The bond angle between two hybrid orbitals is | vedclass

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Question

(A) The relationship between the bond angle $(	heta)$ and the percentage of $s-$character $(s)$ in a hybrid orbital is given by Coulson's formula: $\

Vedclass · Accepted Answer

Between $20 - 21\%$

Vedclass · Answer

(A) The relationship between the bond angle $(	heta)$ and the percentage of $s-$character $(s)$ in a hybrid orbital is given by Coulson's formula: $\cos 	heta = \frac{s}{s-1}$.Given $	heta = 105^{o}$,we have $\cos(105^{o}) = -0.2588$.Substituting this into the formula: $-0.2588 = \frac{s}{s-1}$.$-0.2588(s-1) = s \implies -0.2588s + 0.2588 = s$.$1.2588s = 0.2588 \implies s = \frac{0.2588}{1.2588} \approx 0.2056$.Thus,the percentage of $s-$character is approximately $20.56\%$.However,considering the standard approximation for $sp^{3}$ hybridization where $s-$character is $25\%$ at $109.5^{o}$,as the bond angle decreases due to lone pair repulsion,the $s-$character decreases. For $105^{o}$,the value falls in the range of $20-21\%$. Therefore,option $A$ is the correct choice.

The bond angle between two hybrid orbitals is $105^{o}$. The $\%$ $s-$orbital character of the hybrid orbital is:

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