s - character in ${s p}^{3}$ hybridization is

$\frac{3}{4}$

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$\frac{1}{4}$

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$\frac{1}{2}$

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None of these

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Solution

The correct option is B
$\frac{1}{4}$

Here:

1s orbital + 1p orbital $\Rightarrow$ 2 $s p$ hybrid orbitals

1s orbital + 2p orbital $\Rightarrow$ 3 ${s p}^{2}$ hybrid orbitals

1s orbital + 3p orbital $\Rightarrow$ 4 ${s p}^{3}$ orbital (hybrid)

So,whenever we are getting ${s p}^{3}$ orbitals, we are getting 4 ${s p}^{3}$ orbitals.

s - charcter = $\frac{N u m b e r o f s - o r b i t a l i n v o l v e d i n h y b r i d i z a t i o n}{N u m b e r o f h y b r i d o r b i t a l s}$ = $\frac{1}{4}$

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Similar questions

s - character in ${s p}^{3}$ hybridization is:

s

s p

{s p}^{3}

C H C l_{3}

s p^{3}

‘ s^{'}

[N i C l_{4}]^{2 -}, [N i (C N)_{4}]^{2 -} & [N i (C O)_{4}]

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