What is the value of electron gain enthalpy of `Na^+` if `IE_1` of Na = 5.1 eV?

Options

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

Explanation

To find the value of the electron gain enthalpy of `Na^+`, given that the first ionization energy (IE_1) of sodium (Na) is 5.1 eV, we can follow these steps:

### Step 1: Understand Ionization Energy
Ionization energy is the amount of energy required to remove an electron from a neutral atom in the gas phase. For sodium (Na), the first ionization energy is given as 5.1 eV. This means that to remove one electron from a neutral sodium atom, we need 5.1 eV of energy.

### Step 2: Write the Ionization Reaction
The ionization reaction for sodium can be represented as:
\[ \text{Na (g)} \rightarrow \text{Na}^+ (g) + e^- \]
This reaction shows that when sodium loses one electron, it forms a sodium ion (`Na^+`).

### Step 3: Understand Electron Gain Enthalpy
Electron gain enthalpy is the amount of energy released when an electron is added to a neutral atom in the gas phase. For `Na^+`, the reaction can be represented as:
\[ \text{Na}^+ (g) + e^- \rightarrow \text{Na (g)} \]
This reaction shows that when `Na^+` gains an electron, it becomes neutral sodium (Na).

### Step 4: Relate Ionization Energy and Electron Gain Enthalpy
The electron gain enthalpy for `Na^+` is the reverse of the ionization energy of sodium. Therefore, we can express this relationship as:
\[ \Delta H = -\text{IE}_1 \]
Where:
- \(\Delta H\) is the electron gain enthalpy,
- \(\text{IE}_1\) is the first ionization energy.

### Step 5: Calculate Electron Gain Enthalpy
Given that the first ionization energy of sodium is 5.1 eV, we can substitute this value into our equation:
\[ \Delta H = -5.1 \, \text{eV} \]

### Conclusion
Thus, the value of the electron gain enthalpy of `Na^+` is:
\[ \Delta H = -5.1 \, \text{eV} \]