Question

The state of hybridization of B in $\text{BC}{{\text{l}}_{\text{3}}}$ is :

• A. $s{{p}^{3}}$
• B. $s{{p}^{2}}$
• C. $sp$
• D. $s{{p}^{3}}{{d}^{2}}$

Answer 1

Answer: (B)

$s{{p}^{2}}$

Answer 2

Electronic configuration of boron (At. no. 5) is: $1{{s}^{2}},2{{s}^{2}},2p_{x}^{1}$ There being only one unpaired electron boron should be monovalent. But actually boron is trivalent as in $\text{BC}{{\text{l}}_{\text{3}}}\text{.}$ The formation of three covalent bonds is explained by thinking that the two 2s electrons are unpaired and one of these is excited to 2p state which results in the configuration as: $\text{1}{{\text{s}}^{2}},2{{s}^{1}},p_{x}^{1},2p_{y}^{1}$ The three half-filled orbitals hybridise to give three hybridised $s{{p}^{2}}$ orbitals usually inclined at an angle of $\text{12}{{\text{0}}^{\text{o}}}.$ Hence, hybridisation of boron in $BC{{l}_{3}}$ is $s{{p}^{2}}.$