Hybridisation of boron in diborane $(B_{2} H_{6})$ is:

s p

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s p^{2}

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s p^{3}

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s p^{3} d

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Solution

The correct option is B $s p^{3}$
Hybridisation of boron in $B_{2} H_{6}$ is $s p^{3}$ .

The bonds are neither linear nor in one place.

This type of bond is known as $3 c - 2 e$ type bond. (or banana-bond).

Hence, option $C$ is correct.

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Similar questions

The type of hybridisation of boron in diborane is

(a) sp (b) sp²(c) sp³ (d) dsp²

C (1) H_{2} = C (2) H - C (3) \equiv C H

L i H + {B - -}_{2} H_{6} \to 2 L i B - - H_{4}

The type of hybridisation of boron in diborane is
(a) sp (b) $s p^{2}$ (c) $s p^{3}$ (d) $d s p^{2}$

A l H_{3}

A l H_{4}^{-}

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