Topic 8 · page 29 of 29

Predicting fluorine and astatine

Fluorine and astatine sit at the two ends of Group 7. Apply the trends you know, in the same direction, to predict their state, colour, oxidising power and reactions, and the properties of their compounds.

12 min read8.28

Read first: Silver nitrate test and hydrogen halides

Using the trends 8.28

You have met the trends in chlorine, bromine and iodine. Fluorine is above chlorine in the group and astatine is below iodine. To predict their properties, continue each trend one step in the same direction.

	F₂	Cl₂	Br₂	I₂	At₂ (predicted)
State at room temperature	gas	gas	liquid	solid	solid
Colour	pale yellow	pale green	red-brown	grey-black	black
Melting temperature / °C	−220	−101	−7	114	about 302 (estimate)
Boiling temperature / °C	−188	−34	59	184	about 337 (estimate)
Electronegativity	4.0	3.0	2.8	2.5	about 2.2 (estimate)
Oxidising power of the halogen	strongest	strong	moderate	weak	weakest
Reducing power of the halide ion	weakest	weak	moderate	strong	strongest

Astatine is radioactive and extremely rare, so its values are estimates. Nobody has made a visible sample.

Fluorine 8.28

Fluorine is at the top of the group, so it is:

a pale yellow gas: the smallest molecule, with the fewest electrons, so the weakest London forces and the lowest boiling temperature of the group.
the most electronegative element of all (4.0).
the strongest oxidising agent in the group, with the most reactive atoms. It gains an electron most easily, because the incoming electron enters the shell closest to the nucleus with the least shielding.
able to displace all the other halogens from their halides: $\ce{F2 + 2Cl- -> 2F- + Cl2}$ .
violently reactive with metals, giving ionic fluorides: $\ce{2Na + F2 -> 2NaF}$ .

Compounds of fluorine are predicted in the same way. The fluoride ion is the weakest reducing agent in the group: it is not oxidised by concentrated sulfuric acid at all, so NaF gives HF without any redox change, like NaCl. Silver fluoride is soluble, so there is no precipitate with silver nitrate.

Astatine 8.28

Astatine is at the bottom of the group, so it is:

a black solid, with the highest melting and boiling temperatures, because the At₂ molecule has the most electrons and the strongest London forces.
the least reactive halogen, and the weakest oxidising agent: At₂ would not displace iodide ions: there would be no reaction between At₂ and I⁻.
forming the astatide ion, At⁻, which is the strongest reducing agent in the group. With concentrated sulfuric acid it would reduce sulfur at least as far as iodide does, to H₂S.
giving a precipitate with silver nitrate that is insoluble in ammonia, like AgI.
reacting with hot alkali by disproportionation, giving astatide and astatate(V): $\ce{3At2 + 6NaOH -> 5NaAt + NaAtO3 + 3H2O}$ .
forming hydrogen astatide, HAt, with a very weak H–At bond, which would decompose easily and be the strongest reducing agent of the hydrogen halides.

Worked examplePredicting a reaction of astatine

Predict whether astatine reacts with a solution of sodium iodide, and whether it reacts with sodium metal.

With NaI(aq): astatine is a weaker oxidising agent than iodine, because reactivity falls down Group 7. It cannot take electrons from I⁻. No reaction.
With sodium: every halogen oxidises Group 1 metals, even the weakest. Predict $\ce{2Na(s) + At2(s) -> 2NaAt(s)}$ , sodium astatide, an ionic solid. Na: 0 → +1; At: 0 → −1.

How to answer a prediction question 8.28

MethodPredicting for a halogen you have not met

Find the element's position in the group, and the trend you are asked about.
State the trend, with its reason (electrons, shielding, distance).
Carry the trend on to the new element, and say whether it is greater or less than the one before it.
For a reaction, copy the equation from the nearest halogen, and check the oxidation numbers.

Check yourself

Predict the state and colour of astatine at room temperature, and explain the prediction.
Predict whether fluorine would displace chloride ions from sodium chloride solution, and explain in terms of electrons.
Write the equation for the reaction of astatine with hot sodium hydroxide solution, and state the oxidation numbers of astatine.
Silver nitrate gives a precipitate with astatide ions. Predict whether it dissolves in concentrated ammonia, and explain.
State the trend in oxidising power down Group 7, and place F₂, Cl₂, Br₂, I₂ and At₂ in order.
Why would fluorine's boiling temperature be predicted to be the lowest in the group, and why is hydrogen fluoride an exception to the pattern for the hydrogen halides?

Answers cover these until you have tried the questions

A black solid. Down the group the molecules have more electrons, so London forces are stronger and the melting and boiling temperatures rise, and the colours darken (pale green, red-brown, grey-black). The next step is black.
Yes. Fluorine is a stronger oxidising agent than chlorine: it gains electrons more easily, as the incoming electron enters a shell closer to the nucleus with less shielding. So $\ce{F2 + 2Cl- -> 2F- + Cl2}$ .
$\ce{3At2 + 6NaOH -> 5NaAt + NaAtO3 + 3H2O}$ . Astatine: 0 in At₂, −1 in NaAt and +5 in NaAtO₃ (a disproportionation).
It would not dissolve. The silver halides become less soluble in ammonia down the group: AgCl dissolves in dilute ammonia, AgBr in concentrated ammonia only, and AgI in neither. Silver astatide continues the trend, so it is insoluble in concentrated ammonia.
Oxidising power decreases down the group: F₂ > Cl₂ > Br₂ > I₂ > At₂.
Fluorine has the fewest electrons, so it has the weakest London forces, so the lowest boiling temperature. HF is an exception because it forms hydrogen bonds, which are stronger than the London forces in the other hydrogen halides, so its boiling temperature is out of line (20 °C). It is also only a weak acid in water.