Question 3.67 part B?

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Cowasjee_Sanaea_3E
Posts: 27
Joined: Fri Jul 22, 2016 3:00 am

Question 3.67 part B?

Postby Cowasjee_Sanaea_3E » Wed Oct 19, 2016 10:50 pm

Hi there!

I am confused on the answer to question 3.67 in the textbook which asks:
"in each of these compounds, an atom violates the octet rule. Identify the atom and explain the deviation from the octet rule? B) ClO2"

I would have thought that the answer would have been chlorine in the middle with 3 lone pair electrons, and 4 shared bonds on it while the two oxygens bonded with the chlorine would have 4 lone pair elections (and therefore all three of the atoms would have a formal charge of zero), however the solution manual states that the answer would be chlorine in the middle with 3 lone pair electrons and 2 shared bonds with the two bonded oxygens having 6 paired elections (but in this case Chlorine would have a formal charge of +2 and the oxygens would have a formal charge of -1). how come this is the case? (I understand why chlorine breaks the octet rule, I just don't understand why the structure is this way)?

So in other words, why O-Cl-O and not O=Cl=O?

Thanks in advance!

Belicia Tang 1B
Posts: 32
Joined: Wed Sep 21, 2016 3:00 pm

Re: Question 3.67 part B?

Postby Belicia Tang 1B » Fri Oct 21, 2016 12:11 am

I have the same question! If you calculate the formal charge of Cl in the molecule w/ double bonds, it ends up being 0, while the one w/ single bonds is +2. You would think that the lower formal charge molecule should be stabler... Any thoughts from anyone else?

Audrey_Hall_2I
Posts: 17
Joined: Wed Sep 21, 2016 2:56 pm

Re: Question 3.67 part B?

Postby Audrey_Hall_2I » Fri Oct 21, 2016 5:56 am

We did this example in discussion yesterday and the answer was that it was a resonance hybrid written as with one single bond and one double bond.


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