Percentage $p$ character in ${s p}^{3} d^{2}$ hybridisation is:

33

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75

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50

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25

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Solution

The correct option is C $50$ %
Percentage $P$ character is $s p^{3} d^{2}$
$= \frac{3}{6} \times 100 = 50 %$

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The percentage of p-character in forming P-P bonds in P₄ molecule is:

N H_{3}

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s p^{3}

‘ s^{'}

{s p}^{3} d^{2}

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