Explanation

(b)
Step1. $P - H\$bond energy from bond dissociation energy of $PH_{3}$(g) containing 3 such $P - H$ bonds
$= \frac{228}{3} = 76$ kcal/mol
Step 2. The structure of $P_{2}H_{4}$ is
[image:6_image2.emf]$ie$, it contains 4 $P - H\$bonds and one$\ P - P$ bond, so $P - P\$bond energy can be calculated by $4 \times P - H + P - P\ =$ bond dissociation energy $P_{2}H_{4}$
$\therefore P - P$ bond energy = 335$- 4(76)$
$= 31$ kcal per mol