Which one of the following will show paramagnetism corresponding to 2 unpaired electrons?
(Atomic numbers: Ni = 28, Fe=26)

Options

A. `[FeF_6]^(3-)`
B. `[NiCl_4]^(2-)`
C. `[Fe(CN)_6]^(3-)`
D. `[Ni(CN)_4]^(2-)`

Explanation

To determine which of the given options will show paramagnetism corresponding to 2 unpaired electrons, we need to analyze the electronic configurations of the relevant transition metals and their oxidation states. The two metals we are considering are Iron (Fe) and Nickel (Ni).

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- For Fe in the question, we consider its common oxidation states. The oxidation state of Fe in the complex is +3 (as indicated in the video).
- For Ni in NiCl₄²⁻, the oxidation state is +2.

2. **Write the Electronic Configuration**:
- For Fe (atomic number 26):
- The ground state electronic configuration is [Ar] 3d⁶ 4s².
- For Fe³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration becomes [Ar] 3d⁵.
- For Ni (atomic number 28):
- The ground state electronic configuration is [Ar] 3d⁸ 4s².
- For Ni²⁺, we remove 2 electrons (both from 4s):
- Configuration becomes [Ar] 3d⁸.

3. **Determine the Number of Unpaired Electrons**:
- **For Fe³⁺ (3d⁵)**:
- The 3d subshell has 5 electrons: ↑ ↑ ↑ ↑ ↑ (all unpaired).
- Total unpaired electrons = 5 (paramagnetic).
- **For Ni²⁺ (3d⁸)**:
- The 3d subshell has 8 electrons: ↑ ↑ ↑ ↑ ↑ (the first 6 are singly occupied, and the next 2 will pair up).
- Total unpaired electrons = 2 (paramagnetic).

4. **Conclusion**:
- The only species that shows paramagnetism with 2 unpaired electrons is Ni²⁺ in NiCl₄²⁻.

### Final Answer:
The compound that will show paramagnetism corresponding to 2 unpaired electrons is **Ni²⁺** in **NiCl₄²⁻**.

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