The correct order of increasing s character ( in percentage ) in the hybrid orbitals in below molecules `//` ions is ( assume all hybrid orbitals are exactly equivalent) `:`
`underset(I)(CO_(3)^(2-))` `underset(II)(XeF_(4))` `underset(III)(I_(3)^(-))` `underset(IV)(NCl_(3))` `underset(V)(BeCl_(2)(g))`

Options

A. `IIltIIIltIVltIltV`
B. `IIltIVltIIIltVltI`
C. `IIIltIIltIltVltIV`
D. `IIltIVltIIIltIltV`

Explanation

To determine the correct order of increasing s character in the hybrid orbitals of the given molecules and ions, we will analyze each compound step by step.

### Step 1: Analyze CO₃²⁻ (Carbonate Ion)
- **Hybridization**: The carbonate ion has a trigonal planar structure with three sigma bonds (C-O bonds).
- **Hybridization Type**: The hybridization is sp² (1 s orbital + 2 p orbitals).
- **S Character Calculation**:
\[ \text{S character} = \frac{\text{Number of s orbitals}}{\text{Total number of orbitals}} \times 100 = \frac{1}{3} \times 100 = 33.33\% \]

### Step 2: Analyze XeF₄ (Xenon Tetrafluoride)
- **Hybridization**: XeF₄ has an octahedral structure with four sigma bonds and two lone pairs.
- **Hybridization Type**: The hybridization is sp³d² (1 s orbital + 3 p orbitals + 2 d orbitals).
- **S Character Calculation**:
\[ \text{S character} = \frac{1}{6} \times 100 = 16.67\% \]

### Step 3: Analyze I₃⁻ (Iodide Ion)
- **Hybridization**: The I₃⁻ ion has a linear structure with two sigma bonds and three lone pairs.
- **Hybridization Type**: The hybridization is sp³d (1 s orbital + 3 p orbitals + 1 d orbital).
- **S Character Calculation**:
\[ \text{S character} = \frac{1}{5} \times 100 = 20\% \]

### Step 4: Analyze NCl₃ (Nitrogen Trichloride)
- **Hybridization**: NCl₃ has a tetrahedral structure with three sigma bonds and one lone pair.
- **Hybridization Type**: The hybridization is sp³ (1 s orbital + 3 p orbitals).
- **S Character Calculation**:
\[ \text{S character} = \frac{1}{4} \times 100 = 25\% \]

### Step 5: Analyze BeCl₂ (Beryllium Chloride)
- **Hybridization**: BeCl₂ has a linear structure with two sigma bonds and no lone pairs.
- **Hybridization Type**: The hybridization is sp (1 s orbital + 1 p orbital).
- **S Character Calculation**:
\[ \text{S character} = \frac{1}{2} \times 100 = 50\% \]

### Summary of S Character Percentages
- CO₃²⁻: 33.33%
- XeF₄: 16.67%
- I₃⁻: 20%
- NCl₃: 25%
- BeCl₂: 50%

### Order of Increasing S Character
Now we can arrange the compounds based on their s character percentages:
1. XeF₄: 16.67%
2. I₃⁻: 20%
3. NCl₃: 25%
4. CO₃²⁻: 33.33%
5. BeCl₂: 50%

Thus, the correct order of increasing s character is:
**XeF₄ < I₃⁻ < NCl₃ < CO₃²⁻ < BeCl₂**

### Final Answer
The order is: **II < III < IV < I < V** (or 21435 if numbered).