Question

VBT theory predict Hybridization OR derive Hybridization from VSEPR theory VBT explains bonding with respect to VSEPR - it didn't expect Hybridization H2O SP3 due to vespr not vbt expect second why there is no Hybridization in HF for VBT explaination ?

Answer 1

Here's my answer to what I think you are asking.

Explanation:

Valence bond theory

Valence Bond (VB) theory is based on valence electrons. We use it it help determine the structure of a molecule.

We start with the Lewis structure of a molecule and then explain the formation of bonds by the overlap of atomic orbitals.

If our use of simple atomic orbitals does not explain the observed shape of a molecule, we then resort to the concept of hybridized atomic orbitals.

VB theory and `"HF"`

The Lewis structure of `"HF"` is

The electron configuration of `"H"` is `"1s"`, and the valence electron configuration of `"F"` is `"[He] 2s"^2"2p"_text(x)^2"2p"_text(y)^2"2p"_text(z)`.

Thus, we explain the `"H-F"` bond formation by the overlap of the hydrogen `"1s"` orbital with the half-filled fluorine `"2p"` orbital to form a σ bond.

Since there are only two atoms, the molecule is linear, and we don't have to use hybridization to explain its shape.

VB theory and `"H"_2"O"`

The Lewis structure of `"H"_2"O"` is

The valence electron configuration of `"O"` is `"[He] 2s"^2"2p"_text(x)^2"2p"_text(y)"2p"_text(z)`.

Thus, we could explain the `"OH"` bond formation by the overlap of the hydrogen `"1s"` orbital with the half-filled oxygen `"2p"` orbitals to form two σ bonds.

The problem is that this predicts the `"H-O-H"` bond angle to be 90°, and the observed bond angle is 104.5 °.

VSEPR theory gives a better prediction of the bond angle.

The four electron pairs repel each other to the four corners of a tetrahedron, for which the theoretical bond angle is 109.5°.

The four hybridized `"sp"^3` orbitals also predict a bond angle of 109.5°.

We explain the 5° difference from theoretical as resulting from repulsion by the lone pair electrons.