Tutorial – 4
1) Calculate the moment of inertia (I) and bond length (r) from microwave
spectrum of CO.
Find
r(CO)
First line (J = 0 to J=1 transition) in the rotation spectrum of CO is 3.84235 cm-1.
Calculate the moment of inertia (I) and bond length (r) of CO.
Mass of C = 19.92168 x 10-27 kg
Mass of O = 26.561 x 10-27 kg
Answer
B 
h
2
8 π Ic
for CO
cm
-1
I=
J=0  J=1 (First line)
r2
r
at
I
μ
3.84235 cm-1
2B = 3.84235 cm-1
B = 1.921175 cm-1
=
Mc x mo
=
Mc + mo
=
=
19.92168 x 10-27 x 26.561 x 10-27
19.92168 x
10-27
+ 26.561 x
10-27
[19.92168 x 26.561] x 10-27 x 10-27
[19.92168 + 26.561] x
529.14 x 10-27
kg
46.48
 = 11.384 x 10-27
kg
 = 1.1384 x 10-26 kg
10-27
kg
kg
Answer
B 
I 
h
2
8 π Ic
h
2
8 π Bc
cm
-1
cm
-1
I=
r2
I
r
μ
B = 1.921175 cm-1
B = 192.1175 m-1
6.626 x 10-34 kg m2 s-1
h = 6.626 x 10-34 Kg m2 s-1
c = 2.998×108 m s−1
I=
8 x 9.869 x B m-1 x 2.998×108 m s−1
I Kg m2
0.02799 x 10-42
I=
r2 =
kg m2
192.1175
I = 1.4569 x 10-46
 kg
1.4569 x 10-46
kg m2
r2 =
m2
1.1384 x 10-26
r2 = 1.2797 x 10-20
r = 1.131 x 10-10
r = 1.131 Å
m2
m
µ
B
<
>
µ’
B’
First rotational absorption,
For
=
For
=
Mass of O = 15.9994
Mass of 13C = ??
The effect of isotopic substitution on the energy levels and hence rotational spectrum
of a diatomic molecule such as carbon monoxide
Question-2
Comparison of rotational energy levels of
12CO
and
13CO
Can determine:
(i) isotopic masses accurately, to within 0.02% of other methods for atoms in
gaseous molecules;
(ii) isotopic abundances from the absorption relative intensities.
Question 2: First rotational lines in microwave region for both
13CO are given below:
for
12CO
J=0  J=1 (First line)
at
3.84235 cm-1
for
13CO
J=0  J=1 (First line)
at
3.67337 cm-1
Mass of 12C and
16O
Given : 12C = 12.0000 amu;
16O
= 15.9994 amu
Calculate the atomic weight of 13C
12CO
and
Answer
B 
h
2
8 π Ic
cm
-1
Question & Answer
1) Why vibrational (IR) frequency (ϖos) of triple bond (CC) is higher than C-C ?
Ans:
 
Oscillation Frequency,
m 1m 2
 os 
 os 
m1  m 2
1

2 c
μ
cm-1
1
 (m  m )
2 c
m 1m 2
1
2
k is the force constant (bond strength )
 = reduced mass of atoms
cm-1
The vibrational frequency of a particular bond is increasing with:
 increasing force constant k ( = increasing bond strength)
 decreasing atomic mass
and, therefore, k cc > k c-c [The force constant (k) is proportional to the strength of the covalent
bond linking two atoms]
Here two atoms in both cases are same (carbon) and hence reduced mass () is same. Thus,
ϖos for triple bond (CC) is higher than C-C
Example:
Question & Answer
2) Why vibrational (IR) frequency (ϖos) of N-H (3400 cm-1) is higher than P-H (2350 cm-1)?
Ans:
Oscillation Frequency,
 os 
1

2 c
μ
cm-1
k is the force constant (bond strength )
 = reduced mass of atoms
The vibrational frequency of a particular bond is increasing with:
 increasing force constant k ( = increasing bond strength)
 decreasing atomic mass
Here two atoms in both cases are different (NH Vs PH); P is heavier than N and hence reduced
mass () for P-H bond is higher than that of N-H bond.
Thus, ϖos for N-H bond (3400 cm-1) is higher than P-H (2350 cm-1).
Example:
Question & Answer
3) Why vibrational (IR) frequency (ϖos) of O-H (3600 cm-1) is higher than S-H (2570 cm-1) ?
Ans:
 os 
Oscillation Frequency,
 
m 1m 2
m1  m 2
 os 
1

2 c
μ
1
 (m  m )
2 c
m 1m 2
1
2
cm-1
k is the force constant (bond strength )
 = reduced mass of atoms
cm-1
The vibrational frequency of a particular bond is increasing with:
 increasing force constant k ( = increasing bond strength)
 decreasing atomic mass
Here two atoms in both cases are different (OH Vs SH); S is heavier than O and hence reduced
mass () for S-H bond is higher than that of O-H bond.
Thus, ϖos for O-H bond (3600 cm-1) is higher than S-H (2570 cm-1).
Example:
Question & Answer
4) Why vibrational (IR) frequency (ϖos) of F-H (4000 cm-1) is higher than Cl-H (2890 cm-1)?
Ans:
Oscillation Frequency,
 
m 1m 2
m1  m 2
 os 
 os 
1

2 c
μ
cm-1
1
 (m  m )
2 c
m 1m 2
1
2
k is the force constant (bond strength)
 = reduced mass of atoms
cm-1
The vibrational frequency of a particular bond is increasing with:
 increasing force constant k ( = increasing bond strength)
 decreasing atomic mass
Here two atoms in both cases are different (FH Vs ClH); Cl is heavier than F and hence reduced
mass () for Cl-H bond is higher than that of F-H bond. Thus, ϖos for F-H bond (4000 cm-1) is
higher than Cl-H (2890 cm-1).
Example:
Question & Answer
5) Why vibrational (IR) frequency (os) of CN triple bond is higher than C-N single
bond?
Ans:
 
Oscillation Frequency,
m 1m 2
m1  m 2
 os 
 os 
1

2 c
μ
cm-1
1
 (m  m )
2 c
m 1m 2
1
2
cm-1
k is the force constant (bond strength )
 = reduced mass of atoms
The vibrational frequency of a particular bond is increasing with:
 increasing force constant k ( = increasing bond strength)
 decreasing atomic mass
The equation on the above describes the major factors that influence the stretching frequency
of a covalent bond between two atoms of mass m1 and m2 respectively. The force constant (k)
is proportional to the strength of the covalent bond linking m1 and m2.
In the analogy of a spring, it corresponds to the spring's stiffness. For example, a C=N double
bond is about twice as strong as a C-N single bond, and the C≡N triple bond is similarly stronger
than the double bond. The infrared stretching frequencies of these groups vary in the same
order, ranging from 1100 cm-1 for C-N, to 1660 cm-1 for C=N, to 2220 cm-1 for C≡N.
and, therefore, k CN > k C-N
Here two atoms in both cases are same (c and N) and hence reduced mass () is same. Thus, ϖos
for CN triple bond is higher than C-N single bond.