To determine which of the given compounds have the same magnetic moment, we need to find the number of unpaired electrons in each compound. The magnetic moment (μ) can be calculated using the formula:
\[
μ = \sqrt{n(n + 2)}
\]
where \(n\) is the number of unpaired electrons.
Let's analyze the compounds step by step.
### Step 1: Analyze Chromium (Cr) in +2 Oxidation State
- **Electronic Configuration**: For Cr, the ground state configuration is \( [Ar] 3d^5 4s^1 \). In +2 oxidation state, it loses 2 electrons:
- \(3d^5 4s^0\)
- **Number of Unpaired Electrons**: Since \(3d^5\) has 5 unpaired electrons (due to Hund's rule), \(n = 4\) (as it is a weak ligand, water does not cause pairing).
### Step 2: Analyze Iron (Fe) in +2 Oxidation State
- **Electronic Configuration**: For Fe, the ground state configuration is \( [Ar] 3d^6 4s^2 \). In +2 oxidation state, it loses 2 electrons:
- \(3d^6 4s^0\)
- **Number of Unpaired Electrons**: In \(3d^6\) with a weak ligand, the configuration will have 4 unpaired electrons, so \(n = 4\).
### Step 3: Calculate Magnetic Moment for Cr and Fe
- Using the formula:
\[
μ = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \, \text{BM}
\]
### Step 4: Analyze Cobalt (Co) in +4 Oxidation State (from CoCl6^2-)
- **Oxidation State Calculation**: For Co in CoCl6^2-, we set up the equation:
- \(x - 6 = -2\) → \(x = +4\)
- **Electronic Configuration**: For Co, the ground state configuration is \( [Ar] 3d^7 4s^2 \). In +4 oxidation state, it loses 4 electrons:
- \(3d^5 4s^0\)
- **Number of Unpaired Electrons**: In \(3d^5\), there are 5 unpaired electrons, so \(n = 5\).
### Step 5: Analyze Cobalt (Co) in +2 Oxidation State (from CoCl4^2-)
- **Oxidation State Calculation**: For Co in CoCl4^2-, we set up the equation:
- \(x - 4 = -2\) → \(x = +2\)
- **Electronic Configuration**: For Co in +2 oxidation state:
- \(3d^7 4s^0\)
- **Number of Unpaired Electrons**: In \(3d^7\) with a weak ligand, there are 3 unpaired electrons, so \(n = 3\).
### Step 6: Analyze Cobalt (Co) and Iron (Fe) in +2 Oxidation State with H2O
- **Cobalt**: As calculated, \(n = 3\) (from \(3d^7\)).
- **Iron**: As calculated, \(n = 4\) (from \(3d^6\)).
### Summary of Unpaired Electrons:
1. **Cr +2**: \(n = 4\)
2. **Fe +2**: \(n = 4\)
3. **Co +4**: \(n = 5\)
4. **Co +2**: \(n = 3\)
5. **Fe +2 (H2O)**: \(n = 4\)
### Conclusion:
The compounds that have the same magnetic moment are:
- Cr +2 and Fe +2 (both have \(n = 4\))
### Final Answer:
The compounds with the same magnetic moment are **Cr +2 and Fe +2**.
