= Solution
\b[<Iron> in nonoxidizing <hydrochloric acid> principally gives an <iron(II)> solution; <tin> in oxidizing <aqua regia> principally gives a <tin(IV)> solution.] These are the two preparations paired most directly by “respectively”. Their contrast is clearer when the other acid treatment of each metal is considered as well. “Nitrohydrochloric acid” is the historical name for <aqua regia>, containing nitric and hydrochloric acids.
For <iron> in dilute <hydrochloric acid>,
$$
\boxed{\mathrm{Fe}+2\mathrm{HCl}\longrightarrow\mathrm{FeCl}_2+\mathrm H_2.}
$$
The fresh solution is typically pale green when sufficiently concentrated. <Hydroxide> gives a pale green precipitate of <iron(II) hydroxide>, which darkens and turns brown on exposure to air as oxidation proceeds. <Ammonia> can similarly precipitate a ferrous hydroxide rather than forming a stable soluble excess-ammonia complex under the usual test conditions. <Ferricyanide> gives a deep blue product; the modern structural account of this blue material should not be replaced by an overly literal formula inferred solely from the reagent names. Ferrous solution also reduces a suitable oxidant such as acidic <permanganate>:
$$
\mathrm{MnO}_4^-+5\mathrm{Fe}^{2+}+8\mathrm H^+\longrightarrow\mathrm{Mn}^{2+}+5\mathrm{Fe}^{3+}+4\mathrm H_2\mathrm O.
$$
If instead oxidizing acid treatment produces <iron(III)>, the solution exhibits ferric reactions: <hydroxide> or <Ammonia> gives red-brown <iron(III) hydroxide>; <thiocyanate> gives a strong red colour; and <ferrocyanide> gives <Prussian blue>. Chloride complex formation affects colour and some equilibria, so “yellow” is not a universal iron(III) identity test. Oxidizing <aqua regia> supplies this ferric comparison rather than merely generating more of the same ferrous salt.
<Tin> in ordinary <hydrochloric acid> gives a <tin(II)> solution with hydrogen evolution:
$$
\mathrm{Sn}+2\mathrm{HCl}\longrightarrow\mathrm{SnCl}_2+\mathrm H_2.
$$
Its characteristic property is reducing power. With <mercury(II) chloride>, controlled addition can first form white <mercury(I) chloride> and then, with further reductant, grey metallic <Mercury>:
$$
2\mathrm{HgCl}_2+\mathrm{SnCl}_2\longrightarrow\mathrm{Hg}_2\mathrm{Cl}_2\downarrow+\mathrm{SnCl}_4,
$$
$$
\mathrm{Hg}_2\mathrm{Cl}_2+\mathrm{SnCl}_2\longrightarrow2\mathrm{Hg}\downarrow+\mathrm{SnCl}_4.
$$
The chloride formulae are useful net bookkeeping, not a complete statement of all aqueous chloro-complexes.
In <aqua regia>, <tin> is oxidized predominantly to <tin(IV)>. A representative balance, choosing <nitrogen monoxide> as the nitrate-reduction product, is
$$
\boxed{3\mathrm{Sn}+4\mathrm{HNO}_3+12\mathrm{HCl}\longrightarrow3\mathrm{SnCl}_4+4\mathrm{NO}+8\mathrm H_2\mathrm O.}
$$
Actual gas products depend on the acid conditions. Sufficient <chloride> and acidity retain tin in solution; dilution can cause hydrolysis and precipitation of hydrated <tin(IV) oxide>. This differs from the simple expectation that every dilution leaves unchanged molecular $\mathrm{SnCl}_4$ in water. <Hydrogen sulfide> in suitable acidic conditions gives yellow <tin(IV) sulfide>, compared with brown <tin(II) sulfide> from a stannous solution. <Tin(IV)> does not reproduce the characteristic reducing reaction of fresh <tin(II)> with mercuric chloride; reduction to tin(II) is needed first. Oxidation during storage can therefore change the reactions of a nominally stannous preparation.
Back to article page
