Oxidation number of underlined elements are `N_(2)O_(5),SO_(3)^(2-),NH_(4)^(+)` :

Options

A. `+5,2+,-3`
B. `+6,-2,+3`
C. `+6,+2,-3`
D. `+5,+4,-3`

Answer

To determine the oxidation numbers of the underlined elements in the compounds \( N_2O_5 \), \( SO_3^{2-} \), and \( NH_4^{+} \), we will follow these steps:

### Step 1: Calculate the oxidation number of Nitrogen in \( N_2O_5 \)

1. **Write the oxidation number equation**:
Let the oxidation number of nitrogen be \( X \). The oxidation number of oxygen is typically \(-2\).
The compound is neutral, so:
\[ 2X + 5(-2) = 0 \]

2. **Solve for \( X \)**:
\[ 2X - 10 = 0 \\ 2X = 10 \\ X = 5 \]
Therefore, the oxidation number of nitrogen in \( N_2O_5 \) is **+5**.

### Step 2: Calculate the oxidation number of Sulfur in \( SO_3^{2-} \)

1. **Write the oxidation number equation**:
Let the oxidation number of sulfur be \( X \). The oxidation number of oxygen is \(-2\). The overall charge of the ion is \(-2\):
\[ X + 3(-2) = -2 \]

2. **Solve for \( X \)**:
\[ X - 6 = -2 \\ X = -2 + 6 \\ X = 4 \]
Therefore, the oxidation number of sulfur in \( SO_3^{2-} \) is **+4**.

### Step 3: Calculate the oxidation number of Nitrogen in \( NH_4^{+} \)

1. **Write the oxidation number equation**:
Let the oxidation number of nitrogen be \( X \). The oxidation number of hydrogen is \( +1 \). The overall charge of the ion is \( +1 \):
\[ X + 4(+1) = +1 \]

2. **Solve for \( X \)**:
\[ X + 4 = 1 \\ X = 1 - 4 \\ X = -3 \]
Therefore, the oxidation number of nitrogen in \( NH_4^{+} \) is **-3**.

### Summary of Results:
- The oxidation number of nitrogen in \( N_2O_5 \) is **+5**.
- The oxidation number of sulfur in \( SO_3^{2-} \) is **+4**.
- The oxidation number of nitrogen in \( NH_4^{+} \) is **-3**.

### Final Answer:
- \( N_2O_5 \): +5
- \( SO_3^{2-} \): +4
- \( NH_4^{+} \): -3

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