Why does mercury form polycations?

Question

For example, mercury (I) is $\ce{Hg2^2+}$ and not $\ce{Hg+}$. What causes the stability in covalently bonded $\ce{Hg}$ ions?

Answer 1

I don't have an excellent understanding of the topic, but I will try to write something up.

Polycations are found with the Hg(I) oxidation state. The mercury cation, $\ce{Hg+}$, has a $\mathrm{5d^{10}6s^1}$ configuration. The mercurous cation $\ce{Hg2^2+}$ can then be considered to simply be formed by overlap of the two $\mathrm{6s}$ orbitals on two separate $\ce{Hg+}$ ions. In this sense, the bonding is qualitatively very similar to that in $\ce{H2}$. Note that the $\mathrm{5d^{10}6s^0}$ Hg(II) cation has no chance of forming such covalent bonds.

Generally, going down a group, orbitals become larger, more diffuse, and the overlap between them decreases. As a consequence, $\sigma$-bonding tends to become weaker going down the group. A simple example is provided by the alkali metal dimers:

$$\begin{array}{c|ccc} \text{Element} & \ce{H} & \ce{Li} & \ce{Na} \\ \hline D_\mathrm{0}(\ce{E-E})\text{ / }\pu{kJ mol-1} & 432 & 99 & 71 \\ \hline \text{Reference} & 1 & 2 & 3 \end{array}$$

These dissociation energies are extremely small. For the analogous $\ce{Hg2^2+}$ cation, the situation is not quite as bleak. The Group 12 metals in general have more low-lying and contracted ns orbitals than the corresponding alkali metals, due to the presence of ten poorly-shielding (n-1)d electrons. However, given that zinc does not form similar cations, it is odd that mercury with its larger 6s orbitals should do so.

The most obvious explanation is simply that Zn does not commonly exhibit the +1 oxidation state. The existence of the +1 oxidation state for Hg is directly related to the relativistic stabilisation of the 6s orbital, which increases the first two ionisation energies⁴ of mercury.

$$\begin{array}{c|ccc} \text{Element} & \ce{Zn} & \ce{Cd} & \ce{Hg} \\ \hline I_1\text{ / eV} & 9.393 & 8.992 & 10.44 \\ \hline I_2\text{ / eV} & 17.96 & 16.90 & 18.76 \\ \end{array}$$

However, the question still remains as to why Hg(I) chooses to form polycations instead of existing simply as $\ce{Hg+}$. In fact, relativistic effects have also been implicated in this for a long time: Pyykkö and Desclaux wrote in 1979 that⁵

The relativistic contraction of the half-filled valence s AO makes $\ce{Au2}$ much more strongly bound than $\ce{Ag2}$. Inasmuch as $\ce{Hg2^2+}$ and $\ce{Cd2^2+}$ are isoelectronic with these two molecules, the remarkable stability of the mercurous ion [compared to the Cd analogue] can also be understood.

Therefore, this may most simply be explained as the relativistic contraction of the 6s AOs leading to stronger-than-expected overlap to form a covalent bond. As suggested by the above paragraph, similar effects are seen in the neutral dimers of Group 11 metals:⁶

$$\begin{array}{c|ccc} \text{Element} & \ce{Cu} & \ce{Ag} & \ce{Au} \\ \hline D_\mathrm{0}(\ce{E-E})\text{ / eV} & 1.95 & 1.65 & 2.29 \\ \end{array}$$

The decrease going from Cu to Ag is expected on the basis of weakened overlap, and the dissociation energy of $\ce{Au2}$ would be even lower in the absence of relativistic effects: calculations indicate that relativity contributes roughly $\pu{1 eV}$ to the dissociation energy of $\ce{Au2}$.⁶ Presumably, the effect in $\ce{Hg2^2+}$ is of a similar significance.

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References

1. Frenking, G.; Bickelhaupt, F. M. The EDA Perspective of Chemical Bonding. In The Chemical Bond: Fundamental Aspects of Chemical Bonding; Frenking, G., Shaik, S., Eds.; Wiley: Weinheim, Germany, 2014; p 128.
2. Velasco, R.; Ottinger, Ch.; Zare, R. N. Dissociation Energy of Li2 from Laser‐Excited Fluorescence. J. Chem. Phys. 1969, 51 (12), 5522–5532. DOI: 10.1063/1.1671979. Free version provided by the Zare group here.
3. Jones, K. M.; Maleki, S.; Bize, S.; Lett, P. D.; Williams, C. J.; Richling, H.; Knöckel, H.; Tiemann, E.; Wang, H.; Gould, P. L.; Stwalley, W. C. Direct measurement of the ground-state dissociation energy of Na2. Phys. Rev. A 54, R1006(R). DOI: 10.1103/PhysRevA.54.R1006.
4. Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford UP: Oxford, U.K., 2014; pp 836–7.
5. Pyykko, P.; Desclaux, J. P. Relativity and the periodic system of elements. Acc. Chem. Res. 1979, 12 (8), 276–281. DOI: 10.1021/ar50140a002.
6. Pitzer, K. S. Relativistic effects on chemical properties. Acc. Chem. Res. 1979, 12 (8), 271–276. DOI: 10.1021/ar50140a001.

Answer 2

Only a few $\ce{Hg(I)}$ compounds (also called mercurous compounds) are known which include mercury(I) halides $\ce{Hg2X2}$ and mercury(I) nitrate $\ce{Hg2(NO3)2.2H2O}$ . They contain the ion $\ce{Hg2^2+}$ actually $\ce{(Hg -Hg)^2+}$ and not $\ce{Hg+}$. The two $\ce{Hg}$ atoms are bonded together using the 6s orbital. Zinc and cadmium also forms stable dinuclear ions but are unstable and has been only detected spectroscopically in melts of $\ce{Zn/ZnCl2}$ and $\ce{Cd/CdCl2}$. They are made by reducing mercury(II) salts with metal. Mercury(I) nitrate can be prepared by dissolving $\ce{Hg}$ in $\ce{HNO3}$.

Their standard reduction potential are so close that oxidizing agent like $\ce{HNO3}$ converts $\ce{Hg}$ to $\ce{Hg^2+}$ rather than $\ce{Hg(I)}$.

$\ce{Hg^2+ -> Hg}$ $(E^0 = 0.85\,\rm V)$

$\ce{Hg2^2+ -> Hg}$ $(E^0 = 0.79\,\rm V)$

This shows that $\ce{Hg2^2+}$ is stable to disproportionation by a small margin under standard condition.

$\ce{Hg2^2+ <=> Hg + Hg^2+}$ $(E^0 = 0.13\,\rm V)$

The equilibrium constant can be calculated:-

$$E^0 = \frac{RT}{nF}\cdot\ln K$$

$$K = \frac{[\ce{Hg2^2+}]}{[\ce{Hg^2+}]}=170$$

Thus solution of mercury(I) compounds contains one $\ce{Hg^2+}$ for every $\ce{170}$ $\ce{Hg2^2+}$ and the equilibrium is finely balanced. If any reagent is added to the solution, disproportionation occur.

There are many evidence of mercury(I) having the structure $\ce{(Hg-Hg)^2+}$ like:-

• X-ray diffraction

  Crystal structure of various mercury(I) compounds have been determined by X- ray diffraction. For ex- It was assumed that $\ce{HgCl}$ consist of $\ce{Hg+}$ and $\ce{Cl-}$ but X-ray diffraction showed $\ce{Cl-Hg-Hg-Cl}$ arrangement. Other compound was also proved of having the $\ce{Hg-Hg}$ bond.

• equilibrium constant

Consider the reaction:-

$$\ce{Hg(NO3)2 + Hg -> 2HgNO3}$$

$$\ce{Hg^2+ + Hg -> 2Hg+}$$

Then by law of mass action

$$\frac{\ce{[Hg+]^2}}{\ce{[Hg^2+]}}={\rm constant}$$

Experiment have shown this to be wrong if however:

$$\ce{Hg^2+ + Hg -> Hg2^2+}$$ then $$\frac{\ce{[Hg2]^2+}}{\ce{[Hg^2+]}}={\rm constant}$$

This shows $\ce{Hg2^2+}$ contains$\ce{(Hg-Hg)^2+}$.

• concentration cell

For the cell below, $E^0$ was found to be $0.029\,\rm V$ at $25°\rm C$

Hg/mercury(I) nitrate (0.005M)//mercury(I) nitrate (0.05M) /Hg

$$E = \frac{2.303RT}{nF}\cdot\ln \frac{c_2}{c_1}$$

Putting values,

$$0.029 = \frac{0.059}{n}\cdot\ln \frac{0.05}{0.005}$$

$$n = 2$$ where n = no. of charges on ion. This confirms $\ce{Hg2^2+}$

• Raman spectra

Spectral lines correspond to $\ce{Hg-Hg}$ bond.

• magnetic properties

All mercury(I) compounds are diamagnetic as all electrons are paired in $\ce{(Hg-Hg)^2+}$.

• cryoscopic measurements

Observed depression fits for all mercury(I) compounds containing $\ce{(Hg-Hg)^2+}$ ion.

Also, compounds like $\ce{Hg3(AlCl4)2}$ and $\ce{Hg4(AsF6)2}$ are also discovered and they contains cation with three and four mercury atoms bonded linearly.

Answer 3

All ready answered, but I will give the distilled version: Zn and Cd are most commonly found in the +2 OX State (d10) and can not metal-metal bond in this state. +1 OX state is possible for these elements but is rare due to there low ionization energies, but they do form metal-metal bonds in the +1 OX state.

Hg has relatively high ionization energy due to it having a higher effective nuclear charge from pore shedding from the f electrons (lanthanide contraction). The high ionization energy makes the +1 OX state more stable for Hg, where it is capable of M-M bonding.