The steric number is equal to the number of $\sigma$-bonds + the number of lone pairs of electrons on the central atom. It gives us the number of hybridised orbitals.

It is pretty straight-forward to calculate it, but the problem here is that one must always draw the Lewis Structure before one can actually get to calculating the steric number, and then the number and types of hybrid orbitals. Even that is quite simple for a smaller compound, even like XeF$_6$, but when it comes to complex hydrocarbons, it's a little difficult.

My question is that is there any well-known (or not so well-known, but working) shortcut to doing this, so as to save time? It would be great if anyone could share their ideas and help me out.

Thanks in advance.

  • 1
    $\begingroup$ I’ld like to point out that your first and second sentences contradict themselves. Take for example $\ce{SF4}$. We have two 2-electron-2-centre (2e2c) $\ce{S-F}$ $\sigma$ bonds and one (also 4e3c $\ce{F\bond{...}S\bond{...}F}$ bond. We also have one additional lone pair. The 4e3c bond is also $\sigma$-symmetric. Therefore, we have three or four $\sigma$ bonds — depending on how you count — and thus a steric number of four or five. However, sulphur is $\mathrm{sp^2}$ hybridised, i.e. only three orbitals take part in hybridisation. $\endgroup$
    – Jan
    Commented Jun 18, 2017 at 17:04
  • $\begingroup$ This question was posted before I had learned about the concept of banana bonds, and other special bonds, in which multiple centres are present (such as 4e3c, and 2e3c). Hence, I assumed that in all compounds, steric number equals the number of hybridised orbitals. $\endgroup$
    – Abhigyan
    Commented Jun 19, 2017 at 11:54
  • 1
    $\begingroup$ Interestingly, the steering number is already a gross oversimplification. It always assumes uniform hybridisation patterns, which is likely not true. $\endgroup$ Commented Aug 29, 2023 at 20:47

4 Answers 4


Short answer: no.

The steric number is a property of an atom, not a compound. You need to know what an atom connected to a given atom to know its steric number. For simple compounds, you can usually determine these connections because the formula suggests a central atom and surrounding groups. For hydrocarbons and other organic compounds, you need to consider isomerism. Given the capability of carbon to form complicated bonding patterns, even simple formulas can produce a fair number of isomers with different bonding patterns and steric numbers.

Let's look at some examples.


This formula corresponds to two compounds with the structures shown:

enter image description here

In this case, both compounds have all four carbon atoms with steric number of 4.

it is not always true that a set of hydrocarbon isomers will always have the same steric number for all carbon atoms or even the same set of steric numbers.


This formula corresponds to six isomers:

enter image description here

Note that four of these structures have two carbon atoms with steric number 4 and two carbon atoms with steric number 3. The other two have all four carbon atoms with steric number 4.

Any method to calculate steric number for carbon atoms in an organic compound using just the formula will fail. You must examine the structure.


All right … I found myself a shortcut, and would like to share this in case it is useful for others. However, this formula is applicable to molecules with only one central atom.

Here is how it goes:

  1. Find $N=\frac{V+M \pm I}{2}$, 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). This $N$ is the Steric Number.

  2. Now, find the number of Bond Pairs ($BP$) of electrons, which is equal to the number of atoms surrounding the central atom. However, this is a little difficult for a species like $\ce{H3BO3}$, which is actually $\ce{B(OH)3}$, when written according the IUPAC method of writing the less electronegative atoms first.

  3. Next, find the number of Lone Pairs ($LP$) of electrons, which is equal to $N-BP$.

  4. Now, draw the structure of the atom, using the central atom, drawing the skeleton of the atom using the steric number, and then assigning the Bond Pairs and Lone pairs to the respective bonds/atoms.

That's for an atom with a single central atom.

Now, for a Hydrocarbon, albeit it is not possible to get the shape directly from the molecular formula, it is possible to find its structure and hybridisation if and only if the basic structure of the atom is provided.

  1. For a compound with a single $\sigma$ bond between Carbon atoms, the hybridisation is $sp^3$
  2. For one $\sigma$ and one $\pi$ bond, it is $sp^2$ hybridised, and
  3. For one $\sigma$ and two $\pi$ bonds, it is $sp$ hybridised.

So, essentially, there is no formula for hydrocarbons, but there is a formula for smaller compounds, with a single central atom only.


Here is an alternate method for one central atom molecules which I found in JD LEE adapted by Sudarshan Guha ed 2021.

Motivation: This method assumes that all the corner atoms have complete octet. Number of $\sigma$ bonds $=$ Number of corner atoms.


  1. Define $n$ to be: total valence shell electrons of all atoms + number of negative charge(if any)- number of positive charge (if any).

  2. Write $n$ in the following form: $n = 8j+k $ with $k<8$ and $j>0$ [Remark: this is application of Euclid's division algorithm]

It is now useful to define $Q= 8j$ and $R= k$, it turns out that $Q$ is number of $\sigma$ bond pairs and $R$ is number of non bonding electrons on central atom.

Important point: To make this method work, we have to consider the valence electron of hydrogen as $7$ (because this method assumes that all the corner atoms have complete octet). Otherwise your calculations will go wrong.

  1. The steric number number is given as: $S = Q + \frac{R}{2}$

Hence, from $n$ , we find the non bonding pairs, bond pairs and finally the steric number.

Example calculation:

Consider methane $\ce{CH4}$, due to absence of charges, $n$ is given as the total valence electrons i.e: $4+ 4 \cdot 7=36= 8 \cdot 4$. We see that $Q=4$ , $R=0$ and $S=4$ hence four bond pairs, no lone pairs and tetrahedral geometry.


Show, using the above method, that the steric number of $XeF_5^+$ is $6$

Figure out how to extend this method to the molecule $\ce{HNO3}$


I have been teaching my students the same shortcut by AbhigyanC, but expressed a bit differently. Using the same symbols:

LP = (V-M-I)/2


LP = No. of lone pairs on central atom

V = No. valence electrons brought by central atom

M = No. of hydrogens or halogens bonded to the central atom

I = Charge of the species

It is a rearrangement of the formal charge formula, and uses the following additional observations:

  • Hydrogen always makes single bonds
  • Halogens make single bonds when they are peripheral (at least good enough for General Chemistry)
  • The net charge can be assigned to the central atom because the allowed peripheral atoms do not take on nonzero formal charges

Of course the steric number is: N = M + LP

This shortcut allows me (and any student who adopts it) to simply look at a formula and come up with the VSEPR prediction with a simple mental calculation!


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