HW 4.19 part b and d

[Anna De Schutter - 1A]

Hi!

For the (CH3)2Be molecule, the shape is said to be tetrahedral around the carbon atoms in the solutions manual. Could I also say that the shape of the whole molecule is linear because we have Be in the middle and a CH3 on either side? Or would that be wrong?

Also, for the SnCl2 molecule, I drew the Sn in the middle with each Cl bound to it through a single bond. This gives each element a formal charge of 0, yet doesn't satisfy the octet rule for Sn. Another way to draw this molecule would be to put Sn in the middle with each Cl bound to it through a double bond. This would however give Sn a formal charge of -2 as well as an expanded octet and each Cl a formal charge of +1. Which molecule do we prefer, the one with the single bonds or the one with the double bonds?

Thank you!
Anna De Schutter - section 1A

[Gianna Graziano 1A]

Hi Anna,
b) That would be okay for the lewis structure, but not okay for the VSEPR model. Tetrahedral, with each angel at 109.5 degrees is the correct answer. This is the case because there are 3 bonds, and 0 LP.
d) In every case, you prefer the model that satisfies the octet rule. Formal charge would be ideal at its lowest, but without an octet the molecule would be quite unstable.

[Kaleb Tesfaye 1I]

Hey Gianna
Can you clarify in what you meant when you said "0 LP"?
Also, how we do know which bonds to focus on when we do the VSEPR model. The CH3 molecule is tetrahedaral but the (CH3)2Be molecule is linear. Do we mention the shape and angle for both molecules?

[Rebekah Kaufman 1L]

0 LP means zero lone pairs