I was reading about the Steric Number Formula here.

There, I came to know that the

Steric Number $N=\frac{V+M \pm I}{2}$ where $V = n(\ce{e-})$, the number of valence electrons of central atom, which is equal to the group number according to the old IUPAC system, $M = n(\text{atom})$, the number of monovalent atoms directly bonded to it, and $I$ is the number of positive or negative charges present (subtract it if the charge is positive, and add it if the charge is negative).

I was trying to apply the formula to $\ce{HNO_3}$ with lewis structure:

HNO3 Lewis Structure

Image from here

From this: $N = \frac{5}{2}$ since there is no net charge on the compound and there is no monovalent atom bonded with $\ce{N}$. $N = \frac{5}{2}$ is not possible.

Maybe, I think the charge means, the charge on the central atom, but then again $\ce{N = \frac{4}{2} = 2}$ which is not right. (It's actually $3$)

Also, my friend says that the steric number is the half of

  1. original valence electrons in central atom +
  2. number of monovalent atoms in the entire compound
  3. minus net charge on the entire compound

in which case, the formula would hold right as $V = 5, M = 1$ (as we now consider $\ce{H}$) and $\ce{I = 0}$ which would be $3$ and correct.

I would be glad if some explained what the "charge" represents. If the "charge" represents the net charge on the central atom, then we'd still need to make Lewis Structure and get the charge, then this technique would be not so easy again?


  • 1
    $\begingroup$ I think the simplest formula to use for steric number is - number of sigma bonds + number of lone pairs. So here you'd have number of sigma bonds = 3, lone pairs = 0, so net = 3 $\endgroup$
    – TRC
    Aug 4 at 7:20
  • $\begingroup$ Hi @TRC, I agree with your method - It's correct. But, I want to figure out, what is wrong with the formula that I've stated above? Thanks $\endgroup$
    – MangoPizza
    Aug 4 at 15:11
  • $\begingroup$ There is no real source for the post you have linked. There probably is not a unique definition on how to use that. The whole concept has very, very little value. If you want an easy to understand an follow model that deals with predicting the molecular structure, use VSEPR. But don't fall into the trap of explaining everything with hybridisation. $\endgroup$ Aug 6 at 20:23
  • $\begingroup$ Hi @Martin-マーチン , I agree hybridisation is older than VSEPR, but I want to know hybridisation well because there are several questions in school exams that ask us the hybridisation of molecules. $\endgroup$
    – MangoPizza
    Aug 7 at 6:00
  • 3
    $\begingroup$ Have you considered the $\ce{-OH}$ bond to be a monovalently attached group to the central N atom? considering $\ce{HNO3}$ as $\ce{N(O)2(OH)}$ might help.. $\endgroup$ Aug 7 at 10:28

You are indirectly using the formula already stated by TRC though I am trying to make your formula accurate and unambiguous (especially because of the efforts you have put into the question).


N=Steric Number of atom X (say).

V=original (ie when an atom is isolated (not experimentally possible for all atoms)) valence electrons on the central atom.

M=Number of atoms directly bonded to X through a single bond.

T=Number of atoms directly bonded to X through a triple bond (use the example of $\ce{HCN}$ to check this).

I=Number of positive charges (formal charge) present on X atom (take I negative if the charge is negative).

Don't consider/take into account 'D=Number of atoms directly bonded to X through a double bond.'

Relating it to $\sigma +$lone pair formula:

$$\sigma=M+T+D$$ $$LP _{\text{(Remaining on X after bonding)}} =\frac{V-M-2D-3T-I}{2}$$

As per Buraian's suggestion, you may apply this formula for all the examples available here. Scroll down and find AXE method in that see the second table last row.

  • $\begingroup$ Hi, so in $\ce{HNO_3}$, the steric number of $\ce{N}$ equals $\frac{5 + 0 - 0 - 1}{2} = 2$? It should be $3$? Thanks $\endgroup$
    – MangoPizza
    Aug 10 at 5:57
  • $\begingroup$ @Jay Thanks for the clarification. I wonder if there's a simpler formula for it. After all, we still have to draw the Lewis Diagram to calculate? $\endgroup$
    – MangoPizza
    Aug 11 at 3:02
  • $\begingroup$ @Jay Do you have any idea if the formula that my friend mentioned is correct or not? I have checked the formula on multiple cases and it seems to work fine. Thanks $\endgroup$
    – MangoPizza
    Aug 11 at 10:24
  • 1
    $\begingroup$ @MangoPizza your friend's formula will go incorrect with atoms bonded through triple bond eg:$\ce{HCN, C2H2, N2}$..... Please delete previous comments to keep space. $\endgroup$
    – Jay
    Aug 11 at 10:36
  • 1
    $\begingroup$ I should have read what you said before carefully, I have deleted my answer and reposted it here $\endgroup$
    – Buraian
    Aug 12 at 7:06

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