Naked Science Forum

Non Life Sciences => Chemistry => Topic started by: Indranil on 22/07/2019 16:04:16

Title: In this reaction How many electrons carbon anion has?
Post by: Indranil on 22/07/2019 16:04:16
In this reaction, How many electrons carbon anion has 7 or 8? I mean C- has 7 electrons or 8 electrons and H+ has one proton, how to H+ ion combine with H3C- ion to form CH4 because H+ ion needs 2 electrons? I am confused. Please show the mechanism.
Title: Re: In this reaction How many electrons carbon anion has?
Post by: chiralSPO on 22/07/2019 16:53:53
H+ has no electrons (just one proton--this reaction is a "protonation")
There are a total of 10 electrons in CH3 (6 from the C atom, 1 for each of the 3 H atoms, and 1 extra electron for the negative charge.) Two of these electrons form the inner "core" of the carbon atom (a full 1s orbital), and the remaining 8 electrons are all in the "valence." We can think of there bing 2 electrons in each of the 3 C-H bonds, and another 2 electrons as a lone pair on C.

When the H+ comes close to the electrons in the lone pair, it will be attracted electrostatically, and will get closer and closer until it is trapped and "bonded." The product molecule, CH4 is neutral and has 10 electrons (the proton didn't bring any electrons with it.) The mechanism can be represented by a single curved and double-headed arrow, pointing from the lone pair of the methyl anion (CH3 and going to the proton (H+).
Title: Re: In this reaction How many electrons carbon anion has?
Post by: Indranil on 23/07/2019 05:59:47
H+ has no electrons (just one proton--this reaction is a "protonation")
There are a total of 10 electrons in CH3 (6 from the C atom, 1 for each of the 3 H atoms, and 1 extra electron for the negative charge.)
Where do the carbon get the extra electron for the negetive charge?
Title: Re: In this reaction How many electrons carbon anion has?
Post by: Kryptid on 23/07/2019 06:37:14
Where do the carbon get the extra electron for the negetive charge?

Presumably by taking it from some other atom before it became involved in the reaction that this thread is about.