Problem 65 - A Complete Resource Book in Chemistry for JEE Main

Question

Bond angle in water is (a) \(90^{\circ}\) (b) \(104.5^{\circ}\) (c) \(109.5^{\circ}\) (d) \(120^{\circ}\)

Step-by-Step Solution

Answer

The bond angle in water is (b) \(104.5^{\circ}\).

Step 1: Understanding Molecular Geometry

To determine the bond angle in water, we need to understand its molecular geometry. Water (H₂O) is a molecule with two hydrogen atoms bonded to an oxygen atom. The molecular geometry around the oxygen atom is based on the VSEPR (Valence Shell Electron Pair Repulsion) theory.

Step 2: Applying VSEPR Theory

According to VSEPR theory, the shape of a molecule is determined by the repulsions between electron pairs around the central atom. In water, oxygen has two lone pairs and two bonding pairs (with hydrogen). These pairs will arrange themselves to minimize repulsion, resulting in a bent shape.

Step 3: Calculating Ideal Bond Angle

For a tetrahedral electron pair geometry, the ideal bond angle is approximately 109.5°. However, because of the repulsion between the lone pairs, the bond angle in a water molecule decreases slightly from this ideal angle.

Step 4: Determining the Actual Bond Angle

Due to the repulsion between the two lone pairs being greater than that between the bonding pairs, the bond angle in water is reduced. The measured H-O-H bond angle in water is approximately 104.5°.

Key Concepts

Molecular Geometry

In order to understand the concept of bond angles in water, it's important to first grasp the idea of molecular geometry. Water, with its chemical formula H₂O, is a molecule made up of two hydrogen atoms and one oxygen atom. The arrangement of these atoms is not random; it is influenced by the specific angles and distances between them, which in turn impact the overall shape and properties of the molecule. Water's molecular geometry is characterized by its bent shape. This specific formation is influenced by the presence and arrangement of atoms and electron pairs. To truly understand the shape of the water molecule, we need to delve deeper into the role of electron pairs, which brings us to the next concept.

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a key principle in understanding molecular geometry. This theory posits that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion. This concept is crucial in determining the shape of molecules. In the case of water, the central atom is oxygen. Around this oxygen atom, there are two bonding pairs (shared with hydrogen atoms) and two lone pairs of electrons. The VSEPR theory helps predict that these pairs will position themselves in a manner that reduces repulsive forces, leading to a specific molecular shape. For water, this configuration results in a bent shape rather than a straight line.

Lone Pair Repulsion

Lone pair repulsion is a significant factor in understanding why water doesn't achieve ideal tetrahedral angles. Lone pairs are pairs of valence electrons not involved in bonding. In a water molecule, there are two lone pairs on the oxygen atom. These lone pairs exert a repulsive force on the bonding pairs, actually more so than the bonding pairs exert on each other. As a result, the lone pairs push the hydrogen atoms closer together, decreasing the bond angle from what you would expect in a tetrahedron. This unique interplay is what leads to the bond angle in water shrinking from the ideal tetrahedral angle of 109.5° to around 104.5°.

Tetrahedral Geometry

Understanding tetrahedral geometry provides a basis for comprehending water's bond angle. In a perfect tetrahedral configuration, four bonds spread out from a central atom, spaced symmetrically to minimize repulsion, with bond angles of approximately 109.5°. This is the ideal scenario without lone pair interactions. However, in water, oxygen holds two lone pairs along with two hydrogen bonds. The presence of these lone pairs distorts the ideal tetrahedral geometry. They occupy more space and thus cause the bond angles between the hydrogen atoms to compress. Therefore, instead of maintaining the typical tetrahedral angle, water’s molecule assumes a bent geometry with bond angles adjusted to about 104.5°, highlighting the profound influence of lone pair repulsion.