Question

what is the hybridization of the central atom in the phosphodichloride (pcl₂⁻) anion?

Explanation:

Step1: Determine the number of valence - shell electron pairs

Phosphorus (P) is in Group 5A, so it has 5 valence electrons. Chlorine (Cl) is in Group 7A, and each Cl atom contributes 7 valence electrons. The anion has a - 1 charge.
The number of valence - shell electron pairs ($n$) around the central P atom is calculated using the formula $n=\frac{\text{valence electrons of central atom}+\text{valence electrons from surrounding atoms}+\text{charge of the ion}}{2}$.
For $PCl_{2}^{-}$, the number of valence electrons of P is 5, the number of valence electrons from 2 Cl atoms is $2\times7 = 14$, and the charge of the ion is + 1. So $n=\frac{5 + 14+1}{2}=\frac{20}{2}=4$.

Step2: Determine the hybridization

The number of hybrid orbitals is equal to the number of valence - shell electron pairs around the central atom.
When the number of valence - shell electron pairs ($n = 4$), the hybridization is $sp^{3}$.

Answer:

$sp^{3}$