`IE_(1)` and `IE_(2)` of `Mg` are `178` and `348 kcal mol^(-1)`. The energy required for the reaction `Mg rarr Mg^(2+)+2e^(-)` is

Options

A. `+170 kcal`
B. `+526 kcal`
C. `-170 kcal`
D. `-526 kcal`

Explanation

To find the energy required for the reaction \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \), we need to consider the ionization energies involved in this process.

### Step-by-Step Solution:

1. **Understand the Ionization Energies**:
- The first ionization energy (IE₁) is the energy required to remove the first electron from a neutral magnesium atom (Mg):
\[ \text{Mg} \rightarrow \text{Mg}^+ + e^- \]
- The second ionization energy (IE₂) is the energy required to remove the second electron from the singly charged magnesium ion (Mg⁺):
\[ \text{Mg}^+ \rightarrow \text{Mg}^{2+} + e^- \]

2. **Given Values**:
- \( \text{IE}_1 = 178 \, \text{kcal/mol} \)
- \( \text{IE}_2 = 348 \, \text{kcal/mol} \)

3. **Calculate Total Energy**:
- The total energy required for the reaction \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \) is the sum of the first and second ionization energies:
\[ \text{Total Energy} = \text{IE}_1 + \text{IE}_2 \]
- Substituting the values:
\[ \text{Total Energy} = 178 \, \text{kcal/mol} + 348 \, \text{kcal/mol} = 526 \, \text{kcal/mol} \]

4. **Final Result**:
- The energy required for the reaction \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \) is \( 526 \, \text{kcal/mol} \).