The oxidant number of `Fe` in `Na_(2) [Fe(CN)_(5)NO]` is

Options

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

Explanation

To determine the oxidation number of iron (Fe) in the complex compound \( \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \), we can follow these steps:

### Step 1: Identify the components of the compound
The compound consists of sodium ions (Na\(^+\)), iron (Fe), cyanide ions (CN\(^-\)), and a nitrosyl group (NO).

### Step 2: Assign known oxidation states
- The oxidation state of sodium (Na) is +1.
- The oxidation state of the cyanide ion (CN) is -1.
- The oxidation state of the nitrosyl group (NO) is +1.

### Step 3: Write the overall charge of the complex
The overall charge of the complex ion \( [\text{Fe(CN)}_5\text{NO}] \) is -2, as indicated by the presence of two sodium ions, which contribute a total charge of +2. Therefore, the charge balance gives us:
\[ \text{Charge of } [\text{Fe(CN)}_5\text{NO}] = -2 \]

### Step 4: Set up the equation for oxidation states
Let the oxidation state of iron (Fe) be \( x \). The contribution to the overall charge from the components can be expressed as:
\[ x + 5(-1) + 1 = -2 \]
This equation accounts for the oxidation state of iron, the five cyanide ions, and the nitrosyl group.

### Step 5: Simplify the equation
Substituting the values into the equation gives:
\[ x - 5 + 1 = -2 \]
\[ x - 4 = -2 \]

### Step 6: Solve for \( x \)
Now, solve for \( x \):
\[ x = -2 + 4 \]
\[ x = +2 \]

### Conclusion
The oxidation number of iron (Fe) in \( \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \) is +2.

### Final Answer
The oxidation number of Fe is +2.

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