正文与公式仍可阅读；滚动同步、实验控件和答案验证需要 JavaScript。
The prose and formulas remain readable; interactive figures and checks require JavaScript.

Lesson 6 · 30 min

Halogens and noble gases

Fluorine’s electron affinity is smaller than chlorine’s, so why is it the strongest oxidant? Meet the halogens through energy cycles, bond enthalpies and computed colours, run displacement tests, and see how the “inert” gases bond to fluorine.

In this lesson

Describe how the state, colour and oxidising power of the halogens change down the group, and explain why
Explain with atomisation, electron gain and hydration why F₂ is the strongest oxidant, and point out the anomalous F–F bond
Predict displacement reactions between halogens from standard potentials and identify the halogen formed by the colour of an organic layer
Meet the halogen oxoacids, the interhalogens and the compounds of xenon, and explain the shapes of XeF₂ and XeF₄ with VSEPR

Fluorine is the most ferocious element in chemistry: it sets glass and water alight and even reacts with the noble gas xenon. Yet the data table says that a fluorine atom releases less energy on gaining an electron (its electron affinity) than a chlorine atom does. How can the strongest oxidant not be the atom most eager for an electron?

Predict first

Which halogen atom has the largest electron affinity (releases the most energy on gaining an electron)?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Fluorine Chlorine Bromine Iodine

Down group 17

The halogens exist as diatomic molecules $\ce{X2}$ . As the number of electrons grows, dispersion forces strengthen: fluorine (boiling point −188 °C) and chlorine (−34 °C) are gases, bromine is a deep red-brown liquid boiling at 59 °C, and iodine is a violet-black solid that melts at 114 °C and sublimes readily.

Their colours come from absorbing visible light: the bigger the molecule, the lower its lowest electronic transition, so the absorption moves from the ultraviolet for fluorine (about 285 nm) through 330 nm for chlorine and about 415 nm for bromine to about 520 nm for iodine (it absorbs green, so iodine vapour is violet).

Oxidising power falls down the group: $E^\circ(\ce{X2/X-})$ = +2.866, +1.358, +1.066 and +0.536 V.

\tfrac12\ce{X2} \xrightarrow{\Delta_\mathrm{at}H} \ce{X(g)} \xrightarrow{-EA} \ce{X^-(g)} \xrightarrow{\Delta_\mathrm{hyd}H} \ce{X^-(aq)}

Three steps from halogen to hydrated halide ion: the more negative their sum, the stronger X₂ is as an oxidant.

$\Delta_\mathrm{at}H$: enthalpy of atomisation: half the X–X bond enthalpy (plus vaporisation or sublimation for Br₂ and I₂)
$EA$: electron affinity (energy released on gaining an electron)
$\Delta_\mathrm{hyd}H$: hydration enthalpy of X⁻; more negative for smaller ions

Filling in the numbers (kJ/mol; atomisation / electron gain / hydration → total): fluorine +79.4 / −328.2 / −522.9 → −771.7; chlorine +121.3 / −348.6 / −378.9 → −606.2; bromine +111.9 / −324.5 / −347.9 → −560.6; iodine +106.8 / −295.2 / −305.8 → −494.2. Chlorine leads by 20 kJ/mol on electron affinity but loses to fluorine by 42 kJ/mol and 144 kJ/mol on the other two steps.

The anomalous F–F bond. From chlorine to iodine the X–X bond enthalpy falls as the atoms grow (243 → 193 → 151 kJ/mol), so fluorine “should” be strongest; in fact F–F is only 159 kJ/mol. The F₂ bond is just 141 pm long, and each atom carries three lone pairs: crowded that close, they repel and weaken the bond. In H–F only one partner has lone pairs, so there is no anomaly — H–F (571 kJ/mol) is one of the strongest single bonds.

Computed from thermochemistry, potentials and absorption bands

Trends in the halogens

Gas In water In hexane

very pale yellow 1 bar, 5 cm pathreacts violently with waterreacts violently with water

yellow-green 1 bar, 5 cm path very pale yellow-green (almost colourless) 0.01 M, 1.5 cm path very pale yellow-green 0.01 M, 1.5 cm path

red-brown 0.29 bar, 5 cm path yellow to orange 0.01 M, 1.5 cm path orange to orange-red 0.01 M, 1.5 cm path

violet 0.01 bar, 5 cm path yellow to brown 0.0013 M, 1.5 cm path violet-purple 0.01 M, 1.5 cm path

At room temperature shiny grey-black solid that sublimes to a violet vapour

Molar absorptivity ε(λ) · I₂

Gas
In water
In hexane

Three tabs: the energy cycle, bond enthalpies and colours. On the colour tab compare each halogen as a gas, in water and in hexane: iodine is violet in hexane but yellow-brown in water. Pick a halogen to see how its absorption spectrum shifts with the solvent.

Displacement reactions

A stronger oxidant oxidises the halide of a weaker one to the element:

\ce{Cl2 + 2Br- -> 2Cl- + Br2}\qquad \ce{Cl2 + 2I- -> 2Cl- + I2}\qquad \ce{Br2 + 2I- -> 2Br- + I2}

The reverse reactions ( $\ce{Br2}$ with $\ce{Cl-}$ , $\ce{I2}$ with $\ce{Br-}$ ) do not happen. In water the halogens formed look much alike; shake with hexane and nearly all of the halogen moves into the upper layer, where bromine is orange and iodine violet — unmistakable. In water containing $\ce{I-}$ , iodine also forms brown $\ce{I3-}$ , which is why the water layer is often brown.

Halogen water + halide solution + hexane: colours computed from absorption spectra

Halogen displacement bench

Halogen water (0.010 M X₂, 2 mL)

Halide solution (0.10 M, 2 mL)

1 Pour in the halide solution → 2 Add the halogen water → 3 Add 2 mL hexane → 4 Stopper and shake

Work through steps 1–4 and try all nine combinations; watch the colour of the hexane in the results table. The bench also reports E°cell and K computed from the potentials: every reaction that goes has K above 10⁹, every one that does not has K below 10⁻⁹.

Worked example

Chlorine water and potassium bromide: how complete?

From $E^\circ(\ce{Cl2/Cl-}) = +1.3583$ V and $E^\circ(\ce{Br2/Br-}) = +1.066$ V, find the equilibrium constant of $\ce{Cl2} + \ce{2Br-} \mathrel{\ce{->}} \ce{2Cl-} + \ce{Br2}$ at 25 °C.

Oxoacids and interhalogens

Chlorine dissolves in water and partly disproportionates into hydrochloric and hypochlorous acids: $\ce{Cl2} + \ce{H2O} \mathrel{\ce{<=>}} \ce{HCl} + \ce{HOCl}$ ; HOCl is what disinfects tap water. In cold alkali, $\ce{Cl2} + \ce{2OH-} \mathrel{\ce{->}} \ce{Cl-} + \ce{ClO-} + \ce{H2O}$ gives bleach; in hot concentrated alkali chlorate forms instead: $\ce{3Cl2} + \ce{6OH-} \mathrel{\ce{->}} \ce{5Cl-} + \ce{ClO3-} + \ce{3H2O}$ . The chlorine oxoacids HOCl, $\ce{HClO2}$ , $\ce{HClO3}$ and $\ce{HClO4}$ get stronger as terminal oxygens are added (p $K_\mathrm{a}$ about 7.53, 1.96, −1 and −10); see what makes an acid strong.

Halogens also bond to each other, with the larger halogen in the middle: ClF, $\ce{ClF3}$ (T-shaped), $\ce{BrF5}$ (square pyramidal) and $\ce{IF7}$ (pentagonal bipyramidal) — shapes that VSEPR explains.

Noble gases are not quite inert

The noble gases have full valence shells and very high ionisation energies (helium 2372 kJ/mol), and for decades they were thought to form no compounds. In 1962 Bartlett noticed that the ionisation energy of $\ce{O2}$ (12.07 eV) is almost the same as xenon’s (12.13 eV), and that $\ce{PtF6}$ oxidises $\ce{O2}$ to $\ce{O2+[PtF6]-}$ — so he mixed xenon with $\ce{PtF6}$ and made the first noble-gas compound.

Many xenon compounds are now known: linear $\ce{XeF2}$ (three lone pairs and two bonds on xenon, AX₂E₃), square-planar $\ce{XeF4}$ (AX₄E₂), distorted-octahedral $\ce{XeF6}$ , and the treacherously explosive $\ce{XeO3}$ ( $E^\circ(\ce{XeO3/Xe}) = +2.10$ V, a formidable oxidant). Krypton forms $\ce{KrF2}$ , and in 2000 even an argon compound, HArF, was made at low temperature. The lighter the noble gas, the higher its ionisation energy and the harder it is to oxidise: helium and neon still have no stable neutral compounds.

Valence-shell electron-pair repulsion · 3D

VSEPR shape explorer

Loading 3D view…

Drag to rotate · scroll or pinch to zoom · arrow keys rotate too

Xenon tetrafluoride

Show lone pairs Show bond dipoles Click atoms to measure

VSEPR notation: $\mathrm{AX_{4}E_{2}}$
Electron-domain geometry: Octahedral
Molecular shape: Square planar
Ideal angle: 90°
Measured angle: 90.0°
Bond-dipole sum: non-polar (dipoles cancel)

Rotate XeF₄: its two lone pairs sit opposite each other above and below, leaving the four F atoms in one plane. Then compare XeF₂ with ClF₃ — why do all three lone pairs stay in the equatorial plane?

Concept check

Each pair of solutions is mixed and then shaken with hexane. In which case does a displacement reaction turn the hexane violet?

Bromine water + KCl solution Iodine water + KBr solution Chlorine water + KI solution Chlorine water + KCl solution

Try the calculation

Find $\lg K$ at 25 °C for $\ce{Br2} + \ce{2I-} \mathrel{\ce{->}} \ce{2Br-} + \ce{I2}$ . ( $E^\circ(\ce{Br2/Br-}) = +1.066$ V, $E^\circ(\ce{I2/I-}) = +0.5355$ V)

Key takeaways

From F₂ to I₂ dispersion forces strengthen (gas → liquid → solid), absorption moves to longer wavelengths and oxidising power falls (E° 2.87 → 0.54 V).
Chlorine has the largest electron affinity, but F₂ has a weak F–F bond and F⁻ a huge hydration enthalpy, so ½X₂ → X⁻(aq) is most exothermic for fluorine and F₂ is the strongest oxidant.
A stronger halogen oxidises the halide of a weaker one: Cl₂ displaces Br₂ and I₂, and Br₂ displaces I₂; orange hexane means bromine, violet means iodine.
The heavier noble gases, with lower ionisation energies, bond to fluorine and oxygen: XeF₂ is linear, XeF₄ square planar, and XeO₃ an explosive oxidant.

Progress is stored only in this browser and never uploaded.

Data and references

P. Vanýsek, “Electrochemical series”, CRC Handbook of Chemistry and Physics, 91st ed. (2010).
J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics (1989); D. D. Wagman et al., NBS tables (1982).
K. P. Huber, G. Herzberg, Constants of Diatomic Molecules, Van Nostrand Reinhold (1979).
N. Bartlett, “Xenon hexafluoroplatinate(V) Xe⁺[PtF₆]⁻”, Proc. Chem. Soc. 218 (1962).
L. Khriachtchev et al., “A stable argon compound”, Nature 406, 874 (2000).
N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed. (1997), ch. 17–18 (halogens, noble gases).