Questions: What is the hybridization of the central atom in the (PO2^3-) anion?

Transcript text: What is the hybridization of the central atom in the $\left(\mathrm{PO}_{2}^{3-}\right)$ anion?

Solution

Solution Steps

Step 1: Determine the Central Atom

In the anion $\mathrm{PO}_{2}^{3-}$, phosphorus (P) is the central atom. This is because phosphorus is less electronegative than oxygen and can form multiple bonds.

Step 2: Count the Valence Electrons

Phosphorus has 5 valence electrons, and each oxygen has 6 valence electrons. The anion has a charge of $-3$, which means 3 additional electrons are added. Therefore, the total number of valence electrons is: \[ 5 + 2 \times 6 + 3 = 20 \]

Step 3: Draw the Lewis Structure

To satisfy the octet rule, phosphorus forms double bonds with each oxygen atom. The structure can be represented as: \[ \mathrm{P} = \mathrm{O} - \mathrm{O}^{3-} \] This structure accounts for the 20 valence electrons.

Step 4: Determine the Hybridization

The central phosphorus atom is surrounded by two double bonds and one lone pair. The steric number (number of regions of electron density) is 3, which corresponds to $sp^2$ hybridization.