Amongst the following the incorrect order is :-

Options

A. `IE_(1)(Al) lt IE_(1)(Mg)`
B. `IE_(1)(Na) lt IE_(1) (Mg)`
C. `IE_(2)(Mg) gt IE_(2)(Na)`
D. `IE_(3)(Mg) gt IE_(3)(Al)`

Explanation

To determine the incorrect order of ionization enthalpies among aluminum, magnesium, and sodium, we need to analyze the first, second, and third ionization enthalpies of these elements based on their electronic configurations.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- **Sodium (Na)**: Atomic number 11, electronic configuration is \(1s^2 2s^2 2p^6 3s^1\) or simply \( [Ne] 3s^1 \).
- **Magnesium (Mg)**: Atomic number 12, electronic configuration is \(1s^2 2s^2 2p^6 3s^2\) or \( [Ne] 3s^2 \).
- **Aluminum (Al)**: Atomic number 13, electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^1\) or \( [Ne] 3s^2 3p^1 \).

2. **Analyze the First Ionization Enthalpy**:
- **First Ionization of Sodium (Na)**: Removing one electron from \(3s^1\) results in a noble gas configuration \( [Ne] \). This requires less energy.
- **First Ionization of Magnesium (Mg)**: Removing one electron from \(3s^2\) results in \(3s^1\). This requires more energy than sodium.
- **First Ionization of Aluminum (Al)**: Removing one electron from \(3s^2 3p^1\) results in \(3s^2\). This requires more energy than both sodium and magnesium.

Therefore, the order of first ionization enthalpy is:
\[ \text{IE}_1(\text{Na}) < \text{IE}_1(\text{Al}) < \text{IE}_1(\text{Mg}) \]

3. **Analyze the Second Ionization Enthalpy**:
- **Second Ionization of Sodium (Na)**: After removing one electron, sodium becomes \( [Ne] \). Removing another electron from a noble gas configuration requires a lot of energy.
- **Second Ionization of Magnesium (Mg)**: After removing one electron, magnesium becomes \(3s^1\). Removing another electron from \(3s^1\) results in a noble gas configuration, which requires less energy than removing from sodium.
- **Second Ionization of Aluminum (Al)**: After removing one electron, aluminum becomes \(3s^2\). Removing another electron from \(3s^2\) requires more energy than magnesium.

Therefore, the order of second ionization enthalpy is:
\[ \text{IE}_2(\text{Mg}) < \text{IE}_2(\text{Al}) < \text{IE}_2(\text{Na}) \]

4. **Analyze the Third Ionization Enthalpy**:
- **Third Ionization of Sodium (Na)**: After removing two electrons, sodium has a noble gas configuration, and removing a third electron from a noble gas requires an enormous amount of energy.
- **Third Ionization of Magnesium (Mg)**: After removing two electrons, magnesium becomes \( [Ne] \), and removing a third electron also requires a lot of energy.
- **Third Ionization of Aluminum (Al)**: After removing two electrons, aluminum becomes \(3s^2\) and removing a third electron requires less energy compared to sodium and magnesium.

Therefore, the order of third ionization enthalpy is:
\[ \text{IE}_3(\text{Mg}) > \text{IE}_3(\text{Al}) > \text{IE}_3(\text{Na}) \]

5. **Identify the Incorrect Order**:
From the analysis, the incorrect order is:
\[ \text{IE}_2(\text{Mg}) > \text{IE}_2(\text{Na}) \]
This is incorrect because the second ionization enthalpy of magnesium is less than that of sodium.

### Conclusion:
The incorrect order among the ionization enthalpies is the second ionization enthalpy of magnesium being greater than that of sodium.