E.O.C. Chem/OChem Chap 2

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E.O.C. Chem/OChem Chap 2

Post by taroson » Sun Feb 16, 2020 4:13 am

For standalone question 4, it explains that an atom in a pi bond cannot participate in sp3. Can someone explain why that is?
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Re: E.O.C. Chem/OChem Chap 2

Post by NS_Tutor_Mathias » Tue Feb 18, 2020 2:55 am

The fancy and correct answer is that forming pi bonds requires having two aligned, unhybridized p orbitals between two atoms - nothing else can form a pi bond. Sp3 contains no unhybridized orbitals, so that is out (that is what it is: 4 sp-character mixed orbitals that are overwhelmingly p in character). Sp2 has one unhybridized p orbital and sp has two.

Another way to approach this is to just intuit that orbital hybridization and electronic geometry are kind of co-determined: To have a pi bond, you must have at least a double bond, which forces any central atom with 4 potential ligands to be either sp2 (trig planar) or sp (linear). This view of course explains nothing, but can be easier to remember in a pinch.

While the former explanation is better and more closely aligns with the model, the MCAT in my experience only rarely cares about the mechanical details of hybrizidation to this degree. It can definitely not hurt to familiarize yourself with this though, as it is well within the scope of the exam and may be more heavily emphasized in the future.
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