# How can FeS2 be the formula of iron sulphide and CaC2 be the formula of calcium carbide?

I know that the valency of iron can either be 2 or 3 and so the formulae of the possible sulphides of iron should be $\ce{FeS}$ and $\ce{Fe2S3}$. But I have recently seen the formula $\ce{FeS2}$ used for iron sulphide. How is this possible.

Similarly the formula for Calcium carbide is supposed to be $\ce{Ca2C}$ and not $\ce{CaC2}$. Why do we write $\ce{CaC2}$ for it?

• Because the 'rules' on valences are not firm rules. On the Fe-S binary phase diagram, $FeS_{2}$ exists as a line compound (as do $Fe_{7}S_{8}$, $Fe_{9}S_{10}$, $Fe_{10}S_{11}$, and $Fe_{11}S_{12}$), alongside $Fe_{1-x}S$. Sep 20, 2016 at 14:42

What you call iron sulphide, in my opinion, is more appropriately referred to as or iron disulphide.

If one were to assign oxidation states to each atom, an appropriate description would be $\ce{Fe^2+}$ and $\ce{S_2^2−}$. This formalism recognizes that the sulfur atoms in pyrite occur in pairs with clear $\ce{S–S}$ bonds. These disulphide units can be viewed as derived from hydrogen disulfide, i.e $\ce{H2S2}$ (similar to hydrogen peroxide)

Take a look at the crystal structure (of pyrite):

In the center of the cell a $\ce{S_2^2−}$ pair is seen in yellow. The $\ce{S}$ atoms have bonds with three $\ce{Fe}$ and one other $\ce{S}$ atom.

Contrast this with, $\ce{FeS}$, i.e Iron (II) Sulphide where the counter anion is what you would typically expect $\ce{S^2-}$. This is reflected in its crystal structure, which is the nickel arsenide structure:

The bonding situation is completely different in these two compounds and this is reflected in their chemical formulas.

Similarly, for calcium carbide, the crystal structure is a tetragonally distorted $\ce{NaCl}$ lattice, comprised of discrete $\ce{Ca^2+}$ and discrete $\ce{C_2^-}$ ions.

Also, bear in mind what @Jon Custer said in his comment:

Because the 'rules' on valences are not firm rules. On the $\ce{Fe-S}$ binary phase diagram, $\ce{FeS2}$ exists as a line compound

Also, I'll briefly mention what he probably meant by a line compound:

Vertical lines such as the one seen here for $\ce{FeS2}$ indicate inter metallic compounds, which have precise chemical composition.

References

2. The crystal structures of pyrite and Iron (II) sulphide both come from their respective wikipedia pages here and here
3. The phase diagram comes from: http://www.minsocam.org/msa/collectors_corner/arc/scn1.htm (which further credits it to Ehlers, 1972, after Kullerud, 1967 The Interpretation of Geological Phase Diagrams, Fig. 217, p. 232)
• Persulphide? Disulphide is ok here but per- not really Sep 20, 2016 at 19:31
• @Mithoron yeah, I think you're right. Edited :) Sep 21, 2016 at 1:32
– Karl
Sep 21, 2016 at 3:17
• @Karl I have overloaded OPs with my answers on multiple occasions as well. There is no problem with it, if someone else has the same question and comes to this website looking for help, they may find a more detailed explanation more useful. At the same time, they may find a simpler explanation more useful, such as that which you have provided - there is no better or worse, in such things. The only purpose is to share knowledge and that is what both of you are doing, so I in fact +1'd both answers. Sep 22, 2016 at 2:45
• Good answer. As someone who makes this stuff in a lab, FeS is rarely stoichiometric FeS. It's usually some kind of Fe(1-x)S, having some Fe3+ in it. Sep 22, 2016 at 8:13

There is no point in doubting the composition of these compounds, the measurements don't lie. ;-)

The point is that S or C (or e.g. P) can not only form bonds to the metal, but also between themselves:

$\ce{FeS2}$ for example does not have $\ce{S^2-}$ anions, but $\ce{S2^2-}$, with a covalent bond in it. Add one $\ce{Fe^2+}$ and everything is fine. Sulphur often makes chains of S atoms, for example in vulcanised rubber, or in many sulfides, which are rarely perfectly stochiometric, as Jon mentioned.

Calcium carbide contains pairs of triple-bonded carbon atoms, which are formally $\ce{C2^2-}$, fitting to one $\ce{Ca^2+}$ ion. If you carefully dissolve it in water, you get ethyne.

With oxygen, covalent bonds in inorganic solids are not so common: It doesn't like to form O-O bonds, and so the peroxides (metallic as well as organic) are rather unstable.