Question

3 A conceptually simple model for a chemical bond is to consider the bond as a spring with a certain spring constant. The relationship bet the vibration, the reduced mass () of the two atoms and the spring constant (k) is given by (h-6.626 1034 Js is Planck's constant): tween the energy (E) of h lk E=h,f=- DUg ti rdalii e e increase as the bond character changes from single-, to double-, to triple- bond (assuming that the identity of the two binding atoms remains the same). b) Common ranges of vibrational frequencies are 0.33-0.47 eV for C-H single bonds and 0.19-0.24 eV for C=O double bonds. This seems to contradict the statement in part a) of this problem. Explain how this apparent contradiction can be resolved.

Answer 1

A single bond molecule can be thought of as two molecules by single spring. A double bond molecule can be thought of as two molecules by two springs in parallel. An n-bond molecule can be thought of as two molecules by n springs in parallel. The Equivalent spring constants of n springs of spring constant k each in parallel is . Thus the energy of n bonded molecule becomes

$E_n =\sqrt{n }E$

where E is energy of single bond molecule. Hence we can conclude that the energy increases with the number of bonds n.

b) For C-H bond the reduced mass is $(\frac{1}{6}+1)^{-1}=\frac{6}{7} amu$

For C=O double bond the reduced mass is $(\frac{1}{6}+\frac{1}{8})^{-1}=\frac{24}{7} amu$

Thus the reduced mass of C=O is 4 times larger than C-H bond while spring constant is 2 times larger. Therefore the energy of C=O bond is related to C-H bond energy as follows

$E_{CO}=\sqrt{\frac{2}{4}}E_{CH}=E_{CH}/\sqrt{2}=0.707 E_{CH}$

Thus we see that the CO bonnd energy is less than CH bond energy which is in agreement with the given data.