# Why is the force constant for double and triple bonds approximately double and triple of that of a single bonds?

I am new to spectroscopy, and I am referring to Donald Pavia's Introduction to Spectroscopy. I came across this statement in the book, under the chapter Infrared Spectroscopy:

$K$ is a constant that varies from one bond to another. As a first approximation, the force constants for triple bonds are three times those of single bonds, whereas the force constants for double bonds are twice of those of single bonds.

As far as I know, double bonds are not considered as two single bonds. Double bonds are rather one $\sigma$ bond, and one $\pi$ bond. Neither the bond energies of double bonds are twice the energy of corresponding single bonds, nor is the bond length of double bonds half of the single ones. Then how is the force constant doubles as the bond order doubles?

• IIRC from my undergrad, the 1:2:3 ratio is an approximation based on the two-mass model, which estimates 500 N/m, 1000 N/m, 1500 N/m respectively for single, double, triple C=C bonds . Jun 18, 2017 at 16:30
• @NotEvans. Um whats the two-mass model? Jun 18, 2017 at 16:40
• Force constant is not the same as bond strength. Remember that an approximate measure of bond energy from the bond as a simple harmonic oscillator would be $\frac{1}{2}kx^{2}$ where $x$ is the bond length.
– Zhe
Jun 18, 2017 at 19:35
• @Zhe Ahh, I haven't thought of that. Thanks for your input. Do you want to transform it into an answer so I can accept it? Jun 19, 2017 at 2:15

Force constant is not the same as bond strength. Remember that an approximate measure of bond energy from the bond as a simple harmonic oscillator would be $$E=\frac{1}{2}kx^{2}$$ where $x$ is the bond length. In this case, you would also need to account for differing lengths of single, double, and triple bonds to recover the bond energy trend.
Note that this is not how a quantum mechanical harmonic oscillator is normally described. There, you would have quantized energy levels of $(n+\frac{1}{2})\hbar\omega$, where $\omega = \sqrt{\frac{k}{m}}$. But since the higher order wave functions "concentrate" at the classical turning points, you might still be able to use the classical harmonic oscillator as a model for what is going on, with strong caveats.