2E.5 b

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RRahimtoola1I
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Joined: Fri Aug 09, 2019 12:15 am

2E.5 b

Postby RRahimtoola1I » Wed Nov 13, 2019 7:32 pm

OClO has 19 valence electrons so it's a radical. Do we count the extra electron even though it's not in a pair when determining shape?

nehashetty_2G
Posts: 102
Joined: Thu Jul 25, 2019 12:15 am

Re: 2E.5 b

Postby nehashetty_2G » Wed Nov 13, 2019 7:37 pm

I'm not exactly sure about radicals and shapes, but I know that for 2E.5 when they ask OClO they aren't asking about a molecule ClO2, they are just taking about the bond formed between those three atoms (O, Cl, and O). Hope that helps!

Ruby Richter 2L
Posts: 103
Joined: Thu Jul 25, 2019 12:17 am

Re: 2E.5 b

Postby Ruby Richter 2L » Wed Nov 13, 2019 7:55 pm

I believe question 2E.5 asks about the ClO2 cation, so in total it would have 18 valance electrons due to the positive charge and subtraction of one electron. This would cause the shape to be trigonal planar due to the lone pair on the central Cl atom.

Astrid Lunde 1I
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Re: 2E.5 b

Postby Astrid Lunde 1I » Wed Nov 13, 2019 8:42 pm

The are asking for the ClO2 cation so there are 18 electrons. The shape is trigonal planar and the bond angle is 120 degrees.

KaleenaJezycki_1I
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Re: 2E.5 b

Postby KaleenaJezycki_1I » Wed Nov 13, 2019 9:05 pm

Because it is ClO2+ ion there are 18 valence electrons because the + means to subtract 1 electron. Therefore draw the lewis structure with Cl as the central atom and O atoms connected by double bonds to Cl, then 2 sets of lone electron pairs on each O and 1 set of lone electron pair on the Cl. Then if you look at a VSEPR chart you will see that the shape is angular.


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