Which of the following ions exhibits d-d transitions and paramagnetism as well?

Options

A. `MnO_(4)^(-)`
B. `Cr_(2)O_(7)^(2-)`
C. `CrO_(4)^(2-)`
D. `MnO_(4)^(2-)`

Explanation

To determine which of the given ions exhibits both d-d transitions and paramagnetism, we will analyze each ion step by step.

### Step 1: Analyze MnO4^-
1. **Determine the oxidation state of manganese (Mn):**
- Let the oxidation state of Mn be \( x \).
- The equation is: \( x + 4(-2) = -1 \) (since the overall charge is -1).
- This simplifies to \( x - 8 = -1 \) → \( x = +7 \).

2. **Determine the electronic configuration of Mn in this state:**
- Manganese has a valence electronic configuration of \( [Ar] 4s^2 3d^5 \).
- For Mn in +7 oxidation state, it loses 7 electrons: \( 4s^2 3d^5 \) → \( 4s^0 3d^0 \).
- Therefore, MnO4^- has \( d^0 \) configuration.

3. **Conclusion for MnO4^-:**
- Since there are no d electrons, there can be no d-d transitions or paramagnetism.
- **Result:** Does not exhibit d-d transitions or paramagnetism.

### Step 2: Analyze Cr2O7^2-
1. **Determine the oxidation state of chromium (Cr):**
- Let the oxidation state of Cr be \( x \).
- The equation is: \( 2x + 7(-2) = -2 \).
- This simplifies to \( 2x - 14 = -2 \) → \( 2x = 12 \) → \( x = +6 \).

2. **Determine the electronic configuration of Cr in this state:**
- For Cr in +6 oxidation state: \( 4s^2 3d^5 \) → \( 4s^0 3d^0 \).
- Therefore, Cr2O7^2- has \( d^0 \) configuration.

3. **Conclusion for Cr2O7^2-:**
- Since there are no d electrons, there can be no d-d transitions or paramagnetism.
- **Result:** Does not exhibit d-d transitions or paramagnetism.

### Step 3: Analyze CrO4^2-
1. **Determine the oxidation state of chromium (Cr):**
- Let the oxidation state of Cr be \( x \).
- The equation is: \( x + 4(-2) = -2 \).
- This simplifies to \( x - 8 = -2 \) → \( x = +6 \).

2. **Determine the electronic configuration of Cr in this state:**
- For Cr in +6 oxidation state: \( 4s^2 3d^5 \) → \( 4s^0 3d^0 \).
- Therefore, CrO4^2- has \( d^0 \) configuration.

3. **Conclusion for CrO4^2-:**
- Since there are no d electrons, there can be no d-d transitions or paramagnetism.
- **Result:** Does not exhibit d-d transitions or paramagnetism.

### Step 4: Analyze MnO4^2-
1. **Determine the oxidation state of manganese (Mn):**
- Let the oxidation state of Mn be \( x \).
- The equation is: \( x + 4(-2) = -2 \).
- This simplifies to \( x - 8 = -2 \) → \( x = +6 \).

2. **Determine the electronic configuration of Mn in this state:**
- For Mn in +6 oxidation state: \( 4s^2 3d^5 \) → \( 4s^0 3d^1 \).
- Therefore, MnO4^2- has \( d^1 \) configuration.

3. **Conclusion for MnO4^2-:**
- There is 1 unpaired electron in the d orbital, allowing for d-d transitions.
- Since it has unpaired electrons, it is paramagnetic.
- **Result:** Exhibits both d-d transitions and paramagnetism.

### Final Conclusion:
The ion that exhibits both d-d transitions and paramagnetism is **MnO4^2-**.

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