Under normal conditions which of the given electronic configuration is able to from dispositive ion?

Options

A. Br
B. Cl
C. Mg
D. None of these

Answer

To determine which electronic configuration can form a dipositive ion (a cation with a +2 charge), we need to analyze the given electronic configurations of the elements. A dipositive ion is typically formed by metals, particularly those in groups that have two electrons in their outermost shell.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- We need to look at the electronic configurations of the elements provided in the question. For this example, let's consider bromine (Br), chlorine (Cl), and magnesium (Mg).

2. **Analyze Bromine (Br)**:
- Electronic Configuration: \( \text{Br} = [\text{Ar}] 3d^{10} 4s^2 4p^5 \)
- Bromine is a halogen and has 7 valence electrons. It tends to gain one electron to achieve a stable noble gas configuration (Krypton). Therefore, it forms an anion (Br⁻) rather than a cation.

3. **Analyze Chlorine (Cl)**:
- Electronic Configuration: \( \text{Cl} = [\text{Ne}] 3s^2 3p^5 \)
- Similar to bromine, chlorine has 7 valence electrons and will also gain one electron to form Cl⁻. It does not form a dipositive ion.

4. **Analyze Magnesium (Mg)**:
- Electronic Configuration: \( \text{Mg} = [\text{Ne}] 3s^2 \)
- Magnesium has 2 valence electrons in the 3s subshell. It can easily lose these two electrons to achieve the stable noble gas configuration of Neon. When it loses two electrons, it forms a dipositive ion (Mg²⁺).

5. **Conclusion**:
- Among the elements analyzed, magnesium (Mg) is the only element that can form a dipositive ion under normal conditions.

### Final Answer:
**Magnesium (Mg) is able to form a dipositive ion (Mg²⁺).**

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