the number of d- electrons in `Mn^(2+)` is equal to that of :

Options

A. p-electrons in N
B. s-electron in Na
C. d-electrons in `Fe^(+3)`
D. p-electrons in `O^(-2)`

Answer

To determine the number of d-electrons in \( \text{Mn}^{2+} \) and find which element has the same number of d-electrons, we can follow these steps:

### Step 1: Determine the atomic number of manganese (Mn)
Manganese has an atomic number of 25. This means a neutral manganese atom has 25 electrons.

### Step 2: Write the electronic configuration of neutral manganese
The electronic configuration of neutral manganese (Mn) is:
\[ \text{Mn}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2 \]

### Step 3: Determine the electronic configuration of \( \text{Mn}^{2+} \)
When manganese loses 2 electrons to form \( \text{Mn}^{2+} \), the electrons are removed first from the 4s subshell before the 3d subshell. Therefore, the electronic configuration of \( \text{Mn}^{2+} \) is:
\[ \text{Mn}^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 \]

### Step 4: Count the number of d-electrons in \( \text{Mn}^{2+} \)
From the electronic configuration of \( \text{Mn}^{2+} \), we can see that there are 5 electrons in the 3d subshell. Thus, \( \text{Mn}^{2+} \) has 5 d-electrons.

### Step 5: Compare with other elements
Now we need to find other elements that also have 5 d-electrons in their ions or neutral forms.

1. **Nitrogen (N)**:
- Atomic number: 7
- Electronic configuration: \( 1s^2 2s^2 2p^3 \)
- D-electrons: 0 (not correct)

2. **Sodium (Na)**:
- Atomic number: 11
- Electronic configuration: \( 1s^2 2s^2 2p^6 3s^1 \)
- D-electrons: 0 (not correct)

3. **Iron (Fe)**:
- Atomic number: 26
- Neutral iron has the configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2 \)
- For \( \text{Fe}^{3+} \) (losing 3 electrons):
- Configuration: \( 3d^5 \) (5 d-electrons, correct)

4. **Oxygen (O)**:
- Atomic number: 8
- Electronic configuration: \( 1s^2 2s^2 2p^4 \)
- D-electrons: 0 (not correct)

### Conclusion
The elements that have the same number of d-electrons (5 d-electrons) as \( \text{Mn}^{2+} \) are:
- \( \text{Fe}^{3+} \)

Thus, the answer to the question is that the number of d-electrons in \( \text{Mn}^{2+} \) is equal to that of \( \text{Fe}^{3+} \).