10.3 Reactions of halide ions

. The halide ions lose/give away electrons and become halogen molecules They themselves become oxidised, but they reduce other halogens

.It increases down the group . The size of ions increases down the group as there are more shells . The outer shell gets further from the nucleus, so atomic radii increases . So the attraction between the outer electron and nucleus is less . So it can lose electrons more easily

Solid sodium halides react with concentrated sulphuric acid. The products are different and reflect the reducing power of the halides

. Drops of sulphuric acid are added to sodium chloride. Steamy fumes of hydrogen chloride are seen . Solid product is sodium hydrogensulfate.

sodium chloride + sulfuric acid —> sodium hydrogensulfate + Hydrogen chloride +1 -1 +1,+6,-2 NaCl (s) + H2SO4 (l) –> +1,+1,+6,-2 +1,-1 NaHSO4(s) + HCL(g)

. No oxidation state has changed. The chloride ion is too weak a reducing agent to reduce the sulfur (oxidation state +6) in sulphuric acid. Its an acid base reaction

It produces hydrogen fluoride which is an extremely dangerous gas that can etch glass The fluoride ion is even weaker a reducing agent than chloride ions

You will see steamy fumes of hydrogen bromide, and brown fumes of bromine. Colourless sulphur dioxide is also formed

NaBr(s) + H2S04(l) —-> NaHS04(s) +HBr(g) Sodium hydrogensulfate and hydrogen bromide are formed (in a similar acid base reaction to sodium chloride)

. Bromide ions are strong enough reducing agents to reduce the sulphuric acid to sulphur dioxide. This means the oxidation state of the sulfur is reduced from +6 to +4 and that of the bromine increases from -1 to 0

-1 +6 2HBr- + H2S04(l) —> +4 0 S02(g) + 2H20(l) +Br2(l) .Bromide ion reduces the sulphuric acid forming sulphur dioxide gas and bromine which is orange/brown

Steamy fumes of hydrogen iodide Black solid of iodine Bad egg smell of hydrogen sulphide gas is present. Yellow solid sulphur may be seen Also colourless sulphur dioxide gases also evolved.

NaI(s) + H2S04(l) —> NaHS04(s) + HI (g)

2HI + H2SO4 —> SO2 + 2H2O + I2 SO2 + 6HI —> H2S + 3I2 + 2H2O

Iodide ions are better reducing agents than bromide ions They reduce the sulfur in sulphuric acid even further: . from +6 (making S04) . to 0 (making S8) . and -2( forming H2S) so sulpher dioxide, sulphur and hydrogen sulphide gas are produced.

The oxidation state of sulfur passes from +6 to -2 so goes through state 0 so you see the solid sulfur

. Silver ions in aqueous solution eg silver nitrate, to form precipitate of silver halide .

Cl- + Ag+ –> AgCl(s) White precipitate is formed

. Dilute nitric acid is first added to halide solution to get rid of any soluble carbonate, C03-(aq) or Hydroxide inequalities . Because.. It would interfere with the test by forming insoluble silver carbonate 2Ag+(aq) + C03 2- (aq)–> Ag2C03(s) , or insoluble silver hydroxide Ag+(aq) + 0H-(aq) –> Ag0H(s)

. Add dilute nitric acid HN03 . then add a few drops of silver nitrate solution and halide precipitate forms

White precipitate of silver chloride . It dissolves in dilute ammonia

. cream precipitate of silver bromide . It dissolves in concentrated ammonia

. Pale yellow precipitate of silver iodide .Insoluble in concentrated ammonia

The precipitates formed from silver nitrate solution are too similar in colour to differentiate at times