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Question

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

A
80%
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B
20%
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C
70%
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D
75%
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Solution

The correct option is C 70%
cos θ=1i
cos(104.50)=1i=0.25
Hence, i=4.
Also, Σfp=3
2ii+1+2fp=3
2×45+2fp=3

Where fp represents the fraction of p-charater in lone pair.
fp=710
fp%=70
Thus, the percentage p character in the orbital occupied by the lone pairs in water molecules is 70%.

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