The states of hybridization of boron and oxygen atoms in boric acid $$(H_3BO_3)$$ are respectively

1. Hybridization of Boron ($$\text{B}$$)

• Boron is the central atom and belongs to Group 13, having 3 valence electrons.
• In $$\text{H}_3\text{BO}_3$$, the boron atom forms 3 single covalent bonds ($$\sigma$$-bonds) with three hydroxyl ($$-\text{OH}$$) groups. It has no lone pairs left.
• Steric Number for B: $$3\text{ }\sigma\text{-bonds} + 0\text{ lone pairs} = 3$$
• A steric number of 3 corresponds to a planar geometry with $$sp^2$$ hybridization.

2. Hybridization of Oxygen ($$\text{O}$$)

• Oxygen belongs to Group 16 and has 6 valence electrons.
• Each oxygen atom forms 2 single covalent bonds ($$\sigma$$-bonds)—one with the central boron atom and one with a hydrogen atom. This leaves 4 non-bonding valence electrons, which form 2 lone pairs.
• Steric Number for O: $$2\text{ }\sigma\text{-bonds} + 2\text{ lone pairs} = 4$$
• A steric number of 4 corresponds to a bent geometry derived from a tetrahedral electronic arrangement, signifying $$sp^3$$ hybridization.

Conclusion:

The states of hybridization for the boron and oxygen atoms in boric acid are respectively $$sp^2$$ and $$sp^3$$.

Correct Option: D ($$sp^2$$ and $$sp^3$$)