Question about Opted #93 (ochem)

[meghoo22]

Hey guys!

I was hoping somebody can explain this question/answer to me. On the O-chem section of the opted test, they ask:

93. What is the hybridization of a nitrogen atom if
it forms two σ two π bonds?
A. sp
B. sp2
C. sp3
D. sp3d2

Their answer is "C".

My question is: if Nitrogen is bound with two sigma and two pi bonds, isn't it sp hybridyzed? I assume this is what it would look like X=N=X (X being any atom) Am I not understanding the question correctly or am I just totally off ? Either way, please someone explain this ti me! thanks much 🙂

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[jvelos3]

it forms 4 bonds so therefore it is sp3 hybridized.

to form the 4 bonds an s, p, p, p orbitals had to be hybridized.

 

[Absolute Vision]

I agree with your answer...it seems to be choice A or sp hybridized.
Single bonds = 1 sigma bond; double bond = 1 sigma and 1 pi bond; triple bond = 1 sigma and 2 pi bonds.
To determine the hybridization of an atom, just count the number of atoms to which that atom is bonded plus the number of lone pairs on that atom.
According to your suggested answer nitrogen atom contains two double bonds on each side of the nitrogen atom, so it's sp hybridized ...and each double bond consists of 1 sigma and 1 pi bond, therefore, since there two double bonds, there should be 2 sigma and 2 pi bonds. Unless there's a different way, I believe your answer is correct.

 

[meghoo22]

I must be completely misunderstanding this whole thing then. So the Nitrogen is bound with four different single bonds? (none of them double?) the reason I'm asking is that CH4, for example, has four single sigma (single) bonds and it's sp3. However, once it has a double bond in there, like in COH2, it becomes sp2.

In their discription, they say there are two sigma bonds and two pi bonds. Doesn't this mean two double bonds or one single and one tripple bond? (since each double bonds is made of one sigm and one pi bond,and each tripple bond of one sigma and two pi bonds) By definition, a pi bond can't exist on its own and needs to have a sigma bond associated with it. Either way, this would make the Nitrogen have two bonds...what am I missing here?

Can somebody draw out the structure of this molecule please? I'm even more confused now :/

 

[Absolute Vision]

Based on my post earlier, each of these two nitrogen atoms have a sp hybridization and also contains 2 sigma bonds and 2 pi bonds: =N= or –N=

1) The nitrogen with two double bonds on each side will contain 2 sigma bonds and 2 pi bonds, since each double bond will consist of 1 sigma bond and 1 pi bond).

2) The nitrogen with a single bond on the left side of the atom will contain a sigma bond, while the triple bond on the right side of the atom will contain 1 sigma bond and 2 pi bonds, and therefore the second nitrogen atom will also have 2 sigma bonds and 2 pi bonds.

To determine each hybridization for each nitrogen atom, just count the number of atoms that nitrogen is bonded to plus the lone pairs on the nitrogen.
1) The first nitrogen atom will be bonded to two atoms (one atom for each double bond), so its sp hybridized.
2) Same reasoning as #1.

Is this your reasoning also?

 

[meghoo22]

I completely agree with you Absolute Vision. (I think I started writing my second post as you were writing yours...so when I submitted it you had already posted your reply.) If anyone has an alternate explanation, please post.

What concerns me is that there are a few wrong question/answer sets like this on the opted exam, and I'm not sure if it's a typo or that's the answer they truly count as correct. Kind of disturbing seeing as how thousands of people use this exam as their reference.

 

[jvelos3]

wow...i need to get more sleep...lol

yes the answer is A.

2 sigma bonds and 2 pi bonds...2 sigma bonds are formed by the sp hybridization and the 2 pi bonds are formed by the remaining 2 p orbitals.