Question

Draw lewis structures for CH2Br2, TeF4, BrF5, BeF2

1. which of the molecules is polar
2. which of the molecules violate the octet
3. which of the molecules contain no lone pairs
4. which of the molecules have a sp3d hybridization

Answer 1

Among CH2Br2, TeF4, BrF5, and BeF2, only TeF4 has sp3d hybridization. This hybridization allows for the formation of five electron domains - four bonds and one lone pair.

Step-by-step explanation:

The question asks us to draw Lewis structures for CH2Br2, TeF4, BrF5, and clarify which of the molecules have a sp3d hybridization. When analyzing the hybridization of molecules, we consider the electron domains around the central atom, including bonds and lone pairs:

• CH2Br2 has a central carbon atom with four single bonds, indicating sp3 hybridization.
• TeF4 has a central tellurium atom with four bonds and one lone pair, which is indicative of sp3d hybridization.
• BrF5 has a central bromine atom with five bonds and one lone pair, representing sp3d2 hybridization.
• BeF2 isn't a valid formula for beryllium fluoride, which is BeF2, featuring a linear geometry and sp hybridization.

Therefore, the molecule with sp3d hybridization is TeF4. This hybridization allows for the formation of five electron domains, accommodating both the bonds to fluorine and the lone pair on tellurium.