I was learning about Bent's Rule. I came across a formula

$$\cos\theta = \frac{s}{s-1}$$

I am quite confused about $\theta$. I know that it represents bond angle. But for a compound like $\ce{PCl5}$, which bonds angle has to be considered? For the equatorial bonds, $\theta = 2 \pi/2$ was used and $\theta = \pi/2$ was used for axial bonds. My doubt is that why we are not using $\theta = \pi/2$ for equatorial bonds as they also have form $\pi/2$ angle with with axial bonds.

I asked my mentor about this. He said that $\ce{PCl5}$ has non-equivalent hybridization and $\mathrm{sp^3d}$ is actually $\mathrm{sp^2 + pd}$. The equatorial bonds are $\mathrm{sp^2}$ hybridized and have bond angle as $120^\circ$. The bond angle for axial bonds is the angle formed by the axial bond and the plane formed by chlorine atoms which connected to phosphorus through equatorial bonds.

After this, I became more confused. Can anyone please tell me the meaning of $\theta$ in the formula? Is it really the bond angle? If yes, then which angle has to be considered in $\ce{PCl5}$ and why?


1 Answer 1


TL;DR: $\theta$ is the angle between any two equivalent bonds and using Coulson's theorem for anything beyond the second period is out of its scope.

First things first, Coulson's theorem (the generalization of said formula) cannot be applied in cases of spndm. Therefore using $\ce{PCl5}$ as a comparison is not the right idea. Also, the formula that you've stated is only when finding the bond angle between two equivalent bonds as would be proved below.

This equation is supposedly a simplification of Coulson's theorem that states,

$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$

Here, $\lambda_1$ stands for the hybridization index of one bonded orbital, $\lambda_2$ denotes the hybridization index of the second bonded orbital and $\theta$ signifies the bond angle between the two orbitals. However, this rule as said above can only be applied to spn hybridisation compounds where $\sqrt{n}$ is the hybridisation index. Its formal proof is given in Wikipedia if you are curious and want to find out.

The simplified idea behind Coulson's theorem (as I understand it), is as follows.

Each set of equivalent bonds is made up of a linear combination between s and p orbitals. This is what hybridization is. Given any pair of bonds, we can find the bond angle between the two through their hybridization indexes ($\lambda_1$ and $\lambda_2$) and vice versa. Let's take the example of $\ce{CH4}$,

$\ce{CH4}$ has four equivalent $\ce{C-H}$ bonds and we know that the bond angle is $109.5^\circ$and so according to Coulson's theorem, the hybridisation of the $\ce{C=H}$ bond can be found to be,

$$1+\lambda_\ce{C-H}\cdot\lambda_\ce{C-H}\cos(\theta_\ce{H-C-H}) = 0 $$ $$1 +\lambda_\ce{C-H}^2\cos(\theta_\ce{H-C-H}) = 0 $$ \begin{align} \therefore \lambda_\ce{C-H}^2 &= \frac{-1}{\cos(109.5^\circ)} \\ &\approx 3 \end{align}

Therefore, the $\ce{C-H}$ bonds are sp3 in nature.

Proof of $\cos \theta = \frac{s}{s-1}$

From Coulson's theorem, we have,

$$1 +\lambda_1\lambda_2\cos(\theta) = 0$$

Now, if both bonds are equivalent in nature, we get $\lambda_1 = \lambda_2 = \lambda$, therefore,

$$1 + \lambda^2\cos(\theta) = 0$$

This implies,

$$\cos \theta = \frac{-1}{\lambda^2}$$

Now for such a molecule the hybridisation would be sp$\lambda^2$ and so percentage s-character would be

$$\text{%s character} = \frac{1}{1 + \lambda^2}$$

Solving for $\lambda^2$, we get:

$$\lambda^2 = \frac{1-s}{s}$$

Substituting this in the equation for $\cos \theta$, we get:

$$\cos \theta = \frac{s}{s-1}$$

Note: An assumption that I took here is that the bonds are equivalent. If this is in fact true, then this rule can only be used to find the bond angle between any two equivalent bonds and hence in the case of $\ce{PCl5}$, only the axial-axial ($\ce{P-Cl}$) bonds or equatorial-equatorial $\ce{(P-Cl)}$ bonds can be compared. θ then is the angle between any two equivalent bonds.

  • $\begingroup$ Safdar Faisal is it okay if i ask something in a chat room with you ( related to chemical bonding ) ? I will delete this soon and yeah please invite if possible as i dont know how to create one $\endgroup$
    – Orion_Pax
    Mar 29, 2022 at 19:45

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