Among the following, what is the total number of compounds having `+3` oxidation state of the underlined elements?
`K_(4)ul(P_(2))O_(7)`
`ul(O)F_(2)`
`ul(N)O_(2)^(ө)`

Explanation

To determine the total number of compounds having a +3 oxidation state of the underlined elements in the given compounds, we will analyze each compound one by one.

### Step 1: Analyze the first compound, \( K_4P_2O_7 \)

1. **Identify the oxidation states:**
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of phosphorus (P) be \( x \).

2. **Set up the equation for the neutral compound:**
\[ 4(+1) + 2(x) + 7(-2) = 0 \]
This simplifies to:
\[ 4 + 2x - 14 = 0 \]
\[ 2x - 10 = 0 \]
\[ 2x = 10 \implies x = 5 \]

3. **Conclusion for the first compound:**
- The oxidation state of phosphorus in \( K_4P_2O_7 \) is +5, not +3.

### Step 2: Analyze the second compound, \( OF_2 \)

1. **Identify the oxidation states:**
- Fluorine (F) has an oxidation state of -1.
- Let the oxidation state of oxygen (O) be \( x \).

2. **Set up the equation for the neutral compound:**
\[ x + 2(-1) = 0 \]
This simplifies to:
\[ x - 2 = 0 \implies x = +2 \]

3. **Conclusion for the second compound:**
- The oxidation state of oxygen in \( OF_2 \) is +2, not +3.

### Step 3: Analyze the third compound, \( NO_2^- \)

1. **Identify the oxidation states:**
- Let the oxidation state of nitrogen (N) be \( x \).
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation considering the charge:**
\[ x + 2(-2) = -1 \]
This simplifies to:
\[ x - 4 = -1 \implies x = +3 \]

3. **Conclusion for the third compound:**
- The oxidation state of nitrogen in \( NO_2^- \) is +3.

### Final Conclusion

- Among the three compounds analyzed:
- \( K_4P_2O_7 \): P is +5
- \( OF_2 \): O is +2
- \( NO_2^- \): N is +3

Thus, only **one compound** has an element in the +3 oxidation state, which is \( NO_2^- \).

### Total Number of Compounds with +3 Oxidation State: **1**

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