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单项选择题
A
A
B
B
C
C
D
D

Question

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单项选择题

A

B

C

D

BlackTom AI · Accepted Answer

Oxygen in H2O is sp3 hybridized because it has four electron domains (two bonding pairs and two lone pairs), giving a tetrahedral electron geometry but a bent (V-shaped) molecular shape with an H–O–H angle of about 104.5° (commonly quoted as 105°).

1) Relevant knowledge points and concepts
- Valence electrons and electron domains: count bonding pairs and lone pairs around the central atom.  
- VSEPR theory: electron domains arrange to minimize repulsion; geometry of electron domains determines hybridization.  
- Hybridization: four electron domains → sp3; three → sp2; two → sp.  
- Molecular geometry vs electron geometry: lone pairs occupy space and can change observed bond angles (AXnEm notation: e.g., AX2E2 for water).  
- Lone-pair repulsion > bond-pair repulsion, reducing bond angles from ideal values.

2) Complete reasoning step by step
- Oxygen valence electrons: 6. Each H brings 1 electron for bonding; oxygen forms two O–H bonds (two bonding pairs). Remaining electrons on O form two lone pairs.  
- Count of electron domains around O = 2 bonding pairs + 2 lone pairs = 4 domains. According to VSEPR, 4 domains → tetrahedral electron geometry → sp3 hybridization.  
- Electron-domain arrangement (AX2E2) gives a molecular shape that is bent (V-shaped), not linear or trigonal planar.

3) Intermediate steps / simple "calculation"
- Valence electrons: O(6) + 2×H(1) = 8 electrons → 4 pairs total. Two pairs used for O–H bonds, two remain as lone pairs → 4 regions (sp3).

4) Why the correct answer is correct
- sp3 hybridization follows from 4 domains. Ideal tetrahedral angle is 109.5°, but two lone pairs compress the H–O–H angle to ≈104.5° (rounded to 105°). The molecular shape is bent (V-shaped). This matches option B.

5) Why other options are incorrect
- A: sp2 would imply three domains (trigonal planar) and ~120° angles; 90° is incorrect for sp2 and water does not have only three domains.  
- C: Gives sp3 and V-shaped but uses 109.5° — that is the ideal tetrahedral angle for four bonding regions, not for AX2E2; water’s angle is reduced to ≈105°.  
- D: sp (linear, 180°) corresponds to two domains; oxygen in H2O has four domains, so this is wrong.

6) Conclusion
All evidence (electron-domain count AX2E2, VSEPR prediction, lone-pair compression to ≈105°) confirms option B as correct. Final answer: B.