In which reaction, there is change in oxidation number of N atom?

Options

A. `2NO_(2) iff N_(2)O_(4)`
B. `2NO_(2)+H_(2)O rarr HNO_(3)+HNO_(2)`
C. `NH_(4)OH rarr NH_(4)^(+)+OH^(-)`
D. `N_(2)O_(5)+H_(2)O rarr 2HNO_(3)`

Answer

To determine in which reaction there is a change in the oxidation number of the nitrogen (N) atom, we will analyze each reaction step by step.

### Step 1: Analyze the first reaction
**Reaction:** \( 2NO_2 \rightarrow N_2O_4 \)

- For \( NO_2 \):
- Oxygen (O) has an oxidation number of -2.
- Let the oxidation number of nitrogen (N) be \( x \).
- The equation for \( NO_2 \) is:
\[ x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4 \]

- For \( N_2O_4 \):
- The oxidation number of nitrogen (N) will be \( y \).
- The equation for \( N_2O_4 \) is:
\[ 2y + 4(-2) = 0 \implies 2y - 8 = 0 \implies 2y = 8 \implies y = +4 \]

- **Conclusion:** There is no change in oxidation number (from +4 to +4).

### Step 2: Analyze the second reaction
**Reaction:** \( 2NO_2 + H_2O \rightarrow HNO_3 + HNO_2 \)

- For \( NO_2 \):
- As calculated before, the oxidation number of N is +4.

- For \( HNO_3 \):
- Let the oxidation number of N be \( z \).
- The equation for \( HNO_3 \) is:
\[ z + 1 + 3(-2) = 0 \implies z + 1 - 6 = 0 \implies z - 5 = 0 \implies z = +5 \]

- For \( HNO_2 \):
- Let the oxidation number of N be \( w \).
- The equation for \( HNO_2 \) is:
\[ w + 1 + 2(-2) = 0 \implies w + 1 - 4 = 0 \implies w - 3 = 0 \implies w = +3 \]

- **Conclusion:** There is a change in oxidation number for N in this reaction: from +4 (in \( NO_2 \)) to +5 (in \( HNO_3 \)).

### Step 3: Analyze the third reaction
**Reaction:** \( NH_4OH \rightarrow NH_4^+ + OH^- \)

- For \( NH_4OH \):
- Let the oxidation number of N be \( a \).
- The equation for \( NH_4OH \) is:
\[ a + 4(-1) + (-2) = 0 \implies a - 4 - 2 = 0 \implies a - 6 = 0 \implies a = -3 \]

- For \( NH_4^+ \):
- The oxidation number of N is still \( -3 \).

- **Conclusion:** There is no change in oxidation number (remains -3).

### Step 4: Analyze the fourth reaction
**Reaction:** \( N_2O_5 + H_2O \rightarrow 2HNO_3 \)

- For \( N_2O_5 \):
- Let the oxidation number of N be \( b \).
- The equation for \( N_2O_5 \) is:
\[ 2b + 5(-2) = 0 \implies 2b - 10 = 0 \implies 2b = 10 \implies b = +5 \]

- For \( HNO_3 \):
- As calculated before, the oxidation number of N is +5.

- **Conclusion:** There is no change in oxidation number (remains +5).

### Final Conclusion
The only reaction where there is a change in the oxidation number of nitrogen is:
**Answer:** \( 2NO_2 + H_2O \rightarrow HNO_3 + HNO_2 \) (Change from +4 to +5)

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