The chemical properties of the halogen elements and the hydrogen halides

1. Overview

The halogens are reactive non-metals because each atom needs just one more electron to reach the electron arrangement of a noble gas. This section compares how strongly fluorine, chlorine, bromine and iodine act as oxidising agents, how they react with hydrogen, and how easily the hydrogen halides they form break down when heated. One idea links all three trends: down Group 17 the atoms get bigger, so they attract electrons, including the shared pair in an H-X bond, less strongly.

Key Definitions

• Halogens: the elements of Group 17 of the Periodic Table; their atoms have seven outer-shell electrons, and fluorine, chlorine, bromine and iodine exist as diatomic molecules, X₂, held together by a single covalent bond.
• Oxidising agent: a species that oxidises another species by taking electrons from it; the oxidising agent gains those electrons, so it is itself reduced and its oxidation number decreases.
• Halide ion: the ion with a 1− charge formed when a halogen atom gains one electron, for example Cl⁻, Br⁻ and I⁻.
• Displacement reaction (of the halogens): a redox reaction in which a halogen oxidises the halide ions of a halogen lower in the group, releasing that halogen, for example Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.
• Hydrogen halide: a covalent compound of hydrogen and a halogen with the formula HX, for example HF, HCl, HBr and HI.
• Bond energy: the energy required to break one mole of a particular covalent bond in the gaseous state; it is always positive.
• Bond length: the distance between the nuclei of two atoms joined by a covalent bond.
• Thermal decomposition: the breaking down of a compound into simpler substances by heating, for example 2HI(g) → H₂(g) + I₂(g).
• Thermal stability: how well a compound resists decomposition when it is heated; a more thermally stable compound needs a higher temperature before it decomposes.

Content

The halogens as oxidising agents

A halogen atom has seven electrons in its outer shell (3s² 3p⁵ for chlorine). When a halogen acts as an oxidising agent, each X₂ molecule gains two electrons (X stands for any halogen). In solution it becomes halide ions:

$$\text{X}_2 + 2\text{e}^- \rightarrow 2\text{X}^-$$

The oxidation number of the halogen falls from 0 to −1, so the halogen is reduced. The electrons come from another species, which is oxidised. So the halogen acts as an oxidising agent (oxidation numbers and redox terms are in topic 6.1).

How readily a halogen takes electrons is its oxidising power. It decreases down the group:

F₂ > Cl₂ > Br₂ > I₂ (strongest to weakest oxidising agent)

Why oxidising power decreases down the group:

1. Each halogen has one more occupied shell than the halogen above it, so its atomic radius is larger.
2. An incoming electron enters an outer shell that is further from the nucleus and more shielded by the inner shells.
3. The nuclear charge also increases, but the greater distance and extra shielding outweigh it, so the attraction for the extra electron is weaker.
4. So the atom gains an electron less readily, and the halogen is a weaker oxidising agent.

Fluorine is the strongest oxidising agent in the group. Two things help it: its F-F bond is much weaker than the Cl-Cl bond (158 against 242 kJ mol⁻¹, topic 11.1), so less energy is needed to split F₂ into atoms, and the small fluorine atom forms very strong bonds to other atoms, such as H-F (562 kJ mol⁻¹).

Displacement reactions

A halogen oxidises the halide ions of any halogen below it in the group, releasing that halogen. It cannot oxidise the halide ions of a halogen above it.

Reaction	Ionic equation
chlorine with bromide ions	Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)
chlorine with iodide ions	Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq)
bromine with iodide ions	Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)

In each reaction the halogen added is reduced (0 to −1) and the halide ion is oxidised (−1 to 0). Metal ions such as K⁺ or Na⁺ are spectator ions, so they are left out of the ionic equation. The same reactions, seen from the side of the halide ions as reducing agents, belong to topic 11.3.

Fluorine is not used in these tests, because it reacts with the water itself.

Seeing the result. The halide solutions are colourless, so a released halogen shows up as a colour. Bromine and iodine can look similar in water, so the mixture is often shaken with a little cyclohexane (or hexane). This non-polar solvent does not mix with water and floats on top. Bromine and iodine dissolve much better in it than in water, so they move into the top layer, where their colours are easy to tell apart:

Halogen	Colour in water	Colour in cyclohexane (top layer)
chlorine	very pale green, almost colourless	very pale green, almost colourless
bromine	yellow to orange	orange to red-brown
iodine	yellow-brown to brown	violet (purple)

So an orange top layer shows that bromine is present, and a violet top layer shows iodine. When there is no reaction, the top layer is still coloured: the halogen you added is still there, so you see its colour.

Worked example 1: bromine water and potassium iodide

Question: Bromine water is added to aqueous potassium iodide. The mixture is then shaken with cyclohexane and left to settle. (a) Describe what you would see. (b) Write the ionic equation, with state symbols. (c) Use oxidation numbers to identify the oxidising agent. (d) Explain why bromine water gives no reaction with aqueous potassium chloride.

Step 1: compare oxidising power. Bromine is above iodine, so Br₂ can oxidise I⁻ ions. Bromine is below chlorine, so Br₂ is a weaker oxidising agent than Cl₂ and cannot take electrons from Cl⁻ ions.

Step 2: observations. The potassium iodide solution is colourless. When the yellow-orange bromine water is added, the mixture turns brown as iodine forms. After shaking, two layers separate: the upper cyclohexane layer is violet, and the lower aqueous layer becomes paler.

Step 3: ionic equation. Two iodide ions give up one electron each, and one Br₂ molecule takes two electrons:

Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq)

Check the charges: −2 on the left and −2 on the right. K⁺ is a spectator ion.

Step 4: oxidation numbers. Bromine goes from 0 in Br₂ to −1 in Br⁻ (reduced). Iodine goes from −1 in I⁻ to 0 in I₂ (oxidised).

Answer: (a) the colourless solution turns brown; after shaking, the upper cyclohexane layer is violet. (b) Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq). (c) Br₂ is the oxidising agent: its oxidation number falls from 0 to −1. (d) Br₂ is a weaker oxidising agent than Cl₂, so it cannot oxidise Cl⁻; the mixture keeps the yellow-orange colour of the bromine water, and the cyclohexane layer turns orange.

Reactions with hydrogen

Fluorine, chlorine, bromine and iodine all react with hydrogen to form hydrogen halides:

H₂(g) + X₂(g) → 2HX(g)

The hydrogen halides are colourless, and HCl, HBr and HI are gases that give misty fumes in moist air. The vigour of the reaction falls steeply down the group. Typical conditions and observations:

Halogen	Typical conditions and what you see
fluorine	explodes on mixing, even in the cold and dark
chlorine	explodes in sunlight or ultraviolet light, but reacts only very slowly in the dark at room temperature; the pale green colour disappears
bromine	reacts only when heated, usually with a platinum catalyst, and does not explode; the orange-brown colour of the vapour fades
iodine	reacts slowly even when heated with a platinum catalyst; the violet colour of the vapour fades only partly

With fluorine, chlorine and bromine the reaction goes almost to completion. With iodine the reverse reaction is important at the temperatures used, so the reaction is incomplete and an equilibrium mixture forms: H₂(g) + I₂(g) ⇌ 2HI(g) (equilibrium is in topic 7.1).

Worked example 2: how exothermic is the reaction of bromine with hydrogen?

Question: Use bond energies from the Data Booklet to calculate the enthalpy change for H₂(g) + Br₂(g) → 2HBr(g).

Step 1: bonds broken. One H-H bond and one Br-Br bond: $$436 + 193 = 629\ \text{kJ}$$

Step 2: bonds formed. Two H-Br bonds: $$2 \times 366 = 732\ \text{kJ}$$

Step 3: enthalpy change. Bonds broken minus bonds formed (topic 5.1): $$\Delta H = 629 - 732 = -103\ \text{kJ mol}^{-1}$$

Answer: $\Delta H = -103$ kJ mol⁻¹ for the equation as written, which forms 2 mol of HBr(g). The value is for gaseous bromine: starting from liquid bromine, extra energy is needed to vaporise it, so the reaction would be less exothermic.

Explaining the trend in reactivity with hydrogen

In forming HX the halogen is reduced (0 to −1), so it is again acting as an oxidising agent. Its reactivity falls down the group for the same reason as its oxidising power: a larger atom attracts electrons, including the shared pair in the H-X bond, less strongly.

Bond energies show the same trend in energy terms. The table repeats the Worked example 2 calculation for each halogen, with Data Booklet values ($E$ = bond energy; all values in kJ mol⁻¹; $\Delta H$ is for H₂(g) + X₂(g) → 2HX(g)):

X	$E$(X-X)	$E$(H-X)	$\Delta H$
F	158	562	−530
Cl	242	431	−184
Br	193	366	−103
I	151	299	−11

• The H-X bond gets much weaker down the group, from 562 to 299 kJ mol⁻¹. Much less energy is released when the new bonds form, so the reaction becomes less and less exothermic. For iodine it is only just exothermic, so the reverse reaction (HI splitting up) also happens to a large extent, and an equilibrium mixture forms.
• The X-X bond does not explain the trend from chlorine to iodine. It weakens by 242 − 151 = 91 kJ mol⁻¹, which on its own would make the reaction 91 kJ mol⁻¹ more exothermic. But two H-X bonds weaken by 2 × (431 − 299) = 264 kJ mol⁻¹, which makes it 264 kJ mol⁻¹ less exothermic, nearly three times the effect. Overall the reaction becomes 264 − 91 = 173 kJ mol⁻¹ less exothermic (−184 to −11).

Thermal stability of the hydrogen halides

When a hydrogen halide is heated strongly enough, it splits back into its elements:

2HX(g) → H₂(g) + X₂(g)

Thermal stability decreases down the group: HF > HCl > HBr > HI (most to least stable).

• HF and HCl do not decompose noticeably when heated in the laboratory.
• HBr decomposes only slightly, on strong heating.
• HI decomposes easily: a red-hot wire pushed into a jar of hydrogen iodide produces violet fumes of iodine. The reaction is reversible: 2HI(g) ⇌ H₂(g) + I₂(g).

Explaining the trend with bond strength:

1. Down the group the halogen atom is larger, so the H-X bond is longer.
2. The bonding pair of electrons is further from the halogen nucleus and more shielded, so it is held less strongly.
3. So the H-X bond is weaker: the bond energies are 562 (H-F), 431 (H-Cl), 366 (H-Br) and 299 (H-I) kJ mol⁻¹.
4. Less energy is needed to break the H-X bond, so the molecules split up at a lower temperature, and the compound is less thermally stable.

Decomposition breaks the covalent bond inside each molecule. The hydrogen halides are already gases when they are heated, so the forces between molecules (topic 3.6) play no part here.

Worked example 3: why HF is so much more stable than HBr

Question: (a) Use bond energies to calculate $\Delta H$ for 2HF(g) → H₂(g) + F₂(g). (b) The decomposition 2HBr(g) → H₂(g) + Br₂(g) is the reverse of Worked example 2, so its $\Delta H$ is +103 kJ mol⁻¹. How much of the difference between the two values comes from the H-X bonds, and how much from the X-X bonds? (c) State which compound is more thermally stable, and explain why.

Step 1: bonds broken. Two H-F bonds: $$2 \times 562 = 1124\ \text{kJ}$$

Step 2: bonds formed. One H-H bond and one F-F bond: $$436 + 158 = 594\ \text{kJ}$$

Step 3: enthalpy change. $$\Delta H = 1124 - 594 = +530\ \text{kJ mol}^{-1}$$

Step 4: split the difference. The two values differ by 530 − 103 = 427 kJ mol⁻¹.

• H-X bonds: breaking two H-F bonds takes 2 × (562 − 366) = 392 kJ more than breaking two H-Br bonds.
• X-X bonds: forming F-F gives back 158 kJ, but forming Br-Br gives back 193 kJ, so 35 kJ less is given back for fluorine.
• Check: 392 + 35 = 427 kJ mol⁻¹. The H-X bonds account for 392 ÷ 427 = 92% of the difference.

Answer: (a) $\Delta H = +530$ kJ mol⁻¹. (b) 392 kJ mol⁻¹ (92%) comes from the H-X bonds and 35 kJ mol⁻¹ from the X-X bonds. (c) HF is far more thermally stable, mainly because the H-F bond (562 kJ mol⁻¹) is much stronger than the H-Br bond (366 kJ mol⁻¹): the fluorine atom is smaller, so the bonding pair is closer to its nucleus and held more strongly.

Key Equations

Halogen acting as an oxidising agent (X = any halogen; it gains electrons and is reduced from 0 to −1): $$\text{X}_2 + 2\text{e}^- \rightarrow 2\text{X}^-$$

Displacement (X is above Y in the group, so X₂ oxidises Y⁻): $$\text{X}_2(\text{aq}) + 2\text{Y}^-(\text{aq}) \rightarrow 2\text{X}^-(\text{aq}) + \text{Y}_2(\text{aq})$$

Reaction with hydrogen (an equilibrium, ⇌, for iodine): $$\text{H}_2(\text{g}) + \text{X}_2(\text{g}) \rightarrow 2\text{HX}(\text{g})$$

Enthalpy change from bond energies ($E$ = bond energy in kJ mol⁻¹ from the Data Booklet, for gases; $\Delta H$ in kJ mol⁻¹ for the equation above): $$\Delta H = E(\text{H-H}) + E(\text{X-X}) - 2E(\text{H-X})$$

Thermal decomposition (the reverse reaction, so $\Delta H$ has the same size and the opposite sign): $$2\text{HX}(\text{g}) \rightarrow \text{H}_2(\text{g}) + \text{X}_2(\text{g})$$ $$\Delta H = 2E(\text{H-X}) - E(\text{H-H}) - E(\text{X-X})$$

Common Mistakes to Avoid

• Mistake: mixing up the halogen, its halide ion and its hydrogen halide. Fix: only the element, X₂, is the oxidising agent here. Cl₂ oxidises I⁻ to I₂; chloride ions and HCl cannot oxidise I⁻, because their chlorine is already at −1 and cannot gain another electron.
• Mistake: explaining a displacement only as "the more reactive halogen displaces the less reactive one". Fix: describe the electron transfer: Cl₂ is a stronger oxidising agent than Br₂, so it takes electrons from Br⁻ ions: Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq).
• Mistake: calling iodine in water purple, or expecting a colourless mixture when there is no reaction. Fix: aqueous iodine is yellow-brown to brown; it is violet only in a non-polar solvent such as cyclohexane, or as a vapour. With no reaction the added halogen is still there: bromine water added to KCl(aq) gives an orange cyclohexane layer.
• Mistake: unbalanced ionic equations, such as Cl₂ + I⁻ → Cl⁻ + I₂. Fix: one Cl₂ needs two I⁻: Cl₂(aq) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(aq), with a total charge of −2 on each side.
• Mistake: explaining thermal stability with intermolecular forces. Fix: decomposition breaks the covalent bond inside each molecule. HI decomposes more easily than HCl because the H-I bond (299 kJ mol⁻¹) is weaker than the H-Cl bond (431 kJ mol⁻¹).
• Mistake: stopping at "the bond is longer", or using the X-X bond. Fix: finish the chain: longer bond, weaker bond, less energy needed to break it, less stable. And use H-X: the F-F bond (158 kJ mol⁻¹) is weak, yet HF is the most stable hydrogen halide.

Exam Tips

• Describe the relative reactivity as oxidising agents by giving the order and its direction: "oxidising power decreases from chlorine to bromine to iodine".
• State what is observed with colours before and after, and say which layer: "the colourless solution turns brown; the upper cyclohexane layer turns violet". "Iodine is formed" is a conclusion, not an observation. For chlorine and hydrogen in sunlight, the observation is an explosion.
• Explain reactivity with hydrogen through the halogen: larger atom, weaker attraction for electrons, weaker oxidising agent. You can add that the weaker H-X bond releases less energy, and that the reaction with iodine is reversible. Do not answer with the thermal stability of HX: that is a different question.
• Explain thermal stability as a chain about the H-X bond: larger halogen atom, longer bond, lower bond energy, less energy needed to break it. Name the bond (H-I), not "the bond" or "the halogen".
• Calculate with the bond energies printed in the question when they are given; otherwise use the Data Booklet values. Show bonds broken and bonds formed on separate lines, and give the sign and kJ mol⁻¹.
• Predict for astatine, At, below iodine, by extending each trend one step: At₂ would not oxidise I⁻, and HAt would be less stable than HI. Give a limit (H-At below 299 kJ mol⁻¹), not an invented value.