The first ionisation potential of `Na` is `5.1eV`. The value of electron gain enthalpy of `Na^(+)` will be

Options

A. `-5.1 eV`
B. `-10.2 eV`
C. `+2.55 eV`
D. `-2.55 eV`

Answer

To solve the problem, we need to find the electron gain enthalpy of Na⁺ given that the first ionization potential (IP) of sodium (Na) is 5.1 eV.

### Step-by-Step Solution:

1. **Understanding Ionization Potential (IP)**:
- The first ionization potential is the energy required to remove the most loosely bound electron from a neutral atom in the gas phase. For sodium (Na), this value is given as 5.1 eV. This can be represented by the following equation:
\[ \text{Na (g)} \rightarrow \text{Na}^+ (g) + e^- \quad \Delta H = +5.1 \text{ eV} \]

2. **Understanding Electron Gain Enthalpy (EGE)**:
- The electron gain enthalpy (also known as electron affinity) is the energy change when an electron is added to a neutral atom to form an anion. For Na⁺, the process can be represented as:
\[ \text{Na}^+ (g) + e^- \rightarrow \text{Na (g)} \quad \Delta H = ? \]
- The energy change for this process is the negative of the ionization potential because it involves the reverse process of ionization.

3. **Relating IP and EGE**:
- From the above equations, we can see that the electron gain enthalpy of Na⁺ is the reverse of the ionization potential of Na. Therefore:
\[ \text{EGE of Na}^+ = -\text{IP of Na} \]
- Substituting the value of the ionization potential:
\[ \text{EGE of Na}^+ = -5.1 \text{ eV} \]

4. **Final Answer**:
- Thus, the value of the electron gain enthalpy of Na⁺ is:
\[ \text{EGE of Na}^+ = -5.1 \text{ eV} \]

### Summary:
The electron gain enthalpy of Na⁺ is **-5.1 eV**.