Question

Question #1aba7

Answer 1

Two atoms are `"sp"^3"d"` hybridized and one atom is `"sp"^3` hybridized.

Explanation:

Start by drawing the Lewis structure

We have 5×7 +1 = 36 electrons to place around the five atoms.

We will minimize formal charges by arranging the electrons symmetrically.

Here's what I got.

I put 2 lone pairs on the central atom and 3 lone pairs on each of the other four.

This gives the central atom a formal charge of +1; atoms 2 and 4 get a formal charge of -1; and the terminal atoms have zero formal charge.

Hybridization of central atom

The central atom has 2 bonding pairs and 2 lone pairs (`"AX"_3"E"_2`).

(From Wikimedia Commons)

This corresponds to a tetrahedral electron geometry and a bent molecular geometry (the yellow balls represent the lone pairs).

This corresponds to `"sp"^3` hybridization and a theoretical bond angle of 109.5°.

Hybridization of atoms 2 and 4

Atoms 2 and 4 each have 2 bond pairs and 3 lone pairs (`"AX"_2"E"_3`).

(Adapted from Wikimedia Commons)

This corresponds to a trigonal bipyramidal electron geometry and a linear molecular geometry.

This corresponds to an `"sp"^3"d"` hybridization and a bond angle of 180°.

Thus, two atoms are `"sp"^3"d"` hybridized and one atom is `"sp"^3` hybridized.