Which of the following is paramagnetic ?

Options

A. `[Fe (CN)_6]^(4-)`
B. `[Ni(CO)_4 ]`
C. `[Ni(CN)_4]^(2-)`
D. `[CoF_6]^(3-)`

Answer

To determine which of the given coordination compounds is paramagnetic, we need to analyze the oxidation states and electron configurations of the central metal ions in each compound. Let's go through the steps systematically:

### Step 1: Identify the oxidation state of the central metal ion.
For each coordination compound, we will calculate the oxidation state of the central metal ion based on the charges of the ligands.

### Step 2: Determine the electron configuration of the metal ion.
Once we have the oxidation state, we can determine the electron configuration of the metal ion. This involves knowing the total number of electrons in the d and s orbitals.

### Step 3: Fill the d orbitals according to the ligand strength.
We need to consider whether the ligands are strong field or weak field ligands. Strong field ligands cause pairing of electrons, while weak field ligands do not.

### Step 4: Count the number of unpaired electrons.
After filling the d orbitals, we will count the number of unpaired electrons. If there are unpaired electrons, the species is paramagnetic. If all electrons are paired, it is diamagnetic.

### Step 5: Conclusion
Based on the analysis, we can conclude which compound is paramagnetic.

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### Detailed Solution

1. **First Coordination Compound**:
- Ligand: Cyanide (CN⁻) is a strong field ligand.
- Oxidation state of central atom (Fe):
\[ x + (-1 \times 6) = -4 \implies x = +2 \]
- Electron configuration of Fe²⁺:
\[ [Ar] 3d^6 \]
- Filling the d orbitals (strong field):
\[ \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \]
- Result: All electrons are paired → **Diamagnetic**.

2. **Second Coordination Compound**:
- Ligand: Carbonyl (CO) is a neutral strong field ligand.
- Oxidation state of central atom (Ni):
\[ x = 0 \]
- Electron configuration of Ni:
\[ [Ar] 3d^8 4s^2 \]
- Filling the d orbitals (strong field):
\[ \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \]
- Result: All electrons are paired → **Diamagnetic**.

3. **Third Coordination Compound**:
- Ligand: Cyanide (CN⁻) is a strong field ligand.
- Oxidation state of central atom (Ni):
\[ x + (-1 \times 4) = -2 \implies x = +2 \]
- Electron configuration of Ni²⁺:
\[ [Ar] 3d^8 \]
- Filling the d orbitals (strong field):
\[ \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \]
- Result: All electrons are paired → **Diamagnetic**.

4. **Fourth Coordination Compound**:
- Ligand: Fluoride (F⁻) is a weak field ligand.
- Oxidation state of central atom (Co):
\[ x + (-1 \times 6) = -3 \implies x = +3 \]
- Electron configuration of Co³⁺:
\[ [Ar] 3d^7 \]
- Filling the d orbitals (weak field):
\[ \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \]
- Result: There are 4 unpaired electrons → **Paramagnetic**.

### Conclusion:
The fourth coordination compound (Cobalt with fluoride ligands) is paramagnetic due to the presence of unpaired electrons.

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