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 \( +5 \).

Answer 2

To determine the oxidation numbers of nitrogen in \( \text{NH}_4\text{NO}_3 \), we first need to recognize the oxidation states of other elements in the compound.
In \( \text{NH}_4^+ \), the oxidation state of hydrogen is \( +1 \), and there are 4 hydrogen atoms. Therefore, the total oxidation state for hydrogen is \( 4 \times (+1) = +4 \). The overall charge on the ammonium ion is \( +1 \), so the oxidation state of nitrogen in \( \text{NH}_4^+ \) must be \( -3 \), because: \[ x + 4(+1) = +1 \quad \Rightarrow \quad x = -3 \] Thus, the oxidation state of nitrogen in \( \text{NH}_4^+ \) is \( -3 \). - In \( \text{NO}_3^- \), the oxidation state of oxygen is \( -2 \), and there are 3 oxygen atoms. Therefore, the total oxidation state for oxygen is \( 3 \times (-2) = -6 \). The overall charge on the nitrate ion is \( -1 \), so the oxidation state of nitrogen in \( \text{NO}_3^- \) must be \( +5 \), because: \[ x + 3(-2) = -1 \quad \Rightarrow \quad x - 6 = -1 \quad \Rightarrow \quad x = +5 \] Thus, the oxidation state of nitrogen in \( \text{NO}_3^- \) is \( +5 \). Therefore, the oxidation numbers of nitrogen in \( \text{NH}_4\text{NO}_3 \) are \( -3 \) in \( \text{NH}_4^+ \) and \( +5 \) in \( \text{NO}_3^- \).