Modern inorganic chemistry

198 GROUP IV

Tin

TIN(ll) CHLORIDE

This chloride is prepared by dissolving tin in concentrated hydro-
chloric acid; on cooling, the solution deposits crystals of hydrated
tin(II) chloride, SnCl 2. 2H 2 O ('tin salt'). The anhydrous chloride is
prepared by heating tin in a current of hydrogen chloride:

Sn

Sn + 2HC1 -> + H 2

Cl Cl

The hydrated salt is decomposed by heat:

SnCl 2 .2H 2 O ^ Sn(OH)Cl + HC1 + H 2 O

This reaction proceeds slowly in aqueous solution, so that the

basic salt, Sn(OH)CL is slowly precipitated. Addition of excess

hydrochloric acid gives the acids of formulae HSnCl 3 and H 2 SnCl 4.

Salts of these acids containing the ions SnClJ and SnCl^ (chloro-

stannates(II)) are known.

A solution of tin(II) chloride is a reducing agent. Hence it reduces :

Sn4+ (aq) 4- 2e~ -> Sn^2 + (aq): E^ - 0.15V

mercury(II) chloride, first to the white insoluble mercury(I) chloride

and then, if in excess, to mercury:

2HgCl 2 4- SnCl 2 -> SnCl 4 + Hg 2 Cl 2 |

white

Hg 2 Cl 2 + SnCU -> 2Hg| + SnCl 4

It reduces iron(IIl) to iron(IF) salts:

2Fe3+ 4- Sn2+ -» Sn^4 + + 2Fe^2 +

This provides a method of estimating an iron(III) salt. After
reduction the iron(II) salt is titrated with manganate(VH) solution.
It reduces nitrobenzene (in the presence of hydrochloric acid) to
phenylammonium hydrochloride:

C 6 H 5 NO, + 7HC1 4- BSnCU -> C 6 H 5 NH, .HC1 + 2H.O
4- 3SnCl 4

It reduces phenyl diazonium chloride to phenylhydrazine hydro-
chloride :

[C 6 H 5. N,]C1 + 4HC1 + 2SnCU -^ C 6 H 5 NH. NH,. HC1

4- 2SnCl 4