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I have basically written the entire name except for the oxidation state of $\ce{Co}$.

Mercury tetrathiocyanatocobaltate( )

to know what is the oxidation state of cobalt, I must know the oxidation state of the mercury cation. But mercury can have $+1$ and $+2$ oxidation states. How do I know which one is it in this compound?

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  • $\begingroup$ You don't (that is, unless you know the crystal structure and stuff). In fact it is +2. $\endgroup$ Dec 8 '17 at 13:53
  • $\begingroup$ So, I cannot write the name? It's a question from my homework. $\endgroup$
    – Raknos13
    Dec 8 '17 at 13:56
  • $\begingroup$ Then write (II) and be done with it. I don't see how you were supposed to figure that out. $\endgroup$ Dec 8 '17 at 14:20
  • $\begingroup$ It's Mercury(II) tetrathiocyanatocobaltate(II). $\endgroup$ Dec 8 '17 at 15:43
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    $\begingroup$ It couldn't be Hg (I) Co (III) compound. There would need to be Hg2 group and cobalt is too strong oxidant for it to be present. $\endgroup$
    – Mithoron
    Dec 8 '17 at 18:03
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TL;DR: There is no inner- or outer coordination spheres in this complex. Both cobalt(II) and mercury(II) have nearly ideal tetrahedral coordination environment with $4$ $\ce{N}$ and $4$ $\ce{S}$ atoms, respectively, hence the proper name would be cobalt(II) mercury(II) tetrathiocyanate. Mercury(I) readily undergoes disproportionation, so assuming mercury(II) is the safest option.


Actually, denoting an inner coordination sphere is not correct for this compound. First crystal structure [1] has been assigned to cobalt(II) tetrathiocyanatomercurate(II) $\ce{Co[Hg(SCN)4]}$:

It appears reasonably certain that the $\ce{S}$ atom is attached to the $\ce{Hg}$ in tetrahedral co-ordination, with $\ce{S-Hg-S}$ angles of 120° and 104°.

enter image description here

But two decades later the structure has been refined again, by the group including the author of the original publication. This time the geometry has been established more precisely (ICSD#36062), and it turned out that there is an infinite network of $\ce{Co^2+}$ and $\ce{Hg^2+}$ cations cross-linked via thiocyanate ligands, and the new suggested name was cobalt mercury thiocyanate $\ce{Co(SCN)4Hg}$ [2]:

enter image description here

The combination of two tetrahedrally coordinated atoms, $\ce{Hg}$ and $\ce{Co}$, has produced a most unusual arrangement in which the $\ce{Hg}$ and $\ce{Co}$ atoms are held apart by four spirals, each containing $4$ $\ce{SCN}$ bridges, which are interlinked so that any one $\ce{SCN}$ bridge takes part in eight spirals.

[...] Each such spiral is a spring holding the $\ce{Hg}$ and $\ce{Co}$ atoms apart and straining the bonds in the process. It is almost certainly this strain which flattens the tetrahedral coordination round $\ce{Hg}$ and $\ce{Co}$ in the $c$ direction. If the arrangement could be reproduced mechanically it would probably provide the ideal spring mattress!

References

  1. Jeffery, J. W. Nature 1947, 159 (4044), 610. DOI: 10.1038/159610a0.
  2. Jeffery, J. W.; Rose, K. M. Acta Cryst B 1968, 24 (5), 653–662. DOI: 10.1107/S0567740868002980.
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There are two little clues you can use.

First, mercury(I) isn’t actually present as $\ce{Hg+}$; the cationic species is actually $\ce{Hg2^2+}$ with a $\ce{Hg-Hg}$ bond. Thus, if this were a mercury(I) compound, the sum formula would be $\ce{Hg2[Co(SCN)4]2}$.

This obviously only helps you if the correct formula was supplied as part of the question but the second clue is independent:

Look up a table of standard electrode potentials. There are two redox reactions that can occur, they are reproduced below:

$$\begin{align} \ce{2 Hg^2+ + 2 e- &<=> Hg2^2+}&&&E^0 = \pu{+0.91V}\\[1.0em] \ce{Co^3+ + e- &<=> Co^2+}&&&E^0=\pu{+1.82V}\\ \end{align}$$

The half-cell with a higher standard electrode potential will oxidise a half-cell with a lower standard electrode potential which means that, in solution, cobalt(III) and mercury(I) should react as follows:

$$\ce{2Co^3+ + Hg2^2+ -> 2Co^2+ + 2 Hg^2+}$$

As $$E^0_\text{cell} = E^0_\text{red} - E^0_\text{ox}$$ we get a standard cell potential of $$E^0_\text{cell} = \pu{+1.82V} - (\pu{+0.91V}) = \pu{+0.91V}$$ which is clearly positive, indicating a spontaneous reaction.

Of course, this is using aquacomplexes of both cations rather than whatever structure they would have in the complex you are supposed to name. Different chemical environments will modify a reaction’s electrode potential. However note that the uncorrected potential difference is so large – $\pu{+0.91V}$ – that it will be very hard to stabilise these otherwise incompatible oxidation states in solution.

On the other hand, a compound made up of mercury(II) and cobalt(II) is redox-stable.

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