The Halogens and the Hydrogen Halides

AS · 12 min

The halogens, Group 17, are the most reactive non-metals. Down the group from chlorine to iodine their colour darkens, they become less volatile, and they become weaker oxidising agents. Their compounds with hydrogen, the hydrogen halides, also become less thermally stable down the group. This note covers the physical properties and their explanation in terms of intermolecular forces and bond strength, the halogens as oxidising agents (displacement reactions and their colours), and the reactions with hydrogen. Expect questions in all three AS papers: explanations in Paper 2, colour and trend questions in Paper 1, and displacement observations in Paper 3.

The elements

Every halogen atom has seven outer electrons ( $ns^2 np^5$ ), one short of a noble-gas configuration. The elements exist as diatomic molecules, $\ce{X2}$ , held together by a single covalent bond. Their chemistry is dominated by gaining one electron to form the halide ion, $\ce{X-}$ .

Colours and physical states

halogen	state at room temperature	colour of element	colour of vapour	in water	in a hydrocarbon solvent (e.g. cyclohexane)
fluorine, $\ce{F2}$	gas	pale yellow	pale yellow	(reacts)	(not used)
chlorine, $\ce{Cl2}$	gas	pale yellow-green	pale yellow-green	very pale green (almost colourless)	very pale green (almost colourless)
bromine, $\ce{Br2}$	liquid	red-brown (dark red)	orange-brown	yellow to orange	orange to red
iodine, $\ce{I2}$	solid	grey-black, shiny	purple (violet)	pale brown (brown in $\ce{KI(aq)}$ )	purple (violet)

The colours darken down the group. The syllabus asks specifically for chlorine, bromine and iodine.

Tip

Iodine is only slightly soluble in pure water. It dissolves much better in aqueous potassium iodide, forming a brown solution (containing $\ce{I3-}$ ). In practice "iodine solution" in the laboratory means iodine in $\ce{KI(aq)}$ , which is brown. A positive test for iodine is a blue-black colour with starch.

Volatility

Volatility is how readily a substance turns into a vapour. A volatile substance has a low boiling point.

halogen	$\ce{F2}$	$\ce{Cl2}$	$\ce{Br2}$	$\ce{I2}$
electrons per molecule	18	34	70	106
boiling point / K	85	238	332	457

Volatility decreases down the group (boiling points increase). This is explained by the intermolecular forces.

Key result

Explaining the trend in volatility

The halogens are simple molecular, non-polar substances. Boiling overcomes the instantaneous dipole–induced dipole (id–id) forces between molecules, not the covalent bonds within them.
Down the group the molecules have more electrons (34 in $\ce{Cl2}$ , 70 in $\ce{Br2}$ , 106 in $\ce{I2}$ ).
A larger electron cloud is more easily distorted (polarised), so larger instantaneous dipoles form and induce larger dipoles in neighbouring molecules.
So the id–id forces get stronger, more energy is needed to separate the molecules, and the volatility decreases.

Bond strength

The bond energy of the X–X bond (the energy needed to break one mole of bonds in the gas phase) changes as follows.

bond	F–F	Cl–Cl	Br–Br	I–I
bond energy / kJ mol⁻¹	158	242	193	151

From chlorine to iodine the bond strength decreases. As the atoms get larger, the bonding pair of electrons is further from each nucleus and more shielded, so the attraction between the nuclei and the shared pair is weaker, and less energy is needed to break the bond.

Fluorine is the exception: the F–F bond is weaker than expected. The fluorine atoms are so small that, when they are bonded, the lone pairs on the two atoms are very close together and repel each other strongly, weakening the bond.

Watch out

Do not confuse bond strength with volatility. Iodine has the weakest covalent bond but the highest boiling point, because boiling has nothing to do with the I–I bond: only the forces between molecules are overcome.

The halogens as oxidising agents

A halogen reacts by gaining electrons: $\ce{X2 + 2e- -> 2X-}$ . Gaining electrons is reduction, so the halogen is reduced and acts as an oxidising agent.

Key result

Oxidising power decreases down Group 17: $\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{I2}$ .

Down the group the atoms are larger and more shielded, so an incoming electron is further from the nucleus and attracted less strongly. Electronegativity decreases (F 4.0, Cl 3.0, Br 2.8, I 2.5), and the tendency to gain an electron decreases.

Displacement reactions

A more reactive halogen oxidises the halide ions of a less reactive halogen, displacing it from solution.

	$\ce{KCl(aq)}$	$\ce{KBr(aq)}$	$\ce{KI(aq)}$
add $\ce{Cl2(aq)}$	no reaction	colourless to yellow-orange ( $\ce{Br2}$ formed)	colourless to brown ( $\ce{I2}$ formed)
add $\ce{Br2(aq)}$	no reaction (stays yellow-orange)	no reaction	yellow-orange to brown ( $\ce{I2}$ formed)
add $\ce{I2(aq)}$	no reaction (stays brown)	no reaction	no reaction

The ionic equations:

\ce{Cl2(aq) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq)}

\ce{Cl2(aq) + 2I-(aq) -> 2Cl-(aq) + I2(aq)}

\ce{Br2(aq) + 2I-(aq) -> 2Br-(aq) + I2(aq)}

In aqueous solution bromine and iodine can both look yellow-brown, so the products are often confirmed by shaking with cyclohexane (an organic solvent that does not mix with water). The halogen moves into the upper organic layer, where bromine is orange-red and iodine is purple.

In terms of oxidation numbers, in $\ce{Cl2 + 2Br- -> 2Cl- + Br2}$ chlorine goes from 0 to $-1$ (reduced) and bromine goes from $-1$ to 0 (oxidised).

Reactions with hydrogen

All the halogens react with hydrogen to form hydrogen halides, $\ce{H2 + X2 -> 2HX}$ , but the vigour of the reaction falls sharply down the group.

halogen	reaction with hydrogen	equation
fluorine	explosive, even in the dark and at low temperature	$\ce{H2(g) + F2(g) -> 2HF(g)}$
chlorine	explosive in sunlight (UV light); slow in the dark	$\ce{H2(g) + Cl2(g) -> 2HCl(g)}$
bromine	needs heating (about 300 °C) with a platinum catalyst; less vigorous	$\ce{H2(g) + Br2(g) -> 2HBr(g)}$
iodine	slow even when heated; reversible, incomplete	$\ce{H2(g) + I2(g) <=> 2HI(g)}$

Explaining the relative reactivity

The reactivity decreases down the group for two connected reasons.

The halogen becomes a weaker oxidising agent (less attraction for electrons, as above).
The H–X bond formed gets weaker down the group, so less energy is released on forming the hydrogen halide and the reaction is less exothermic.

Using bond energies (H–H 436 kJ mol⁻¹), the enthalpy change for $\ce{H2 + X2 -> 2HX}$ is:

reaction	bonds broken / kJ	bonds formed / kJ	$\Delta H$ / kJ mol⁻¹
$\ce{H2 + F2}$	$436 + 158 = 594$	$2 \times 562 = 1124$	$-530$
$\ce{H2 + Cl2}$	$436 + 242 = 678$	$2 \times 431 = 862$	$-184$
$\ce{H2 + Br2}$	$436 + 193 = 629$	$2 \times 366 = 732$	$-103$
$\ce{H2 + I2}$	$436 + 151 = 587$	$2 \times 299 = 598$	$-11$

The reaction with iodine is only just exothermic, which is why it is reversible and incomplete.

Thermal stability of the hydrogen halides

bond	H–F	H–Cl	H–Br	H–I
bond energy / kJ mol⁻¹	562	431	366	299

Key result

Thermal stability of the hydrogen halides decreases down the group.

The H–X bond gets longer and weaker from H–F to H–I, because the halogen atom gets larger and the shared pair is further from the halogen nucleus and more shielded. Less energy is needed to break the H–X bond, so the hydrogen halide decomposes more easily on heating.

Observations:

$\ce{HF}$ and $\ce{HCl}$ do not decompose at ordinary high temperatures (HCl is stable even at about 1500 °C).
$\ce{HBr}$ decomposes slightly on strong heating, giving a little brown bromine vapour.
$\ce{HI}$ decomposes readily: a red-hot wire or glass rod plunged into hydrogen iodide produces purple fumes of iodine. $\ce{2HI(g) -> H2(g) + I2(g)}$

Worked examples

Explaining volatility

Explain why bromine is a liquid at room temperature but chlorine is a gas.

Solution

Both are simple molecular substances; boiling overcomes the instantaneous dipole–induced dipole forces between molecules. A $\ce{Br2}$ molecule has 70 electrons compared with 34 in $\ce{Cl2}$ . Its larger electron cloud is more easily polarised, so it forms larger instantaneous dipoles and stronger id–id forces. More energy is needed to separate bromine molecules, so bromine has a higher boiling point (332 K, above room temperature) than chlorine (238 K).

Displacement and oxidation numbers

Chlorine water is added to aqueous potassium iodide, and the mixture is then shaken with cyclohexane. Describe what is seen, write the ionic equation, and identify the oxidising agent using oxidation numbers.

Solution

The colourless solution turns brown; after shaking with cyclohexane, the upper (organic) layer is purple.

\ce{Cl2(aq) + 2I-(aq) -> 2Cl-(aq) + I2(aq)}

Chlorine: $0 \to -1$ , reduced. Iodine: $-1 \to 0$ , oxidised. Chlorine is the oxidising agent: it gains electrons from the iodide ions.

Bond energies and reactivity with hydrogen

Use the bond energies H–H 436, Cl–Cl 242, H–Cl 431, I–I 151 and H–I 299 kJ mol⁻¹ to calculate $\Delta H$ for the formation of 2 mol of HCl and of HI from the elements in the gas phase. Use your answers to explain the difference in the reactions of chlorine and iodine with hydrogen.

Solution

$\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$

HCl: $(436 + 242) - (2 \times 431) = 678 - 862 = -184\ \text{kJ mol}^{-1}$

HI: $(436 + 151) - (2 \times 299) = 587 - 598 = -11\ \text{kJ mol}^{-1}$

The formation of HCl is strongly exothermic, because the H–Cl bond is much stronger than the H–I bond. The reaction of hydrogen with chlorine is explosive in UV light and goes to completion; the reaction with iodine releases very little energy, is slow and reversible, and does not go to completion.

Exam-style: thermal stability

A hot glass rod is placed into separate gas jars of hydrogen chloride and hydrogen iodide. Describe and explain the observations.

Solution

Hydrogen chloride: no visible change; HCl does not decompose.

Hydrogen iodide: purple fumes form: $\ce{2HI(g) -> H2(g) + I2(g)}$ .

Explanation: the H–I bond (299 kJ mol⁻¹) is much weaker than the H–Cl bond (431 kJ mol⁻¹), because the iodine atom is larger, so the bonding pair is further from the iodine nucleus and more shielded, and is held less strongly. Less energy is needed to break the H–I bond, so HI is less thermally stable.

Exam-hard: predicting astatine

Astatine, At, is below iodine in Group 17.

(a) Predict the physical state and colour of astatine at room temperature.

(b) Predict whether astatine would react with aqueous sodium iodide, explaining your answer.

(d) A student predicts that the At–At bond is stronger than the I–I bond "because astatine has more electrons". Comment on this.

Solution

(a) Solid (volatility decreases down the group as id–id forces increase with the number of electrons); black or very dark (colours darken down the group).

(b) No reaction. Astatine is a weaker oxidising agent than iodine: its atoms are larger and more shielded, so it attracts electrons less strongly and cannot oxidise $\ce{I-}$ to $\ce{I2}$ .

(c) HAt is less thermally stable than HI. The At atom is larger, so the H–At bond is longer and weaker (bonding pair further from the At nucleus and more shielded), and less energy is needed to break it.

(d) The prediction is wrong. Bond strength decreases from Cl–Cl to I–I because the larger atoms hold the shared pair less strongly; the number of electrons is relevant to intermolecular (id–id) forces, not to the strength of the covalent bond. The At–At bond is expected to be weaker than I–I.

Watch out

Iodine solid is grey-black, not purple. Purple is the colour of the vapour and of iodine in cyclohexane.
Bromine is red-brown as a liquid; in water it is yellow to orange.
The trend in volatility is explained by id–id forces between molecules, not by bond strength. "Iodine has a higher boiling point because the I–I bond is stronger" is wrong twice.
In displacement reactions the halogen is reduced (it is the oxidising agent) and the halide ion is oxidised.
$\ce{H2 + I2 <=> 2HI}$ is an equilibrium. Use the reversible arrow.

Exam tip

Colour questions: be precise and use the syllabus words. "Brown" for iodine in aqueous solution; "orange" or "yellow" for bromine water; "purple" for iodine in an organic solvent.
Explaining volatility (about three marks): id–id forces; more electrons down the group; stronger forces need more energy to overcome.
Explaining thermal stability of HX: bond length increases, bond energy decreases, less energy needed to break H–X. Quote the data booklet values if they are given.
In "explain the relative reactivity with hydrogen", you can use oxidising power and H–X bond strength. Both are credited.

Practical skills

Displacement reactions are done in test-tubes: add about $1\ \text{cm}^3$ of aqueous chlorine, bromine or iodine to $1\ \text{cm}^3$ of aqueous potassium chloride, bromide or iodide, record the colour, then add about $1\ \text{cm}^3$ of cyclohexane, stopper, shake and allow the layers to separate. Record the colour of each layer. Work in a fume cupboard or well-ventilated area (chlorine and bromine are toxic). Paper 3 rewards precise colour descriptions and comparisons, for example "pale yellow solution turns orange-brown".

Summary

Halogens are $\ce{X2}$ molecules; atoms have seven outer electrons.
Colours darken down the group: $\ce{Cl2}$ pale yellow-green gas, $\ce{Br2}$ red-brown liquid, $\ce{I2}$ grey-black solid (purple vapour).
Volatility decreases down the group: more electrons, stronger id–id forces.
X–X bond strength decreases from Cl to I (larger atoms); F–F is weak because of lone-pair repulsion.
Oxidising power decreases down the group: $\ce{Cl2}$ displaces $\ce{Br2}$ and $\ce{I2}$ ; $\ce{Br2}$ displaces $\ce{I2}$ .
Reaction with $\ce{H2}$ becomes less vigorous down the group; $\ce{H2 + I2 <=> 2HI}$ is reversible.
Thermal stability of HX decreases down the group because the H–X bond gets weaker; HI decomposes to purple iodine vapour with a hot rod.

Practice

Question

State the colour and physical state of chlorine, bromine and iodine at room temperature.
Explain why the boiling point of iodine is higher than that of chlorine.
Write the ionic equation for the reaction of bromine water with aqueous potassium iodide, and describe the colour change.
Explain why bromine water does not react with aqueous potassium chloride.
State and explain the trend in bond energy from Cl–Cl to I–I, and explain why the F–F bond is weaker than the Cl–Cl bond.
$240\ \text{cm}^3$ of chlorine gas, measured at room conditions, is bubbled into excess aqueous potassium iodide. Calculate the mass of iodine formed. ( $A_r$ : I 126.9)
Use bond energies (H–H 436, Br–Br 193, H–Br 366 kJ mol⁻¹) to calculate the enthalpy change of $\ce{H2(g) + Br2(g) -> 2HBr(g)}$ .
Arrange HF, HCl, HBr and HI in order of increasing thermal stability and explain the order.
A colourless solution contains either potassium bromide or potassium iodide. Describe how you could use chlorine water and cyclohexane to identify the halide, giving the observations and equation for each possibility.
$0.508\ \text{g}$ of iodine is formed when an excess of aqueous bromine is added to a solution of potassium iodide. Calculate the mass of potassium iodide in the original solution, and explain why the reaction would not take place if aqueous iodine were added to potassium bromide. ( $A_r$ : K 39.1, I 126.9)

Answers

Chlorine: pale yellow-green gas. Bromine: red-brown liquid. Iodine: grey-black (shiny) solid.
Both are simple molecular; boiling overcomes id–id forces between molecules. $\ce{I2}$ has more electrons (106) than $\ce{Cl2}$ (34), so its electron cloud is more polarisable, the instantaneous and induced dipoles are larger and the id–id forces stronger. More energy is needed to separate the molecules.
$\ce{Br2(aq) + 2I-(aq) -> 2Br-(aq) + I2(aq)}$ . Yellow-orange (bromine water) turns brown.
Bromine is a weaker oxidising agent than chlorine. Its atoms are larger and more shielded, so it has a weaker attraction for electrons and cannot remove electrons from chloride ions; chloride cannot be oxidised to chlorine by bromine.
Bond energy decreases from Cl–Cl (242) through Br–Br (193) to I–I (151 kJ mol⁻¹). The atoms get larger, so the bonding pair is further from the nuclei and more shielded, and is attracted less strongly. F–F (158) is weaker than Cl–Cl because the very small fluorine atoms bring the lone pairs on the two atoms close together, and their repulsion weakens the bond.
$n(\ce{Cl2}) = 0.240 / 24.0 = 0.0100\ \text{mol}$ . $\ce{Cl2 + 2I- -> 2Cl- + I2}$ : $n(\ce{I2}) = 0.0100\ \text{mol}$ . $M(\ce{I2}) = 253.8$ ; mass $= 2.54\ \text{g}$ .
$\Delta H = (436 + 193) - 2(366) = 629 - 732 = -103\ \text{kJ mol}^{-1}$ .
$\ce{HI} < \ce{HBr} < \ce{HCl} < \ce{HF}$ . From HF to HI the halogen atom gets larger, the H–X bond gets longer, and the bonding pair is further from the halogen nucleus and more shielded, so the bond energy decreases (562, 431, 366, 299 kJ mol⁻¹). Weaker bonds break more easily on heating, so stability decreases.
Add chlorine water to a sample, then add cyclohexane, shake and let the layers separate. Bromide: solution turns yellow-orange; cyclohexane layer orange-red. $\ce{Cl2 + 2Br- -> 2Cl- + Br2}$ . Iodide: solution turns brown; cyclohexane layer purple. $\ce{Cl2 + 2I- -> 2Cl- + I2}$ .
$n(\ce{I2}) = 0.508 / 253.8 = 2.002 \times 10^{-3}\ \text{mol}$ . $\ce{Br2 + 2I- -> 2Br- + I2}$ , so $n(\ce{KI}) = 4.003 \times 10^{-3}\ \text{mol}$ . $M(\ce{KI}) = 166.0$ ; mass $= 0.665\ \text{g}$ . Iodine is a weaker oxidising agent than bromine (larger atom, weaker attraction for electrons), so it cannot oxidise bromide ions to bromine.

The elements

Colours and physical states

Volatility

Bond strength

The halogens as oxidising agents

Displacement reactions

Reactions with hydrogen

Explaining the relative reactivity

Thermal stability of the hydrogen halides

Worked examples

Practice

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