Write the electronic configuration of the following ions.

Options

A. `Mn^(2+)`
B. `Fe^(3+)`
C. `Ni^(2+)`
D. `Cr^(3+)`

Answer

To write the electronic configuration of the given ions, we first need to understand the electronic configuration of the neutral atoms and then adjust for the charge of the ions.

### Step-by-Step Solution:

1. **Manganese Ion (Mn²⁺)**:
- Neutral manganese (Mn) has an atomic number of 25.
- Its electronic configuration is:
\[ \text{Mn: } [\text{Ar}] 3d^5 4s^2 \]
- For Mn²⁺, we remove 2 electrons. Electrons are removed first from the 4s orbital:
\[ \text{Mn}^{2+}: [\text{Ar}] 3d^5 4s^0 \]

2. **Iron Ion (Fe³⁺)**:
- Neutral iron (Fe) has an atomic number of 26.
- Its electronic configuration is:
\[ \text{Fe: } [\text{Ar}] 3d^6 4s^2 \]
- For Fe³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
\[ \text{Fe}^{3+}: [\text{Ar}] 3d^5 4s^0 \]

3. **Nickel Ion (Ni²⁺)**:
- Neutral nickel (Ni) has an atomic number of 28.
- Its electronic configuration is:
\[ \text{Ni: } [\text{Ar}] 3d^8 4s^2 \]
- For Ni²⁺, we remove 2 electrons from the 4s orbital:
\[ \text{Ni}^{2+}: [\text{Ar}] 3d^8 4s^0 \]

4. **Chromium Ion (Cr³⁺)**:
- Neutral chromium (Cr) has an atomic number of 24.
- Its electronic configuration is:
\[ \text{Cr: } [\text{Ar}] 3d^5 4s^1 \]
- For Cr³⁺, we remove 3 electrons (1 from 4s and 2 from 3d):
\[ \text{Cr}^{3+}: [\text{Ar}] 3d^3 4s^0 \]

### Final Electronic Configurations:
- Mn²⁺: \([Ar] 3d^5 4s^0\)
- Fe³⁺: \([Ar] 3d^5 4s^0\)
- Ni²⁺: \([Ar] 3d^8 4s^0\)
- Cr³⁺: \([Ar] 3d^3 4s^0\)