Question

Calculate the percentage of $p-$ character of the orbitals of $‘B^{\prime }$ atom involved in bonding with $H$-atom in $B{H}_3$ molecule.

• A. $66.66\%$
• B. $33.33\%$
• C. $50\%$
• D. $75\%$

Answer 1

The correct option is A $66.66\%$

The ground state electronic configuration of $B$ atom is: $1s^{2}2s^{2}2p^1$

The excited state electronic configuration of $B$ atom is: $1s^{2}2s^{1}2p^2$

Thus, here $2s$ orbital and two $2p$ orbitals undergoes hybridisation to give three $s{p}^2$ hybridised orbitals. Hence, boron in $B{H}_3$ is $s{p}^2$ hybridised.

Thus, the percentage of $p-$ character of the hybridised orbital of $‘B^{\prime }$ in $B{H}_3$ molecule is $\frac{2}{3}\times 100=66.66\%$