Which of the following comp. is coloured and has unpaired electron :

Options

A. `CuF_2`
B. `K_2Cr_2O_7`
C. `KMnO_4`
D. `K_4[Fe(CN)_6]`

Answer

To determine which of the given compounds is colored and has unpaired electrons, we need to analyze the oxidation states and electronic configurations of the compounds. Let's go through the analysis step by step.

### Step 1: Analyze Copper(II) Fluoride (CuF2)
1. **Determine the oxidation state of Copper (Cu)**:
- In CuF2, fluorine (F) has an oxidation state of -1. Since there are two fluorine atoms, the total contribution is -2.
- Let the oxidation state of Cu be \( x \):
\[ x + 2(-1) = 0 \implies x - 2 = 0 \implies x = +2 \]

2. **Find the electronic configuration of Cu in +2 state**:
- Copper (Cu) has an atomic number of 29. Its electronic configuration is:
\[ [Ar] 3d^{10} 4s^1 \]
- In the +2 oxidation state, it loses two electrons, typically from the 4s and then from the 3d:
\[ 3d^9 \]

3. **Identify unpaired electrons**:
- The configuration \( 3d^9 \) indicates that there is one unpaired electron.

4. **Determine the color**:
- CuF2 is known to be colored (specifically, it appears blue or green).

### Step 2: Analyze Potassium Dichromate (K2Cr2O7)
1. **Determine the oxidation state of Chromium (Cr)**:
- The formula can be simplified to Cr2O7^2-. The total oxidation state must equal -2:
\[ 2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]

2. **Find the electronic configuration of Cr in +6 state**:
- Chromium (Cr) has an atomic number of 24. Its electronic configuration is:
\[ [Ar] 3d^5 4s^1 \]
- In the +6 oxidation state, it loses 6 electrons:
\[ 3d^0 \]

3. **Identify unpaired electrons**:
- The configuration \( 3d^0 \) indicates that there are no unpaired electrons.

### Step 3: Analyze Potassium Permanganate (KMnO4)
1. **Determine the oxidation state of Manganese (Mn)**:
- The formula can be simplified to MnO4^-. The total oxidation state must equal -1:
\[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \]

2. **Find the electronic configuration of Mn in +7 state**:
- Manganese (Mn) has an atomic number of 25. Its electronic configuration is:
\[ [Ar] 3d^5 4s^2 \]
- In the +7 oxidation state, it loses 7 electrons:
\[ 3d^0 \]

3. **Identify unpaired electrons**:
- The configuration \( 3d^0 \) indicates that there are no unpaired electrons.

### Step 4: Analyze Ferricyanide (Fe(CN)6^4-)
1. **Determine the oxidation state of Iron (Fe)**:
- The formula can be simplified to Fe(CN)6^4-. The total oxidation state must equal -4:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]

2. **Find the electronic configuration of Fe in +2 state**:
- Iron (Fe) has an atomic number of 26. Its electronic configuration is:
\[ [Ar] 3d^6 4s^2 \]
- In the +2 oxidation state, it loses two electrons (typically from 4s):
\[ 3d^6 \]

3. **Identify unpaired electrons**:
- The configuration \( 3d^6 \) can have unpaired electrons depending on the ligand. Since CN^- is a strong field ligand, it causes pairing:
- The paired configuration would be \( 3d^6 \) with no unpaired electrons.

### Conclusion:
From the analysis, only **CuF2** has an unpaired electron and is colored. Therefore, the answer is **CuF2**.