In solid ammonia, each `NH_(3)` molecule has six other `NH_(3)` molecules as nearest neighbours. `DeltaH` sublimation of `NH_(3)` at the melting point is `30.8 k J mol^(-1)`, and the estimated `DeltaH` sublimation in the absence of hydrogen bonding is `14.4 kJ mol^(-1)`. the strength of a hydrogen bond is `NH_(3)` is

Options

A. `5.47 kJ " mol"^(-1)`
B. `10.93 kJ " mol"^(-1)`
C. `16.40 kJ " mol"^(-1)`
D. `-16.4 kJ " mol"^(-1)`

Explanation

To find the strength of a hydrogen bond in solid ammonia (NH₃), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- ΔH (sublimation of NH₃ at melting point) = 30.8 kJ/mol
- ΔH (sublimation in absence of hydrogen bonding) = 14.4 kJ/mol

2. **Calculate the Contribution of Hydrogen Bonds**:
- The contribution of hydrogen bonding to the sublimation enthalpy can be calculated by subtracting the estimated ΔH in the absence of hydrogen bonding from the actual ΔH at the melting point.
\[ \Delta H_{\text{hydrogen bonding}} = \Delta H_{\text{sublimation}} - \Delta H_{\text{no hydrogen bonding}} \]
\[ \Delta H_{\text{hydrogen bonding}} = 30.8 \, \text{kJ/mol} - 14.4 \, \text{kJ/mol} = 16.4 \, \text{kJ/mol} \]

3. **Determine the Number of Hydrogen Bonds**:
- Each NH₃ molecule forms 3 hydrogen bonds with neighboring NH₃ molecules. Since each molecule has 6 nearest neighbors, we can consider that the effective number of hydrogen bonds per molecule is 3.

4. **Calculate the Strength of Each Hydrogen Bond**:
- To find the strength of each hydrogen bond, divide the total contribution of hydrogen bonding by the number of hydrogen bonds per molecule.
\[ \text{Strength of each hydrogen bond} = \frac{\Delta H_{\text{hydrogen bonding}}}{\text{Number of hydrogen bonds}} \]
\[ \text{Strength of each hydrogen bond} = \frac{16.4 \, \text{kJ/mol}}{3} \approx 5.47 \, \text{kJ/mol} \]

5. **Final Result**:
- The strength of a hydrogen bond in solid ammonia is approximately **5.47 kJ/mol**.