The molecular shape and bond angle of water are respectively

Analysis

The question asks for the molecular geometry (shape) and the specific bond angle of a water molecule ($H_2O$). To determine this, we look at the hybridization and the electron pair arrangement of the oxygen atom.

Concept: VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory helps us predict the shape of molecules based on the repulsion between electron pairs around a central atom.

1. Electron Configuration: Oxygen has 6 valence electrons. In $H_2O$, it forms two single covalent bonds with two hydrogen atoms.
2. Electron Pairs: Oxygen uses 2 electrons for bonding (bonding pairs) and retains 4 electrons as 2 lone pairs.
3. Geometry: According to VSEPR, 4 electron pairs (2 bonding + 2 lone) arrange themselves in a tetrahedral electron geometry to minimize repulsion.
4. Molecular Shape: Because the two lone pairs exert a stronger repulsive force than the bonding pairs, they "push" the hydrogen atoms closer together. This distortion changes the shape from tetrahedral to bent (or V-shaped).
5. Bond Angle: While a perfect tetrahedron has an angle of 109.5°, the strong repulsion from the two lone pairs on the oxygen atom compresses the $H-O-H$ bond angle to approximately 104.5° to 105°.

Step-by-Step Solution

• Step 1: Identify the central atom (Oxygen).
• Step 2: Count the electron domains. There are 2 bonding pairs and 2 lone pairs, totaling 4 domains.
• Step 3: Determine the shape. Due to the 2 lone pairs, the molecule is not linear or tetrahedral; it is bent.
• Step 4: Determine the angle. The repulsion from the lone pairs reduces the ideal tetrahedral angle (109.5°) to approximately 105°.

Comparing this to the options provided:

• A is incorrect (that is the shape of $CO_2$).
• B is incorrect (109.5° is the angle for a perfect tetrahedron like $CH_4$).
• C is incorrect (the molecule is not tetrahedral in shape).
• D matches our derivation.

Exam Tip

Always remember the "Lone Pair Rule": Lone pairs take up more space than bonding pairs. Every time you have a lone pair on the central atom, the bond angle will be slightly smaller than the "ideal" angle for that geometry (e.g., $NH_3$ is 107°, $H_2O$ is 105°).