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When drawing a lewis structure with a double/triple bond, how does one know to give that structure a double bond instead of just a single bond with more "." around it? For example, with H2CO, how do you know to draw it with a double bond between the C and O and 4 "." around the O, instead of a single bond between the C and O and 6 "." around the O?
You know to draw a double bond between the C and the O because Carbons always has to have 4 bonds around it, and since the only other elements are Hydrogen who can only have one bond, then the bond needs to be between Carbon and Oxygen.
The student above me is absolutely right. The key to this example is not so much focusing on the Oxygen in the molecule but rather the Carbon. You could have 6 electrons around the oxygen, however, you would need to satisfy the octet around the carbon first. Since Carbon would only have three pairs of shared electrons (3 bonds), you would need to add a double bond between the oxygen and the carbon. Hope this helps.
In order to determine whether or not you use triple/double bonds instead of more lone pairs you must calculate the number of valence electrons. For example, for O2 you have 2(6 e) = 12 e. If you drew the lewis structure with more lone pairs, there would be too many electrons so instead you add a double bond o=o with four valence electrons on each o to get a total of 12 e.
H2CO has a total of 12 electrons. If you were to include a single bond for O instead of a double bond, you would result in 14e- instead of 12e-. Another thing to keep in mind is the formal charge. For instance if you had a single bond on the O atom, the FC would be 6-(6+1)=-1 while a double bond would would have a formal charge of 6-(4+2)=0. Since the FC of the double bond is 0, it is more stable than having a single bond.
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