Help w/ 4.19 (6th edition)

(Polar molecules, Non-polar molecules, etc.)

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KylieY_3B
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

Help w/ 4.19 (6th edition)

Postby KylieY_3B » Mon Nov 19, 2018 11:15 pm

For part b, I've been having difficulty with part b, which is the molecule (CH3)2Be. I understand the basic molecular shape and geometry, but I'm not sure I understand the Lewis structure. From what I've looked up, Be has single bonds with Carbon but no lone pairs. How is this possible if Be doesn't have an octet? I could be wrong about the Lewis structure.

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: Help w/ 4.19 (6th edition)

Postby ChathuriGunasekera1D » Tue Nov 20, 2018 8:14 am

Be has 4 electrons total, and from that has 2 valence electrons. If of course cannot follow the octet rule because it doesn't have enough electrons. So, it can only form one bond with each carbon which uses up its two valence electrons. The structure will look like 3HC-Be-CH3 but the Hs will be spread out.

Sean Reyes 1J
Posts: 67
Joined: Fri Sep 28, 2018 12:24 am

Re: Help w/ 4.19 (6th edition)

Postby Sean Reyes 1J » Tue Nov 20, 2018 6:58 pm

Be is one of the exceptions to the octet rule along with hydrogen, helium, and lithium.
Because beryllium does not have enough electrons at all to form an octet, it would only be able to from a single bond with each of the carbons.
Additonally,if you used formal charge, the formula would be:
2 - ( 0 + (4/2)), which would equal a charge of 0.


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