A : `K_(4)[Fe(CN)_(6)]` and `K_(3) [Fe(CN)_(6)]` have different magnetic moment..
.R: Magnetic moment is decided by the number ofunpaired electron and both have different number of unpaired electrons.

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1)
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).
C. ) If Assertion is true statement but Reason is false, then mark (3).
D. If both Assertion and Reason are false statements, then mark (4)

Answer

To analyze the assertion and reason provided in the question, we can break down the solution into clear steps:

### Step 1: Identify the Compounds
The two compounds given are \( K_4[Fe(CN)_6] \) and \( K_3[Fe(CN)_6] \). We need to determine the oxidation states of iron in both compounds.

### Step 2: Calculate the Oxidation State of Iron in \( K_4[Fe(CN)_6] \)
1. **Dissociate the compound**:
\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]
2. **Set up the equation for oxidation state**:
Let the oxidation state of iron be \( x \). The cyanide ion \( CN^- \) has a charge of -1, and there are 6 cyanide ions:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]

### Step 3: Determine the Electron Configuration of \( Fe^{2+} \)
- The atomic number of iron is 26, so its electron configuration is \( [Ar] 3d^6 4s^2 \).
- For \( Fe^{2+} \), the configuration becomes \( [Ar] 3d^6 \).

### Step 4: Analyze the Electron Pairing
- Since cyanide (\( CN^- \)) is a strong field ligand, it causes pairing of electrons in the \( d \) orbitals.
- The \( 3d^6 \) configuration will lead to all electrons being paired:
\[ \text{Electron configuration: } \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \quad (0 \text{ unpaired electrons}) \]

### Step 5: Calculate the Magnetic Moment for \( K_4[Fe(CN)_6] \)
- The formula for magnetic moment (\( \mu \)) is:
\[ \mu = \sqrt{n(n+2)} \]
where \( n \) is the number of unpaired electrons.
- Since \( n = 0 \):
\[ \mu = \sqrt{0(0+2)} = 0 \, \text{Bohr magneton} \]

### Step 6: Calculate the Oxidation State of Iron in \( K_3[Fe(CN)_6] \)
1. **Dissociate the compound**:
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]
2. **Set up the equation for oxidation state**:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

### Step 7: Determine the Electron Configuration of \( Fe^{3+} \)
- For \( Fe^{3+} \), the configuration becomes \( [Ar] 3d^5 \).

### Step 8: Analyze the Electron Pairing for \( Fe^{3+} \)
- The \( 3d^5 \) configuration will have all five electrons unpaired due to the strong field ligand:
\[ \text{Electron configuration: } \uparrow \uparrow \uparrow \uparrow \uparrow \quad (5 \text{ unpaired electrons}) \]

### Step 9: Calculate the Magnetic Moment for \( K_3[Fe(CN)_6] \)
- Using the formula:
\[ n = 5 \implies \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{Bohr magneton} \]

### Conclusion
- The assertion is true: \( K_4[Fe(CN)_6] \) and \( K_3[Fe(CN)_6] \) have different magnetic moments (0 and 5.92 Bohr magneton, respectively).
- The reason is also true: magnetic moments depend on the number of unpaired electrons, which differ in these two complexes.