The oxidation number of nitrogen in `(N_(2)H_(5))^(+)` is

Options

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

Explanation

To find the oxidation number of nitrogen in the ion \( (N_2H_5)^{+} \), we can follow these steps:

### Step 1: Assign variables
Let the oxidation number of nitrogen be \( x \). Since there are two nitrogen atoms in the compound, the contribution from nitrogen will be \( 2x \).

### Step 2: Consider the contribution from hydrogen
Each hydrogen atom has an oxidation number of \( +1 \). Since there are five hydrogen atoms, their total contribution will be \( 5 \times (+1) = +5 \).

### Step 3: Set up the equation
The overall charge of the ion \( (N_2H_5)^{+} \) is \( +1 \). Therefore, we can set up the equation based on the contributions from nitrogen and hydrogen:

\[ 2x + 5 = +1 \]

### Step 4: Solve for \( x \)
Now, we can solve the equation for \( x \):

\[ 2x + 5 = 1 \]
\[ 2x = 1 - 5 \]
\[ 2x = -4 \]
\[ x = -2 \]

### Step 5: Determine the oxidation number for each nitrogen
Since there are two nitrogen atoms, the oxidation number of each nitrogen atom is \( -2 \).

### Conclusion
The oxidation number of nitrogen in \( (N_2H_5)^{+} \) is \( -2 \).

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