Hw 6.5

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Edmund Zhi 2B
Posts: 118
Joined: Sat Jul 20, 2019 12:16 am

Hw 6.5

Postby Edmund Zhi 2B » Wed Dec 04, 2019 3:27 pm

In the reaction, H2O2 + SO3 -> H2SO5, how do you identify which one is the lewis acid and which one is the lewis base? I know what the definitions are, but how do you determine which of the reactions is the electron "donor" and which one is the "acceptor" since they merge into one molecule?

sarahwu3a
Posts: 50
Joined: Sat Aug 17, 2019 12:17 am

Re: Hw 6.5

Postby sarahwu3a » Thu Dec 05, 2019 2:21 am

In the SO3 molecule, the sulfur atom has space to accept another electron pair so that in the molecule there are 4 regions of electron density in the molecule. Thus, as the electron pair acceptor it is the Lewis acid.

Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am

Re: Hw 6.5

Postby Anisha Chandra 1K » Thu Dec 05, 2019 11:04 pm

In the solutions manual, why is SO3 drawn as having 1 double bond and 2 single bonds instead of 3 double bonds?

DTingey_1C
Posts: 55
Joined: Fri Aug 30, 2019 12:16 am

Re: Hw 6.5

Postby DTingey_1C » Fri Dec 06, 2019 3:44 pm

Yes isn't SO3 with one double bond more unstable than SO3 with 3 double bonds? Maybe this is because it wouldn't act as a lewis acid anymore.

Junxi Feng 3B
Posts: 52
Joined: Sat Sep 14, 2019 12:17 am

Re: Hw 6.5

Postby Junxi Feng 3B » Fri Dec 06, 2019 4:15 pm

I think the best way to determine is to draw out all the lewis structures. And you can see the lewis structure of the final product is still drawn around the sulfur atom, meaning it accepts the electrons. Therefore, it's the Lewis acid.


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