The correct order of increasing s-character (in percentage) in the hybrid orbitals of following molecules/ions is :-
`(A) CO_(3)^(-2) " "(B) NCl_(3)" "(C ) BeCl_(2)`

Options

A. `B lt A lt C`
B. `A lt B lt C`
C. `C lt A lt B`
D. `C lt B lt A`

Answer

To determine the correct order of increasing s-character in the hybrid orbitals of the given molecules/ions (CO₃²⁻, NCl₃, and BeCl₂), we need to analyze the hybridization of each central atom in these compounds.

### Step 1: Analyze CO₃²⁻ (Carbonate Ion)
- The carbonate ion has the formula CO₃²⁻. The central atom is carbon (C), which is bonded to three oxygen atoms (O).
- Since carbon is connected to three atoms, the hybridization is **sp²**.
- The percentage of s-character in sp² hybridization is calculated as follows:
\[ \text{s-character} = \frac{1}{3} \times 100\% = 33.3\% \]

### Step 2: Analyze NCl₃ (Nitrogen Trichloride)
- The nitrogen trichloride molecule has the formula NCl₃. The central atom is nitrogen (N), which is bonded to three chlorine atoms (Cl) and has one lone pair.
- Since nitrogen is connected to four regions of electron density (three bond pairs and one lone pair), the hybridization is **sp³**.
- The percentage of s-character in sp³ hybridization is calculated as follows:
\[ \text{s-character} = \frac{1}{4} \times 100\% = 25\% \]

### Step 3: Analyze BeCl₂ (Beryllium Dichloride)
- The beryllium dichloride molecule has the formula BeCl₂. The central atom is beryllium (Be), which is bonded to two chlorine atoms (Cl).
- Since beryllium is connected to two atoms and has no lone pairs, the hybridization is **sp**.
- The percentage of s-character in sp hybridization is calculated as follows:
\[ \text{s-character} = \frac{1}{2} \times 100\% = 50\% \]

### Step 4: Compare the s-character percentages
- Now we can summarize the s-character percentages:
- CO₃²⁻: 33.3% (sp²)
- NCl₃: 25% (sp³)
- BeCl₂: 50% (sp)

### Step 5: Order the compounds based on increasing s-character
- The order of increasing s-character is:
\[ \text{NCl}_3 < \text{CO}_3^{2-} < \text{BeCl}_2 \]

### Final Answer
Thus, the correct order of increasing s-character (in percentage) in the hybrid orbitals of the given molecules/ions is:
**NCl₃ < CO₃²⁻ < BeCl₂**

---