`Be^(2+)` is isoelectronic with which of the following ions ?

Options

A. `H^+`
B. `Li^+`
C. `Na^+`
D. `Mg^(2+)`

Explanation

To determine which ion is isoelectronic with \( \text{Be}^{2+} \), we need to follow these steps:

### Step 1: Determine the Atomic Number and Electron Configuration of Beryllium
- The atomic number of beryllium (Be) is 4. This means that a neutral beryllium atom has 4 electrons.
- The electron configuration of neutral beryllium is \( 1s^2 2s^2 \).

### Step 2: Calculate the Number of Electrons in \( \text{Be}^{2+} \)
- The \( \text{Be}^{2+} \) ion is formed by removing 2 electrons from the neutral beryllium atom.
- Therefore, \( \text{Be}^{2+} \) has \( 4 - 2 = 2 \) electrons.

### Step 3: Identify the Electron Configuration of \( \text{Be}^{2+} \)
- With 2 electrons, the electron configuration for \( \text{Be}^{2+} \) is \( 1s^2 \).

### Step 4: Analyze the Given Ions
Now we will analyze the given ions to see which one has the same number of electrons (2 electrons):

1. **\( \text{H}^{+} \)**:
- Atomic number = 1 (1 electron)
- After losing 1 electron, \( \text{H}^{+} \) has 0 electrons.

2. **\( \text{Li}^{+} \)**:
- Atomic number = 3 (3 electrons)
- After losing 1 electron, \( \text{Li}^{+} \) has \( 3 - 1 = 2 \) electrons.

3. **\( \text{Na}^{+} \)**:
- Atomic number = 11 (11 electrons)
- After losing 1 electron, \( \text{Na}^{+} \) has \( 11 - 1 = 10 \) electrons.

4. **\( \text{Mg}^{2+} \)**:
- Atomic number = 12 (12 electrons)
- After losing 2 electrons, \( \text{Mg}^{2+} \) has \( 12 - 2 = 10 \) electrons.

### Step 5: Conclusion
- The only ion that has the same number of electrons (2 electrons) as \( \text{Be}^{2+} \) is \( \text{Li}^{+} \).
- Therefore, \( \text{Li}^{+} \) is isoelectronic with \( \text{Be}^{2+} \).

### Final Answer
**\( \text{Li}^{+} \) is isoelectronic with \( \text{Be}^{2+} \).**

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