Question #73746

Jun 15, 2016

Actually it doesn't. In the planar molecule the $2 s$ orbital on oxygen mixed with only the two $2 p$ orbitals in the same plane to make the bonds.

Explanation:

A diagram of the molecular orbitals is given here. Note that the $2 {p}_{z}$ remains unchanged.

https://en.wikipedia.org/wiki/Molecular_orbital_diagram

Jun 15, 2016

It is a consequence of both Hund's rule of maximum multiplicity (which gives oxygen two unpaired electrons) and achievement of maximum separation of the orbitals.

Explanation:

Following Hunds rule electronic configuration of oxygen is $1 {s}^{2} 2 {s}^{2} 2 p {x}^{2} 2 p {y}^{1} 2 p {z}^{1}$.

Oxygen, therefore, has two unpaired electrons which can bond with the two hydrogen atoms. In order to obtain maximum separation of the electron orbitals from the 3 atoms, bonding is achieved by the formation of four $s {p}^{3}$ hybrid orbitals. However, two of these contain only non-bonded pairs, so the overall shape looks more like a "bent" V than a normal tetrahedron. Repulsion from the remaining lone pair on the oxygen atom means the bond angle is reduced to around 105 degrees..