Percent s character of the central atom in PBr3

The percent s character of the central atom in PBr3 can be calculated using the formula:

% s character = 100 × (1 - cosθ/4)

where θ is the bond angle in degrees.

Given that the bond angle in PBr3 is 101°, we can plug this value into the formula to get:

% s character = 100 × (1 - cos101/4)
= 0.96 or 1 (rounded off)

Therefore, the percent s character of the central atom in PBr3 is approximately 1.

Explanation

When a central atom is involved in a covalent bond, it contributes to the bonding by sharing its valence electrons. These electrons can be classified into s and p orbitals, which have different shapes and energies.

The percent s character of an orbital is a measure of the amount of s orbital character it contains. This value ranges from 0 to 100%, with 0% indicating pure p character and 100% indicating pure s character.

The bond angle in a molecule depends on the hybridization of the central atom and the number of bonding and non-bonding electron pairs around it. In PBr3, the central atom is phosphorus, which has five valence electrons. To form four bonds with bromine atoms, one of the valence electrons is promoted to the 3d orbital, and the remaining four electrons are hybridized to form four sp3 orbitals. The bond angle in this case is approximately 109.5°, which is the ideal tetrahedral angle.

However, due to the larger size of the bromine atoms, there is some repulsion between them, which causes the bond angle to decrease slightly to 101°. This decrease in bond angle indicates that the orbitals involved in bonding have more p character than s character.

Using the formula mentioned earlier, we can calculate the percent s character of the sp3 hybrid orbitals in PBr3 to be approximately 1%. This means that these orbitals have predominantly p character, which is consistent with the decreased bond angle observed in the molecule.