This question already has an answer here:

Chlorine can be $sp^3d^3$ hybridized.

If so, it can form $\ce{ClH_7}$ and then chlorine, being more electronegative, will have (-7) as its oxidation number. But we know that the oxidation number of $\ce{Cl}$ lies between (-1) and (+7).



marked as duplicate by ron, jerepierre, bon, Martin - マーチン Jun 24 '15 at 18:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Your question is based on a skewed premise, as gsurfer04 already pointed out. Rather than invoking d-orbitals, which are usually rather far away in energy from the corresponding p-orbitals (in fact, usually higher in energy than the next shell’s s-orbital, so for chlorine 3p → 4s → 3d), these hypervalent halogen compounds are considered to derive from multi-centre bonds.

For details in the bonding mechanism, I am going to forward you to this excellent answer.

I haven’t seen any multi-hydrogen chloride yet. In fact, when discussing interhalogens like $\ce{ClF3}$, our professor in the introductory chemistry course went as far as saying even $\ce{H3Cl}$ is impossible. I’ve forgotten the exact reasoning (and unfortunately it’s not explicitly stated in the web material any more) but I believe it to be along the lines of hydrogen’s s-orbitals not being able to form the outside parts of a four-electron three-centre bond.


First of all, $d$ orbitals play no part in hypervalency.

If you look at existing hypervalent molecules, you'll see that the central atom is always less electronegative. As H is less electronegative than $\ce{Cl}$, $\ce{ClH7}$ will be very unstable.

  • 6
    $\begingroup$ The central atom isn't always less electron negative pubs.acs.org/doi/abs/10.1021/ja00293a020 and pubs.acs.org/doi/pdf/10.1021/ja961081v $\endgroup$ – DavePhD Jun 10 '15 at 15:04
  • $\begingroup$ @DavePhD Fascinating find on the triprotonated methane! Thanks for the link. $\endgroup$ – Nicolau Saker Neto Jun 10 '15 at 17:07
  • $\begingroup$ @DavePhD Although the central atom isn't necessarily less electronegative, in order for the molecule to have 3c-4e bonds, theoretically, the peripheral atoms should be electronegative atoms to withdraw electron density away from the central atom. After all, two of the electrons of the 4 electrons in the 3c-4e bond becomes a lone pair on the peripheral atoms. $\endgroup$ – Tan Yong Boon Nov 26 '17 at 3:13

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