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Advanced Level Inorganic Chemistry Periodic Table Revision Notes Part 9. Group 7/17 The Halogens

9.4 Heating halide salts with conc. sulphuric acid & 9.5 Tests for halogens and halide ions

The reaction of fluorides, chlorides, bromides and iodides with hot concentrated sulfuric acid is fully described and the observations fully explained. The chemical test methods for detecting and identifying halogens and the corresponding halide ions are described.

PLEASE NOTE KS4 Science GCSE/IGCSE/O Level GROUP 7 HALOGENS NOTES are on a separate webpage

INORGANIC Part 9 Group 7/17 Halogens sub-index: 9.1 Introduction, trends & Group 7/17 data * 9.2 Halogen displacement reaction and reactivity trend  * 9.3 Reactions of halogens with other elements * 9.4 Reaction between halide salts and conc. sulfuric acid * 9.5 Tests for halogens and halide ions * 9.6 Extraction of halogens from natural sources * 9.7 Uses of halogens & compounds * 9.8 Oxidation & Reduction - more on redox reactions of halogens & halide ions * 9.9 Volumetric analysis - titrations involving halogens or halide ions * 9.10 Ozone, CFC's and halogen organic chemistry links * 9.11 Chemical bonding in halogen compounds * 9.12 Miscellaneous aspects of halogen chemistry

Advanced Level Inorganic Chemistry Periodic Table Index * Part 1 Periodic Table history * Part 2 Electron configurations, spectroscopy, hydrogen spectrum, ionisation energies * Part 3 Period 1 survey H to He * Part 4 Period 2 survey Li to Ne * Part 5 Period 3 survey Na to Ar * Part 6 Period 4 survey K to Kr and important trends down a group * Part 7 s-block Groups 1/2 Alkali Metals/Alkaline Earth Metals * Part 8  p-block Groups 3/13 to 0/18 * Part 9 Group 7/17 The Halogens * Part 10 3d block elements & Transition Metal Series * Part 11 Group & Series data & periodicity plots * All 11 Parts have their own sub-indexes near the top of the pages


9.4 The effect of heating halide salts with concentrated sulphuric acid

  • If salts such as sodium chloride, sodium bromide and sodium iodide are heated with conc. sulfuric acid, no simple pattern is observed, each reaction gives different products irrespective of the halogen element/compound formed. Similar results are obtained with potassium salts (just swap Na with a K) and most other metal salts of the halogens.

    • Conc. sulphuric acid is a viscous covalent liquid which can act as a mild oxidising agent with reducing agents.

    • In each case the metal ion is a spectator ion.

    • In writing the equations, it doesn't really matter whether you show the formation of sodium sulphate (Na2SO4) or sodium hydrogensulphate (NaHSO4), I think the former is generally more convenient.

  • (1) Sodium chloride

    • The less volatile sulphuric acid displaces the more 'volatile' colourless hydrogen chloride gas and leaving a white residue of sodium hydrogensulphate or sodium sulphate.

    • NaCl(s) + H2SO4(l) ==> NaHSO4(s) + HCl(g)

    • or 2NaCl(s) + H2SO4(l) ==> Na2SO4(s) + 2HCl(g)

    • or Cl-(s) + H2SO4(l) ==> HSO4-(s) + HCl(g)

    • Since it is not a redox reaction, non of the elements involved changes in oxidation state.

    • Sodium fluoride would behave in a similar way i.e. hydrogen fluoride gas, HF, would be released.

    • These are 'displacement' reaction, NOT a redox reaction.

    • However, in the case of bromides and iodides, the hydrogen halides are still formed, other more significant redox reactions take places in which the halide ions are oxidised to the halogen and the conc. sulfuric acid undergoes reduction. These redox reactions are now described below.

    • Note:

      • Other chloride salts e.g. potassium chloride etc. behave in the same way.

      • Sodium fluoride or potassium fluoride salts also behave in the same way e.g.

        • KF(s) + H2SO4(l) ==> KHSO4(s) + HF(g)

        • The hydrogen fluoride gas released reacts with glass and will etch it.

  • Advanced Inorganic Chemistry Page Index and Links(2) Sodium bromide

    • With hot conc. sulphuric acid, the bromide ion in solid salts, can act as a reducing agent and the sulphuric acid is reduced to sulphur dioxide (sulfur(IV) oxide) and the bromide ion oxidised to free bromine. You observe the orange-brown vapour of the bromine as well as hydrogen bromide fumes and a white residue of sodium sulfate is formed.

    • This is a redox reaction.

    • 2NaBr(s) + 2H2SO4(l) ==> NaHSO4(s) + Br2(g) + SO2(g) + 2H2O(l)

    • The half-cell reactions are:

      • (i) 2Br- ==> Br2 + 2e- (oxidation, 2 electron loss)

      • (ii) H2SO4 + 2H+ + 2e- ==> SO2 + 2H2O (reduction, 2 electron gain)

      • (i) balances (ii) in terms of electron/oxidation number change, so the ionic redox equation is

    • 2Br- + H2SO4 + 2H+ ==> Br2 + SO2 + 2H2O

    • The bromide ion is oxidised to bromine, half of the 'sulphate' essentially remains unchanged, but the other molecule of sulphuric acid is reduced to sulfur dioxide (sulphur dioxide, sulphur(IV) oxide).

    • Redox analysis

    • The oxidation state of bromine changes from -1 in Br-, to 0 in Br2 (oxidation, increase in oxidation state).

    • The oxidation state of sulphur changes from +6 in H2SO4, to +4 in SO2 (reduction, decrease in oxidation state).

    • The sulphur undergoes 2 'units' of oxidation number decrease, hence for each of 2 bromide ions, an oxidation number increase of 1 per bromine atom.

      • NOTE: You can write the bromide oxidation reaction as if hydrogen bromide is first formed and then oxidised to bromine and the concentrated sulphuric acid reduced to sulfur(IV) oxide (sulfur dioxide).

      • NaBr(s) + H2SO4(l) ==> NaHSO4(s) + HBr(g)

      • 2HBr(g) + H2SO4(l) ==> Br2(g) + SO2(g) + 2H2O(l)

      • I don't know how much of the HBr survives the oxidation?

  • Advanced Inorganic Chemistry Page Index and Links(3) Sodium iodide

    • With hot conc. sulphuric acid, the iodide ion in solid salts acts as an even stronger reducing agent than bromide, and the sulphuric acid is reduced to hydrogen sulphide gas and the iodide ion oxidised to free iodine. You observe the purple vapour of iodine, the strong smell of hydrogen sulphide (rotten eggs!) as well as hydrogen iodide fumes and a white residue of sodium sulfate is formed. The iodine will crystallise out on the upper cooler parts of the test tube.

    • This is a redox reaction.

    • 8NaI(s) + 5H2SO4(l) ==> 4Na2SO4(s) + 4I2(g/s) + H2S(g) + 4H2O(l)

    • The half-cell reactions are:

      • (i) 2I- ==> I2 + 2e- (oxidation, 2 electron loss)

      • (ii) H2SO4 + 8H+ + 8e- ==> SO2 + 4H2O (reduction, 8 electrons gained)

      • 4 x (i) balances (ii) in terms of electron/oxidation number change, so the ionic redox equation is

    • 8I- + H2SO4 + 8H+ ==> 4I2 + H2S + 4H2O

    • The iodide ion is oxidised to iodine, 4/5ths of the 'sulphate' essentially remains unchanged, but the 5th molecule of sulfuric acid is reduced to hydrogen sulphide.

    • Apart from the smell, hydrogen sulfide gives a black precipitate with lead ethanoate paper.

    • Redox analysis

    • The oxidation state of iodine changes from -1 in I-, to 0 in I2 (oxidation, increase in oxidation state).

    • The oxidation state of sulphur changes from +6 in H2SO4, to -2 in H2S (reduction, decrease in oxidation state).

    • The sulphur undergoes 8 'units' of oxidation number decrease, hence for each of 8 iodide ions, an oxidation number increase of 1 per iodine atom.

      • NOTE: You can write the iodide oxidation reaction as if hydrogen iodide is first formed and then oxidised to iodine and the concentrated sulphuric acid reduced to hydrogen sulfide (hydrogen sulphide).

      • NaI(s) + H2SO4(l) ==> NaHSO4(s) + HI(g)

      • 8HI(g) + H2SO4(l) ==> 4I2(g/s) + H2S(g) + 4H2O(l)

      • I don't know how much of the HI survives the oxidation?

  • (4) Note the increasing reducing power as you descend the Group 7 Halogens, i.e. the halide ion is increasingly more easily oxidised.

    • This is reflected in the oxidation potentials for the half cell halogen/halide ion

    • EθF2(aq)/F-(aq) = +2.87V > EθCl2(aq)/Cl-(aq) = +1.36V > EθBr2(aq)/Br-(aq) = +1.09V > EθI2(aq)/I-(aq) = +0.54V

    • The more positive the half-cell potential, the stronger the oxidising power (of the halogen)

      • and the weaker the reducing power of the corresponding halide ion.

      • Oxidising power: F2 > Cl2 > Br2 > I2

      • Reduction power: I- > Br- > Cl- > F-

Advanced Inorganic Chemistry Page Index and Links


 

9.5 Tests for halogens and halide Ions

Test for halogen Test method Test observations Test chemistry and comments
Chlorine gas Cl2

A pungent green gas.

(i) Apply damp blue litmus. (Can use red litmus and just see bleaching effect.)

(ii) A drop silver nitrate on the end of a glass rod into the gas.

(i) litmus turns red and then is bleached white.

(ii) White precipitate.

(i) Non-metal, is acid in aqueous solution and a powerful oxidising agent

(ii) It forms a small amount of chloride ion in water, so gives a positive result for the chloride test.

Bromine Br2 (l or aq)

A dark red liquid - orange-brown fumes, yellow-orange aqueous solution. The other common orange-brown gas is nitrogen dioxide

(i) Shake with a liquid alkene.

(ii) Mix with silver nitrate solution.

(ii) Decolourised. See alkene test.

(ii) Cream ppt. of silver bromide as in the test for bromide.

(i) Forms a colourless organic dibromo-compound

>C=C< + Br2 ==> >CBr-CBr<

(ii) Ag+(aq) + Br-(aq) ==> AgBr(s) 

 Any soluble bromide gives a silver bromide precipitate.

Iodine (i) solid

or (ii) solution

A very dark solid

(i) Gently heat the dark coloured solid.

(ii) Test aqueous solution or solid with starch solution.

(i) Gives brilliant purple vapour.

(ii) A blue black colour.

(i) Iodine forms a distinctive coloured vapour.

(ii) Forms a blue-black complex with starch and in biology the test is used to detect starch with iodine solution.

Advanced Inorganic Chemistry Page Index and LinksTests for Halide Ions

In test (i) the silver nitrate is acidified with dilute nitric acid to prevent the precipitation of other non-halide silver salts.

Test for halide ion Test method Observations Test chemistry and comments
Fluoride Ion  F-

Fluoride and hydrogen fluoride gas are harmful, irritating and corrosive substances.

(i) If the suspected fluoride is soluble add dilute nitric acid and silver nitrate solution.

(ii) You can warm a solid fluoride with conc. sulphuric acid and hold in the fumes (ONLY!) a glass rod with a drop of water on the end.

(i) There is NO precipitate!

(ii) Look for etching effects on the surface of the glass rod.

(i) Silver fluoride, AgF, is moderately soluble so this test proves little except that it isn't chloride, bromide and iodide!

(ii) Hydrogen fluoride gas is produced by displacement

F- + H2SO4 ==> HSO4- + HF which reacts with the glass silica to form silicic acid, silicon oxyfluoride, silicon fluoride. The chemistry is messy and complex BUT the glass rod is clearly etched.

Chloride ion  Cl-

If the solution also contains the sulphate ion, you test with barium ions 1st, filter off any barium sulphate precipitate and then test for chloride ion. This is because silver sulphate is also ~insoluble, so the two precipitates of silver sulfate and silver chloride could not be distinguished

(i) If the chloride is soluble, add dilute nitric acid and silver nitrate solution.

(ii) If insoluble salt, add conc. sulphuric acid, warm if necessary then test gas as for HCl.

(iii) Add lead(II) nitrate solution. Not a very specific test - test (i) is best.

(i) white precipitate of silver chloride soluble in dilute ammonia.

(ii) You get nasty fumes of hydrogen chloride which turn blue litmus red and give a white precipitate with silver nitrate solution.

(iii) A white ppt. of lead(II) chloride is formed.

(i) Ag+(aq) + Cl-(aq) ==> AgCl(s)

Any soluble silver salt + any soluble chloride  gives a white silver chloride precipitate, that darkens in light.

(ii) Cl-(s) + H2SO4(l) ==> HSO4-(s) + HCl(g) ,

then Ag+(aq) + Cl-(aq) ==> AgCl(s)

(iii) Pb2+(aq) + 2Cl-(aq) ==> PbCl2(s)

Bromide ion

Br-

(i) If bromide soluble, add dilute nitric acid and silver nitrate solution.

(ii) If insoluble salt, add conc. sulphuric acid, warm if necessary.

(iii) Add lead(II) nitrate solution. Not a very specific test - test (i) is best.

(i) Cream precipitate of silver bromide, only soluble in concentrated ammonia.

(ii) Orange vapour of bromine and pungent fumes of SO2

(iii) A white ppt. of lead(II) bromide is formed.

(i) Ag+(aq) + Br-(aq) ==> AgBr(s)

Any soluble silver salt + any soluble bromide gives a cream silver bromide precipitate.

(ii) The bromide ion is oxidised to bromine and the sulphuric acid is reduced to sulphur dioxide. (Test for sulphur dioxide -  potassium dichromate(VI) paper changes from orange to green)

(iii) Pb2+(aq) + 2Br-(aq) ==> PbBr2(s)

Iodide ion

I- 

(i) If iodide soluble, add dilute nitric acid and silver nitrate solution.

(ii) If insoluble salt can heat with conc. sulphuric acid, (ii) get purple fumes of iodine and very smelly hydrogen sulphide.

(iii) If iodide soluble, add lead(II) nitrate solution.

(i) Yellow precipitate of silver iodide insoluble in concentrated ammonia.

(ii) purple vapour and rotten egg smell!, (iii) a yellow precipitate forms

(iii) Yellow precipitate of lead(II) iodide. Not too definitive -Test (i) best.

(i) Ag+(aq) + I-(aq) ==> AgI(s) , any soluble silver salt + any soluble iodide  ==> silver iodide precipitate,

(ii) iodide ion is oxidised to iodine and the sulphuric acid is reduced to 'rotten eggs' smelly hydrogen sulphide,

(iii) insoluble lead(II) iodide formed

Pb2+(aq) + 2I-(aq) ==> PbI2(s)

The solubility tend for AgX in ammonia solution is (AgF water soluble) AgCl >> AgBr >> AgI

Advanced Inorganic Chemistry Page Index and Links


PLEASE NOTE KS4 Science GCSE/IGCSE/O Level GROUP 7 HALOGENS NOTES are on a separate webpage

WHAT NEXT?

INORGANIC Part 9 Group 7/17 Halogens sub-index: 9.1 Introduction, trends & Group 7/17 data * 9.2 Halogen displacement reaction and reactivity trend  * 9.3 Reactions of halogens with other elements * 9.4 Reaction between halide salts and conc. sulfuric acid * 9.5 Tests for halogens and halide ions * 9.6 Extraction of halogens from natural sources * 9.7 Uses of halogens & compounds * 9.8 Oxidation & Reduction - more on redox reactions of halogens & halide ions * 9.9 Volumetric analysis - titrations involving halogens or halide ions * 9.10 Ozone, CFC's and halogen organic chemistry links * 9.11 Chemical bonding in halogen compounds * 9.12 Miscellaneous aspects of halogen chemistry


keywords phrases formula oxidation states balanced symbol equations: NaCl(s) + H2SO4(l) ==> NaHSO4(s) + HCl(g) 2NaCl(s) + H2SO4(l) ==> Na2SO4(s) + 2HCl(g) Cl-(s) + H2SO4(l) ==> HSO4-(s) + HCl(g) KF(s) + H2SO4(l) ==> KHSO4(s) + HF(g) 2NaBr(s) + 2H2SO4(l) ==> NaHSO4(s) + Br2(g) + SO2(g) + 2H2O(l) H2SO4 + 2H+ + 2e- ==> SO2 + 2H2O 2Br- + H2SO4 + 2H+ ==> Br2 + SO2 + 2H2O NaBr(s) + H2SO4(l) ==> NaHSO4(s) + HBr(g) 2HBr(g) + H2SO4(l) ==> Br2(g) + SO2(g) + 2H2O(l) 8NaI(s) + 5H2SO4(l) ==> 4Na2SO4(s) + 4I2(g/s) + H2S(g) + 4H2O(l) 8I- + H2SO4 + 8H+ ==> 4I2 + H2S + 4H2O NaI(s) + H2SO4(l) ==> NaHSO4(s) + HI(g) 8HI(g) + H2SO4(l) ==> 4I2(g/s) + H2S(g) + 4H2O(l)


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