Isostructural chemical compounds have similar chemical structures.

Chemical Structure

A chemical structure determination includes a chemist's specifying the molecular geometry and, when feasible and necessary, the electronic structure of the target molecule or other solid. Molecular geometry refers to the spatial arrangement of atoms in a molecule and the chemical bonds that hold the atoms together, and can be represented using structural formulae and by molecular models;[citation needed] complete electronic structure descriptions include specifying the occupation of a molecule's molecular orbitals.

My Doubt

Given that, while practicing, I encountered a question asking whether $\ce{NO3-}$ and $\ce{SO3}$ are isostructural or not. If we go as per the geometry, they are isostructural ($\ce{sp^2}$ hybridization). But if we consider the electronic structure, the former has bond order 4/3, the latter has 2 (assuming $\ce{S=O}$), and so may be, they are not isostructural. (In addition, they also have different total number of electrons.) This (that they are not isostructural) is the answer.

So, this question. Is the term defined or is it used without any strict definition?

  • $\begingroup$ Bonding is identical between those two. $\endgroup$
    – Mithoron
    May 4, 2017 at 16:31
  • $\begingroup$ I did read somewhere that in $\ce{SO_3}$, the bond order is 4/3, but am unsure of it, as well as unable to find the source. (Does formal charge +2 on S indicates this? Moreover, the diagrams I find over the internet containg three S=O bonds.) Therefore, I did mention "assuming S=O". The answer is given as 'not isostructural'. (The question intends to ask: Is the answer wrong? Can the definitions be worded better? Is there anything more to it?) $\endgroup$
    – digikar
    May 4, 2017 at 17:13
  • $\begingroup$ Ad question you mention you have encountered.. this type of questions not focused on the problem matter, but rather on vague terminology, and are surely made by evil professors. In the answer I would describe all my train of thought. $\endgroup$
    – mykhal
    May 5, 2017 at 9:34

1 Answer 1


Two molecules are isostructural if they have a similar number of atoms arranged in a similar structure (spatial coordinates). In your example, both $\ce{NO3^-}$ and $\ce{SO3}$ have same number of atoms, and they have similar spatial arrangement as well. So they are isostructural. Sometimes using hybridization of the central atom works in finding out, but not always. For example $\ce{SCl3}$ and $\ce{BF3}$ are isostructural, even though the former is $\text{sp}^3\text{d}$, while the latter is $\text{sp}^2$.

  • $\begingroup$ Unable to 'simply' google the structure for SCl3; do the lone pairs occupy the axial orbitals? If yes, then is there any explanation why this is against the [atleast-as-I-know-it] usual configuration of lone pairs occupying equatorial position? $\endgroup$
    – digikar
    May 4, 2017 at 14:35
  • $\begingroup$ Lone pairs occupy axial positions because they're they face least repulsions. $\endgroup$ May 4, 2017 at 15:44
  • $\begingroup$ My coursebook gives the same reason for the structure of ClF3, just that the lone pairs are equatorial (because they face less repulsion)! How do these two structures differ then? $\endgroup$
    – digikar
    May 4, 2017 at 15:54
  • $\begingroup$ $\ce{ClF3}$ isnt isostructural with $\ce{SCl3}$. Central atoms have different number of electrons. $\endgroup$ May 4, 2017 at 15:57
  • 4
    $\begingroup$ This is incorrect. (1) $\ce{SCl3}$ does not exist as far as I can tell. (2) In a structure with 3 bp and 2 lp, lone pairs are best placed equatorial, because the total energy is dictated not just by lp-lp repulsions but also lp-bp repulsions. In the T-shaped structure (i.e. both lp equatorial) the lp-lp repulsion is slightly larger, but lp-bp repulsions are minimised. $\endgroup$ May 5, 2017 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.