The oxidation number of iron in sodium nitroprusside is

Options

A. `+3`
B. `+4`
C. `+2`
D. `+1`

Explanation

To find the oxidation number of iron in sodium nitroprusside (Na2[Fe(CN)5(NO)]), we can follow these steps:

### Step 1: Write the formula of sodium nitroprusside
The formula of sodium nitroprusside is Na2[Fe(CN)5(NO)].

### Step 2: Identify the components of the compound
- Sodium (Na) is a Group 1 element and has an oxidation number of +1.
- The cyanide ion (CN) has a charge of -1.
- The nitrosyl ion (NO) has a charge of +1.

### Step 3: Set up the equation for the oxidation state of iron
Let the oxidation number of iron (Fe) be \( x \). The overall charge of the complex ion [Fe(CN)5(NO)] is neutral, so we can set up the equation:

\[ 2(+1) + x + 5(-1) + (+1) = 0 \]

### Step 4: Simplify the equation
Now, simplify the equation:

\[ 2 + x - 5 + 1 = 0 \]

This simplifies to:

\[ x - 2 = 0 \]

### Step 5: Solve for \( x \)
Now, solve for \( x \):

\[ x = 2 \]

### Conclusion
The oxidation number of iron in sodium nitroprusside is +2.

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