The hybridisation of the central atom in `ICI_(2)^(o+)` is .

Options

A. `sp^(3)`
B. `sp^(2)`
C. `sp^(2)`
D. `sp`

Explanation

To determine the hybridization of the central atom in the molecule ICl₂⁺, we will follow these steps:

### Step 1: Identify the central atom and its valence electrons
The central atom in ICl₂⁺ is iodine (I). Iodine is in group 17 of the periodic table and has 7 valence electrons.

### Step 2: Count the number of monovalent species
In ICl₂⁺, there are 2 chlorine (Cl) atoms. Chlorine is also a halogen and is monovalent (it forms one bond). Therefore, the number of monovalent species is 2.

### Step 3: Account for the charge of the molecule
The molecule has a positive charge (⁺). This means we will subtract 1 from our total count of electrons.

### Step 4: Calculate the steric number
The steric number (SN) is calculated using the formula:
\[ \text{Steric Number} = \frac{1}{2} \times (\text{Valence Electrons of Central Atom} + \text{Number of Monovalent Species} - \text{Cationic Charge}) \]

Substituting the values:
- Valence electrons of iodine = 7
- Number of monovalent species = 2
- Cationic charge = 1

So,
\[ \text{Steric Number} = \frac{1}{2} \times (7 + 2 - 1) \]
\[ \text{Steric Number} = \frac{1}{2} \times 8 = 4 \]

### Step 5: Determine the hybridization
A steric number of 4 corresponds to sp³ hybridization. This means that the central atom (iodine) uses four hybrid orbitals to form bonds.

### Conclusion
The hybridization of the central atom in ICl₂⁺ is **sp³**.

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