Question

9. Describe:

a. Electron-domain geometry

b. Molecular geometry (Note: Describe and draw the shape)

c. Hybridization of the central atom

d. Polarity

in the following molecules:

1. TeCl₂
2. Ethane: CH₃CH₃ (Describe a, b, c for each carbon; Describe d for the whole molecule).

Answer 1

Solution:

Part A) TeCl2:

The central Te atom has 6 valence electrons. When the two Cl atoms bond to the central atom, they each contribute one electron to form two sigma bonds. This leaves 4 electrons on the central atoms in the form of 2 lone electron pairs. It has sp3 hybrization. Hence,, a tetrahedron shape with two electron pairs and two atoms, creating a bent molecular shape. The bent shape of molecule makes BeCl2 polar.

1) Electron geometry = Tetrahedral

2) Molecular geometry = Bent shaped

3) Hybridization on central atom = sp3

4) Polarity = Polar

Part B) CH3-CH3:

The each central C atom has 4 valence electrons and they binds to 3H and one C atoms which contribute four sigma bonds. It has sp3 hybrization. Hence, Carbon having with four electron pairs creating a tetrahedral molecular shape. Thus, molecule becomes nonpolar.

1) Electron geometry = Tetrahedral

2) Molecular geometry = Tetrahedral

3) Hybridization on central atom = sp3

4) Polarity = Nonpolar