Question

Why is C--O--C bond angle in ethers slightly greater than tetrahedral angle?

Hint:

In ethers: \[ \boxed{Oxygen\ is\ sp^3\ hybridised} \] \[ \boxed{2\ bond\ pairs + 2\ lone\ pairs} \] Lone pair repulsion increases the bond angle slightly above the normal tetrahedral angle. Remember: \[ \boxed{\text{More lone pair repulsion} \Rightarrow \text{larger bond angle}} \]

Answer 1

Concept: According to the valence shell electron pair repulsion (VSEPR) theory, the shape of molecules depends on the repulsion between electron pairs present around the central atom. In ethers, oxygen is the central atom: \[ R-O-R' \] The oxygen atom is $sp^3$ hybridised and contains four electron pairs: \[ \boxed{2 \text{ bond pairs} + 2 \text{ lone pairs}} \] The arrangement of these electron pairs is approximately tetrahedral. \[ \text{Ideal tetrahedral angle}=109.5^\circ \] However, lone pairs occupy more space than bond pairs because they are attracted only by the central atom and are not shared with another atom. Therefore, the repulsion order is: \[ \boxed{\text{Lone pair--lone pair} \gt \text{lone pair--bond pair} \gt \text{bond pair--bond pair}} \] The two lone pairs on oxygen repel the C--O bond pairs strongly and push the two alkyl groups farther apart. As a result, the C--O--C bond angle becomes slightly larger than the tetrahedral angle. \[ \boxed{\angle C-O-C \approx 111.7^\circ} \]

Final Answer: The C--O--C bond angle in ethers is slightly greater than $109.5^\circ$ because oxygen contains two lone pairs of electrons. The strong repulsion between lone pairs and bond pairs pushes the two C--O bonds apart, increasing the bond angle. \[ \boxed{\angle C-O-C = 111.7^\circ \gt 109.5^\circ} \]