`Na_(2)S+Na_(2)[Fe(CN)_(5)NO]rarrNa_(4)[Fe(NH)_(5)NOS],` oxidation number of Fe in reactant ( complex ) and product ( complex) are `:`

Options

A. `2,1`
B. `2,2`
C. `2,3`
D. `3,3`

Explanation

To determine the oxidation number of iron (Fe) in the reactant and product complexes of the given reaction, we can follow these steps:

### Step 1: Identify the Reactants and Products
The reaction given is:
\[ \text{Na}_2\text{S} + \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \rightarrow \text{Na}_4[\text{Fe(NH)}_5\text{NOS}] \]

Here, the reactant complex is sodium nitroprusside, \(\text{Na}_2[\text{Fe(CN)}_5\text{NO}]\), and the product complex is \(\text{Na}_4[\text{Fe(NH)}_5\text{NOS}]\).

### Step 2: Calculate the Oxidation Number of Fe in the Reactant
For the reactant complex \(\text{Na}_2[\text{Fe(CN)}_5\text{NO}]\):
- Let the oxidation state of Fe be \(X\).
- The contribution from sodium (Na) is \(2 \times (+1) = +2\).
- The contribution from cyanide (CN) is \(5 \times (-1) = -5\).
- The contribution from nitrosyl (NO) is \(+1\).

Setting up the equation for the oxidation state:
\[ 2 + X + 5(-1) + 1 = 0 \]
\[ 2 + X - 5 + 1 = 0 \]
\[ X - 2 = 0 \implies X = +2 \]

### Step 3: Calculate the Oxidation Number of Fe in the Product
For the product complex \(\text{Na}_4[\text{Fe(NH)}_5\text{NOS}]\):
- Let the oxidation state of Fe be \(Y\).
- The contribution from sodium (Na) is \(4 \times (+1) = +4\).
- The contribution from five amine groups (NH) is \(5 \times 0 = 0\) (NH is neutral).
- The contribution from NOS is \(-1\).

Setting up the equation for the oxidation state:
\[ 4 + Y + 0 - 1 = 0 \]
\[ 4 + Y - 1 = 0 \]
\[ Y + 3 = 0 \implies Y = +2 \]

### Conclusion
The oxidation number of Fe in both the reactant complex \(\text{Na}_2[\text{Fe(CN)}_5\text{NO}]\) and the product complex \(\text{Na}_4[\text{Fe(NH)}_5\text{NOS}]\) is +2.

### Final Answer
The oxidation number of Fe in the reactant (complex) is +2, and in the product (complex) is also +2.

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