White phosphorus is a tetra atomic solid `P_(4) (s)` at room temperature. Find average `(P-P)` bond enthalpy (in KJ/mol).
`{:("Given",DeltaH_("sublimation") "of" P_(4)(s)=59 KJ//mol,),(,DeltaH_("atomizatiom") "of" P_(4)(s) = 1265 KJ//mol,):}`

Options

A. 201
B. 1206
C. `301.5`
D. `210.83`

Explanation

To find the average (P-P) bond enthalpy for white phosphorus (P₄), we can follow these steps:

### Step 1: Understand the given data
We are given:
- ΔH (sublimation) of P₄(s) = 59 kJ/mol
- ΔH (atomization) of P₄(s) = 1265 kJ/mol

### Step 2: Identify the total number of P-P bonds in P₄
The structure of P₄ consists of a tetrahedral arrangement of four phosphorus atoms. Each phosphorus atom is bonded to three other phosphorus atoms, forming a total of 6 P-P bonds.

### Step 3: Use the formula for average bond enthalpy
The average bond enthalpy can be calculated using the following formula:

\[ \text{Average P-P bond enthalpy} = \frac{\Delta H_{\text{atomization}} - \Delta H_{\text{sublimation}}}{\text{Number of P-P bonds}} \]

### Step 4: Substitute the values into the formula
Substituting the values we have:

\[ \text{Average P-P bond enthalpy} = \frac{1265 \, \text{kJ/mol} - 59 \, \text{kJ/mol}}{6} \]

### Step 5: Calculate the difference
First, calculate the difference between the atomization and sublimation enthalpies:

\[ 1265 \, \text{kJ/mol} - 59 \, \text{kJ/mol} = 1206 \, \text{kJ/mol} \]

### Step 6: Divide by the number of bonds
Now, divide this result by the number of P-P bonds (which is 6):

\[ \text{Average P-P bond enthalpy} = \frac{1206 \, \text{kJ/mol}}{6} = 201 \, \text{kJ/mol} \]

### Conclusion
Thus, the average (P-P) bond enthalpy in P₄ is **201 kJ/mol**.

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