The heat of atomisation of `PH_(3)(g)` and `P_(2)H_(4)(g)` are `954 kJ mol^(-1)` and `1485 kJ mol^(-1)` respectively. The `P-P` bond energy in `kJ mol^(-1)` is

Options

A. 213
B. 426
C. 318
D. 1272

Explanation

To find the P-P bond energy based on the given heats of atomization for PH₃(g) and P₂H₄(g), we can follow these steps:

### Step 1: Understand the Heat of Atomization
The heat of atomization is the energy required to break all bonds in one mole of a compound to form individual atoms in the gas phase.

### Step 2: Analyze PH₃(g)
For PH₃(g):
- The heat of atomization is given as 954 kJ/mol.
- In PH₃, there are 3 P-H bonds.

To find the energy required to break one P-H bond:
\[ \text{Energy per P-H bond} = \frac{\text{Total energy for PH}_3}{\text{Number of P-H bonds}} = \frac{954 \text{ kJ/mol}}{3} = 318 \text{ kJ/mol} \]

### Step 3: Analyze P₂H₄(g)
For P₂H₄(g):
- The heat of atomization is given as 1485 kJ/mol.
- In P₂H₄, there are 4 P-H bonds and 1 P-P bond.

Let \( x \) be the energy of the P-P bond. The total energy to break all bonds in P₂H₄ is the sum of the energy required to break the 4 P-H bonds and the 1 P-P bond:
\[ \text{Total energy for P}_2\text{H}_4 = 4 \times \text{(Energy per P-H bond)} + \text{(Energy of P-P bond)} \]
Substituting the known values:
\[ 1485 \text{ kJ/mol} = 4 \times 318 \text{ kJ/mol} + x \]

### Step 4: Calculate the Energy of P-P Bond
Calculate the total energy for breaking the 4 P-H bonds:
\[ 4 \times 318 = 1272 \text{ kJ/mol} \]
Now substitute this back into the equation:
\[ 1485 = 1272 + x \]
Solving for \( x \):
\[ x = 1485 - 1272 = 213 \text{ kJ/mol} \]

### Conclusion
The P-P bond energy is 213 kJ/mol.