The hybridisation scheme for the central atom includes a d-orbital contribution in:

Options

A. `I_(3)^(-)`
B. `PCl_(3)`
C. `NO_(3)^(-)`
D. `H_(2)Se`

Explanation

To determine which hybridization scheme for the central atom includes a d-orbital contribution, we will calculate the steric number for each option provided. The formula for calculating the steric number (S) is:

\[ S = \frac{1}{2} \times (\text{Valence electrons of central atom}) + (\text{Monovalent atoms attached}) - (\text{Cationic charge}) + (\text{Anionic charge}) \]

Let's analyze each option step by step:

### Step 1: Analyze I3⁻
1. **Identify the central atom**: Iodine (I)
2. **Valence electrons of iodine**: 7
3. **Monovalent atoms attached**: 2 (the two other iodine atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 1

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \times 7 + 2 - 0 + 1 = \frac{7}{2} + 2 + 1 = 3.5 + 2 + 1 = 6.5 \]

Since we round to the nearest whole number, we get \( S = 5 \).

This corresponds to **sp³d hybridization**, which includes d-orbital contribution.

### Step 2: Analyze PCl3
1. **Identify the central atom**: Phosphorus (P)
2. **Valence electrons of phosphorus**: 5
3. **Monovalent atoms attached**: 3 (the three chlorine atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 0

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \times 5 + 3 - 0 + 0 = \frac{5}{2} + 3 = 2.5 + 3 = 5.5 \]

Rounding gives us \( S = 4 \).

This corresponds to **sp³ hybridization**, which does not include d-orbital contribution.

### Step 3: Analyze NO3⁻
1. **Identify the central atom**: Nitrogen (N)
2. **Valence electrons of nitrogen**: 5
3. **Monovalent atoms attached**: 3 (the three oxygen atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 1

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \times 5 + 3 - 0 + 1 = \frac{5}{2} + 3 + 1 = 2.5 + 3 + 1 = 6.5 \]

Rounding gives us \( S = 3 \).

This corresponds to **sp² hybridization**, which does not include d-orbital contribution.

### Step 4: Analyze H2Se
1. **Identify the central atom**: Selenium (Se)
2. **Valence electrons of selenium**: 6
3. **Monovalent atoms attached**: 2 (the two hydrogen atoms)
4. **Cationic charge**: 0
5. **Anionic charge**: 0

Now, substitute these values into the formula:

\[ S = \frac{1}{2} \times 6 + 2 - 0 + 0 = 3 + 2 = 5 \]

This corresponds to **sp³ hybridization**, which does not include d-orbital contribution.

### Conclusion
From the analysis above, only **I3⁻** has a hybridization scheme that includes d-orbital contribution, specifically **sp³d hybridization**. The other options (PCl3, NO3⁻, and H2Se) do not involve d-orbital contributions.

### Final Answer
The hybridization scheme for the central atom that includes a d-orbital contribution is found in **I3⁻**.