Question

The oxidation number of nitrogen atom in MH₄NO₃ are

• A. -3,-3
• B. +5,+5
• C. -3,+5
• D. +3,-5

Answer 1

The Correct Option is C

The compound \( \text{NH}_4\text{NO}_3 \) consists of two nitrogen atoms: one in the ammonium ion (\( \text{NH}_4^+ \)) and one in the nitrate ion (\( \text{NO}_3^- \)).
1. Oxidation state of nitrogen in ammonium ion (\( \text{NH}_4^+ \)): The ammonium ion consists of one nitrogen atom and four hydrogen atoms. The oxidation state of hydrogen is \( +1 \). Since the total charge of the ammonium ion is \( +1 \), we can find the oxidation state of nitrogen (\( x \)) by the equation: \[ x + 4(+1) = +1 \] Solving for \( x \): \[ x + 4 = 1 \quad \Rightarrow \quad x = -3 \] Therefore, the oxidation number of nitrogen in \( \text{NH}_4^+ \) is \( -3 \).
2. Oxidation state of nitrogen in nitrate ion (\( \text{NO}_3^- \)): The nitrate ion consists of one nitrogen atom and three oxygen atoms. The oxidation state of oxygen is \( -2 \). Since the total charge of the nitrate ion is \( -1 \), we can find the oxidation state of nitrogen (\( x \)) by the equation: \[ x + 3(-2) = -1 \] Solving for \( x \): \[ x - 6 = -1 \quad \Rightarrow \quad x = +5 \] Therefore, the oxidation number of nitrogen in \( \text{NO}_3^- \) is \( +5 \).

Thus, the oxidation numbers of nitrogen in \( \text{NH}_4\text{NO}_3 \) are \( -3 \) and \( +5 \).