4B.12-4B.14 - Halogen Reactions And Halide Tests

Chlorine Oxidation Numbers

Chlorine can look like several different chemicals in one topic: a poisonous green gas, a water-treatment reagent, bleach, chloride salts and chlorate(V) salts. The unifying idea is not the name of the product; it is the oxidation number of chlorine.

In elemental chlorine, $\ce{Cl2}$, each chlorine atom has oxidation number $0$. In chloride, $\ce{Cl-}$, chlorine is $-1$. In chlorate(I), $\ce{ClO-}$, chlorine is $+1$ because oxygen is $-2$ and the overall ion charge is $-1$. In chlorate(V), $\ce{ClO3-}$, chlorine is $+5$ because three oxygens contribute $-6$, so chlorine must be $+5$ to give the $-1$ ion charge.

[DIAGRAM: asset_name: Edexcel A Level Chemistry L022 Chlorine Disproportionation Oxidation Numbers; asset_slug: edexcel_a_level_chemistry_l022_chlorine_disproportionation_oxidation_numbers; recommended_method: deterministic_code; description: Water and cold alkali give 0 to −1/+1; hot alkali gives 0 to −1/+5 with a 5:1 chloride:chlorate(V) ratio. Pale 1600 by 900 diagram rendered with Pillow; technical geometry and Unicode chemical labels inspected.]

The diagram writes the $+1$ chlorine product as $\ce{HClO}$ / $\ce{ClO-}$ because water gives chloric(I) acid, $\ce{HClO}$, while alkaline conditions give the chlorate(I) ion, $\ce{ClO-}$.

If the same element from a single starting species is both oxidised and reduced in the same reaction, the reaction is called disproportionation. Chlorine does this in water and in alkali:

• 0 → -1 is reduction because chlorine gains electrons.
• 0 → +1 or 0 → +5 is oxidation because chlorine loses electrons.

Worked example: identify the redox change in cold dilute sodium hydroxide

For the ionic equation:

$\ce{Cl2 + 2OH^- -> Cl^- + ClO- + H2O}$

Start with oxidation numbers:

$\ce{Cl}$ in $\ce{Cl2}$ = 0

$\ce{Cl}$ in $\ce{Cl-}$ = -1

$\ce{Cl}$ in $\ce{ClO-}$ = +1

One chlorine atom is reduced from $0$ to $-1$; the other is oxidised from $0$ to $+1$. The same element, chlorine, has moved in two opposite redox directions, so this is disproportionation.

Chlorine With Water And Alkali

The important chlorine reactions in this lesson are not isolated facts. They are three versions of the same redox story, with the conditions deciding which oxygen-containing chlorine product is formed.

Chlorine with water

Chlorine reacts reversibly with water:

$\ce{Cl2 + H2O ⇌ HCl + HClO}$

or, showing ions in acidic solution:

$\ce{Cl2 + H2O ⇌ H^+ + Cl^- + HClO}$

The chlorine in $\ce{HCl}$ is $-1$; the chlorine in $\ce{HClO}$ is $+1$. This is disproportionation. The $\ce{HClO}$ formed is chloric(I) acid, often called hypochlorous acid, and it is the species mainly responsible for killing microorganisms in water treatment.

Water treatment therefore needs chemical judgement, not just recall. Chlorine is useful because it forms oxidising species that disinfect water. It must be controlled because chlorine and its oxidising products are hazardous at high concentration and can react with organic material in water.

Chlorine with cold dilute sodium hydroxide: bleach

Cold, dilute aqueous sodium hydroxide gives sodium chloride and sodium chlorate(I), commonly called sodium hypochlorite:

$\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}$

The ionic equation makes the redox pattern clearer:

$\ce{Cl2 + 2OH^- -> Cl^- + ClO- + H2O}$

$\ce{NaClO}$ is the active compound in many bleaches. Its chlorine can be reduced from +1 to −1 in chloride, accepting electrons from another species; this explains its oxidising action.

Chlorine with hot alkali

Hot alkali pushes the oxidation further to chlorate(V):

The ionic version is:

Here most chlorine atoms are reduced to $\ce{Cl-}$, while one chlorine atom in every six is oxidised to $\ce{ClO3-}$. That is why the balanced equation contains $\ce{5Cl^-}$ for every $\ce{ClO3-}$.

Common-error contrast

Do not describe the cold alkali and hot alkali reactions as simple neutralisation. Sodium hydroxide is present, but the key chemistry is the redox change in chlorine. A neutralisation answer would not explain why $\ce{Cl-}$, $\ce{ClO-}$ or $\ce{ClO3-}$ appear.

Guided practice

Bromine can show analogous reactions. Complete the cold dilute alkali pattern for $\ce{Br2}$:

$\ce{Br2 + 2OH- ->}$ $\boxed{?}$ + $\boxed{?}$ + $\ce{H2O}$

Answer:

$\ce{Br2 + 2OH^- -> Br^- + BrO^- + H2O}$

The oxidation numbers are the same pattern as chlorine: bromine is reduced from $0$ to $-1$ and oxidised from $0$ to $+1$.

Halogens With Metals

When a halogen reacts with a Group 1 or Group 2 metal, the halogen acts as an oxidising agent. The metal atoms lose electrons and form cations; the halogen molecules gain electrons and form halide ions.

For a Group 1 metal:

$\ce{2Na + Cl2 -> 2NaCl}$

Oxidation numbers:

$\ce{Na}$: 0 → +1

$\ce{Cl}$: 0 → -1

Sodium has been oxidised because its oxidation number increases. Chlorine has been reduced because its oxidation number decreases. Chlorine is therefore the oxidising agent.

For a Group 2 metal:

$\ce{Mg + Cl2 -> MgCl2}$

Oxidation numbers:

$\ce{Mg}$: 0 → +2

$\ce{Cl}$: 0 → -1

The same redox roles apply. Magnesium is oxidised, chlorine is reduced, and chlorine is the oxidising agent.

The analogous equations use the same structure with other halogens:

$\ce{2K + Br2 -> 2KBr}$

$\ce{Ca + I2 -> CaI2}$

The halogen trend from the previous lesson still matters. Fluorine is the strongest oxidising halogen; iodine is weaker; astatine is predicted to be weaker still. Here, though, the key is the oxidation-number explanation inside the metal-halogen reaction.

Worked example: calcium and bromine

Write the equation for calcium reacting with bromine and identify the oxidising agent.

Equation:

$\ce{Ca + Br2 -> CaBr2}$

Calcium changes from $0$ to $+2$, so calcium is oxidised. Bromine changes from $0$ to $-1$, so bromine is reduced. Bromine is the oxidising agent because it causes calcium to lose electrons.

Hydrogen Halides And Sulfuric Acid

Solid Group 1 halides react with concentrated sulfuric acid to produce hydrogen halides. This is also used to show the trend in reducing ability of the hydrogen halides.

The first step is an acid-base reaction:

$\ce{NaCl + H2SO4 -> NaHSO4 + HCl}$

$\ce{NaBr + H2SO4 -> NaHSO4 + HBr}$

$\ce{NaI + H2SO4 -> NaHSO4 + HI}$

For fluoride and chloride, this first step is normally the main reaction in this context. Hydrogen fluoride and hydrogen chloride are not strong enough reducing agents to reduce concentrated sulfuric acid under these conditions.

Hydrogen bromide and hydrogen iodide are stronger reducing agents. They can reduce sulfur in sulfuric acid while the halide ion is oxidised to the halogen.

For bromide:

$\ce{H2SO4 + 2HBr -> Br2 + SO2 + 2H2O}$

The bromine changes from $-1$ in $\ce{HBr}$ to $0$ in $\ce{Br2}$, so bromide is oxidised. Sulfur changes from $+6$ in $\ce{H2SO4}$ to $+4$ in $\ce{SO2}$, so sulfur is reduced. This shows that $\ce{HBr}$ is a reducing agent.

For iodide, reduction can go further. Possible sulfur-containing products include $\ce{SO2}$, $\ce{S}$ and $\ce{H2S}$:

$\ce{H2SO4 + 2HI -> I2 + SO2 + 2H2O}$

$\ce{H2SO4 + 6HI -> 3I2 + S + 4H2O}$

$\ce{H2SO4 + 8HI -> 4I2 + H2S + 4H2O}$

The pattern is:

$\ce{HF}$ < $\ce{HCl}$ < $\ce{HBr}$ < $\ce{HI}$

in increasing reducing ability.

The reason is linked to the hydrogen-halogen bond and the halide ion. Down Group 7, the $\ce{H-X}$ bond becomes weaker and the halide ion is larger, so it is more easily oxidised to $\ce{X2}$. That is why iodide is a much stronger reducing agent than chloride.

The observations depend on the product: hydrogen halides give steamy acidic fumes; bromine gives orange-brown vapour; iodine gives dark solid or purple vapour; sulfur is yellow. Sulfur dioxide and hydrogen sulfide are harmful gases, so use a supervised microscale procedure in a fume cupboard. Do not smell gases directly. These observations support the equations; they are not permission to identify a gas by inhalation.

Worked example: explaining a bromide result

A student adds concentrated sulfuric acid to solid potassium bromide. Brown fumes and a choking acidic gas are observed. Explain the redox chemistry.

The brown fumes indicate $\ce{Br2}$. Bromide ions in $\ce{HBr}$ have been oxidised from $-1$ to $0$. Sulfur in sulfuric acid has been reduced from $+6$ to $+4$ in $\ce{SO2}$. Therefore hydrogen bromide is acting as a reducing agent, and the result shows that reducing ability has increased compared with hydrogen chloride.

Common-error contrast

Do not say "bromide is a stronger oxidising agent down the group". The halide ion is the reducing species. The halogen molecule, such as $\ce{Br2}$, is the oxidising species. Down the group, halogen molecules become weaker oxidising agents, while halide ions become stronger reducing agents.

Halide Tests And Hydrogen Halides

The silver nitrate test turns invisible aqueous halide ions into visible evidence. Aqueous $\ce{Cl-}$, $\ce{Br-}$ and $\ce{I-}$ react with silver ions to form insoluble silver halide precipitates:

$\ce{Ag+ + Cl^- -> AgCl(s)}$

$\ce{Ag+ + Br^- -> AgBr(s)}$

$\ce{Ag+ + I- -> AgI(s)}$

[DIAGRAM: asset_name: Edexcel A Level Chemistry L022 Halide Precipitation Ammonia Tests; asset_slug: edexcel_a_level_chemistry_l022_halide_precipitation_ammonia_tests; recommended_method: deterministic_code; description: AgCl dissolves in dilute ammonia; AgBr requires concentrated ammonia; AgI remains in both. Pale 1600 by 900 diagram rendered with Pillow; technical geometry and Unicode chemical labels inspected.]

The usual observations are:

The ammonia step is useful because the precipitate colours can be hard to distinguish, especially cream versus pale yellow. Ammonia forms soluble diamminesilver(I) complexes with $\ce{AgCl}$ and, less readily, with $\ce{AgBr}$:

$\ce{AgI}$ is too insoluble for ammonia to dissolve in this test.

For an unknown sample, acidify with dilute nitric acid before adding silver nitrate. Do not use hydrochloric acid, because it adds chloride and can cause a false positive. This removes ions such as carbonate that could also form precipitates with silver ions. The specified halide evidence in this lesson, however, is the silver nitrate precipitate followed by the ammonia solubility test.

Hydrogen halides also have two simple reactions to know.

With ammonia:

$\ce{HCl(g) + NH3(g) -> NH4Cl(s)}$

$\ce{HBr(g) + NH3(g) -> NH4Br(s)}$

$\ce{HI(g) + NH3(g) -> NH4I(s)}$

The solid ammonium halides form white fumes or a white smoke when the gases meet.

With water, hydrogen halides form acids:

The key representation shift is from gas molecule to aqueous ions. A hydrogen halide gas is covalent, but in water it donates a proton and produces hydrated hydrogen ions and halide ions.

Hydrogen fluoride follows the proton-transfer pattern but is a weak acid in water:

The equilibrium arrow shows partial ionisation. With ammonia it forms ammonium fluoride, $\ce{HF + NH3 -> NH4F}$. This contrasts with the essentially complete ionisation of HCl, HBr and HI in dilute aqueous solution.

Worked example: deducing a halide

An unknown aqueous halide is acidified with dilute nitric acid. Silver nitrate solution gives a cream precipitate. Dilute ammonia has little effect, but concentrated ammonia dissolves the precipitate. Deduce the halide ion.

The cream precipitate suggests $\ce{AgBr}$. The fact that it dissolves only in concentrated ammonia confirms $\ce{Br-}$. The unknown contains bromide ions.

Fluorine And Astatine Predictions

Questions can ask about fluorine and astatine even when the main taught reactions use chlorine, bromine and iodine. The safe method is to state the relevant trend first, then apply it to the new element.

Use this prediction frame:

position in Group 7

→ atomic radius, shielding and electronegativity

→ oxidising ability of $\ce{X2}$ or reducing ability of $\ce{X-}$

→ predicted reaction or observation

Fluorine predictions

Fluorine is a gas at room temperature and has lower melting and boiling temperatures than chlorine because its small molecules have weaker London forces.

Fluorine is at the top of Group 7. It is very small and very electronegative, so $\ce{F2}$ is the strongest oxidising halogen. Fluoride ions, $\ce{F-}$, are the weakest reducing halide ions.

Consequences:

• $\ce{F2}$ reacts extremely readily as an oxidising agent.
• $\ce{F-}$ is not expected to reduce concentrated sulfuric acid.
• $\ce{HF}$ is not expected to show the reducing behaviour seen for $\ce{HBr}$ and $\ce{HI}$.
• $\ce{AgF}$ is soluble, so fluoride does not fit the same silver nitrate precipitate pattern as chloride, bromide and iodide.
• Fluorine is normally assigned oxidation number $-1$ in its compounds, so positive fluorine oxidation states analogous to $\ce{ClO-}$ or $\ce{ClO3-}$ are not a good prediction.

A useful contrast is fluorine with water. Chlorine disproportionates in water to give chloride and chloric(I) acid, but fluorine is such a strong oxidising agent that it oxidises water:

$\ce{2F2 + 2H2O -> 4HF + O2}$

Astatine predictions

From the physical trend, astatine is predicted to be solid at room temperature with higher melting and boiling temperatures than iodine, as its larger electron clouds give stronger London forces. These are trend-based predictions, not observations from a routine school experiment.

Astatine is below iodine. It is predicted to have a larger atomic radius, lower electronegativity and weaker oxidising ability than iodine. Astatide ions, $\ce{At-}$, would be predicted to be very strong reducing agents.

Consequences:

• $\ce{At2}$ would be a weaker oxidising agent than $\ce{I2}$.
• $\ce{At-}$ and $\ce{HAt}$ would be predicted to reduce concentrated sulfuric acid at least as readily as iodide and hydrogen iodide.
• Silver astatide, $\ce{AgAt}$, would be predicted to be a very insoluble silver halide, likely less soluble than $\ce{AgI}$.
• Astatine is rare and radioactive, so A level predictions are trend arguments rather than routine bench observations.

Worked example: predict a silver nitrate result for astatide

Question: Predict what would happen when aqueous silver nitrate is added to a solution containing $\ce{At-}$, then aqueous ammonia is added.

Trend route: from $\ce{Cl-}$ to $\ce{Br-}$ to $\ce{I-}$, silver halides become less soluble and dissolve less readily in ammonia. Astatide is below iodide, so $\ce{AgAt}$ is predicted to be even less soluble than $\ce{AgI}$. Therefore a precipitate would be expected, and it would probably be insoluble in ammonia.

Recap

For this lesson, a strong answer does not list isolated observations. It moves from equation to oxidation number, from precipitate to confirmatory ammonia evidence, or from position in Group 7 to a justified prediction. That is the difference between remembering halogen facts and using halogen chemistry.