In my textbook it was written that if you put alkali solution into Hg2+ solution you won't get Hg(OH)2 but you get HgO.
And this is because Hg has high electronegativity and this makes polarization of O-H in virtual Hg(OH)2 big enough to make dehydration of Hg(OH)2.

I couldn't understand the reason well. Dehydration seems like a kind of neutralization because it emit H2O and makes salt(HgO), but neutralization between the same thing(in this case Hg(OH)2 )is usually not favored (reaction like 2HCO3-->H2O+CO2+CO32- is not favored.) Why Hg(OH)2 dehydration favored ? Electronegativity of Hg is 2.0(Pauling) and 2.2(H) so it is smaller in Hg actually.

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    – Poutnik
    Sep 25, 2022 at 5:33
  • 2
    $\begingroup$ The core 5d orbitals of Hg are relatively close in energy to the valence oxygen orbitals and repel them. $\endgroup$ Sep 27, 2022 at 1:46
  • $\begingroup$ @KanghunKim how does that explain formation of HgO over Hg(OH)2? In both of them oxygen is binding to Hg. $\endgroup$
    – Harshil
    Sep 27, 2022 at 3:48
  • $\begingroup$ An Hg can barely mind binding to one oxygen(HgO is actually only kinetically stable)- it just cannot withstand binding to TWO oxygens. $\endgroup$ Sep 27, 2022 at 14:30
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    $\begingroup$ Suppose that you get an upset stomach to apples. Eat one apple and you're barely trying not to throw it up(but it gets wasted as diarrhea at some time). This is HgO. Eat two apples and you throw half up(Hg(OH)2 -> HgO + H2O), later you excrete the other half as diarrhea. Not a metaphor that agrees with the table, but you get the idea. $\endgroup$ Sep 27, 2022 at 14:33

1 Answer 1


The filled core 5d orbitals of mercury(II), were they valence, would be antibonding with respect to the filled valence 2s and 2p orbitals of oxygen(-II).

For the lighter elements this effect, which technically exists in every species with filled core orbitals, could be ignored- however it is not that straightforward for the heavier elements, especially the later-period post-transition metals, due to relativistic effects that stabilise the valence s orbital and destabilise the core d orbitals.

The only ways to remedy it is to either form strong covalent bonds (using the low-energy unfilled 6s orbital of mercury(II)) to increase the bonds with the ligands, or to form dispersive interactions (using the relatively high-energy filled 5d orbitals of mercury(II)) with the ligands to decrease the antibonds with the ligands. The former requires high-energy ligand HOMO's, while the latter requires diffuse ligand HOMO's- oxygen(-II) has neither. (Sulfur(-II) has both, which explains why HgS is one of the most insoluble substances in water)

In short, HgO is unstable due to the "unfortunate" combination of core-valence repulsion (effected by relativistic effects) and poor bonding overlap. Of course, similar effects occur in ZnO and CdO as well, but HgO is the most unstable of these, in fact being the only thermodynamically unstable of the three, due to the very fact that relativistic effects get stronger when the element gets heavier.

All this is about how mercury(II) literally wants to "expel" oxygen(-II) when forced to eat it. Eat one oxygen(-II) and mercury(II) tries to keep it down(eventually expelling it as diarrhea)- eat two oxygen(-II)'s and mercury(II) ends up immediately throwing one of them up(no comments shall be made on the fate of the other one). This explains the kinetic (never mind the thermodynamic) instability of Hg(OH)2.

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    $\begingroup$ Could you explain what you mean by the first paragraph? Molecular orbitals are bonding or antibonding, not atomic ones. Do you mean higher in energy? If so that's not the same as antibonding. $\endgroup$
    – Ian Bush
    Oct 1, 2022 at 15:33
  • $\begingroup$ "Were they valence", some of the Hg 5d orbitals would end up being the predominant parts of filled antibonding orbitals. $\endgroup$ Oct 11, 2022 at 1:36

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