`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 will use the provided first and second ionization energies of magnesium, which are given as follows:

- \( \text{IE}_1 = 178 \, \text{kcal mol}^{-1} \)
- \( \text{IE}_2 = 348 \, \text{kcal mol}^{-1} \)

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction involves the ionization of magnesium to form \( \text{Mg}^{2+} \) by losing two electrons. This can be represented in two steps:
- Step 1: \( \text{Mg} \rightarrow \text{Mg}^+ + e^- \) (First ionization)
- Step 2: \( \text{Mg}^+ \rightarrow \text{Mg}^{2+} + e^- \) (Second ionization)

2. **Identify the Ionization Energies**:
- The energy required for the first ionization (to form \( \text{Mg}^+ \)) is \( \text{IE}_1 = 178 \, \text{kcal mol}^{-1} \).
- The energy required for the second ionization (to form \( \text{Mg}^{2+} \)) is \( \text{IE}_2 = 348 \, \text{kcal mol}^{-1} \).

3. **Calculate Total Energy Required**:
To find the total energy required for the reaction \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \), we need to add the energies of the two ionization processes:
\[ \Delta H = \text{IE}_1 + \text{IE}_2 \]
Substituting the values:
\[ \Delta H = 178 \, \text{kcal mol}^{-1} + 348 \, \text{kcal mol}^{-1} \]

4. **Perform the Addition**:
\[ \Delta H = 526 \, \text{kcal mol}^{-1} \]

5. **Conclusion**:
The energy required for the reaction \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \) is \( 526 \, \text{kcal mol}^{-1} \).

### Final Answer:
The energy required for the reaction is \( 526 \, \text{kcal mol}^{-1} \).

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