In a water molecule, the O atom is A. sp2 hybridized (trigonal), the bond angle between the O-H bonds is 90o, and the overall geometry of the molecule is V-shaped B. sp3 hybridized (tetrahedral), the bond angle between the O-H bonds is 105o, and the overall geometry of the molecule is V-shaped C. sp3 hybridized (tetrahedral), the bond angle between the O-H bonds is 109.5o, and the overall geometry of the molecule is V-shaped D. sp hybridized (linear), the bond angle between the O-H bonds is 180o, and the overall geometry of the molecule is linear

Question

BlackTom AI · Accepted Answer

Oxygen in H2O has a steric number of 4 (two O–H bonds + two lone pairs), so the O atom is sp3 hybridized and the electron-pair geometry is tetrahedral; the molecular shape (positions of atoms) is bent or V-shaped with a reduced H–O–H angle (~104.5°, often rounded to 105°).

1) Relevant concepts
- Valence electrons and Lewis structure determine bonding and lone pairs.
- Steric number = number of σ bonds + number of lone pairs → determines hybridization.
- VSEPR: lone pairs occupy more space than bonding pairs, so lone pair–bonding pair repulsions compress bond angles.
- Ideal tetrahedral angle = 109.5°; lone pairs reduce observed bond angles.

2) Step-by-step reasoning and intermediate steps
- Valence electrons: O has 6, each H has 1 → total electrons = 6 + 1 + 1 = 8 electrons (4 pairs).
- Two O–H bonds use 2 electron pairs; remaining 2 pairs are lone pairs on O.
- Steric number = 2 bonds + 2 lone pairs = 4 → hybridization = sp3 (four hybrid orbitals).
- Electron geometry = tetrahedral (arrangement of four electron pairs).
- Molecular geometry (considering only atoms) = bent (V-shaped) because two positions are occupied by lone pairs.
- Ideal tetrahedral angle (109.5°) is decreased because lone pair–bond pair repulsion > bond pair–bond pair repulsion. This pushes the H atoms closer, producing the observed H–O–H angle ≈ 104.5°, commonly given as 105°.

3) Why the correct answer is correct
- Option B states sp3 hybridization (tetrahedral electron geometry), an H–O–H angle ≈ 105°, and a V-shaped molecular geometry — exactly matching the VSEPR prediction and experimental bond angle (~104.5°).

4) Why the other choices are incorrect
- A: sp2/trigonal would give ~120° and cannot give the observed bent shape with two lone pairs; 90° is also incorrect.
- C: sp3 is right but 109.5° applies to four bonding pairs (perfect tetrahedron). Water’s lone pairs reduce the angle, so 109.5° is not correct.
- D: sp (linear, 180°) is inconsistent with the two lone pairs and observed bent shape.

Conclusion: The correct choice is B.

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