-
1 1
0 0
0 0
+ 1/2 - 1/2
2 2
0 0
0 0
+1/2 - 1/2
2 2 2 2 . 2 2
1 1
1 1 1 1
-1 -1 0 0 1 1
0 0
0 ■ 0
+ 1/2 -1/2
1 1 1 1
-1 -1
0 0 1 1
+ 1/2 -1/2 . +1/2 -1/2 +1/2 - 1/2
-2 -2 -1 ~ -4 0 ?0 1 * 1 2.
+ 1/2 -1/2 + 1/2 - 1/2 + 1/2 -1/2 + 1/2 -1/2 + 1/2
.
2
is2
2
2?
2(K)
•
+ 1/2 -1/2 ■ + 1/2 -1/2 + 1/2 - 1/2
6 .
2/
2
3s2
8(£)
-
-
3 3
-
3 3 3 3 3 3 3 3 3 3 3. 3 3 3 3 3
1 1 2 2 2 2
.
2 2 2 2 2 2
2
-
>
3p6
6 •
V -
•
-
10
3J10
18(Af)
■
.
-1/2
Hence the maximum number of electrons in an orbit is 2n2. This formula is applicable only to closed shells. The outermost shell cannot have more than eight electrons.
The distribution of electrons in the various shells generally follows the rules given below. (f) The maximum number of electrons possible in a main shell is 2n where n is the principal quantum number for the shell. , (ii) The outermost main shell cannot contain more ectrons (iii) The “last but one” main shell cannot contain more than' elSh^ ec^
(iv) The last shell cannot contain more than two electrons c°ntain more than nine electrons until the “last but two” shells has its full capacity of electrons.
Some Examples of Electron Configuration Hydrogen (Z = 1) £4, „ , „ „ The hydrogen atom has only one electron occupying the r-subshell of the K-shell for which n = 1. This shell can have a maximum of two electrons. Since the shell is incomplete hydrogen is chemically very active and forms new compounds on chemical reaction wit
suitable elements. This
single electron exists either in the m, = + | state or m,=-1 state. Symbolically, the electron configuration of hydrogen is represented as Is1.
Helium (Z = 2) This atom has two electrons which fill the first orbit or AT-shell. The two electrons in this shell are in the m = +— state and m = --7 state (having opposite spins). 2 15 2 The shell is complete and hence helium is inert element. The electronic configuration of helium is represented as k2. The number 4 refers to the principal
quantum number 1. The symbol s refers to the subshell I = 0. The number of electrons 2 is fixed at the top of 5. Carbon (Z = 6) Carbon atom has 6 electrons. Two electrons fill the first main shell. The other four electrons are
in the second main shell. Of these four electrons, two are in the s-subshell and the other two are in the p-subshell. The electronic configuration is represented as lx2 2s2 2p2.
Neon (Z = 10) Two electrons fill the first main shell. Eight electrons fill the second main shell. This is another inert element.
Mendelyeev, a Russian scientist in 1869 showed that the elements when arranged according to the
magnitude of atomic weights, displayed marked periodicity in their properties. Elements were also
found to exhibit periodicity in various physical properties such as atomic volume, expansion coefficient, electrical conductivity and melting point. Based on these observations, the law of periodic table states that the properties of the elements are periodic functions of their atomic weights. Mendelyeev arranged all the elements in the order of their increasing atomic weights in horizontal
rows in such a way that the elements having similar properties come directly under one another in the
Nitrogen Oxygen
7
Scandium
Titanium
Vanadium Chromium
Manganese
21
22
23 24
25
Ti V Cr Mn
47.90 50.94 52.00 54.94
44.96
K Ca
Potassium Calcium
18 19 20
Sc
35.45 39.95 39.10 40.08
Cl Ar
28.09 30.99 32.06
16 17
Si P
22.99 24.31 26.98
12.01 14.01 16.00 19.00 20.18
M N
0
9
6 6
2
6 6 6 6 6
2 2
2 2 2 2 2 2 2 2
4 5 6 6 6
2 2 2 2 2
6 6 6 6 6
2 2 2 2
6
6 6
3
2 2 2 2 2
2
2 2 2 2 2
6
2
1 2 2'1 2 2’
8
2
6 6 6 6
3 4 5 6
2
1
,7
2 2 2 2
2 2 2 2
2
1 2 2
6
11
1 2 3 5 5
2 1 2
2
2
-1 -2
10 12
13 14
15 16 19
s
20
p
P
21
d 22
s
Q
18 22 20
16
14 16
12 12 14
7 8 10 10
6
6
4 5
2
23
Number of neutrons
(Contd,)
30
28 28
26
24
________________________ 20
18
f
CHAPTER 2
17
spspdspdfspd
L
Distribution of electrons within the shells
2 2 2 2
2 2 2 2
2
2
2 2
6.939
9.012 10.81
1 2
5
J
K
1.008 4.003
4
•
Atomic weight
S
Silicon Phosphorus
14 15
Mg Al
Na
Ne
0 F
N
C
Be B
H He Li
Symbol
Sulphur Chlorine Argon
Aluminium
Sodium Magnesium
12 13
11
IP
9
Fluorine Neon
Carbon
6
8
Boron
4 5
Hydrogen Helium Lithium Beryllium
2
1
1 2 3
Name of element
Atomic numbeir
Table 2.2 Electronic configuration of elements
Element
Atame of
Atomic weight
106.4 107.9 112.4 114.8 118.7
Pd Ag Cd In Sa
10 10 10 10 10
2 2 6 2 6 22626 2 2 6 2 6 22626 22626
2 6 26 2 6 26 26
10 10 10 10 10
16
-1 2 21 22
264-1 2 6 5-1 2 6 6 1 267-1 268-1
Palladium Silver Cadmium Indium Tin
10 10 10 10 10
22626 2 2 6 2 6 2 2 6 2 6 22626 22626
46 47 48 49 50
92.9! 95.94 (99) 101.1 102.9
Niobium Molybdenum Technetium Ruthenium Rhodium
41 42 43 44 45
Nb Mo Tc* Ru Rh
26 2 6 t 2 6 2 261-2 2 6 2 2
10 10 10 10 !0
22626 22626 2 2 6 2 6 22626 22626
83.80 83.47 87,62 88 91 91.22
31 32 33 34 35
Kr Rb Sr Y Zr
Iron Cobalt Nickel Copper Zinc
26 27 28 29 30
Krypton Rubidium Svomtum Yttrium Zirconium
15
36 37 38 39 40
14
2 1 2 2 23 24 2 5
13
10 10 10 10 16
12
2 2 6 2 6 2 2 6 2 6 22626 226 2 6 2 2 6 2 6
11
69.72 72.59 7492 7896 79.91
10
Ga Ge As Se Br
9
Gallium Germanium Arsenic Selenium Bromine
8
2262662 2 2 6 2 6 7 2 2262682 2 2 6 2 6 10 1 2 2 6 2 6 10 2
7
17
jg
O spdf
55.85 58.93 58.71 63.54 65.37
6
L M N sspspdspdf
K
Distribution of Electrons within the shells
3_________ 4_________ 5
Symbol
(ComdJ
Ke Co N: Cu Za
J___________ 2
Atomic Number
Table 2, B
0
r™ bn (n + 1) > am (m + 1) r0"
i.e.,
bn(n + 1) > am (m + 1 )r0'' m Substituting for rQ from equation (3.6), we get
i.e.,
bn (n + 1) > am (m +
(3.6a)
n>m
The forces acting between the atoms are mostly electrostatic in nature and are determined
essentially by the extent to which the wave functions of the outer electrons are perturbed by the presence of other atoms at close proximity.
Wil
The energy corresponding to the equilibrium position (r =
, denoted by t/(r„) is called the bonding
energy or the energy of cohesion of the molecule. This is the energy required to dissociate the two atoms of the molecule (AB) into an infinite separation. This energy is also known as the energy of dissociation.
It is obtained as follows:
Equation (3.4) is
U(r) will be minimum when r = r0.
Interat&nic Forces and Bondings in Solids
=-*
Thus
dU
Hence
51
(3.7)
=0 q
Thus
0
a
fo
Thus substituting this m equation (3 J)* we gel ri
U
H.Jbi
y
—
=-
171
a
(3.8)
In conclusion, it may thus be said that the forces of repulsion are the result of interpenetration of outer
electronic shells between atoms, ions and molecules when they approach one another each in interatomic or iftiermolecuiar distances The forces of attraction are due to interaction between outer electrons* of the two atoms, resulting in the formation of a sufficiently stable aggregate which can be considered as an indepen
dent molecular species. Thus all stable arrangements of atoms in solids we such that the potential energy is a minimum. This is one way of ex plaining the cohesion of atoms in solid aggregates. Dissociation may occur as a result of strong electric fields and mechanical strains or at high temperature. For example. 4.4 eV of energy is required to break one (H-Cl) bond, or 420 x 101 kJ/k mol.
V.
BONDING IN SOLIDS
From experience one knows that solids are usually moderately strong and slightly elastic structures. Thus the individual atoms must be held together in solids by interatomic forces or bonds However, in addition to these attractive forces there must be a repulsive force because solids are not easily compressed The attrac tive forces between the constituent particles in solids are basically electrostatic in origin and the classifica
tion of the different types of bonding is strongly dependent on the electronic structure of the atoms concerned, and hence directly related to the periodic lable. A solid is composed of billions of atoms packed closely together and the characteristic properties of
this state of matter can be accounted for by their proximity and the forces of attractions that hold the atoms together. The importance of these attractions is evident when we consider a piece of copper wire of which each gram contains (6.025 x 10*^)763,54 atoms under normal conditions; the forces of attraction that bind these atoms are very strong If thU is not true, the atoms will easily disintegrate or crumble or the metal will
deform under small loads. According to the strength and directionality, chemical bonds are grouped into primary and srrandary.
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52
Solid Stare Physics
Primary bonds by virtue of their nature arc interatomic bonds, whereas secondary bonds are mrermolecular
bonds. The attractive forces in primary bonds are directly associated with rhe valence electrons. The outer shell, which contains rhe valence electrons, is in a high energy state and hence relatively unstable. If it can
acquire more electrons to bring the total upto eight or lose all its electrons to another, it become stable. This is how atomic or primary bonds are formed- In fact, each bond is a direct consequence of the exchange or
sharing of valence electrons. The study of the solid structure reveals that there are three strong principal types of primary bonds:
ionict covalent and metallic. These bonds are distinguished on the basis of the positions assumed by the
bond electrons during the formation of the bond ran der WWr and hydrogen bonds are typical examples of secondary bonds and they result from inter molecular attracnon. The bonding energy or the cohesive energy
is defined as the energy of formation of one kmol of a substance from its atoms nr ions. It is equal but opposite in sign to the energy of dissociation of the substance. The bond energy can be calculated as the
energy of the atoms or urns at the equilibrium spacing in the crystal structure, using the state of infinite sep aration of the atoms or ions as the zero potential. The strength of a bond is best measured by the energy
required to break it, that is, the amount of heat which must be supplied to vaporize the solid (infinite separation) and hence separate the constituent atoms. The melting points and the boiling points of the ele
ments are dependent on the strength of the bond. Generally, the stronger the bond, the higher are the melting arul boiling points.
VI. IONIC BONDING Perhaps ionic or heteropolar bonding is the simplest type of chemical bonding to visualize since it is almost
totally electrostatic in nature. It occurs between electropositive elements (metals; i.e. those elements on the
left side of the periodic table) and electronegative elements (non-metals; i.e,, those on the right side of the periodic table), NaCI ami MgO are examples of solids in which ionic bonding dominated The criterion for ionic bonding is the difference in electronegativity (the tendency to acquire elec
trons). An ionic bond is really the attractive force existing between a positive and a negative ion when they
axe brought into close proximity, The^e ions, of course, are formed when the atoms involved lose or gain electrons in order to stabilise their outer shell electron configuration.
A typical example of an Ionic bond is the bond between ihe positive sodium ton and the negative chlorine ion in sodium chlonde. The white soft metal Na has the following electron configuration;
K(2)Lt8)M(l), This atom has a low ionisation energy and hence easily loses an electron; the other atom chlorine [K(2)L(8)M(7)J< has a high electron affinity and strongly tends to acquire an electron. Suppose a chlorine atom and a sodium atom approach each other. This occurs when sodium bums in an atmosphere of
chlorine. The sodium gives up its valence electron to the chlorine, each of the resulting ions then has a stable
filled shell of outer electrons, and a strong electrostatic attraction is set up that bonds the Na" cation and the O anion into a very stable molecule (NaCI) at the equilibrium spacing. The reaction is represented by
Na+CL^Na*+CT-*NaCI Since chlorine exists as molecules, the chemical reaction must be written as 2Na + CU -* 2Na* + 2CT -> 2NaCl To release the valence electrons, a kmd of free sodium atoms requires an expenditure of energy, / - 496 x IO3 kJ (this quantity is the first ionisation energy per kmol of Na atoms). Acquisition of the elec trons by a kmoi of free chlorine atoms results in a release of energy,
* 549 x ICPkL Thus, to initiate a
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Interatomic
To release the valence electrons, a kmol of free sodium atoms requires an expenditure of energy, I = 496 x 103 kJ (this quantity is the first ionisation energy per kmol of Na atoms). Acquisition of the
electrons by a kmol of free chlorine atoms results in a release of energy, Ea = 349 x 103 kJ. Thus, to
initiate a reaction between a kmol of free neutral sodium atoms and a kmol of free neutral chlorine
atoms requires a net expenditure of (/ - E() = (496 - 349) x 103 = 147 x 103 kJ. Once started, the reaction proceeds vigorously with the evolution of light and heat because of the sizeable energy decrease
resulting from ionic bonding. The final product usually is a large number of salt crystallites. Other examples of ionic crystals are 2 Mg + O2---------- > 2 Mg+++ 2 O—---------- > 2 MgO
Mg + Cl2---------- > Mg++ + 2 Cl"---------- > MgCl2 The different steps showing energy absorption and liberation in the formation of NaCl is given
in Table 3.1. The total energy on the right side is much greater than that on the left side and hence a considerable amount of energy is given out during the formation of NaCi crystal. This appears as heat
Table 3.1 Energy absorbed and energy released in the formation of NaCl to
1.
Energy released in the formation of NaCl
> Energy absorbed in the formation of NaCl Separation of chlorine atoms of the diatomic
Completion of Af-shell of chlorine
CL molecule
•
2.
Separation of sodium atoms of solid sodium
3.
Separation of electron from sodium atom
Lattice energy of sodium chloride • —
NaCl is one of the best examples of ionic compound and let the sodium and chlorine atoms be free at
infinite distance of separation. The energy required to remove the outer electron from the Na atom
- (ionisation energy of sodium atom), leaving it as an Na+ ion is 5.1 eV. That is Na + 5.1 eV---------- > Na+ + e~
The electron affinity of chlorine is 3.6 eV. Thus, when the removed electron from sodium atom
is added to chlorine atom, 3.6 eV of energy is released and the chlorine atom becomes negatively charged. Hence Cl + «r---------- > Cl"+ 3.6 eV
Thus a net energy of (5.1 - 3.6) = 1.5 eV is spent in creating a positive sodium ion and a negative chlorine ion at infinity. Thus
Na + Cl + 1.5 eV---------- > Na+ + Cl"
What happens when the .electrostatic attraction between Na+ and Cl- ions brings them together to the equilibrium spacing r0 = 0.24 nm? At the equilibrium position, the potential energy will be
minimum and the energy released in the formation of NaCl molecule is called the bond energy ol the molecule and it is obtained as follows:
CHAPTER 3
and light during the chemical reaction.
54
S&lid Slate Physics
(c) Ionic molecule Fifl. 3.3
Schcmaiic representation of the formation of an Ionic molecule of sodium chloride.
(3.9) (1.602x10 ”)
_(47t)(8.85x 10"ls}(2.4x 10’’°),
________ (1.602X 1O~‘V________
.4n(8.85x 10’2:x 2.4)(1.602 x IO1’).
eV
*-6eV Thus the energy released in the formation of NaCl molecuk starting from neutral Na and Ct atoms having
zero potential energies is (5- i - 3,6 - 6) ~ - 45 eV+ Schematically Na+ + CF
Na + Cl
+ 4J eV -> NaO
This is the energy released. Thus the entire process evolves an energy of 6
15
45 eV. This means that to
dissociate a NaCI molecule into Na and Cl iorw, ft requires an energy of 4,5 eV,
Bond lengths and bond energies of some diatomic tonic molecules arc given in Tabic 3,B.
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Interatomic Forces and Bondings in Solids
55
Bond lenglhs and bond energies of some diatomic ionic molecules
Molecule KC1 UH NaF NaCl Nai NaBr
VIII.
Bond length (nnj 0.267 0360 0i£5 0236 0 271 0250
Bond energy (in eV) 4.4 2.5 4.7 4.3 3.1 3J
CALCULATION OF LATTICE ENERGY OF IONIC CRYSTALS
The lattice energy of an ionic solid will differ from the bond energy of diatomic ionic molecules since, in rhe
former case there will be interactions between more than two ions* The coAeu've energy of an ionic crystal is rhe energy that would be liberated by the formation of the crystal from individual neutral atoms. Cohesive energy is usually expressed in eV/atom, or eV/molecute or in kJ/kmol The bond energy of a molecule held
together by an ionic bond is not the same as the coterfw energy of the crystal, because in rhe crystal each ion interacts with all the other ions present and not just with one. two or three of the opposite sign* Some*
times rhe lattice energy, rather than the cohesive energy, is presented; the lattice energy is that energy
evolved when a crystal is formed from individual ions, rather than from individual atoms. Let us now consider the case of NaCl which is one of the best examples of an ionk campmind. Ionic crystals are strongly bound w ith binding energies (cohesive energies) of about 5-10 eV per molecule This
is the energy required to dissociate the lattice into positive and negative ions at infinite separation. Two
common types of structures found in ionic crystals are NaCl structure which is face-centred cubic and CsCI which is body-centred cubic* The calculation of binding energy of ionic crystals generally requires a
knowledge of forces acting between the constituent particles. In the classical Bom Madelung theory of ionic crystals. it is assumed that the electrons are transferred from electropositive atoms (Na. Ca, K, Mg) to electronegative atoms (0, F, Cl). The stability of an ionic
crystal depends un the balancing of alleast three forces; the electrostatic, or Coulomb forces between the
iem which give a resultant attraction falling off with the square of the distance; van def Waals forces of attraction diminishing according to the seventh power of the distance (usually neglected); and the interionic
repulsive forces falling off still more rapidly with distance. The resultant of the attractive and repulsive
forces is to lead to an equilibrium position of minimum potential energy; i.e.; of greatest stability as shown tn Fig. 3*4.
Far two ions of charges fore this energy is -
Z&e2 Z/ and Zjg separated by a distance r, the attractive energy is ~ 4.n e \ and there-
—----- - if both rhe atoms are respectively mmo valent, divalent and 4n r
trivalcnt. For the whole crystal, the Coulomb potential energy may be written as -
. This term represents
the net Coulomb potential energy of any one ion due to the presence of all other similar and dissimilar ions
present in the crystal The minus sign shows that the irci Coulomb energy is attractive. The constant A is known as Madelung constant.
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_ —/-1 r -. This tcnn represents the net Coulomb potential energy ol any one ion due in ,L 4* fA r ‘ 'he
presence of all other similar and dissimilar ions present in the crystal. The minus sign shows that the net coulomb energy is attractive. The constant A is known as Madelung constant.
Rg. 3.4 Schematic representation of lattice energy of NaCI as a function of interatomic spacing To prevent the lattice from collapsing, there must also be repulsive forces between the ions.
These repulsive forces become more noticeable when the electron shells of neighbouring ions begin to overlap, and they increase strongly in this region with decreasing values of r, That is, this repulsive
force arises from the interaction of the electron clouds surroundings an atom. Bom, in his early work, made the simple assumption that the repulsive energy due to the overlap of the outer electron shells
between two ions is inversely proportional to some power of the distance r or equal to btf1. Focusing
our attention again on one particular ion, we may thus write for the repulsive energy of this ion, due to the presence of all other ions in the crystal as, Bld*. n is called repulsive exponent.
Now the total energy of one ion due to the presence of all others, is given by AZiZ2e2
B
47teor
r"
U(r) = -
(3.10)
For the univalent alkali halides
Ae2 U(r) = - ----------4n e0 r
W n
(3.11)
The total energy per kmol of the crystal is
£ n
U(r) = N
Ae2 (3.12)
4n e0 r
The P.E. will be minimum at the equilibrium spacing r0. (KT
Thus
. dr,
, -N
Ae2 or
4k e0 r2 ~
A
L4TC eo r0
Btl
fo
A£ 4tt e r2
Bn +l
=0
(3.13)
r»
Ae2
Ae2^'1
.
(3.14)
4ne0 n
Substituting this value of H in equation (3.12) and putting r = rt). we get the total equilibrium energy per kmol °f *hc crystal. Thus
lU]rsru=U() = - ^AAel
+ N^Ae2r~1 47t£on
4
dV
drjdvldV
(3.21a)
dry d I dU dV