Question

Calculate the % p-character in the orbital occupied by the lone pairs in water molecules :
[Given : $\angle HOH$ is ${104.5}^{\circ }$ and cos $\left(104.5\right)=-0.25$]

• A. $80$%
• B. $20$%
• C. $70$%
• D. $75$%

Answer 1

The correct option is C $70$%

cos $\theta =\frac{-1}{i}$
$cos\left({104.5}^0\right)=\frac{-1}{i}=-0.25$
Hence, $i=4$.
Also, $\Sigma f_p=3$
$\frac{2i}{i+1}+2f_p^{\prime }=3$
$\frac{2\times 4}{5}+2f_p^{\prime }=3$

Where $f_p^{\prime }$ represents the fraction of p-charater in lone pair.
$f_p^{\prime }=\frac{7}{10}$
$f_p^{\prime }\%=70$
Thus, the percentage p character in the orbital occupied by the lone pairs in water molecules is 70%.