The heats of atomization of `PH_(3)(g)` and `P_(2)H_(4)(g)` are 954 kJ `"mol"^(-1)` and `1485kJ"mol"^(-1)` respectivel. 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 using the heats of atomization of `PH3(g)` and `P2H4(g)`, we can follow these steps:

### Step 1: Understand the heats of atomization
The heat of atomization is the energy required to break a compound into its constituent atoms in the gaseous state.

- For `PH3(g)`, the heat of atomization is given as 954 kJ/mol.
- For `P2H4(g)`, the heat of atomization is given as 1485 kJ/mol.

### Step 2: Calculate the bond energy of P-H in `PH3`
`PH3` has three P-H bonds. Therefore, the energy required to break one P-H bond can be calculated as follows:

\[ \text{Energy per P-H bond} = \frac{\text{Heat of atomization of } PH3}{\text{Number of P-H bonds}} = \frac{954 \text{ kJ/mol}}{3} = 318 \text{ kJ/mol} \]

### Step 3: Set up the equation for `P2H4`
In `P2H4`, there are four P-H bonds and one P-P bond. The total heat of atomization can be expressed as:

\[ \text{Heat of atomization of } P2H4 = \text{Energy for breaking 4 P-H bonds} + \text{Energy for breaking 1 P-P bond} \]

Substituting the known values:

\[ 1485 \text{ kJ/mol} = 4 \times (\text{Energy per P-H bond}) + (\text{Energy for P-P bond}) \]

Substituting the value of the energy per P-H bond:

\[ 1485 \text{ kJ/mol} = 4 \times 318 \text{ kJ/mol} + (\text{Energy for P-P bond}) \]

### Step 4: Calculate the P-P bond energy
Now, we can solve for the P-P bond energy:

\[ 1485 \text{ kJ/mol} = 1272 \text{ kJ/mol} + (\text{Energy for P-P bond}) \]

Subtracting 1272 kJ/mol from both sides gives:

\[ \text{Energy for P-P bond} = 1485 \text{ kJ/mol} - 1272 \text{ kJ/mol} = 213 \text{ kJ/mol} \]

### Final Answer
The P-P bond energy is **213 kJ/mol**.

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