The % p-character in the orbitals forming C-H bond in CH4 is

The % p-character in the orbitals forming $C - H$ bond in ${CH}_{4}$ is

A
25
B
50
C
75
D
100

Solution:

Steric no. = Sigma bond+ Lone pair

$= 3 + 1 = 4$

The % p-character,

$= \frac{3}{4} \times 100 = 75 %$

So,

$s p^{3} \to 75 %$

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