Sulphide ion reacts with `[Fe(CN)_(5)NO]` to form a purple coloured compound (X). In this reaction oxidation state of iron .

Options

A. changes from +2 to +3
B. changes from +3 to +2
C. changes from +2 to +4
D. does not change.

Answer

To determine the oxidation state of iron in the reaction between the sulfide ion and the complex `[Fe(CN)_(5)NO]`, we can follow these steps:

### Step 1: Identify the components in the reaction
- The sulfide ion is represented as \( S^{2-} \).
- The complex is \( [Fe(CN)_{5}NO] \).

### Step 2: Determine the charge of the complex
- The overall charge of the complex is neutral.
- The cyanide ion \( CN^{-} \) has a charge of \(-1\). Since there are 5 cyanide ions, their total charge is \( 5 \times (-1) = -5 \).
- The nitrosyl ion \( NO \) can be considered as \( NO^{+} \) in this context, contributing a charge of \( +1 \).
- Therefore, the total charge from the ligands is \( -5 + 1 = -4 \).
- To balance this, the iron must have a charge of \( +4 \) to make the overall charge of the complex neutral.

### Step 3: Calculate the oxidation state of iron
- The oxidation state of iron in the complex can be calculated as follows:
\[ \text{Charge of iron} + \text{Charge from ligands} = 0 \]
\[ x + (-4) = 0 \implies x = +4 \]
- Thus, the oxidation state of iron in the complex \( [Fe(CN)_{5}NO] \) is \( +4 \).

### Step 4: Analyze the reaction with sulfide ion
- When the sulfide ion \( S^{2-} \) reacts with the complex, it forms a new compound \( X \).
- The new compound formed will have the iron still in the same oxidation state, as the sulfide ion does not oxidize or reduce the iron.

### Step 5: Conclusion
- The oxidation state of iron in the reaction remains unchanged at \( +3 \) after the reaction with the sulfide ion.

### Final Answer
The oxidation state of iron in the reaction is \( +3 \).

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