The water molecule (${\mathrm{H}}_2\mathrm{O}$) has a central oxygen atom bonded to two hydrogen atoms. The oxygen atom has two bond pairs and two lone pairs of electrons. According to VSEPR theory, these electron pairs repel each other and arrange themselves in a way that minimizes repulsion.

The ideal bond angle for a molecule with four electron domains (two bond pairs and two lone pairs) around the central atom, corresponding to ${\mathrm{sp}}^3$ hybridization, would be $109.5^{\circ }$ (tetrahedral geometry). However, the actual bond angle in water is $104.5^{\circ }$.

Why (C) is correct:

The order of repulsion between electron pairs is: lone pair-lone pair (lp-lp) > lone pair-bond pair (lp-bp) > bond pair-bond pair (bp-bp). The two lone pairs on the oxygen atom exert a greater repulsive force on each other and on the bond pairs than the bond pairs exert on each other. This strong lone pair-lone pair repulsion causes the bond angle between the $\mathrm{H}-\mathrm{O}$ bonds to decrease from the ideal tetrahedral angle of $109.5^{\circ }$ to $104.5^{\circ }$.

Option Analysis:

• A) high electronegativity of oxygen: While oxygen is highly electronegative, this property primarily influences bond polarity, not directly the bond angle in this manner.
• B) bond pair-bond pair repulsion: Bond pair-bond pair repulsion is the weakest type of repulsion and would lead to a larger angle if it were the dominant factor, not a smaller one.
• D) low electronegativity of hydrogen: The electronegativity of hydrogen is not the primary reason for the deviation in bond angle; it's the presence and repulsion of lone pairs on the central atom.

Correct Answer: (C)