The state of hybridisation of atoms in boric acid `(H_(3)BO_(3))` is .

Options

A. `sp^(3)`
B. `sp^(2)`
C. `sp`
D. None of these

Explanation

To determine the state of hybridization of atoms in boric acid (H₃BO₃), we can follow these steps:

### Step 1: Identify the central atom
In boric acid, the central atom is boron (B). It is surrounded by three hydrogen (H) atoms and one oxygen (O) atom.

### Step 2: Determine the valence electrons
Boron has 3 valence electrons, and each hydrogen has 1 valence electron. The oxygen atom has 6 valence electrons. Therefore, the total number of valence electrons in H₃BO₃ is:
- Boron: 3
- Hydrogen: 3 (1 from each of the 3 H atoms)
- Oxygen: 6
- Total = 3 + 3 + 6 = 12 valence electrons

### Step 3: Draw the Lewis structure
In the Lewis structure of H₃BO₃, boron is at the center, bonded to three hydrogen atoms and one oxygen atom. The oxygen atom has two lone pairs of electrons. The structure can be represented as:
```
H
|
H - B - O
|
H
```
The oxygen atom has two lone pairs, making it sp³ hybridized.

### Step 4: Determine the hybridization of boron
Boron forms three sigma bonds (one with each hydrogen and one with oxygen). To accommodate these three bonds, boron undergoes sp² hybridization, which involves mixing one s orbital and two p orbitals.

### Step 5: Determine the hybridization of oxygen
The oxygen atom forms one sigma bond with boron and has two lone pairs. This arrangement leads to sp³ hybridization, as it involves mixing one s orbital and three p orbitals.

### Conclusion
In boric acid (H₃BO₃):
- The hybridization of boron is sp².
- The hybridization of oxygen is sp³.

### Final Answer
The state of hybridization of atoms in boric acid (H₃BO₃) is:
- Boron: sp²
- Oxygen: sp³