Halogens

Class 11 Chemistry, unit 12, book pages 245–250

Practise 39 MCQs Take the unit quiz Review 15 cards

Halogens are the group 17 elements, from fluorine to astatine and oganesson, and they are all reactive non-metals that exist as diatomic molecules. This unit describes how their colour, volatility and bond strength change down the group, how this sets their reactivity as oxidizing agents and with hydrogen, and how the halide ions behave as reducing agents. It ends with the silver nitrate test, the reactions of chlorine with sodium hydroxide and water, and the use of chlorine in water purification.

Learning outcomes (11)

Describe the colours and the trends in volatility of chlorine, bromine, and iodine.
Describe the trend in bond strength of halogen molecules.
Interpret the volatility of the elements in terms of instantaneous dipole-induced dipole forces.
Describe the relative reactivity of halogen elements as oxidizing agents.
Describe the reactions of elements with hydrogen and explain their relative reactivity in these reactions.
Describe the relative thermal stabilities of hydrogen halides and explain these in terms of bond strength.
Describe the relative reactivity of halide ions as reducing agents.
Explain the reactions of halide ions with aqueous silver nitrate and concentrated sulphuric acid.
Describe the reactions of halides with aqueous silver ions followed by aqueous ammonia.
Interpret the reaction of chlorine with cold and hot aqueous sodium hydroxide as disproportion reactions.
Explain the use of chlorine in water purification, including the production of the active species HOCl and ClO⁻ which kill bacteria.

12.1Halogens

Halogens are the elements of group 17 of the periodic table (Group VII-A in the older numbering). The name comes from the Greek words halos, meaning salt, and gen, meaning former.

The group contains fluorine, chlorine, bromine, iodine, astatine and oganesson. Astatine is radioactive and oganesson is an artificial radioactive element, so little is known about their properties. All halogens are reactive non-metals, and all of them exist as diatomic molecules: $F X_{2}$ , $Cl X_{2}$ , $Br X_{2}$ and $I X_{2}$ .

Colour and state of the first four halogens
Halogen	Formula	Colour and state at room temperature
Fluorine	$F X_{2}$	Pale yellow gas
Chlorine	$Cl X_{2}$	Greenish yellow gas
Bromine	$Br X_{2}$	Reddish brown liquid
Iodine	$I X_{2}$	Shiny greyish black solid

12.1.1Volatility of halogens

As you move down the group, the halogens become less volatile and darker. The decrease in volatility is due to the increase in shell size.

Halogen molecules are non-polar. The forces between them are instantaneous dipole-induced dipole forces, and the strength of these forces increases as the size of the halogen molecule increases.

Why the state changes down the group

Halogen molecules are non-polar, so the molecules attract each other only by instantaneous dipole-induced dipole forces.
A larger molecule has stronger forces of this kind, so more energy is needed to separate the molecules.
A substance that needs more energy to separate its molecules is less volatile, so it stays a liquid or a solid at room temperature.
Applied to the halogens: $F X_{2}$ and $Cl X_{2}$ are gases, $Br X_{2}$ is a liquid and $I X_{2}$ is a solid at room temperature.

12.1.2Bond strength of halogens

As you go down the group, the bond energy of the halogen molecules decreases steadily from chlorine to iodine. Fluorine does not follow this pattern: its bond energy is lower than that of chlorine. Its small size causes inter-electronic repulsions, which makes $F X_{2}$ less stable.

Bond energies of the halogen molecules
Halogen molecule	Bond energy ( $kJ mo l^{- 1}$ )
$F - F$	156
$Cl - Cl$	243
$Br - Br$	193
$I - I$	151

Worked example: bond energy graph and energy to break bonds

Plot the bond energies of

F X_{2}

Cl X_{2}

Br X_{2}

and

I X_{2}

(

156

243

193

and

151 kJ mo l^{- 1}

). Describe the pattern, then find the energy needed to break all the bonds in

2.0 mol

Br X_{2}

and in

0.50 mol

Cl X_{2}

Given

$n (Br X_{2}) = 2.0 mol$
$n (Cl X_{2}) = 0.50 mol$

To find

\text{energy to break all the bonds}

F X_{2}

Cl X_{2}

Br X_{2}

I X_{2}

100

150

200

250

300

halogen molecule

E (kJ mo l^{- 1})

156

243

193

151

Bond energy of the halogen molecules from the book's table: it rises from

F X_{2}

Cl X_{2}

and then falls to

I X_{2}

Describe the pattern. Bond energy falls steadily from $Cl X_{2}$ to $I X_{2}$ , but $F X_{2}$ breaks the trend: it is lower than $Cl X_{2}$ because the small $F$ atoms repel each other's electrons, which makes the molecule less stable.
Each mole of $Br X_{2}$ has one mole of Br-Br bonds, so the energy is the amount times the bond energy.
$E = n \times E_{bond} = 2.0 mol \times 193 kJ mo l^{- 1} = 386 kJ$
The same for chlorine.
$E = 0.50 mol \times 243 kJ mo l^{- 1} = 121.5 kJ$
Do not say that bond energy falls from fluorine to iodine; the order is $Cl X_{2} > Br X_{2} > F X_{2} > I X_{2}$ .

386 kJ for Br X_{2}, 121.5 kJ for Cl X_{2}

12.2ReactivityOften examined

The reactivity of a halogen depends on how easily its atoms gain an electron to form a halide ion ( $F X^{-}$ , $Cl X^{-}$ , $Br X^{-}$ or $I X^{-}$ ). Fluorine has the highest tendency to gain an electron, so it is the most reactive halogen.

As you go down the group, the electronegativity of the halogens decreases, and so does their reactivity. Because halogens have a strong tendency to gain electrons, they are strong oxidizing agents, and this oxidizing power also decreases down the group. The order of decreasing oxidizing power is:

F X_{2} > Cl X_{2} > Br X_{2} > I X_{2}

A stronger oxidizing agent can oxidize, and so displace, the halide ion of a weaker halogen from its aqueous solution. Fluorine can oxidize and displace all the other halide ions:

Displacement by fluorine

Fluorine displaces chloride ions from potassium chloride solution, liberating chlorine gas.
$F X_{2} (g) + 2 KCl (aq) 2 KF (aq) + Cl X_{2} (g)$
Fluorine displaces bromide ions from potassium bromide solution, liberating liquid bromine.
$F X_{2} (g) + 2 KBr (aq) 2 KF (aq) + Br X_{2} (l)$
Fluorine displaces iodide ions from sodium iodide solution, liberating solid iodine.
$F X_{2} (g) + 2 NaI (aq) 2 NaF (aq) + I X_{2} (s)$

Similarly, chlorine can oxidize bromide and iodide ions. Iodine cannot oxidize any halide ion.

Worked example: which halogen displacements can occur

Decide which of these reactions take place and write the balanced equation for each one that does: (a)

Cl X_{2}

with

KBr

solution, (b)

Br X_{2}

with

KCl

solution, (c)

Br X_{2}

with

NaI

solution, (d)

I X_{2}

with

KBr

solution.

Use the rule: a halogen displaces the halide ion of a weaker oxidizing agent. The oxidizing power order is $F X_{2} > Cl X_{2} > Br X_{2} > I X_{2}$ .
$Cl X_{2} > Br X_{2} > I X_{2}$
(a) $Cl X_{2}$ is stronger than $Br X_{2}$ , so it oxidizes bromide ions and reddish brown bromine is formed. Cl goes from $0$ to $- 1$ and Br from $- 1$ to $0$ .
$Cl X_{2} (g) + 2 KBr (aq) 2 KCl (aq) + Br X_{2} (l)$
(b) $Br X_{2}$ is weaker than $Cl X_{2}$ , so it cannot oxidize chloride ions. No reaction.
(c) $Br X_{2}$ is stronger than $I X_{2}$ , so it displaces iodide ions and iodine is formed.
$Br X_{2} (l) + 2 NaI (aq) 2 NaBr (aq) + I X_{2} (s)$
(d) $I X_{2}$ is the weakest oxidizing agent here and cannot oxidize bromide ions. No reaction. A common mistake is to decide by position in the equation instead of by the order of oxidizing power.

Displacement grid. A halogen displaces only the halide ions below it in the oxidizing power order; these cells (accent) show the reactions that occur.

(a) and (c) occur; (b) and (d) do not

12.2.1Reactions of halogens with hydrogenOften examined

Halogens react with hydrogen to form hydrogen halides, where X stands for F, Cl, Br or I:

H X_{2} + X X_{2} 2 HX

As the bond energy of the halogens decreases down the group, the atoms become larger and the forces between the molecules weaker. A halogen atom then attracts hydrogen atoms less strongly, so the reaction is less vigorous. The relative reactivity therefore decreases from fluorine to iodine:

F X_{2} > Cl X_{2} > Br X_{2} > I X_{2}

Reactions of the halogens with hydrogen, from fluorine to iodine
Halogen	Reaction with hydrogen
Fluorine	Combines explosively with hydrogen even in the cold and in darkness, to form hydrogen fluoride gas.
Chlorine	Hydrogen and chlorine explode when exposed to sunlight or a flame, to form hydrogen chloride gas.
Bromine	Bromine vapour and hydrogen produce a small explosion when ignited, and hydrogen bromide gas is formed.
Iodine	Combines only partially with hydrogen, even on continued heating. An equilibrium forms between hydrogen, iodine and hydrogen iodide gas.

12.2.2Thermal stability of hydrogen halides

The strength of the hydrogen-halogen bond depends on the electronegativity difference between the hydrogen and halogen atoms. A larger electronegativity difference gives a stronger bond.

Why thermal stability falls from HF to HI

Going from HF to HI, the electronegativity of the halogen decreases, so the difference in electronegativity from hydrogen gets smaller.
A smaller difference gives a weaker H-X bond. The bonds weaken gradually from HCl to HBr to HI.
Weaker bonds need less energy to break, so the energy needed to break H-X decreases in this order:
$HF > HCl > HBr > HI$
A hydrogen halide with the stronger bond is more stable to heat, so the thermal stability decreases from HF to HI.

12.2.3Acidic strength of halogen acids

The strength of the H-X bond affects how easily a halogen acid donates its proton. The H-F bond is the strongest and the H-I bond is the weakest, so $HI$ loses a proton most easily.

Order of acidic strength

A weaker H-X bond breaks more easily, so the acid gives up its proton more readily.
The H-X bond weakens from HF to HI, so the acidic strength increases in this order:
$HI > HBr > HCl > HF$
So HI is the strongest of the halogen acids and HF is the weakest.

12.2.4Reactivity of halide ionsOften examined

The reducing power of an anion depends on how easily it can lose electrons. Down the group, the attraction between the nucleus and the valence electrons decreases, so the halide ions lose electrons more easily. The reducing power of the halide ions therefore increases down the group: $I X^{-}$ is the strongest reducing agent and $F X^{-}$ the weakest.

a) With concentrated sulphuric acid

Fluoride and chloride ions are weak reducing agents, so they cannot reduce concentrated sulphuric acid. They only form the hydrogen halide:

NaF + H X_{2} SO X_{4} NaHSO X_{4} + HF

NaCl + H X_{2} SO X_{4} NaHSO X_{4} + HCl

Bromide ions are strong enough reducing agents to reduce concentrated sulphuric acid to sulphur dioxide:

NaBr + H X_{2} SO X_{4} NaHSO X_{4} + HBr

2 HBr + H X_{2} SO X_{4} Br X_{2} + SO X_{2} + 2 H X_{2} O

Iodide ions are even stronger reducing agents. They reduce concentrated sulphuric acid to hydrogen sulphide, and iodine is liberated:

NaI + H X_{2} SO X_{4} NaHSO X_{4} + HI

8 HI + H X_{2} SO X_{4} 4 I X_{2} + H X_{2} S + 4 H X_{2} O

Balancing the iodide reaction

Each iodine atom changes from -1 in HI to 0 in $I X_{2}$ , so eight electrons are lost by eight iodine atoms.
The sulphur atom changes from +6 in H2SO4 to -2 in H2S, so it gains eight electrons.
The electrons lost equal the electrons gained, so one H2SO4 reacts with eight HI and the equation balances.

Worked example: how far each halide ion reduces concentrated sulphuric acid

Concentrated sulphuric acid is added to solid

NaF

NaCl

NaBr

and

NaI

. Find the sulphur product with each, give the oxidation number of sulphur in it, and explain the trend.

F X^{-}

Cl X^{-}

Br X^{-}

I X^{-}

−4

−2

halide ion

n_{ox} (S)

H X_{2} SO X_{4}

H X_{2} SO X_{4}

SO X_{2}

H X_{2} S

Oxidation number of sulphur in the sulphur-containing product when concentrated sulphuric acid meets each halide. The more the acid is reduced, the stronger the reducing agent.

$F X^{-}$ and $Cl X^{-}$ are weak reducing agents. They only form the hydrogen halide, so sulphur stays at $+ 6$ in $H X_{2} SO X_{4}$ .
$NaCl + H X_{2} SO X_{4} NaHSO X_{4} + HCl$
$Br X^{-}$ is stronger and reduces the acid to $SO X_{2}$ . Sulphur goes from $+ 6$ to $+ 4$ , gaining 2 electrons, which two bromide ions supply (each $- 1 \to 0$ ).
$2 HBr + H X_{2} SO X_{4} Br X_{2} + SO X_{2} + 2 H X_{2} O$
$I X^{-}$ is the strongest and reduces the acid to $H X_{2} S$ : sulphur goes from $+ 6$ to $- 2$ , gaining 8 electrons from eight iodide ions.
$8 HI + H X_{2} SO X_{4} 4 I X_{2} + H X_{2} S + 4 H X_{2} O$
Trend: down the group the nucleus holds the valence electrons less strongly, so the halide ion loses electrons more easily. A stronger reducing agent reduces sulphur further.

+ 6, + 6, + 4, - 2 for F X^{-}, Cl X^{-}, Br X^{-}, I X^{-}

b) With silver nitrate solution

Adding aqueous silver nitrate to a solution of a halide gives a coloured precipitate of the silver halide. This is the silver nitrate test for halide ions. The nitrate ion is a spectator ion and is left out of the ionic equation:

Ag X^{+} (aq) + X X^{-} (aq) AgX (s)

Silver nitrate test for halide ions
Halide ion	Colour of the silver halide precipitate	Behaviour in aqueous ammonia
Chloride, $Cl X^{-}$	White	Remains undissolved
Bromide, $Br X^{-}$	Cream or off-white	Partially dissolves
Iodide, $I X^{-}$	Yellow	Remains undissolved

Worked example: identifying halide ions with silver nitrate

Three colourless solutions X, Y and Z are each treated with aqueous silver nitrate. X gives a white precipitate that stays in aqueous ammonia, Y gives a cream precipitate that partly dissolves in ammonia, and Z gives a yellow precipitate. (a) Identify the halide ion in each and write the ionic equations. (b) What mass of precipitate forms from

0.010 mol

of X's halide ion? (

Ag = 108

Cl = 35.5

)

Given

$n (Cl X^{-}) = 0.010 mol$

To find

\text{halide ions; }m(\ce{AgCl})

Match the colour with the table: white is chloride, cream is bromide and yellow is iodide. The ammonia test agrees for Y: only the bromide precipitate partly dissolves. So X is $Cl X^{-}$ , Y is $Br X^{-}$ and Z is $I X^{-}$ .
Write the ionic equations. The nitrate ion is a spectator ion, so it is left out.
$Ag X^{+} (aq) + Cl X^{-} (aq) AgCl (s) Ag X^{+} (aq) + Br X^{-} (aq) AgBr (s) Ag X^{+} (aq) + I X^{-} (aq) AgI (s)$
(b) One $Cl X^{-}$ gives one $AgCl$ , so $0.010 mol$ of $AgCl$ forms. Its molar mass is $108 + 35.5 = 143.5 g mo l^{- 1}$ .
$m = 0.010 mol \times 143.5 g mo l^{- 1} = 1.435 g \approx 1.4 g$
Use two observations, the colour and the effect of ammonia, since white and yellow precipitates are told apart by colour and cream is separated by its partial dissolving.

X = Cl X^{-}, Y = Br X^{-}, Z = I X^{-}; m (AgCl) = 1.4 g

c) Reaction with sodium hydroxide

Chlorine reacts with sodium hydroxide in two ways. The product depends on the temperature and the concentration of the NaOH.

In cold, dilute NaOH, chlorine forms sodium hypochlorite, NaOCl:

Cl X_{2} + 2 NaOH NaOCl + NaCl + H X_{2} O

In hot, concentrated NaOH, chlorine forms sodium chlorate, $NaClO X_{3}$ :

3 Cl X_{2} + 6 NaOH NaClO X_{3} + 5 NaCl + 3 H X_{2} O

Chlorine in cold dilute sodium hydroxide is both oxidized (to NaOCl) and reduced (to NaCl).

Reading the hot reaction

$3 Cl X_{2}$ contains six chlorine atoms.
One chlorine atom ends up in $NaClO X_{3}$ , so it is oxidized.
The other five chlorine atoms end up in $NaCl$ , so they are reduced.
The electrons lost by the one oxidized atom equal the electrons gained by the five reduced atoms, which is why the equation balances in this 1 to 5 ratio.

Worked example: electron balance in chlorine and hot concentrated sodium hydroxide

Show that

3 Cl X_{2} + 6 NaOH NaClO X_{3} + 5 NaCl + 3 H X_{2} O

is a disproportionation reaction and that the electrons lost equal the electrons gained.

Oxidation number of Cl in $Cl X_{2}$ is $0$ . In $NaClO X_{3}$ : $(+ 1) + x + 3 (- 2) = 0$ , so $x = + 5$ . In $NaCl$ : $(+ 1) + x = 0$ , so $x = - 1$ .
$Cl X_{2} : 0, NaClO X_{3} : + 5, NaCl : - 1$
$3 Cl X_{2}$ has 6 Cl atoms. One atom goes $0 \to + 5$ and loses 5 electrons: oxidised.
$1 \times 5 = 5 e^{-} lost$
The other 5 atoms go $0 \to - 1$ and each gains 1 electron: reduced.
$5 \times 1 = 5 e^{-} gained$
The electrons lost (5) equal those gained (5), which explains the ratio of 1 $NaClO X_{3}$ to 5 $NaCl$ . One element, chlorine, is both oxidised and reduced, so it is a disproportionation reaction. In cold dilute NaOH the same idea gives NaOCl ( $+ 1$ ) and NaCl ( $- 1$ ) in a 1 to 1 ratio.

Chlorine in hot concentrated sodium hydroxide: the electrons lost by one atom (5) equal those gained by the other five.

5 e^{-} lost = 5 e^{-} gained

In both reactions chlorine is oxidized and reduced at the same time. A reaction in which one element is both oxidized and reduced is called a self oxidation-reduction reaction, or a disproportionation reaction.

Disproportionation reaction: A self oxidation-reduction reaction, in which one element is both oxidized and reduced. Chlorine in cold or hot sodium hydroxide is an example.

d) Reaction of chlorine with water

Chlorine dissolves in water to form hydrochloric acid and hypochlorous acid, $HOCl$ :

Cl X_{2} + H X_{2} O HCl + HOCl

Hypochlorous acid is very reactive. It decomposes to produce atomic oxygen, $[O]$ , which can kill bacteria present in water. For this reason chlorine is used to purify water by killing bacteria, and the process is called chlorination.

HOCl HCl + [O]

In sunlight, or on standing, hypochlorous acid decomposes to give oxygen gas:

2 HOCl 2 HCl + O X_{2}

Key terms

Halogen: An element of group 17: fluorine, chlorine, bromine, iodine, astatine or oganesson. All halogens are reactive non-metals that exist as diatomic molecules.
Diatomic molecule: A molecule made of two atoms, such as \ce{Cl2} or \ce{I2}.
Volatility: The ease with which a substance changes into vapour. Halogens become less volatile down the group.
Instantaneous dipole-induced dipole forces: Weak attractions between non-polar molecules, caused by a momentary uneven spread of electrons. They grow stronger as the molecule gets larger.
Bond energy: The energy needed to break a bond. The bond energies of the halogen molecules are given in kJ per mol.
Electronegativity: The tendency of an atom to attract the shared electrons of a bond. It decreases down the group, and the difference from hydrogen governs the H-X bond strength.
Oxidizing power: The ability of a halogen to gain electrons and oxidize another species. It decreases from fluorine to iodine.
Halide ion: The negative ion of a halogen, such as \ce{F-}, \ce{Cl-}, \ce{Br-} or \ce{I-}.
Reducing power: The ability of a species to lose electrons. The reducing power of a halide ion increases down the group.
Thermal stability: The resistance of a hydrogen halide to breaking down on heating. It decreases from HF to HI.
Acidic strength: How easily an acid donates its proton. HI is the strongest halogen acid and HF the weakest.
Silver nitrate test: A test for halide ions: adding aqueous silver nitrate forms a coloured precipitate, and its colour and its solubility in aqueous ammonia identify the halide.
Disproportionation reaction: A self oxidation-reduction reaction in which one element is both oxidized and reduced, as in chlorine with sodium hydroxide.
Hypochlorous acid: HOCl, formed when chlorine dissolves in water. It decomposes to give atomic oxygen, which kills bacteria.
Chlorination: The purification of water by adding chlorine, which kills bacteria.

Multiple choice questionsp. 249–250

Which halogen has greenish yellow colour?

A\ce{F2}
B\ce{Cl2}Answer
C\ce{Br2}
D\ce{I2}

Why

Chlorine, \ce{Cl2}, is a greenish yellow gas. Fluorine is a pale yellow gas, bromine is a reddish brown liquid and iodine is a shiny greyish black solid, so none of the other three is greenish yellow.

Which halogen exists in liquid state?

A\ce{F2}
B\ce{Cl2}
C\ce{Br2}Answer
D\ce{I2}

Why

Bromine, \ce{Br2}, is a reddish brown liquid. Fluorine and chlorine are gases, and iodine is a solid.

Volatility falls down the group because the instantaneous dipole-induced dipole forces between the non-polar molecules get stronger as the molecules grow larger. So \ce{Br2} is a liquid and \ce{I2} is a solid.

iii

Which of the following is the strongest reducing agent?

A\ce{F-}
B\ce{Br-}
C\ce{Cl-}
D\ce{I-}Answer

Why

The reducing power of a halide ion depends on how easily it loses an electron. Going down the group the outer electrons are further from the nucleus and held less strongly, so the ion loses electrons more easily. Iodide, \ce{I-}, is therefore the strongest reducing agent and fluoride, \ce{F-}, the weakest (option A is the weakest, not the strongest).

This matches the book's observation that \ce{F-} and \ce{Cl-} are too weak to reduce concentrated \ce{H2SO4}, while \ce{I-} reduces it to \ce{H2S}.

Oxidizing power of which halogen is highest?

A\ce{F2}Answer
B\ce{Cl2}
C\ce{Br2}
D\ce{I2}

Why

Oxidizing power is the tendency of a halogen to gain electrons. It decreases down the group, \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}, so fluorine has the highest oxidizing power. Iodine, \ce{I2}, has the lowest.

Which is strongest acid?

A\ce{HF}
B\ce{HCl}
C\ce{HBr}
D\ce{HI}Answer

Why

Acid strength depends on how easily the H-X bond breaks to release a proton. The H-X bond gets weaker from \ce{HF} to \ce{HI}, so \ce{HI} loses its proton most easily and is the strongest acid. \ce{HF} has the strongest bond and is the weakest of the four.

Short questionsp. 250

i3 marks

Give the order of oxidizing power of halogens.

Solution

The oxidizing power of the halogens decreases down the group: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}. Fluorine is the strongest oxidizing agent and iodine the weakest.

Oxidizing power is the tendency to gain electrons. As the atoms become larger, the electronegativity falls and the nucleus attracts an incoming electron less strongly, so the tendency to gain electrons, and so the oxidizing power, falls down the group.

This is shown by displacement: \ce{F2} oxidizes and displaces all the halide ions, \ce{Cl2} oxidizes \ce{Br-} and \ce{I-}, but \ce{I2} cannot oxidize any halide ion.

ii3 marks

How does reactivity of halogens with hydrogen vary?

Solution

Reactivity with hydrogen decreases from fluorine to iodine: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}. Each halogen forms the hydrogen halide, \ce{H2 + X2 -> 2HX} with X = F, Cl, Br, I.

Fluorine combines explosively with hydrogen even in the cold and in darkness. Chlorine explodes in sunlight or on heating with a flame. Bromine gives only a small explosion when ignited. Iodine combines only partially, even on continued heating, and an equilibrium is set up with \ce{HI}.

The reason is the atomic size and bond energy: down the group the atoms get larger and the halogen bond energy falls, so the halogen atoms attract hydrogen atoms less effectively and the reactions become less vigorous.

Note. The book writes the general equation as \ce{H2 + X2 -> HX}, without the coefficient 2 on HX. The balanced form is shown here.

iii3 marks

Why is the strength of \ce{HF} different from \ce{HI}?

Solution

The acidic strength of a hydrogen halide depends on the strength of its H-X bond. An acid is stronger when the bond is weaker, because the proton is then released more easily. The electronegativity difference between hydrogen and the halogen decreases from F to I, so the H-X bond becomes weaker from \ce{HF} to \ce{HI}.

The H-F bond is the strongest, so \ce{HF} loses its proton least easily and is the weakest acid. The H-I bond is the weakest, so \ce{HI} loses its proton most easily and is the strongest acid. The order of acidic strength is \ce{HI} > \ce{HBr} > \ce{HCl} > \ce{HF}.

So the difference is that the H-F bond is strong, while the H-I bond is weak.

Note. The book gives this bond-strength order. In aqueous solution, \ce{HF} is a weak acid while \ce{HCl}, \ce{HBr} and \ce{HI} are strong acids, so the book's order is a simplification based on bond strength.

iv3 marks

Write reactions of sulphuric acid with \ce{NaCl} and \ce{NaBr}.

Solution

With sodium chloride: concentrated sulphuric acid displaces hydrogen chloride from its salt. The chloride ion is too weak a reducing agent to reduce the acid, so no redox reaction takes place:

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

With sodium bromide: the first step is the same and gives hydrogen bromide. The bromide ion is a strong enough reducing agent to reduce the concentrated acid to sulphur dioxide, and bromine is set free:

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

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

Note. The book prints 2\ce{HBr} + \ce{H2SO4} -> \ce{Br2} + \ce{SO2} + \ce{H2O}, which is not balanced (4 H and 4 O on the left, 2 H and 3 O on the right). The balanced equation has 2\ce{H2O}, as shown here.

v3 marks

What is a disproportion reaction?

Solution

A disproportion reaction is a redox reaction in which one element is at the same time oxidized and reduced. Such reactions are also called self-oxidation-reduction reactions.

Chlorine reacts with cold dilute sodium hydroxide in this way. In \ce{Cl2} the oxidation number of chlorine is 0. In \ce{NaCl} it is -1 (reduced) and in \ce{NaOCl} it is +1 (oxidized):

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

With hot concentrated sodium hydroxide the same change gives \ce{NaClO3}, in which chlorine is +5.

52026 Annual I, Set D, Q.2 (iii) OR3 marks

How does sulphuric acid react with halide ions?

Solution

The reducing power of the halide ions decides the reaction with concentrated sulphuric acid.

Fluoride and chloride ions are weak reducing agents, so they cannot reduce the acid. They only give the hydrogen halide by acid displacement:

\ce{NaF + H2SO4 -> NaHSO4 + HF}\qquad \ce{NaCl + H2SO4 -> NaHSO4 + HCl}

Bromide ions are strong enough to reduce the concentrated acid to sulphur dioxide, and bromine is formed:

\begin{gathered}\ce{NaBr + H2SO4 -> NaHSO4 + HBr}\\ \ce{2HBr + H2SO4 -> Br2 + SO2 + 2H2O}\end{gathered}

Iodide ions are even stronger reducing agents. They reduce the acid to hydrogen sulphide, and iodine is formed:

\ce{NaI + H2SO4 -> NaHSO4 + HI}\qquad \ce{8HI + H2SO4 -> 4I2 + H2S + 4H2O}

Note. The book prints 8\ce{HI} + 4\ce{H2SO4} -> 4\ce{I2} + 4\ce{H2S} + 4\ce{H2O}. This is not balanced: it has 16 O on the left and 4 O on the right. The balanced equation with one molecule of \ce{H2SO4} is shown here.

63 marks

Compare the strength of halogen acids.

Solution

The acidic strength of the hydrogen halides increases from \ce{HF} to \ce{HI}: \ce{HF} < \ce{HCl} < \ce{HBr} < \ce{HI}.

Acid	Strength of H-X bond	Acidic strength
\ce{HF}	Strongest	Weakest
\ce{HCl}	Strong	Stronger than \ce{HF}
\ce{HBr}	Weaker	Stronger than \ce{HCl}
\ce{HI}	Weakest	Strongest

The electronegativity difference between hydrogen and the halogen decreases down the group, so the H-X bond weakens and the proton is released more easily.

Note. The book's order \ce{HI} > \ce{HBr} > \ce{HCl} > \ce{HF} is the order of H-X bond strength. In aqueous solution \ce{HF} is in fact a weak acid, while \ce{HCl}, \ce{HBr} and \ce{HI} are strong acids, so the book's order is a simplification.

93 marks

Interpret the reactivity of halogens as oxidizing agents.

Solution

Halogens are strong oxidizing agents because each atom easily gains one electron to form a halide ion, \ce{X2 + 2e- -> 2X-}. Each halogen is reduced in the process.

Their oxidizing power falls down the group, because the electronegativity falls and the atoms get larger, so the nucleus attracts an incoming electron less strongly. Fluorine has the highest tendency to gain an electron and is the strongest oxidizing agent.

This shows in displacement reactions, where a halogen oxidizes and displaces the halide ion of a halogen below it in the group:

\begin{gathered}\ce{F2(g) + 2KCl(aq) -> 2KF(aq) + Cl2(g)}\\ \ce{F2(g) + 2NaI(aq) -> 2NaF(aq) + I2(s)}\end{gathered}

Long questionsp. 250

35 marks

Explain the reactions of chlorine with sodium hydroxide.

Solution

Principle

Chlorine undergoes a disproportion reaction with sodium hydroxide. The same element is both oxidized and reduced, and the product depends on the temperature and the concentration of the alkali.

Cold and dilute sodium hydroxide

When chlorine gas is passed through cold, dilute aqueous \ce{NaOH}, sodium hypochlorite (\ce{NaOCl}) and sodium chloride form:

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

Hot and concentrated sodium hydroxide

When chlorine is passed through hot, concentrated aqueous \ce{NaOH}, sodium chlorate (\ce{NaClO3}) and sodium chloride form:

\ce{3Cl2 + 6NaOH -> NaClO3 + 5NaCl + 3H2O}

Check with oxidation numbers

Species	Oxidation number of Cl	Change for each Cl atom
\ce{Cl2}	0	Starting point
\ce{NaCl}	-1	Reduced: gains 1 electron
\ce{NaOCl}	+1	Oxidized: loses 1 electron
\ce{NaClO3}	+5	Oxidized: loses 5 electrons

In the cold reaction one chlorine atom is reduced and one is oxidized by one electron each, so the electrons balance. In the hot reaction five chlorine atoms are reduced, gaining 5 electrons in total, and one chlorine atom is oxidized, losing 5 electrons, so the electrons again balance.

Both reactions are therefore disproportion reactions, also called self-oxidation-reduction reactions.

42025 Annual I, Set B, Q.2 (ii) OR5 marks

Describe the silver nitrate test for the identification of halide ions.

Solution

Principle

The silver nitrate test identifies a halide ion from the colour of the silver halide precipitate formed when a solution of the halide is treated with aqueous silver nitrate. The general reaction is:

\ce{X^- (aq) + AgNO3 (aq) -> AgX (s) + NO3^- (aq)}

Procedure

Take a solution of the halide salt in a test tube.
Add aqueous silver nitrate and observe the colour of the precipitate.
Add aqueous ammonia and observe whether the precipitate dissolves.

Observations

Halide ion	Colour of \ce{AgX} precipitate	Effect of aqueous ammonia
\ce{Cl-}	White	Remains undissolved
\ce{Br-}	Cream or off-white	Partly dissolves
\ce{I-}	Yellow	Remains undissolved

The test identifies the halide from the colour of the precipitate and from whether it dissolves in aqueous ammonia. Chloride gives a white precipitate, bromide a cream precipitate that partly dissolves, and iodide a yellow precipitate that does not dissolve.

Note. Misprint in the book's text: it says the white precipitate of AgCl remains undissolved in aqueous ammonia. In fact AgCl dissolves in excess dilute aqueous ammonia, forming the complex \ce{[Ag(NH3)2]+}. The answer above follows the book's observations; check the wording with your teacher before using it in an exam.

76 marks

Evaluate the reactivity of halogens.

Solution

Introduction

The reactivity of a halogen is its tendency to gain an electron and form a halide ion. Three pieces of evidence from the book (oxidizing power, reaction with hydrogen and displacement) are used to judge it.

1. Oxidizing power

Fluorine has the highest tendency to gain electrons, so it is the strongest oxidizing agent: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}. \ce{F2} oxidizes and displaces all the halide ions, \ce{Cl2} oxidizes \ce{Br-} and \ce{I-}, and \ce{I2} oxidizes no halide ion.

2. Reaction with hydrogen

Fluorine reacts explosively with hydrogen even in the dark, chlorine explodes in sunlight or on heating, bromine gives only a small explosion when ignited, and iodine reacts only partially, setting up an equilibrium. The reactions become less vigorous down the group.

3. Atomic size and electronegativity

The atoms grow larger down the group and the electronegativity falls, so the nucleus attracts an incoming electron less strongly. This explains the trend in both tests.

Evaluation

Fluorine is the most reactive halogen: its reactions are the most vigorous and it reacts even in the dark.
Iodine is the least reactive: it oxidizes no halide ion and reacts with hydrogen only partially.
Bromine lies between chlorine and iodine in both tests, so the trend is consistent across the evidence and the conclusion is reliable.

Overall, the reactivity of the halogens decreases from fluorine to iodine.

\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}

Note. The book's text asks which halogen is least reactive as 'Bromine or iodine', which is ambiguous. Iodine is the least reactive, as shown above.

86 marks

Discuss trends in (a) volatility of halogens (b) strength in halogen molecules.

Solution

(a) Volatility of halogens

Volatility decreases down the group. Fluorine and chlorine are gases, bromine is a reddish brown liquid and iodine is a shiny greyish black solid, so volatility falls from \ce{F2} to \ce{I2}.

Halogen molecules are non-polar, so the forces between them are instantaneous dipole-induced dipole forces. These forces get stronger as the molecule grows larger, because it has more electrons and a larger shell. More energy is then needed to separate the molecules, so the halogen is less volatile.

Hence \ce{Br2} is a liquid and \ce{I2} is a solid at room temperature, and the halogens become darker as you go down the group.

(b) Strength in halogen molecules

Bond energies of halogen molecules (book table).
Halogen molecule	Bond energy (kJ/mol)
\ce{F2}	156
\ce{Cl2}	243
\ce{Br2}	193
\ce{I2}	151

From chlorine to iodine the bond energy decreases steadily, because the atoms are larger and the shared pair of electrons is held less firmly. So \ce{Cl2} is the strongest molecule of those compared, and \ce{I2} the weakest.

Fluorine has a lower bond energy than chlorine. Although it is smaller, its small size brings the lone pairs close together, so the inter-electronic repulsions weaken the F-F bond and make \ce{F2} less stable.

Overall, the bond strength order is \ce{Cl2} > \ce{Br2} > \ce{F2} > \ce{I2}.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 111 mark

Which of the following is possible?

A\ce{Br2 + 2KF -> F2 + 2KBr}
B\ce{Cl2 + 2KF -> F2 + 2KCl}
C\ce{F2 + 2KBr -> 2KF + Br2}Answer
D\ce{I2 + 2KBr -> Br2 + 2KI}

Why

A halogen can displace another halogen from its halide only if it is the stronger oxidizing agent. The oxidizing power falls down the group: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}.

\ce{F2 + 2KBr -> 2KF + Br2}

Fluorine is stronger than bromine, so \ce{F2} oxidizes \ce{Br-} to \ce{Br2} (option C). The other reactions go the wrong way, because a weaker halogen cannot oxidize the halide ion of a stronger one: \ce{Br2} and \ce{Cl2} cannot oxidize \ce{F-} (A and B), and \ce{I2} cannot oxidize \ce{Br-} (D).

2025 Annual I, Set B, Q.2 (ii) OR1+1+1 = 3 marks

Write down the chemical tests with equations to identify chlorides, bromides and iodides present in aqueous solution.

Solution

The test is the silver nitrate test. Aqueous silver nitrate is added to the aqueous halide solution, and the halide ion gives a coloured precipitate of the silver halide. The nitrate ion is a spectator ion.

\ce{Ag+(aq) + X-(aq) -> AgX(s)}

Chloride (1 mark)

\ce{NaCl + AgNO3 -> AgCl\downarrow + NaNO3}

A white precipitate of \ce{AgCl}.

Bromide (1 mark)

\ce{NaBr + AgNO3 -> AgBr\downarrow + NaNO3}

A cream (off-white) precipitate of \ce{AgBr}, which partly dissolves in aqueous ammonia.

Iodide (1 mark)

\ce{NaI + AgNO3 -> AgI\downarrow + NaNO3}

A yellow precipitate of \ce{AgI}, which does not dissolve in aqueous ammonia.

Halide ion	Colour of \ce{AgX}	Behaviour in aqueous ammonia (as in the book)
\ce{Cl-}	White	Remains undissolved
\ce{Br-}	Cream	Partly dissolves
\ce{I-}	Yellow	Remains undissolved

2026 Annual I, Set B, Q.2 (iii) OR3 marks

State three factors affecting reducing power of halide ions.

Solution

The reducing power of a halide ion is its tendency to lose an electron (to change to the halogen, \ce{2X- -> X2 + 2e-}). Going down the group it increases: \ce{I-} > \ce{Br-} > \ce{Cl-} > \ce{F-}. Three factors explain this.

Size of the ion. The halide ion becomes larger down the group (\ce{F-} is the smallest, \ce{I-} the largest), so the outer electron is farther from the nucleus and is lost more easily.
Attraction between the nucleus and the valence electrons. The number of shells (and so the shielding) increases down the group, so the nucleus attracts the outer electrons less strongly and the ion loses an electron more easily.
Electronegativity. It decreases down the group, so the ion holds its electrons less tightly. \ce{F-} holds its electron most strongly and is the weakest reducing agent; \ce{I-} is the strongest.

Evidence: with concentrated \ce{H2SO4}, \ce{F-} and \ce{Cl-} cannot reduce it, \ce{Br-} reduces it to \ce{SO2} and \ce{I-}, the strongest reducing agent, reduces it further to \ce{H2S}.

Note. The book states the cause as the decreasing attraction between the nucleus and the valence electrons; size and electronegativity are the other points that explain the same trend.

2026 Annual I, Set D, MCQ 111 mark

Which of the following is possible?

A\ce{Br2 + 2KF -> F2 + 2KBr}
B\ce{Cl2 + 2KF -> F2 + 2KCl}
C\ce{F2 + 2KBr -> 2KF + Br2}Answer
D\ce{I2 + 2KBr -> Br2 + 2KI}

Why

A halogen can displace another halogen from its halide only if it is the stronger oxidizing agent. The oxidizing power falls down the group: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}.

\ce{F2 + 2KBr -> 2KF + Br2}

2026 Annual I, Set D, Q.2 (iii) OR1+2 = 3 marks

Explain the reactivity of \ce{F-}, \ce{Cl-} and \ce{Br-} with concentrated \ce{H2SO4}.

Solution

The reactivity of a halide ion with concentrated sulphuric acid depends on its reducing power, which increases from \ce{F-} to \ce{I-}.

Fluoride and chloride (1 mark)

\ce{F-} and \ce{Cl-} are weak reducing agents and cannot reduce concentrated \ce{H2SO4}. They only form the hydrogen halide.

\ce{NaF + H2SO4 -> NaHSO4 + HF}

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

Bromide (2 marks)

\ce{Br-} is a stronger reducing agent. \ce{HBr} formed first is oxidized to bromine, and \ce{H2SO4} is reduced to \ce{SO2} (sulphur changes from +6 to +4).

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

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

2025 Annual I, Set D, Q.53+3 = 6 marks

Explain the relative reactivity of halogens: (i) as oxidizing agents (ii) in their reaction with \ce{H2}, by giving suitable examples.

Solution

Reactivity is the tendency of a halogen atom to gain an electron and form the halide ion. Fluorine has the highest tendency, and the electronegativity and reactivity fall down the group.

(i) Relative reactivity as oxidizing agents (3 marks)

Halogens gain electrons readily, so they are strong oxidizing agents. The oxidizing power decreases down the group because the atoms become larger and the electronegativity decreases.

\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}

A stronger oxidizing agent oxidizes, and so displaces, the halide ion of a weaker halogen from its solution.

\ce{F2(g) + 2KCl(aq) -> 2KF(aq) + Cl2(g)} (fluorine displaces chloride, bromide and iodide; for example \ce{F2(g) + 2KBr(aq) -> 2KF(aq) + Br2(l)})
\ce{Cl2 + 2KBr -> 2KCl + Br2} and \ce{Cl2 + 2KI -> 2KCl + I2} (chlorine displaces bromide and iodide)
\ce{Br2 + 2KI -> 2KBr + I2} (bromine displaces iodide only)
\ce{I2} does not oxidize any halide ion.

(ii) Reaction with hydrogen (3 marks)

Halogens react with hydrogen to form hydrogen halides.

\ce{H2 + X2 -> 2HX}

As the halogen atoms become larger down the group, they attract hydrogen atoms less strongly and the reaction becomes less vigorous: \ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}.

Halogen	Reaction with hydrogen
Fluorine	Combines explosively even in the cold and in the dark: \ce{H2 + F2 -> 2HF}
Chlorine	Explodes in sunlight or with a flame: \ce{H2 + Cl2 -> 2HCl}
Bromine	Small explosion when ignited: \ce{H2 + Br2 -> 2HBr}
Iodine	Combines only partially even on continued heating; an equilibrium forms: \ce{H2 + I2 <=> 2HI}

So reactivity in both respects decreases from fluorine to iodine.

Exam focus

Explain the reactivity of halide ions \ce{F-}, \ce{Cl-}, \ce{Br-} and \ce{I-} with concentrated \ce{H2SO4}, with equations. (short)
Asked in 2026 (Set D Q.2 (iii) OR; Set B Q.2 (iii) OR on the factors of reducing power), and book short questions 4 and 6.
Go to answer
Explain the relative reactivity of halogens as oxidizing agents and in their reaction with hydrogen, with examples. (long)
A 6-mark Section C question in 2025 (Set D Q.5, split 3+3); book short questions 2 and 8 and long question 3.
Go to answer
Decide which halogen displacement reaction is possible (for example \ce{F2 + 2KBr -> 2KF + Br2}). (mcq)
Asked in both 2026 sets (Set B and Set D MCQ 11).
Go to answer
Describe the silver nitrate test to identify chloride, bromide and iodide ions, with colours and equations. (short)
Asked in 2025 (Set B Q.2 (ii) OR) and book long question 2.
Go to answer
State the three factors (and the trend) that make reducing power of halide ions increase down the group. (short)
Asked in 2026 (Set B Q.2 (iii) OR).
Go to answer
Explain the reactions of chlorine with cold dilute and hot concentrated \ce{NaOH}, and define a disproportionation reaction. (long)
Book long question 1 and short question 5; not asked in the 2025 or 2026 papers.
Go to answer
Compare the trends in volatility, bond strength, thermal stability and acidic strength of halogens and hydrogen halides. (long)
Book short questions 1, 3 and 7, MCQ 5 and long question 4; not asked directly in the 2025 or 2026 papers.
Go to answer