Tutorial 21112
4. The microwave spectrum of 1H19F gives series of lines
separated by 40.5 cm-1. Calculate the moment of
inertia and the internuclear distance for the molecule.
•
•
•
•
•
•
h = 6.626 × 10-34 Js
c= 2.99792 x 108 ms-1
2B = 40.5 cm-1
B = 20.25 X 102 m-1
 h 
B 2 
 8 IC 
34
6.626 x 10 JS
B 2
8
1
8 I x 2.99793 x 10 ms
34
2 1
6.626 x 10 Kgm s
I 2
8
1
8 x 2.99793 x 10 B ms
34
6.626 x 10 kgm
I 2
8
2
1
8 x 2.99793 x 10 x 20.25 x 10 m
I  1.3822 x10
 47
2
kgm
I  r
2
 m H mF
I 
 m  mH
 F
mH
 2
r


• mH = 1.0X10-3 kg / 6.023 x 1023
• mF = 19.0X10-3 kg / 6.023 x 1023
μ = 1.577 X 10-27 kg
r2 = 0.8764 x 10-20 m2
r = 0.9362 x 10-10 m
r = 0.9362 Ǻ
mF
5.
Calculate the separation between the rotational lines of the
spectrum of gaseous 1H35Cl and the bond length of HCl, given
that its moment of inertia is 2.8 x 10 -47 kg m2.
 h 
B 2 
 8 IC 
6.626 x 10 34 JS
B 2
8 I x 2.99793 x 108 ms 1
6.626 x 10 34 Kgm 2 s 1
B 2
8 x 2.8 x 10 -47 kg m 2 x 2.99793 x 108 B ms 1
6.626 x 10 34 kgm
B 2
8 x 2.99793 x 108 x 20.25 x 10 2 m 1
 2 B  ................. m 1
I  r 2
mH = 1.0X10-3 kg / 6.023 x 1023
•
mCl = 35.0X10-3 kg / 6.023 x 1023
μ = 1.614 X 10-27 kg
r2 = 1.734 x 10-20 m2
r = 1.734 x 10-10 m
r = 1.734 Ǻ
 mH mCl
I  
 mCl  mH
 2
r

6. The microwave absorption of nitrogen oxide (NO)
showed line at the following wavenumbers:
• 3.405 cm-1
6.810 cm-1
10.215 cm-1
13.620 cm-1
•
Calculate the bond length of NO.
•
N= 14.0 g and O = 16.0 g
•
•
•
•
2B = 3.405 cm-1
B = 1.7025 X 102 m-1
I = 16. 4402 x 10-47 Kgm2
r = 1.151 Ǻ
7. The rotational spectrum of 79Br19F shows a series of
equidistant lines spaced 0.71433 cm-1 apart. Calculate
the rotational constant B, and hence the moment of
inertia and the bond length of the molecule.
8.
What is the zero-point energy of a diatomic molecule with a
fundamental vibration frequency of 3 x 10 14 Hz?
1

Ev   v  hosc
2

Ev=0= ½ h ω osc
•
h = 6.626 × 10-34 Js
•
Ev=0= ½ x (6.626 × 10
•
Zero point energy = 9.94 x 10-20 J
•
-34
Js) x (3 x 10 14 s-1)
9. Calculate the fundamental vibration wave number of
gaseous 1H35Cl, given that the force constant for the
bond stretching is 516 Nm-1.
1
 osc 
2C
k

k = 516 Nm-1
μ = 1.614 X 10-27 kg
C= 2.998 X 108 ms-1
ωosc = 2990 cm-1
11. The force constants of the CO group and that of CN groups
are 1230 Nm-1 and 1750 Nm-1 respectively. Calculate the
infrared frequency in wavenumbers at which an absorption
band could be observed for CO and CN groups.
1
 osc 
2C
•
k

C=O
μCO = 11.11 X 10-27 kg
C=N
μCN = 10.736 X 10-27 kg
ωosc = 1766 cm-1
ωosc = 2143 cm-1
12.
•
•
Analysis of the fundamental, 1st overtone and 2nd overtone
bands of HCl leads to the following equation for the vibrational
enegy levels in the ground state.
ν = 2990.6 (v + 1/2) - 52.63 (v + 1/2)2 cm-1
v is the vibrational quantum number
(i) calculate the zero-point energy
(ii) Find the separation between successive energy levels v and
v+1 in terms of v.
(iii) calculate the wavenumber of the fundamental band
(i) V =2990.6 (0+ 1/2)-52.63 (0 +1/2)2 cm-1
= 1482.1 cm-1
E=hcv
find E
• (ii) separation between successive energy levels
V cm-1 = {2990.6 (v + 3/2) - 52.63 (v + 3/2)2 cm-1}
– {2990.6 (v + 1/2) - 52.63 (v + 1/2)2 cm-1}
= 2990.6 -105.26 (v+1)
(iii) Fundamental band
V cm-1 = 2990.6 -105.26 (0+ 1)
V = 2885.3 cm-1
13. The fundamental vibration is observed at 2650 cm-1 for 1H80Br,
Calculate the shift in wavenumber if the sample analyzed
contains 2H80Br.
The fundamental vibration frequency is given by
1
 osc 
2C
•
•
k

Force constant k needs to be assume as a constant for both
molecules
 osc,1HBr
2
Therefore
2650 cm-1
 osc, HBr
2

HBr
 HBr
1
 osc, HBr  1886 cm
1
2
• Therefore the shift is 764 cm-1.
• Force Constant: A constant of proportionality for
springs which indicates the relative stiffness or
elasticity of the spring when it is stretched within its
elastic limits.
• SI unit of measure is Nm-1.
2S 1
LJ
15. Two electrons are in the 2p3d configuration. In
what spectroscopic state would you expect these
to be found?
•
•
•
•
l1= 1, l2= 2, L = 1, 2, 3 and
s1= ½ , s2= ½ S = 0, 1
So we have 1P, 1D, 1F
•
•
Consider the J values:
1P , 1D , 1F , 3 P
3D
3F
,
,
1
2
3
0, 1, 2
1, 2, 3
2, 3, 4,
, 3P, 3D, 3F
16. Two electrons are in the 4d5p
configuration. In what are the possible
spectral terms can be formed from this
state? Give complete descriptions,
including J value.
•
1P , 1D , 1F , 3P
3D
3F
,
,
1
2
3
0, 1, 2
1, 2, 3
2, 3, 4,