To determine the hybridisation of the oxygen atom in H₂O, we need to consider the number of sigma bonds and lone pairs around the central atom.

Step 1: Determine the central atom and its valence electrons.
In H₂O, oxygen (O) is the central atom. Oxygen has 6 valence electrons.

Step 2: Determine the number of sigma bonds.
Oxygen forms two single bonds with two hydrogen atoms. Each single bond is a sigma bond. So, there are 2 sigma bonds.

Step 3: Determine the number of lone pairs.
After forming two bonds, oxygen uses 2 of its 6 valence electrons. The remaining 4 electrons form 2 lone pairs (4 electrons / 2 electrons per pair = 2 lone pairs).

Step 4: Calculate the steric number.
Steric number = (Number of sigma bonds) + (Number of lone pairs)
Steric number = 2 + 2 = 4

Step 5: Determine the hybridisation based on the steric number.
A steric number of 4 corresponds to sp³ hybridisation.

Option Analysis:

• A) sp²: This hybridisation corresponds to a steric number of 3 (e.g., in BF₃ or ethene). Oxygen in H₂O has a steric number of 4.
• B) sp: This hybridisation corresponds to a steric number of 2 (e.g., in BeCl₂ or ethyne). Oxygen in H₂O has a steric number of 4.
• C) sp³d: This hybridisation corresponds to a steric number of 5 (e.g., in PCl₅). Oxygen in H₂O has a steric number of 4.
• D) sp³: This hybridisation corresponds to a steric number of 4, which matches the oxygen atom in H₂O.

Correct Answer: (D)