The heats of atomization of PH₃(g) and P₂H₄(g) are 954 kJ mol^-1 and 1485 kJ mol^-1 respectively. The P—P bond energy in kJ mol^-1 is

In ${\mathrm{PH}}_3\left(\mathrm{g}\right)$, energy required to break 3 P-H bonds = 954 kJ mol^-1 

$\therefore$Energy required to break 1 P —H bond$=\frac{954}{3}=318{\mathrm{kJmol}}^{-1}$

In P₂H₄(g), energy of 1 P—P bond + 4 P—H bonds = 1485 kJ mol^-1

$\because \text{ Energy of }1\mathrm{P}-\mathrm{H}\text{ bond }=318\mathrm{kJ}{\mathrm{mol}}^{-1}$

$\therefore \text{ Energy of }4\mathrm{P}-\mathrm{H}\text{ bond }=318\times 4$

                                                 = 1272 kJ mol^-1

Thus, the P — P bond energy = 1485 -1272

                                                 = 213 kJ mol^-1