3
$\begingroup$

HCl is a covalent compound.

I do not understand why HCl cannot exist as an ionic compound with the combination of hydrogen ions and chloride ions. Is it a matter of electronegativity?

I have once wondered if it is possible to obtain HCl crystals from hydrochloric acid, since I was told that HCl(g) ionizes in water to produce hydrogen and chloride ions. However, I have overlooked the fact that the so-called hydrogen ions exist as hydronium ions instead of pure hydrogen ions.

Does anyone have some thoughts on the reason why HCl should be covalent, and whether it is possible to prepare HCl salt?

$\endgroup$
7
  • 2
    $\begingroup$ H+ as a naked ion in a bulk compound is quite definitely impossible. Hydronium is another thing; solid salts of H3O+ are known, but it seems this is not what you want. $\endgroup$ Nov 15, 2023 at 11:47
  • 3
    $\begingroup$ H+ alone is not like other ions = atoms with nonzero net charge. It is just a proton, an elementary particle with size roughly 10^5 smaller than atom, with intensity of electrostatic field roughly 10^10 stronger than e.g. for Li+. Any free proton encountering Cl- ion would form HCl. $\endgroup$
    – Poutnik
    Nov 15, 2023 at 12:04
  • 2
    $\begingroup$ chemistry.stackexchange.com/questions/73254 $\endgroup$
    – andselisk
    Nov 15, 2023 at 12:14
  • 1
    $\begingroup$ The crystalline state «does not mind» if the content of a unit cell are neutral atoms (as in metals), molecules (water as ice, iodine), ions (NaCl), or a mixture of them ($\ce{CuSO4 * 5 H2O}$). Simplified: the crystalline state only is one form (or in case of polymorphism, a couple of forms) where objects are arranged in a somewhat regular pattern that once you know the unit cell, you equally know the properties «here and thousands of unit cells away» (because of periodicity, in either 1D, 2D / tiling, 2.5D, or 3D / space). It still can become a bit more complicated down the road, though. $\endgroup$
    – Buttonwood
    Nov 15, 2023 at 12:58
  • 3
    $\begingroup$ Pure HCl is a gas at room temperature and its melting point is ~-114° . The pure compound is a covalent molecule as a gas and as a solid, apparently. It doesn't exist as an ionic compound. The reasons why it doesn't form an ionic at room temperature is because it is a gas; in the solid state explanations are likely more complex but possibly because "naked" protons are very unfavourable isolated and always react to avid being isolated. $\endgroup$
    – matt_black
    Nov 15, 2023 at 14:41

1 Answer 1

5
$\begingroup$

As has been mentioned, a compound with a bare hydrogen cation cannot exist. Salts derived from strong acids generally contain hydrogen ion covalently bonded to a neutral molecule, so the charged species is multiatomic.

Ironically, the best approximation to an ionically bonded hydrogen cation in a stable compound is not derived from an acid at all. Inverse sodium hydride[1], described as a section of this Wikipedia article, contains natride ion ($\ce{Na^-}$) combined with a protonated $3^6$-adamazane ($\ce{C18H36N4 * H+}$). The hydrogen ion represented by $\ce{H+}$ is not covalently bonded to any one nitrogen atom, but tetrahedrally coordinated to all four of them like the central metal ion in a typical chelate complex. The hydrogen-nitrogen interactions must be a combination of a delocalized covalent bond to the $1s$ orbital of the hydrogen with electrostatic attraction.

Reference

  1. Redko, M. Y.; Vlassa, M.; Jackson, J. E.; Misiolek, A. W.; Huang, R. H.; Dye, J. L. "Inverse Sodium Hydride": A Crystalline Salt that Contains $\ce{H+}$ and $\ce{Na−}$. J. Am. Chem. Soc. 2002, 124, 5928–5929. doi: 10.1021/ja025655+. PMID 12022811.
$\endgroup$
3
  • 2
    $\begingroup$ Note, one of the nice details of mhchem is the following: instead of a lengthy $\ce{C_{6}H_{12}O_{6}}$ subscripts in glucose's Hill formula simply type as $\ce{C6H12O6}$ to yield $\ce{C6H12O6}$. $\endgroup$
    – Buttonwood
    Nov 15, 2023 at 21:04
  • 1
    $\begingroup$ You may be right, but in computer programming class I learned to put extra parentheses to make things doubly clear (e.g. (3×6)+5 versus 3×6+5). Same principle here. $\endgroup$ Nov 15, 2023 at 21:17
  • 1
    $\begingroup$ This requires three additional characters per subscript – in a simple Hill formula, in absence of any charges? My preference is to use curly braces in expressions (test page) of higher complexity. On the other hand, yes, especially in a hurry late arvo, «explicit is better than implicit» (PEP20 in Zen of Python) has its point (an echo of implicit none in Fortran?). Even calculators can be a cause errors here if not used well (see photo in Wikipedia's order of operations, for instance). $\endgroup$
    – Buttonwood
    Nov 16, 2023 at 19:00

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.