Question

The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in ${\left[B{\left(OH\right)}_4\right]}^{-}$ and the geometry of the complex are respectively.

• A. ${sp}^3$, tetrahedral
• B. ${sp}^3$, square planar
• C. ${sp}^3{d}^2$, octahedral
• D. ${dsp}^3$, square planar

Answer 1

The correct option is A ${sp}^3$, tetrahedral

Solution:- (A) $s{p}^3$, tetrahedral

Boron has di electronic configuration:

$1s^{2}2s^{2}2{p_x}^{1}2{p_y}^{0}2{p_z}^0$

In the excited state, $2s$-orbital electrons are impaired and one electron is shifted to a $p$-orbital. Now, hybridisation occurs between one $s$- and three $p$-orbitals to give $s{p}^3$ hybridisation and tetrahedral geometry.