Can someone please explain how these compounds are formed/how the 3c 2e comes into existence.

  1. Polymeric structure of BeCl2 - 3c 4e
  2. BeH2 - 3c 2e
  3. CH5+

Please explain in terms of the hybridization of the central atom and how the electrons are shared etc. Thanks!

My approach -

I know that in B2H6 there is a banana bond (3c 2e), but in BH3 it makes more sense - both B go into sp3 hybridization, and one hybrid orbital of each is empty where some overlapping takes place due to the H bonded with the other B, thus causing the electron pair to jump from one B to another. Thus named banana bond. With CH5 I don't understand how C is able to hybridize in a way such that it can have a banana bond. I read that it is of the form CH3 and H2 and 3c 2e bonding in between them, but I don't understand how it takes place. For BeCl2 there are 2 Be having 2 Cls each and also coordinate bond by other 2 chlorines. Is that right? But for BeH2, how can there be a 3c 2e bond? I mean based on the same principle, there must be a bond between Be, H that can jump to the other Be with same H, but Be doesn't have enough electrons to do this. Thus my confusion.

  • $\begingroup$ This looks like a homework question. $\endgroup$
    – Jan
    Commented Dec 7, 2015 at 14:03
  • $\begingroup$ No, really, I am more interested in how a 3c 2e bond is formed than getting these answers. I want to know how the "banana bond" comes into existence in such compounds. It is the concept of 3c 2e that I'm interested in. Please edit the question if you want. $\endgroup$
    – Shodai
    Commented Dec 7, 2015 at 14:09
  • $\begingroup$ Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. Don't worry, they're not banned. But, we require a minimal effort. You can just tell us what you did that didn't get answers. We'll take it from there. $\endgroup$
    – M.A.R.
    Commented Dec 7, 2015 at 14:16
  • $\begingroup$ Ok I am editing my question now $\endgroup$
    – Shodai
    Commented Dec 7, 2015 at 14:26


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