Among the following compounds the one that is polar and has the central atom with `sp^2`-hybridisation is

Options

A. `H_2 CO_3`
B. `SiF_4`
C. `BF_3`
D. `HCIO_2`

Explanation

To determine which of the given compounds is polar and has the central atom with sp² hybridization, we will analyze each compound step by step.

### Step 1: Identify the Compounds
The compounds given are:
1. H₂CO₃ (Carbonic Acid)
2. SiF₄ (Silicon Tetrafluoride)
3. BF₃ (Boron Trifluoride)
4. HClO₂ (Chlorous Acid)

### Step 2: Determine Polarity of Each Compound
- **H₂CO₃**: This compound is polar. The carbon atom is bonded to three oxygen atoms (one with a double bond and two with single bonds), leading to an unequal distribution of charge.

- **SiF₄**: This compound is nonpolar. Although silicon has four fluorine atoms attached, the symmetrical tetrahedral shape cancels out the dipole moments.

- **BF₃**: This compound is also nonpolar. Boron trifluoride has a trigonal planar structure, which is symmetrical, leading to the cancellation of dipole moments.

- **HClO₂**: This compound is polar. The chlorine atom is bonded to two oxygen atoms and one hydrogen atom, leading to an asymmetrical charge distribution.

### Step 3: Determine Hybridization of the Central Atom
Now, we will find the hybridization of the central atom in each compound.

1. **H₂CO₃**:
- Carbon (C) has 4 valence electrons.
- It is bonded to 3 oxygen atoms and 2 hydrogen atoms.
- Hybridization = (Number of valence electrons + Number of monovalent atoms) / 2
- Hybridization = (4 + 2) / 2 = 3 → sp² hybridization.

2. **SiF₄**:
- Silicon (Si) has 4 valence electrons.
- It is bonded to 4 fluorine atoms.
- Hybridization = (4 + 4) / 2 = 4 → sp³ hybridization.

3. **BF₃**:
- Boron (B) has 3 valence electrons.
- It is bonded to 3 fluorine atoms.
- Hybridization = (3 + 3) / 2 = 3 → sp² hybridization.

4. **HClO₂**:
- Chlorine (Cl) has 7 valence electrons.
- It is bonded to 2 oxygen atoms and 1 hydrogen atom.
- Hybridization = (7 + 1) / 2 = 4 → sp³ hybridization.

### Step 4: Summary of Findings
- **H₂CO₃**: Polar, sp² hybridization.
- **SiF₄**: Nonpolar, sp³ hybridization.
- **BF₃**: Nonpolar, sp² hybridization.
- **HClO₂**: Polar, sp³ hybridization.

### Conclusion
Among the compounds analyzed, **H₂CO₃** is the only compound that is polar and has a central atom with sp² hybridization.

### Final Answer
The answer is **H₂CO₃**.