hybridization - Determine the bond angle in a compound

How can we find the bond angle between 3 atoms in a compound? Please specify an equation which can be used for all compounds.

Answer

In carbon compounds Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.

$$1+\lambda_{i} \lambda_{j} \cos(\theta_{ij})=0$$

where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

Let's look at some examples.

• In the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.



• In the case of the methyl carbocation ($\ce{CH3^{+}}$) the $\ce{C-H}$ bonds are $\ce{sp^2}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt2$. In this case, using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 120 degrees.


• Finally, in the case of acetylene the $\ce{C-H}$ bond is $\ce{sp}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt1$. In this case, using Coulson's theorem we find that the $\ce{C-C-H}$ bond angle is 180 degrees.

Conversely, if the bond angle is known, the hybridization can be determined.