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

Answer

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

### Step 1: Identify the oxidation state of sodium (Na)
Sodium (Na) is an alkali metal, and its oxidation state is always +1. Since there are 4 sodium atoms in the complex, their total contribution to the oxidation state is:
\[ 4 \times (+1) = +4 \]

### Step 2: Identify the oxidation states of the ligands
1. The oxidation state of the cyanide ion (\( \text{CN} \)) is -1. Since there are 5 cyanide ions, their total contribution is:
\[ 5 \times (-1) = -5 \]
2. The oxidation state of the nitroso group (\( \text{NOS} \)) is also -1.

### Step 3: Set up the equation for the oxidation state of Fe
Let the oxidation state of Fe be \( x \). The overall charge of the complex is neutral (0), so we can set up the equation:
\[ \text{Total charge} = \text{Charge from Na} + \text{Charge from Fe} + \text{Charge from CN} + \text{Charge from NOS} \]
Substituting the known values:
\[ 0 = (+4) + x + (-5) + (-1) \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 0 = 4 + x - 5 - 1 \]
\[ 0 = x - 2 \]

### Step 5: Solve for x
Rearranging the equation gives:
\[ x = +2 \]

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

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