To solve the problem, we need to analyze the reactions of Fe²⁺ and Fe³⁺ with excess cyanide (CN⁻) and identify the complexes formed, denoted as P and Q, respectively.
### Step-by-Step Solution:
1. **Identify the Complexes Formed:**
- For the reaction of Fe²⁺ with excess CN⁻:
\[
\text{Fe}^{2+} + \text{CN}^- \rightarrow \text{P}
\]
The complex formed is \(\text{K}_4[\text{Fe(CN)}_6]\), which is known as potassium ferrocyanide.
- For the reaction of Fe³⁺ with excess CN⁻:
\[
\text{Fe}^{3+} + \text{CN}^- \rightarrow \text{Q}
\]
The complex formed is \(\text{K}_3[\text{Fe(CN)}_6]\), which is known as potassium ferricyanide.
2. **Determine the Color of the Complexes:**
- The complex P (\(\text{K}_4[\text{Fe(CN)}_6]\)) is typically yellow or light green.
- The complex Q (\(\text{K}_3[\text{Fe(CN)}_6]\)) is known to form a deep blue color, which is referred to as Prussian blue.
3. **Hybridization of the Complexes:**
- Both complexes involve the coordination of cyanide, which is a strong field ligand.
- For Fe²⁺ (\(d^6\) configuration), the hybridization is \(d^2sp^3\).
- For Fe³⁺ (\(d^5\) configuration), the hybridization is also \(d^2sp^3\).
4. **Magnetic Properties:**
- Complex P (from Fe²⁺) has no unpaired electrons, making it diamagnetic.
- Complex Q (from Fe³⁺) has one unpaired electron, making it paramagnetic.
5. **Comparison of Magnetic Moments:**
- Since complex P is diamagnetic (no unpaired electrons), and complex Q is paramagnetic (one unpaired electron), the magnetic moment of Q will be greater than that of P.
### Conclusion:
Based on the analysis, we can summarize the findings about complexes P and Q:
- **P** is \(\text{K}_4[\text{Fe(CN)}_6]\) (yellow/light green, diamagnetic).
- **Q** is \(\text{K}_3[\text{Fe(CN)}_6]\) (deep blue, paramagnetic, Prussian blue).
Thus, the correct statement regarding P and Q is that Q gives the Prussian blue color, while P does not.
