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 compound \( Na_2[Fe(CN)_5NO] \), we can follow these steps:

### Step 1: Identify the components of the compound
The compound consists of:
- Sodium (Na)
- Iron (Fe)
- Cyanide (CN)
- Nitric oxide (NO)

### Step 2: Assign known oxidation states
- The oxidation state of sodium (Na) is +1.
- The oxidation state of cyanide (CN) is -1.
- The oxidation state of nitric oxide (NO) is typically +1.

### Step 3: Set up the equation
Let the oxidation state of iron (Fe) be \( X \). The compound is neutral, so the sum of the oxidation states must equal zero.

The formula can be expressed as:
\[ 2(\text{Na}) + X + 5(\text{CN}) + (\text{NO}) = 0 \]

Substituting the known values:
\[ 2(+1) + X + 5(-1) + (+1) = 0 \]

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

### Step 5: Solve for \( X \)
Now, solve for \( X \):
\[ X = 2 \]

### Step 6: Conclusion
Thus, the oxidation number of iron (Fe) in \( Na_2[Fe(CN)_5NO] \) is +2.