Question

What is hybridisation of boron in $\left[B\right(OH{)}_4{]}^{-}$?

• A. $s{p}^2$
• B. $s{p}^3$
• C. $s{p}^3{d}^2$
• D. None of these

Answer 1

The correct option is B $s{p}^3$

Hybridisation of boron in $\left[B\right(OH{)}_4{]}^{-}$ is $s{p}^3$

Hybridisation is $s{p}^3$ (Refer to Image 1)

electronic configuration of boron $\longrightarrow1s^{2}2s^{2}2p^1$

(Refer to Image 2)

electronic configuration of boron in first excited state

(Refer to Image 3)

Boron forms $4s{p}^3$ hybrid orbitals

The unpaired orbitals forms covalent bond with $3-OH$ groups

The empty orbital forms co-ordinate covalent with $-O{H}^{-}$ group

The boron atom in $B(OH{)}_3$ is electron deficient: it has 6 rather than 8 electrons in its valence shell. It acts as a Lewis acid by readily accepting the lone pair from the oxygen in a water molecule to go from sp2 to sp3 hybridisation.