Which hybridization results in non-planar orbitals?

s p

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s p^{2}

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s p^{3}

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d s p^{2}

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Solution

The correct option is C $s p^{3}$
Consider the geometry defined by the hybridization:-
A) $s p$ - Linear
B) $s p^{2}$ - Trigonal planar
C) $s p^{3}$ - Tetrahedral
D) $d s p^{2}$ - Square Planar.
Hence, $s p^{3}$ hybridisation results in non-planar orbitals.

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Similar questions

The correct order of hybridization of the central atom in the following species $N H_{3}, [P t C l_{4}]^{2 -}$ , $P C l_{5}$ and $B C l_{3}$ is

s p, s p^{2}, s p^{3}

C H_{4}, C_{2} H_{6}, a n d C_{2} H_{4}

Question 03

The types of hybrid orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ respectively are expected to be

(a) $s p, s p^{3} a n d s p^{2}$

(b) $s p, s p^{2} a n d s p^{3}$

(d) $s p^{2}, s p^{3} a n d s p$

s p^{3}, s p^{2}, s p

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