Question

According to valence bond theory bond angle in water should be \(90^{\circ} .\) But experimental value is \(104.5^{\circ}\). This is due to (1) repulsion between positively charged hydrogen atoms (2) the presence of lone pair of electrons in oxygen (3) repulsion between two bonds is very high (4) None

Short answer

The bond angle in water is larger due to the lone pairs on the oxygen atom (Option 2).

Step-by-step solution

- Understand the Problem

The question is about the discrepancy between the bond angle in water predicted by valence bond theory (\(90^{\circ}\)) and the experimentally observed bond angle (\(104.5^{\circ}\)).

- Review the Options

Examine the provided options to understand what could cause the bond angle in water to be different from the predicted angle.

- Analyze Repulsion in Water Molecule

Consider the structure of the water molecule, specifically how lone pairs on the oxygen atom influence bond angles. Lone pairs create greater repulsion compared to bond pairs.

- Identify the Correct Explanation

Identify that the presence of lone pairs on oxygen creates a larger repulsion force. This repulsion decreases the H-O-H bond angle from the theoretical value.

- Select the Correct Option

Given the detailed analysis, the correct choice is (2) the presence of lone pair of electrons in oxygen, which explains why the bond angle is observed to be larger than the predicted angle.

Key concepts

bond angles

In chemistry, bond angles are critical in understanding molecular structure. These angles are formed between atoms within a molecule. Specifically, for a water molecule, valence bond theory initially predicts a bond angle of \(90^{\circ}\).
This is because valence bond theory might not fully consider certain electron repulsion forces.

However, real-life measurements show that the bond angle in a water molecule is actually \(104.5^{\circ}\). This slight disparity raises questions about the factors influencing bond angles.

water molecule structure

The structure of the water molecule (H\(_2\)O) is V-shaped, or angular. Each hydrogen atom is bonded to the oxygen atom via a sigma bond. Oxygen, with its 6 valence electrons, forms two bonds with hydrogen, sharing one electron with each.
This leaves two lone pairs on the oxygen atom.

• The oxygen atom thus adopts a tetrahedral geometry, with four pairs of electrons (two bonding pairs and two lone pairs) around it.
• Though tetrahedral arrangements have bond angles of around \(109.5^{\circ}\), lone pair repulsion makes the H-O-H bond angle slightly smaller than expected.

Hence, without lone pairs, the bond angles would likely be closer to \(109.5^{\circ}\).

lone pair repulsion

Lone pairs of electrons exert a greater repulsive force than bonding pairs. This is due to the lone pair electrons being localized solely on one atom, creating higher electron density.

In water, the lone pairs (on oxygen) push the bonding pairs closer together, reducing the bond angle.

• Lone pairs occupy more space surrounding the central atom, forcing bonding pairs to be closer.
• While bonding pairs share electrons between atoms, their repulsion is lesser compared to the lone pairs.

Thus, the theoretical \(90^{\circ}\) angle, ignoring lone pair repulsion, adjusts to \(104.5^{\circ}\) experimentally.

experimental vs theoretical values

The theoretical value for bond angles in water is largely derived from simplified models that do not account for all inter-electron forces. Valence bond theory predicts a \(90^{\circ}\) angle assuming only bonding interactions.

However, experimental values often differ as they reflect real-world complexity. Certain factors include:

• Lone pair-bond pair repulsion, as seen in water's \(104.5^{\circ}\) bond angle.
• Physical conditions like temperature and measurement precision.

These discrepancies highlight the importance of experimental validation in chemistry, ensuring theoretical models align with observed data.