Same number of unpaired electron is observed in which of the following complexes
(a) `[MnCl_(6)]^(3-)` (b) `[Fe(CN)_(6)]^(3-)`
( c) `[CoF_(6)]^(3-)` (d) `[Ni(NH_(3))_(6)]^(2+)`

Options

A. a and b
B. a & c
C. b & d
D. c & d

Answer

To determine which of the given complexes has the same number of unpaired electrons, we will analyze each complex step by step.

### Step 1: Determine the oxidation state of the metal in each complex.

1. **For `[MnCl6]^(3-)`:**
- Let the oxidation state of Mn be \( x \).
- The oxidation state of Cl is \(-1\).
- The equation becomes:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- Thus, Mn is in the +3 oxidation state.

2. **For `[Fe(CN)6]^(3-)`:**
- Let the oxidation state of Fe be \( y \).
- The oxidation state of CN is \(-1\).
- The equation becomes:
\[ y + 6(-1) = -3 \implies y - 6 = -3 \implies y = +3 \]
- Thus, Fe is in the +3 oxidation state.

3. **For `[CoF6]^(3-)`:**
- Let the oxidation state of Co be \( z \).
- The oxidation state of F is \(-1\).
- The equation becomes:
\[ z + 6(-1) = -3 \implies z - 6 = -3 \implies z = +3 \]
- Thus, Co is in the +3 oxidation state.

4. **For `[Ni(NH3)6]^(2+)`:**
- Let the oxidation state of Ni be \( w \).
- The oxidation state of NH3 is 0.
- The equation becomes:
\[ w + 6(0) = +2 \implies w = +2 \]
- Thus, Ni is in the +2 oxidation state.

### Step 2: Write the electronic configuration of each metal in its oxidation state.

1. **For Mn in +3 state:**
- Atomic number of Mn = 25.
- Electronic configuration: \( [Ar] 3d^5 4s^2 \).
- For Mn\(^{3+}\): Remove 3 electrons (2 from 4s and 1 from 3d).
- Configuration: \( [Ar] 3d^4 \).
- Unpaired electrons: 4.

2. **For Fe in +3 state:**
- Atomic number of Fe = 26.
- Electronic configuration: \( [Ar] 3d^6 4s^2 \).
- For Fe\(^{3+}\): Remove 3 electrons (2 from 4s and 1 from 3d).
- Configuration: \( [Ar] 3d^5 \).
- Unpaired electrons: 5 (due to strong field ligand CN causing pairing).

3. **For Co in +3 state:**
- Atomic number of Co = 27.
- Electronic configuration: \( [Ar] 3d^7 4s^2 \).
- For Co\(^{3+}\): Remove 3 electrons (2 from 4s and 1 from 3d).
- Configuration: \( [Ar] 3d^6 \).
- Unpaired electrons: 4 (due to weak field ligand F).

4. **For Ni in +2 state:**
- Atomic number of Ni = 28.
- Electronic configuration: \( [Ar] 3d^8 4s^2 \).
- For Ni\(^{2+}\): Remove 2 electrons (2 from 4s).
- Configuration: \( [Ar] 3d^8 \).
- Unpaired electrons: 2.

### Step 3: Summarize the number of unpaired electrons for each complex.

- **[MnCl6]^(3-)**: 4 unpaired electrons.
- **[Fe(CN)6]^(3-)**: 5 unpaired electrons.
- **[CoF6]^(3-)**: 4 unpaired electrons.
- **[Ni(NH3)6]^(2+)**: 2 unpaired electrons.

### Conclusion:
The complexes with the same number of unpaired electrons are:
- **[MnCl6]^(3-)** and **[CoF6]^(3-)**, both having **4 unpaired electrons**.

### Final Answer:
The correct options are **(a)** and **(c)**.

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