Among the following, what is the total number of compounds having `+3` oxidation state of the underlined elements?
`Naul(Au)Cl_(4)`
`ul(I)Cl`
`ul(N)O_(2)^(ө)`

Explanation

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

### Step 1: Analyze NaAuCl₄
- The underlined element is Au (Gold).
- Let the oxidation state of Au be \( x \).
- The oxidation states of other elements are:
- Na: +1
- Cl: -1 (there are 4 Cl atoms, so total contribution from Cl is -4)

The equation for the oxidation states in the compound is:
\[ +1 + x + 4(-1) = 0 \]
This simplifies to:
\[ +1 + x - 4 = 0 \]
\[ x - 3 = 0 \]
\[ x = +3 \]

**Conclusion:** Au in NaAuCl₄ has an oxidation state of +3.

### Step 2: Analyze ICl
- The underlined element is I (Iodine).
- Let the oxidation state of I be \( x \).
- The oxidation state of Cl is -1.

The equation for the oxidation states in the compound is:
\[ x + (-1) = 0 \]
This simplifies to:
\[ x - 1 = 0 \]
\[ x = +1 \]

**Conclusion:** I in ICl has an oxidation state of +1, not +3.

### Step 3: Analyze NO₂⁻
- The underlined element is N (Nitrogen).
- Let the oxidation state of N be \( x \).
- The oxidation states of O is -2 (there are 2 O atoms, so total contribution from O is -4).

The equation for the oxidation states in the compound is:
\[ x + 2(-2) = -1 \]
This simplifies to:
\[ x - 4 = -1 \]
\[ x = +3 \]

**Conclusion:** N in NO₂⁻ has an oxidation state of +3.

### Final Count of Compounds with +3 Oxidation State
- NaAuCl₄: +3 (Au)
- ICl: +1 (I)
- NO₂⁻: +3 (N)

**Total number of compounds with +3 oxidation state:** 2 (NaAuCl₄ and NO₂⁻)

### Final Answer
The total number of compounds having a +3 oxidation state of the underlined elements is **2**.

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