Question

The II-O-II bond angle in \(\mathrm{II}_{2} \mathrm{O}\) is \(104.5^{\circ}\) and not \(109^{\circ} 28^{\prime}\) because of (1) high electronegativity of oxygen (2) bond pair-bond pair repulsion (3) Lonc pair-lone pair repulsion (4) Lonc pair-bond pair repulsion

Short answer

The bond angle in \(\text{H}_{2}\text{O}\) is \[104.5^{\text{circ}}\] because of lone pair-lone pair repulsion (option 3).

Step-by-step solution

Identify the Geometry

Recognize that the molecule \(\text{H}_{2}\text{O}\) has a bent or V-shaped molecular geometry.

Determine Ideal Bond Angle

Understand that the ideal bond angle for a tetrahedral arrangement without lone pairs is \[109.5^{\text{circ}}\].

Consider Lone Pair Effects

Hydrogen peroxide (\(\text{H}_{2}\text{O}\)) has two lone pairs on the oxygen atom. These lone pairs repel more strongly than bonding pairs, reducing the bond angle.

Examine Repulsions

Lone pair-lone pair repulsion is stronger than bond pair-bond pair and lone pair-bond pair repulsion, causing the bond angle to decrease from the ideal tetrahedral angle.

Confirm the Correct Answer

Given the information, the bond angle in \(\text{H}_{2}\text{O}\) is \[104.5^{\text{circ}}\], primarily due to lone pair-lone pair repulsion. Therefore, the correct answer is option (3).

Key concepts

Molecular Geometry

The molecular geometry of a molecule describes the three-dimensional arrangement of its atoms. For \(\text{H}_{2}\text{O}\), the molecular geometry is bent or V-shaped. This is because the oxygen atom in water forms two bonds with hydrogen atoms and has two lone pairs of electrons. According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, a molecule will arrange its electron pairs to minimize repulsion. In water's case, the two lone pairs and two bond pairs spread out to least interact with each other. This results in a bent shape with a bond angle less than the ideal tetrahedral angle.

Lone Pair Repulsion

Lone pair repulsion refers to the repulsive force between the non-bonding pairs of electrons in a molecule. These lone pairs are located on the oxygen atom in \(\text{H}_{2}\text{O}\). Lone pairs occupy more space around the central atom than bonding pairs, causing greater repulsion. This increased repulsion between lone pairs forces the bonding pairs closer together. As a result, the bond angle in \(\text{H}_{2}\text{O}\) is reduced from the ideal tetrahedral angle of 109.5° to 104.5°.

Bond Pair Repulsion

Bond pair repulsion involves the repulsive force between bonding electron pairs in a molecule. Although bond pair repulsion is present, it is not as strong as lone pair repulsion. In \(\text{H}_{2}\text{O}\), the bond pairs between the oxygen and hydrogen atoms have a repulsive interaction. This repulsion is less intense compared to lone pair-lone pair and lone pair-bond pair interactions. Thus, while bond pair repulsion does affect the structure of the molecule, it is not the primary reason for the reduced bond angle.

Electronegativity Effects

Electronegativity is the ability of an atom to attract electrons towards itself. Oxygen, with its high electronegativity, attracts electrons more strongly than hydrogen. This results in a partial negative charge on the oxygen and a partial positive charge on the hydrogens. While the electronegativity of oxygen influences bond polarity and the molecule’s overall dipole moment, it does not directly cause the reduction of the bond angle in \(\text{H}_{2}\text{O}\). The primary factor responsible for the bond angle of 104.5° is the repulsion between lone pairs on the oxygen atom, not the difference in electronegativity.