Use VSEPR theory to predict the molecular geometry of H₂Se.

Question

Quizplus · Accepted Answer

First, let's determine the Lewis structure of the molecule.

H |- S - B12 - S - B0 - Se
   |         |         |
  U1       U1      U1

Each of the letters represent an atom, the numbers represent the number of valence electrons available for each atom, and the dashes represent covalent bonds. The lone pairs on each atom are represented by U1.

Counting the total number of valence electrons, we get: 
1 (H) + 6 (S) + 12 (B12) + 6 (B0) + 34 (Se) = 59

We need to arrange these electrons to form bonds and lone pairs around each atom in the molecule.

Starting with the central atom, B12, we can form single bonds with the two adjacent S atoms, using up 4 electrons:

H |- S - B12 - S - B0 - Se
   |   ||        |         |
  U1U1     U1  U1

Each S atom can form one more single bond with a neighboring B0 atom, using up 2 more electrons:

H |- S - B12 - S - B0 - Se
   |   ||   |   ||        |
  U1U1     U1    U1U1

Finally, the remaining electrons can be placed as lone pairs:

H |- S - B12 - S - B0 - Se
   |   ||   |   ||   |
  U1U1     U1    U1U1

Each atom in the molecule is now surrounded by a full octet of valence electrons, except for the central B12 atom, which has only 6 electrons around it. This suggests that the molecule is bent, with a bond angle less than 180 degrees at the B12 atom.

Therefore, the correct answer is A) bent.

Use VSEPR Theory to Predict the Molecular Geometry of H2Se

Use VSEPR Theory to Predict the Molecular Geometry of H₂Se