By Sunil Bhardwaj

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TLC

Consider a diatomic molecule AB. Two atoms A and B are connected by an elastic spring. A is fixed at one end and B is kept hanging at their normal equilibrium distance \(r\). When the spring is stretched by an external force \(f\) and then released, the equilibrium distance is disturbed, and its length increases.

The opposing force restores back the spring to its original equilibrium distance. If the displaced distance is \(x\), then according to Hooks Law, $$f \propto (-x) \qquad or \qquad f = -Kx\qquad ...(1) $$ The negative sign indicates the direction of force in opposite direction. where \(K\), the proportionality constant is called the FORCE CONSTANT. Its unit is Newton/metre. This force constant is the measure of the strength of the bond. Higher the force constant, stronger is the bond.

Thus, the atom B performs simple harmonic motion along the axis of the bond. When bond atoms A and B are vibrating and oscillating simultaneously then the vibrational frequency \((\upsilon )\) is equal to oscillating frequency \((\omega )\), which is given by the relation, $$\omega = \upsilon = \frac { 1 }{ 2\pi } \sqrt { \frac { K }{ \mu } } \qquad ...(2) $$ In terms of wave numbers the oscillation frequency \(\overline { v }\) or \(\overline { \omega } \) is given by, $$\overline { \omega } = \overline { v } = \frac { v }{ c } = \frac { 1 }{ 2\pi c } \sqrt { \frac { K }{ \mu } } $$ On rearranging equation, $$\boxed { K = 4{ \pi }^{ 2 }{ c }^{ 2 }{ \overline { \omega } }^{ 2 }\mu } $$ This is the expression for the force constant. where \(c\) is the velocity of light \(= 3 \times { 10 }^{ 8 }\) m/s
\(\mu\) is the reduced mass \(= \left( \frac { { m }_{ 1 } \times { m }_{ 2 } }{ { m }_{ 1 } + { m }_{ 2 } } \right) \)
\(\overline { \omega } \) wave number \(= \frac { 1 }{ \lambda } \) i.e. reciprocal of wavelength.

This force constant helps to know the strength of the bond and the molecular stability. It has the following significance:

1. Higher the force constant, stronger the bond and greater the bond energy. The force constant of HF is about three times greater than that of HI. Thus, HF bond energy is greater than HI and HF cannot be easily oxidised.

2. Molecules with two or more bonds between the atoms have greater stability than those with single bonds. Thus, the molecules like NO, CO and C2H2 are more stable.

3. Force constants for different compounds where oxygen and carbon linked with double bond and triple bond are \(12.5 \times { 10 }^{ 2 } N{ m }^{ -1 }\) and \(18.6 \times { 10 }^{ 2 } N{ m }^{ -1 }\) respectively. But for carbon dioxide it is about \(15.2 \times { 10 }^{ 2 } N{ m }^{ -1 }\) This value is in between the above two values. It suggests that there is a resonance in carbon dioxide molecule.

The force constant of HI molecule is \(2.9 \times { 10 }^{ 2 } N{ m }^{ -1 }\). At what frequency will the origin of fundamental bond appear? (Given : H=1, I = 127)

We have force constant \(K = 2.9 \times { 10 }^{ 2 } N{ m }^{ -1 }\)
Mass of H = 1 a.m.u.
Mass of I = 127 a.m.u. $$\therefore reduced \ mass \ \mu = \left( \frac { { m }_{ 1 }{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) $$ $$ = \left( \frac { 1 \times 127 }{ 1 + 127 } \right) $$ $$ = \frac { 127 }{ 128 } = 0.992 a.m.u. $$ $$ = 0.992 \times \left( 1.6 \times { 10 }^{ -27 } kg \right) $$ $$ = 1.588 \times { 10 }^{ -27 } kg $$ Now frequency of fundamental band is $$ \overline { \upsilon } = \frac { 1 }{ 2\pi c } \sqrt { \frac { K }{ \mu } } $$ $$ = \frac { 1 }{ 2 \times 3.14 \times \left( 3 \times { 10 }^{ 8 } m/s \right) } \sqrt { \frac { 2.9 \times { 10 }^{ 2 } N{ m }^{ -1 } }{ 1.588 \times { 10 }^{ -27 } kg } } $$ $$ = \frac { 1 }{ 18.84 \times { 10 }^{ 8 } } \sqrt { 1.826 \times { 10 }^{ 29 } } $$ $$ = \frac { 1 }{ 18.84 \times { 10 }^{ 8 } } \times 4.273 \times { 10 }^{ 14 } $$ $$ = 2.268 \times { 10 }^{ 5 } m $$

The force constant of the H-F bond is \(9.7 \times { 10 }^{ 2 } N{ m }^{ -1 }\). Calculate the vibrational frequency of the molecules. (Given : H=1, F = 19)

We have force constant \(K = 9.7 \times { 10 }^{ 2 } N{ m }^{ -1 }\)
Mass of H = 1 a.m.u.
Mass of F = 19 a.m.u. $$\therefore reduced \ mass \ \mu = \left( \frac { { m }_{ 1 }{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right) $$ $$ = \left( \frac { 1 \times 19 }{ 1 + 19 } \right) $$ $$ = \frac { 19 }{ 20 } = 0.95 a.m.u. $$ $$ = 0.95 \times \left( 1.6 \times { 10 }^{ -27 } kg \right) $$ $$ = 1.52 \times { 10 }^{ -27 } kg $$ Now frequency of fundamental band is $$ \overline { \upsilon } = \frac { 1 }{ 2\pi c } \sqrt { \frac { K }{ \mu } } $$ $$ = \frac { 1 }{ 2 \times 3.14 \times \left( 3 \times { 10 }^{ 8 } m/s \right) } \sqrt { \frac { 9.7 \times { 10 }^{ 2 } N{ m }^{ -1 } }{ 1.52 \times { 10 }^{ -27 } kg } } $$ $$ = \frac { 1 }{ 18.84 \times { 10 }^{ 8 } } \sqrt { 6.382 \times { 10 }^{ 29 } } $$ $$ = \frac { 1 }{ 18.84 \times { 10 }^{ 8 } } \times 7.988 \times { 10 }^{ 14 } $$ $$ = 4.240 \times { 10 }^{ 5 } m $$

MCQ on Spectroscopy from Physical Chemistry

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Prof. Gianfranco Coletti

Shared publicly - 2019-08-23 00:00:00

Don’t want your columns to simply stack in some grid tiers? Use a combination of different classes for each tier as needed. See the example below for a better idea of how it all works.

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For grids that are the same from the smallest of devices to the largest, use the .col and .col-* classes. Specify a numbered class when you need a particularly sized column; otherwise, feel free to stick to

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