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Double bonds vs. lone pairs

Posted: Sun Nov 04, 2018 7:56 pm
by KylieY_3B
When drawing a Lewis dot structure, how do we know when to draw a molecule with double bonds or when to draw in lone pairs?

For example, CH2Cl2 has 20 valence electrons, but how do we know whether to draw Carbon and Chlorine with double bonds or to draw in lone pairs for chlorine?

Re: Double bonds vs. lone pairs

Posted: Sun Nov 04, 2018 8:02 pm
by Netpheel Wang 4L
We know what kind of bond to make when calculating formal charge of each atom. The Lewis structure with the lowest total formal charge is what is ideally wanted.

Re: Double bonds vs. lone pairs

Posted: Sun Nov 04, 2018 8:06 pm
by 405112316
Before even calculating formal charge, we have to find a structure that contains all 20 valence electrons. Carbon is in the middle because it has 4 potential bonding sites. We know the both hydrogens have to bond to carbon with a single bond because hydrogens can only have two electrons in their valence shell. If we were to make a double bond with one chlorine, we would have to move hydrogen to attach to that chlorine (otherwise carbon would have 10 valence electrons); we would only have accounted for 18 valence electrons.

Re: Double bonds vs. lone pairs

Posted: Sun Nov 04, 2018 8:18 pm
by Ronald Thompson 1F
Count the total valence electrons and arrange them in a ways that satisfies the elements' octect.

Re: Double bonds vs. lone pairs

Posted: Sun Nov 04, 2018 8:21 pm
by Dong Hyun Lee 4E
You want to first make sure that you have the total number of valence electrons of all elements in that compound and at least one bond per connected element. When that is done, you want to make sure every element fits the octet rule (except the exceptions) and then check for the least ammount of formal charge. Only then you can choose whether you need double bonds or lone pairs.

Re: Double bonds vs. lone pairs

Posted: Sun Nov 04, 2018 8:56 pm
by daniella_knight1I
Arrange the structure with single bonds and calculate the formal charge, then try double bonding and calculate it again. You want to have the formal charge as close to 0 as possible.