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

Step 1: Hybridisation of oxygen in ethers.
In ethers ($R-O-R'$), oxygen is $sp^3$ hybridised with two bond pairs (to two carbons) and two lone pairs. Pure $sp^3$ geometry gives an ideal angle of $109.5^\circ$.
Step 2: Lone pairs exert greater repulsion.
Lone pairs are held only by the oxygen nucleus, occupy more angular space than bond pairs, and exert stronger repulsions. The repulsion order is: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.
Step 3: Effect on C-O-C angle.
The two lone pairs on oxygen repel both $C-O$ bond pairs outward, forcing the two carbon atoms apart and expanding the $C-O-C$ angle beyond the tetrahedral value. \[ \boxed{\angle C-O-C \approx 111.7^\circ > 109.5^\circ \text{ due to lone pair repulsion}} \]