Hybridisation for a molecule is given by: $$\frac{1}{2}(V + H - C + A)$$ Where,

  • V = Number of valance electrons in central atom
  • H = Number of surrounding monovalent atoms
  • C = Cationic charge
  • A = Anionic charge

  • If this expression is 2 then its $sp$ hybridised, 3 then $sp^2$ hybridised and so on.

    However, I am not able to understand why this works.

    • 1
      $\begingroup$ duplicate of chemistry.stackexchange.com/questions/4399 $\endgroup$
      – MaxW
      Dec 20, 2015 at 16:54
    • 4
      $\begingroup$ @MaxW No, it’s not. $\endgroup$
      – Jan
      Dec 20, 2015 at 17:01
    • 4
      $\begingroup$ @MaxW I've already read that question before asking. $\endgroup$
      – dark32
      Dec 21, 2015 at 9:46

    1 Answer 1


    Since you understand why it works for monovalent surrounding atoms, I will assume that we have a molecule $\ce{MA_{$m$} B_{$n$}}$ where $\ce{M}$ is the central atom, $\ce{A}$ represents the monovalent atoms which may or may not be identical, and $\ce{B}$ represents not necessarily identical atoms which are polyvalent. First up, there are some assumptions to be made:

    1. All "surrounding atoms" are directly bonded to the central atom $\ce{M}$, and with no other atom;
    2. All atoms form as many covalent bonds as is indicated by valency;

    Now say the $n\ \ce{B}$ atoms are absent. Let the steric number be $x$,
    Then, $x=0.5\times (V+m-C+A)$
    Now add the $\ce{B}$ atoms. Now the central atom forms $n$ more sigma bonds, but if the sigma bonds were not present, these would exist as lone pairs, and hence already counted in $V$. Take for example, $\ce{PCl_3}$ and $\ce{POCl_3}$. For $\ce{PCl_3}$ you would get $0.5\times (5+3)=4$ and for $\ce{POCl_3}$ you would get $0.5\times (5+3)=4$. The reason the $\ce{O}$ made no difference is because phosphorus uses two of its electrons to bond with both of which are nothing but valence electrons of phosphorus, only present as a lone pair in one case and a sigma bond in the other. Also, this formula does not work when you have triple bonds, say for $\ce{HCN}$. As to how it works for mono valent atoms being the only surrounding atoms, it is simple.The number of sigma bonds is the number of mono valent atoms, and thus by adding that to the number of lone pairs we get the hybridization. Say you have $ n $ lone pairs and $s$ sigma bonds.You can see that $V=2n+s$, and that $s=H$. So $0.5*(H+V)=n+s$, as required.

    • $\begingroup$ Please format chemical expressions using the \ce{...} syntax. Also please remember to put a space behind a full stop. $\endgroup$
      – Jan
      Dec 21, 2015 at 12:47
    • $\begingroup$ I think that I didn't (or incorrectly) understand why hybridisation formula works for monovalent atoms too. I am sorry for that, I'll change the question too. $\endgroup$
      – dark32
      Dec 22, 2015 at 15:34
    • $\begingroup$ @xXLemonProgrammerXx I've edited my answer. $\endgroup$ Dec 22, 2015 at 16:00

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