In `CO_(3)^(2-), CH_(4)` and `B Cl_(3)` hybridisation of carbon atoms are respectively.

Options

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

Explanation

To determine the hybridization of carbon atoms in the given compounds CO₃²⁻, CH₄, and BCl₃, we will analyze each compound step by step.

### Step 1: Analyze CO₃²⁻ (Carbonate Ion)
1. **Determine the structure**: The carbonate ion (CO₃²⁻) consists of one carbon atom centrally bonded to three oxygen atoms.
2. **Count the regions of electron density**: The carbon atom forms three sigma bonds with the oxygen atoms and has one lone pair of electrons.
3. **Hybridization**: The presence of three sigma bonds indicates that the carbon atom is sp² hybridized. This is because sp² hybridization involves one s orbital and two p orbitals, resulting in three sp² hybrid orbitals.
4. **Geometry**: The geometry of CO₃²⁻ is trigonal planar due to the sp² hybridization.

### Step 2: Analyze CH₄ (Methane)
1. **Determine the structure**: The methane molecule (CH₄) consists of one carbon atom bonded to four hydrogen atoms.
2. **Count the regions of electron density**: The carbon atom forms four sigma bonds with the hydrogen atoms and has no lone pairs.
3. **Hybridization**: The presence of four sigma bonds indicates that the carbon atom is sp³ hybridized. This is because sp³ hybridization involves one s orbital and three p orbitals, resulting in four sp³ hybrid orbitals.
4. **Geometry**: The geometry of CH₄ is tetrahedral due to the sp³ hybridization.

### Step 3: Analyze BCl₃ (Boron Trichloride)
1. **Determine the structure**: The boron trichloride molecule (BCl₃) consists of one boron atom bonded to three chlorine atoms.
2. **Count the regions of electron density**: The boron atom forms three sigma bonds with the chlorine atoms and has no lone pairs.
3. **Hybridization**: The presence of three sigma bonds indicates that the boron atom is sp² hybridized. This is because sp² hybridization involves one s orbital and two p orbitals, resulting in three sp² hybrid orbitals.
4. **Geometry**: The geometry of BCl₃ is trigonal planar due to the sp² hybridization.

### Summary of Hybridizations
- **CO₃²⁻**: sp² hybridized
- **CH₄**: sp³ hybridized
- **BCl₃**: sp² hybridized

### Final Answer
The hybridization of carbon atoms in CO₃²⁻, CH₄, and BCl₃ are respectively: **sp², sp³, and sp²**.

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