Oxidation number of Fe in violet coloured complex `Na_(4)[Fe(CN)_(5)(NOS)]` is :

Options

A. `0`
B. `2`
C. `3`
D. `4`

Explanation

To determine the oxidation number of Fe in the complex \( \text{Na}_4[\text{Fe(CN)}_5(\text{NOS})] \), we can follow these steps:

### Step 1: Identify the charges of the components
- Sodium (Na) has a charge of +1. Since there are 4 sodium ions, the total charge contributed by sodium is:
\[ 4 \times (+1) = +4 \]

### Step 2: Determine the charges of the ligands
- The cyanide ion (CN) has a charge of -1. Since there are 5 cyanide ions, the total charge contributed by cyanide is:
\[ 5 \times (-1) = -5 \]
- The NOS ligand is also considered to have a charge of -1. Therefore, the total charge contributed by NOS is:
\[ -1 \]

### Step 3: Calculate the total charge from ligands
- Adding the charges from the cyanide and NOS ligands gives:
\[ -5 + (-1) = -6 \]

### Step 4: Set up the equation for the oxidation state of Fe
- Let the oxidation state of Fe be \( x \). The overall charge of the complex ion must equal the sum of the charges from sodium, the ligands, and the oxidation state of Fe:
\[ +4 + x + (-6) = 0 \]

### Step 5: Solve for the oxidation state of Fe
- Rearranging the equation gives:
\[ x - 2 = 0 \]
\[ x = +2 \]

### Conclusion
The oxidation number of Fe in the complex \( \text{Na}_4[\text{Fe(CN)}_5(\text{NOS})] \) is \( +2 \).