Question

Phosphine ( PH₃ )reacts with borane ( BH₃ ) as follows:

$\mathit{P}{\mathit{H}}_{\mathbf{3}}\mathbf{+}\mathit{B}{\mathit{H}}_{\mathbf{3}}\mathbf{\to }\mathit{P}{\mathit{H}}_{\mathbf{3}}\mathbf{-}\mathit{B}{\mathit{H}}_{\mathbf{3}}$

1. Which of the illustrations below depicts the change, if any, in the orbital hybridization of P during this reaction?
2. Which depicts the change, if any, in the orbital hybridization of B

Short answer

In hybridization, the number of hybrid orbitals formed is equal to the number of electron groups bonded to the central atom.

Step-by-step solution

For the orbital hybridization P during the reaction

The P atom PH₃ has 1 lone pair and 3 bond pairsmean it has four electron groups that form SP³hybridization. In the product, the P atom is also surrounded by four electron groups and formshybridization.Hence, in the whole reaction, the hybridization of Pan atom is sp³.

Therefore, illustration B is correct.$\mathit{s}{\mathit{p}}^{\mathbf{3}}\mathbf{\to }\mathit{s}{\mathit{p}}^{\mathbf{3}}$

For the orbital hybridization of B during the reaction

The B atom BH₃ has 3 bond pairsmeans it has three electron groupsthat form sp² hybridization. In the product, the B atom is surrounded by four electron groups and formsSP³hybridization. Hence, in the reaction, the hybridization of the B atom changes from sp²to SP³.

Therefore, illustration A is correct.$\mathit{s}{\mathit{p}}^{\mathbf{2}}\mathbf{\to }\mathit{s}{\mathit{p}}^{\mathbf{3}}$