Question

Calculate the percentage of p-character in the orbital occupied by the lone pairs in water molecules : [Given:∠HOH is 104.5∘ and cos (104.5)= − 0.25]

Answer 1

Given Data:

$\angle \mathrm{HOH}=104.5°$

$\cos (104.5°)=-0.25$

Applying the formula to find p-character:

$$\begin{gathered} \cos \mathrm{\theta }=\frac{-1}{\mathrm{i}} \\ \cos (104.5°)=\frac{-1}{\mathrm{i}}=-0.25 \\ \mathrm{i}=\frac{1}{0.25}=4 \\ \mathrm{Now},\sum _{}{\mathrm{f}}_{\mathrm{p}}=3 \\ \frac{2\mathrm{x}\mathrm{i}}{\mathrm{i}+1}+2{{\mathrm{f}}^{,}}_{\mathrm{p}}=3\mathrm{where}{{\mathrm{f}}^{,}}_{\mathrm{p}}=\mathrm{fraction}\mathrm{of}\mathrm{p}-\mathrm{character}\mathrm{in}\mathrm{lone}\mathrm{pair} \\ Puttingi=4 \\ \frac{2x4}{4+1}+2{f^{,}}_p=3 \\ \mathrm{On}\mathrm{solving},\mathrm{we}\mathrm{get} \\ {f^{,}}_{p\mathit{}\mathit{=}\mathit{}}0.7 \\ \mathrm{So},\mathrm{percentage}\mathrm{of}\mathrm{p}-\mathrm{character}\mathrm{in}\mathrm{the}\mathrm{orbital}=0.7\mathrm{x}100=70\% \end{gathered}$$

Therefore, %p-character in the orbital occupied by lone pairs in water molecules is 70%.