Fall 2011 Midterm 6C and fall 2012 midterm 6A


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Shannon Han 2B
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Joined: Fri Sep 26, 2014 2:02 pm

Fall 2011 Midterm 6C and fall 2012 midterm 6A

Postby Shannon Han 2B » Sun Nov 09, 2014 2:21 pm

I'm confused by these two problems about the hybridization of each atom. I missed all of them except for the hybridization of the carbon atoms because it always forms 4 bonds. Can anybody explain the answers to these?

K Honeychurch 1K
Posts: 27
Joined: Fri Sep 26, 2014 2:02 pm

Re: Fall 2011 Midterm 6C and fall 2012 midterm 6A

Postby K Honeychurch 1K » Sun Nov 09, 2014 8:37 pm

The hybridization is determined by the number of regions of high electron density. Each bond (single, double, or triple) and each electron pair (or even a single electron if it is unpaired) is a region of high electron density.
2 regions= sp
3 regions= sp2
4 regions= sp3
5 regions= sp3d
6 regions= sp3d2

Fall 2011 Midterm 6C-
1. Each carbon atom has a single bond to H, a single bond to C, and a double bond to C. That is 3 regions which means sp2 hybridization.
2. The N has two single bonds to hydrogens, a single bond to a C, and a lone pair. 4 regions= sp3.
3. The last O is bonded to a H and to a C, and has 2 lone pairs. 4 regions= sp3
4. The C connected to 2 O atoms is double bonded to an O, single bonded to another O, and single bonded to a C. 3 regions= sp2
Hopefully you can work through 6A now!

tchar96_1F
Posts: 14
Joined: Fri Sep 26, 2014 2:02 pm

Re: Fall 2011 Midterm 6C and fall 2012 midterm 6A

Postby tchar96_1F » Tue Nov 11, 2014 5:25 pm

I don't understand how we're supposed to know where the lone pairs are if we're not given the chemical formula. Could you explain how you know where to put the lone pairs? Thank you!

K Honeychurch 1K
Posts: 27
Joined: Fri Sep 26, 2014 2:02 pm

Re: Fall 2011 Midterm 6C and fall 2012 midterm 6A

Postby K Honeychurch 1K » Sun Nov 16, 2014 10:13 pm

Sorry I didn't see this sooner! I don't know if this is irrelevant now since it's after the midterm...but I guess there's always the final...
Each atom must have a completed octet (8 electrons around it either in bonds or lone pairs) so when an atom does not have 4 bonds it must have a lone pair or two or three. For example, the N in part 2 of the 2011 midterm question is single bonded to three atoms. That would only be 6 electrons around it. For the N to have a full octet it must have a lone pair. So, even if the drawing does not show the lone pairs, you have to know if there are any and where they are.


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