In how may of the following compounds an elements has fractional oxidation number `KO_(2), N_(3)H, O_(2)F_(2), NaN_(3),Ni(CO)_(4),Na_(2)S_(4)O_(6),N_(2)H_(6)^(++), KI_(3)`.

Answer

To determine how many of the given compounds have elements with fractional oxidation numbers, we will analyze each compound step by step.

### Step 1: Analyze KO₂
- **Compound**: KO₂
- **Oxidation State Calculation**:
- Let the oxidation state of K be +1.
- Let the oxidation state of O in O₂ be x. Since there are two O atoms, we have:
\[ +1 + 2x = 0 \]
\[ 2x = -1 \]
\[ x = -\frac{1}{2} \]
- **Conclusion**: The oxidation state of O is -1/2 (fractional).

### Step 2: Analyze N₃H
- **Compound**: N₃H
- **Oxidation State Calculation**:
- Let the oxidation state of N be x.
- For H, it is +1. Therefore:
\[ 3x + 1 = 0 \]
\[ 3x = -1 \]
\[ x = -\frac{1}{3} \]
- **Conclusion**: The oxidation state of N is -1/3 (fractional).

### Step 3: Analyze O₂F₂
- **Compound**: O₂F₂
- **Oxidation State Calculation**:
- Let the oxidation state of O be x and F be y. For F, it is -1.
- Therefore:
\[ 2x + 2(-1) = 0 \]
\[ 2x - 2 = 0 \]
\[ 2x = 2 \]
\[ x = +1 \]
- **Conclusion**: The oxidation state of O is +1 (not fractional) and F is -1 (not fractional).

### Step 4: Analyze NaN₃
- **Compound**: NaN₃
- **Oxidation State Calculation**:
- Let the oxidation state of N be x. For Na, it is +1. Therefore:
\[ +1 + 3x = 0 \]
\[ 3x = -1 \]
\[ x = -\frac{1}{3} \]
- **Conclusion**: The oxidation state of N is -1/3 (fractional).

### Step 5: Analyze Ni(CO)₄
- **Compound**: Ni(CO)₄
- **Oxidation State Calculation**:
- Let the oxidation state of Ni be x. For CO, it is neutral (0).
- Therefore:
\[ x + 4(0) = 0 \]
\[ x = 0 \]
- **Conclusion**: The oxidation state of Ni is 0 (not fractional).

### Step 6: Analyze Na₂S₄O₆
- **Compound**: Na₂S₄O₆
- **Oxidation State Calculation**:
- Let the oxidation state of S be x. For Na, it is +1 and for O, it is -2.
- Therefore:
\[ 2(+1) + 4x + 6(-2) = 0 \]
\[ 2 + 4x - 12 = 0 \]
\[ 4x = 10 \]
\[ x = \frac{10}{4} = 2.5 \]
- **Conclusion**: The oxidation state of S is 2.5 (fractional).

### Step 7: Analyze N₂H₆²⁺
- **Compound**: N₂H₆²⁺
- **Oxidation State Calculation**:
- Let the oxidation state of N be x. For H, it is +1.
- Therefore:
\[ 2x + 6(+1) + 2 = 0 \]
\[ 2x + 6 + 2 = 0 \]
\[ 2x + 8 = 0 \]
\[ 2x = -8 \]
\[ x = -4 \]
- **Conclusion**: The oxidation state of N is -4 (not fractional).

### Step 8: Analyze KI₃
- **Compound**: KI₃
- **Oxidation State Calculation**:
- Let the oxidation state of I be x. For K, it is +1.
- Therefore:
\[ +1 + 3x = 0 \]
\[ 3x = -1 \]
\[ x = -\frac{1}{3} \]
- **Conclusion**: The oxidation state of I is -1/3 (fractional).

### Final Count of Compounds with Fractional Oxidation Numbers
- The compounds with fractional oxidation numbers are:
1. KO₂
2. N₃H
3. NaN₃
4. Na₂S₄O₆
5. KI₃

Thus, there are **5 compounds** with fractional oxidation numbers.

### Summary of Steps:
1. Analyze KO₂: O has -1/2 (fractional).
2. Analyze N₃H: N has -1/3 (fractional).
3. Analyze O₂F₂: No fractional oxidation states.
4. Analyze NaN₃: N has -1/3 (fractional).
5. Analyze Ni(CO)₄: No fractional oxidation states.
6. Analyze Na₂S₄O₆: S has 2.5 (fractional).
7. Analyze N₂H₆²⁺: No fractional oxidation states.
8. Analyze KI₃: I has -1/3 (fractional).