In the reaction between mercury and nitric acid, mercury(II) nitrate only forms when concentrated nitric acid is heated up and added to it, if the acid is dilute, mercury(I) nitrate forms.

I understand heat must be added to make the nitric acid properly oxidise with the mercury, but how does the concentration affect the reaction? Is it due to extra water content in dilute nitric acid? How does that affect it?

  • $\begingroup$ My guess is that with concentration nitric acid, the reaction first proceed with the formation of the +1 salt but then oxidises to the +2 salt owing to the excess of acid present as compared to mercury. $\endgroup$ Oct 20 at 3:01
  • $\begingroup$ Oxidation Hg(I)->Hg(II) has higher standard redox potential(data page) than for Hg(0)->Hg(I), so HNO3 may need high enough concentration ( and therefore oxidative power/redox potential) for the former to be preferred. $\endgroup$
    – Poutnik
    Oct 20 at 8:03

Why does mercury in dilute HNO3 give mercurous nitrate while hot concentrated HNO3 produces mercuric nitrate?

“Mercury dissolves in oxidizing acids, producing either Hg${^{2+}}$ or Hg$_2$$^{2+}$, depending on which reagent (mercury or e.g., nitric acid) is in excess.” Ref 1

Wikipedia says “Mercuric nitrate can be reacted with elemental mercury to form mercurous nitrate…(and)…If the solution is boiled or exposed to light, mercury(I) nitrate undergoes a disproportionation reaction yielding elemental mercury and mercury(II) nitrate:” Ref 2

So, dilute acid is simply acid that is insufficient to oxidize all the mercury to mercuric; excess mercury reduces mercuric to mercurous nitrate by disproportionation. The equilibrium seems labile enough. In my simplified imagination, a mercuric ion just picks up a mercury atom and brings mercury nuclei even closer together than in the metal. The Hg-Hg bond length in mercurous nitrate is 254 pm (Ref 3) and 253 pm in mercurous chloride vs 300 pm in the metal. Ref 4

enter image description hereenter image description here

Figure 1. Structure of mercurous nitrate

enter image description here

Figure 2. Structure of mercurous chloride

Mercury will dissolve in nitric acid because the oxidation potential of nitric acid is higher than necessary to dissolve metallic mercury. Electrochemical potentials (Ref 5) give a more quantitative picture of the possibilities (see eq 4):

eq 1: Hg${^{2+}}$ + 2e- = Hgº ...................................0.851V

eq 2: NO$^-_3$ + 3 H$^+$ + 2e- = HNO$_2$ + H$_2$O ..........0.94V

eq 3: 2 Hg${^{2+}}$ + 2e- = Hg$_2$$^{2+}$..............................0.905V

Combining equations 1 and 2 gives equation 4:

eq 4: Hgº + NO$^-_3$ + 3 H$^+$ +2e- = HNO2 + H2O + Hg${^{2+}}$ +2e- ......0.089V

The first product of dissolution of mercury will be mercuric ion. Hg${^{2+}}$ is a powerful oxidant, capable of oxidizing metallic mercury. Combine equations 1 and 3; cancel the electrons on both sides:

eq 5: 2 Hg${^{2+}}$ + Hgº + 2e- = Hg$_2$$^{2+}$ + Hg${^{2+}}$ + 2e- ............0.054V

So mercuric ion in the presence of excess mercury (or insufficient HNO3) is reduced to mercurous, but the driving forces are small. The labile reaction is well known. Excerpts from Ref 6 give equations for “excess acid” and “excess mercury”:

"eq 6: 3 Hgº + 2 NO$^-_3$ + 8 H$^+$ = 3 Hg${^{2+}}$ + 2 NO + 4 H2O (excess acid) and

eq 7: 6 Hgº + 2 NO$^-_3$ + 8 H$^+$ = 3 Hg$_2$$^{2+}$ + 2 NO + 4 H2O (excess mercury).

…the value for the potential of the reaction (Hg${^{2+}}$ + Hgº = Hg$_2$$^{2+}$) corresponds to an equilibrium concentration of Hg$_2$$^{2+}$ at 25ºC only 166 times that of Hg${^{2+}}$; hence the equilibrium is easily reversed…”(by solubility variations) “…reducing agents first reduce mercuric ion to mercurous…and…most reducing agents capable of reducing mercuric ion will, in excess, reduce the mercurous ion to mercury as a second step…”

The binuclear structure of mercurous ion is readily accepted, but it was proved meticulously, in five steps, in Ref 7. Interesting observations include that mercurous compounds are diamagnetic, ruling out Hg$^+$ with an unpaired electron. Electrical conductance of solutions of mercurous salts is consistent with a divalent mercury species. Equilibrium experiments between mercuric ion and mercury metal yield a constant only when the mercurous product is dimeric. The reference also mentions the different oxidations of mercury metal due to excess or deficiency of either reagent. But of course, X-ray diffraction is a clincher.

Interestingly, the ground state electronic configuration of neutral mercury is [Xe].4f14.5d10.6s2 Ref 8. The f electrons are supposed to be poor shielders, so the ionization potential of mercury is a little over a volt higher than might be expected. Whatever is going on (like resonance) appears to make metal-metal bonding very attractive.


Ref 2.https://en.wikipedia.org/wiki/Mercury(I)_nitrate

Ref 3.http://nopr.niscair.res.in/bitstream/123456789/11011/1/IJCA%2050A(2)%20137-140.pdf

Ref 4. https://en.wikipedia.org/wiki/Mercury(I)_chloride

Ref 5: CRC Handbook, p D134, 62 ed.

Ref 6: Reference Book of Inorganic Chemistry, W.M. Latimer and J.H. Hildebrand, 3rd ed, Macmillan Co., 1951, pp 140-141

Ref 7. Advanced Inorganic Chemistry, 2nd ed, F. A. Cotton and G. Wilkinson, Interscience /John Wiley, 1962, p611ff

Ref 8. https://www.webelements.com/mercury/

  • $\begingroup$ Is there anywhere explicitly stated an excess of metallic mercury is required for forming mercurous instead of mercuric nitrate with dilute acid ? (unless I have missed it in the answer) I ask as notes I have read focus on acid dilution, not excess of mercury. $\endgroup$
    – Poutnik
    Oct 23 at 9:08
  • $\begingroup$ @Poutnik: The answer is way too long; it really consists of about 7 answers. The one you are looking for is illustrated in equation 7 from Ref 6. Lines 3 and 4 in my answer use the word "excess" rather than "dilute" but is not clear on whether this is REQUIRED. $\endgroup$ Oct 23 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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