why bond angle of alcohols are less than tetrahedral

Question

shapes

 

Answer 1

 

The lone pairs do not "take up" more space.  They are electrons which are quite tiny and are restricted to (in the water example below) the same sized sp³ hybridized orbitals as the electrons of the oxygen-hydrogen sigma bonds. In actuality, the lone pairs are a "more concentrated" negative moiety, therefore REPEL the valance electrons of the other substituents more strongly.  

Answer 2

Emily provided a good answer. All I would add is to think of a similar, but more common example-water.

With H₂O, the O atom has two hydrogen atoms, and two lone pairs of electrons. These loan pairs take up

more space than bonding pairs, forcing the hydrogen atoms closer, so the H-O-H bond angle is about 104.5 ^o

Answer 3

In a "perfect tetrahedral" (like CH₄), all four hydrogen groups are distributed around the carbon atom equally-- leading to the bond angle of 109.5 deg.

 

However, in the case of the oxygen atom in an alcohol (I assume that's the bond angle you are looking at), the substituents on the oxygen are a carbon atom, a hydrogen atom, and two lone pairs. These lone pairs take up more space than a bond would and therefore distort the "perfect tetrahedral". Since the lone pairs need more room they push the hydrogen and the carbon closer together causing the angle between these two atoms to be less than 109.5.

 

Hope this helps!

Emily