Properties and Uses of Group 18 Elements

Lessons

Senior 4 · Chemistry

Properties and Uses of Group 18 Elements

3 lessons

Lesson 2 of 3

Reactivity and Inertness of the Noble Gases

25 min

Seven o'clock on a Friday evening

For seventy years every textbook called Group 18 the "inert gases" and Group 0, "valency zero". In 1962 Neil Bartlett, a young chemist at the University of British Columbia in Canada, found that platinum hexafluoride, $\mathrm{PtF_6}$ , is so strong an oxidising agent that it takes an electron from an oxygen molecule, forming the red solid $\mathrm{O_2^+[PtF_6]^-}$ . He then noticed that the energy needed to remove an electron from $\mathrm{O_2}$ (1175 kJ mol⁻¹) is almost the same as the first ionisation energy of xenon (1170 kJ mol⁻¹). At about 7 p.m. on Friday 23 March 1962 he let red $\mathrm{PtF_6}$ vapour meet colourless xenon. A deep-yellow solid formed at once, with the overall composition $\mathrm{XePtF_6}$ . A "noble" gas had reacted. This lesson explains why the noble gases are so unreactive, why the heavier ones are not completely so, and what their compounds look like.

Definition

Inert: having no tendency to react at all. The noble gases were once called the "inert gases"; the name was dropped after 1962 because xenon, krypton and radon form compounds.

Noble gas compound: a compound in which a noble gas atom is chemically bonded to another atom. Example: xenon tetrafluoride, $\mathrm{XeF_4}$ , in which xenon has the oxidation state +4.

Hypervalent (expanded octet): describing a central atom with more than eight electrons in its outer shell in a compound. Example: xenon in $\mathrm{XeF_2}$ has five electron pairs around it.

Why the noble gases are so unreactive

An atom reacts by losing, gaining or sharing electrons. A noble gas atom resists all three:

Losing electrons is very costly. Each noble gas has the highest first ionisation energy in its period (Figure 13.3). The outer electrons are in a full shell, close to a large effective nuclear charge.
Gaining electrons gives nothing back. An extra electron would have to go into a new, outer shell, well shielded from the nucleus. The noble gases do not form stable negative ions.
Sharing electrons needs unpaired electrons. All the electrons are paired in full s and p sub-shells. To form covalent bonds an atom must first promote electrons into a higher, empty sub-shell, and only the strongest bonds repay that cost.

Figure 13.3: First ionisation energies of hydrogen to calcium. Every peak is a noble gas; every sharp drop after a peak is an alkali metal, starting a new shell.

Why xenon reacts but helium does not

Down the group the outer electrons are further from the nucleus and better shielded, so ionisation energy falls from 2372 kJ mol⁻¹ (He) to 1170 kJ mol⁻¹ (Xe) and 1037 kJ mol⁻¹ (Rn). The large xenon atom also has empty 5d orbitals not far above its 5p orbitals, so, in the simple model used at this level, electrons can be promoted to give unpaired electrons for bonding. (Modern bonding theory describes these bonds differently, but it agrees on the main point.) Even so, xenon reacts only with the two most electronegative elements, fluorine and oxygen, which form bonds strong enough to pay for disturbing the full shell.

Table 13.3: Known compounds of the noble gases

Element	First IE / kJ mol⁻¹	Compounds
He, Ne	2372, 2081	no stable neutral compounds known
Ar	1521	argon fluorohydride, HArF, made in 2000 and stable only at about 40 K
Kr	1351	krypton difluoride, $\mathrm{KrF_2}$ (1963), and a few others
Xe	1170	about five hundred: $\mathrm{XeF_2}$ , $\mathrm{XeF_4}$ , $\mathrm{XeF_6}$ , $\mathrm{XeO_3}$ , $\mathrm{XeO_4}$ and many more
Rn	1037	$\mathrm{RnF_2}$ ; radon reacts more easily than xenon, but its radioactivity makes it hard to study

The xenon fluorides

Xenon and fluorine combine directly. Which fluoride forms depends on the ratio of the gases and the conditions:

\mathrm{Xe(g) + F_2(g) \rightarrow XeF_2(s)} \qquad \text{(xenon in excess; heat, or ultraviolet light or even sunlight)}

\mathrm{Xe(g) + 2F_2(g) \rightarrow XeF_4(s)} \qquad \text{(Xe : F}_2 = 1 : 5\text{, 400 °C, nickel vessel)}

\mathrm{Xe(g) + 3F_2(g) \rightarrow XeF_6(s)} \qquad \text{(large excess of F}_2\text{, about 300 °C and 60 atm)}

All three are colourless crystalline solids, formed exothermically, and all are powerful oxidising and fluorinating agents: xenon's oxidation state falls back to 0 when they react. The nickel vessel does not catalyse the reaction; a thin layer of nickel fluoride protects the metal from attack. $\mathrm{XeF_6}$ reacts with water to give xenon trioxide, a dangerously explosive solid: $\mathrm{XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF}$

Shapes from VSEPR

The shapes follow from counting electron pairs around xenon, as in Unit 4. Xenon brings 8 outer electrons; each fluorine supplies 1 to its bond.

Compound	Oxidation state of Xe	Bonding pairs	Lone pairs	Shape
$\mathrm{XeF_2}$	+2	2	3	linear
$\mathrm{XeF_4}$	+4	4	2	square planar
$\mathrm{XeF_6}$	+6	6	1	distorted octahedral
$\mathrm{XeO_3}$	+6	3 (double bonds)	1	trigonal pyramidal

Figure 13.4: The shapes of $\mathrm{XeF_2}$ and $\mathrm{XeF_4}$ . Lone pairs repel most strongly, so they take the positions furthest apart: around the middle of $\mathrm{XeF_2}$ , and above and below the square of $\mathrm{XeF_4}$ .

Worked example 1: the shape of $\mathrm{XeF_4}$

Use VSEPR theory to predict the shape and bond angle of $\mathrm{XeF_4}$ , and give the oxidation state of xenon.

Solution. Electrons around Xe: 8 (its own) + 4 (one from each F) = 12, which is 6 pairs. Four pairs bond to F atoms; the other two are lone pairs. Six pairs point to the corners of an octahedron. The two lone pairs repel most, so they sit opposite each other (above and below), leaving the four F atoms in one plane: the molecule is square planar, with F–Xe–F angles of 90°. Fluorine is always −1, so $x + 4 (- 1) = 0$ and $x = + 4$ .

Worked example 2: identifying a xenon fluoride

A colourless xenon fluoride contains 53.5% xenon by mass. Find its empirical formula. $A_{r}$ : Xe = 131.29, F = 19.00.

Solution. In 100 g there are 53.5 g Xe and 46.5 g F. $n(\mathrm{Xe}) = \frac{53.5}{131.29} = 0.4075\ \mathrm{mol} \qquad n(\mathrm{F}) = \frac{46.5}{19.00} = 2.447\ \mathrm{mol}$ $\frac{n(\mathrm{F})}{n(\mathrm{Xe})} = \frac{2.447}{0.4075} = 6.01 \approx 6$ The formula is $\mathrm{XeF_6}$ . Check: $\frac{131.29}{245.29} \times 100 = 53.5\%$ . It was made with a large excess of fluorine at high pressure.

Warning

Common mistakes. (1) Saying the noble gases "cannot react": xenon, krypton and radon form compounds, which is why "inert" was dropped. (2) Expecting helium to be the most reactive because it is the smallest: it has the highest ionisation energy of all elements and is the least reactive. (3) Drawing $\mathrm{XeF_2}$ as bent like water: its three lone pairs sit around the middle, so it is linear. (4) Forgetting lone pairs when naming a shape: $\mathrm{XeF_4}$ has six electron pairs but a square planar shape.

Safety. Fluorine and the xenon fluorides are far too dangerous for a school laboratory: they attack glass and skin, and with water they release hydrogen fluoride, which causes deep burns. Xenon trioxide explodes. These compounds are made only in specialised laboratories.

Tip

"Big, and only with F or O." A noble gas reacts only if it is big enough for its outer electrons to be pulled away (Kr, Xe, Rn), and only by the most electronegative elements (fluorine and oxygen).

Periodic Videos: Xenon Bonds (7 min). A professor at the University of Nottingham recalls stories from the chemistry of “unreactive” xenon. Watch for which elements xenon actually bonds to.

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Practice

Give three reasons, in terms of electrons, why argon does not react with sodium or chlorine.
Explain why the first ionisation energy of xenon (1170 kJ mol⁻¹) made Bartlett think that xenon might react with $\mathrm{PtF_6}$ .
Explain why xenon forms compounds but neon does not.
Give the oxidation state of xenon in $\mathrm{XeF_2}$ , $\mathrm{XeO_3}$ and $\mathrm{XeO_4}$ .
Use VSEPR theory to explain why $\mathrm{XeF_2}$ is linear.
Calculate the percentage by mass of fluorine in $\mathrm{XeF_2}$ .
0.500 g of xenon is converted completely into $\mathrm{XeF_4}$ . Calculate the mass of $\mathrm{XeF_4}$ formed.
A krypton fluoride contains 68.8% krypton by mass. Find its formula ( $A_{r}$ : Kr = 83.80, F = 19.00) and predict its shape.

Show answersHide answers

Its outer shell ( $3s^2\,3p^6$ ) is full; its first ionisation energy (1521 kJ mol⁻¹) is very high, so it will not lose an electron; an extra electron would have to enter a new, well-shielded shell, so it will not gain one; it has no unpaired electrons to share without costly promotion.
$\mathrm{PtF_6}$ can remove an electron from $\mathrm{O_2}$ , which needs 1175 kJ mol⁻¹. Removing one from xenon needs almost the same energy (1170), so $\mathrm{PtF_6}$ should oxidise xenon too.
Xenon's outer electrons are far from the nucleus and well shielded (IE 1170 kJ mol⁻¹), and it has empty 5d orbitals close in energy. Neon's electrons are close to the nucleus (IE 2081 kJ mol⁻¹) and it has no low-lying empty orbitals in its outer shell, so no bond is strong enough to repay the energy.
+2; +6; +8.
Xe has 8 + 2 = 10 electrons, 5 pairs: 2 bonding and 3 lone pairs, in a trigonal bipyramid. The 3 lone pairs take the equatorial positions (most room), so the two F atoms are axial, 180° apart.
$M(\mathrm{XeF_2}) = 169.29$ ; % F = $\frac{38.00}{169.29} \times 100 = 22.4\%$ .
$n = \frac{0.500}{131.29} = 3.81 \times 10^{-3}\ \mathrm{mol}$ ; mass $= 3.81 \times 10^{-3} \times 207.29 = 0.789\ \mathrm{g}$ .
Kr: $\frac{68.8}{83.80} = 0.821$ ; F: $\frac{31.2}{19.00} = 1.642$ ; ratio 1 : 2, so $\mathrm{KrF_2}$ . Like $\mathrm{XeF_2}$ , 2 bonding + 3 lone pairs: linear.

Quick check

Which noble gas forms the largest number of known compounds?

Key points

Noble gases are unreactive because they have full outer shells, the highest ionisation energy in each period, no tendency to gain electrons and no unpaired electrons.
Reactivity increases down the group as ionisation energy falls: He and Ne form no stable neutral compounds; Kr, Xe and Rn react with fluorine (and Xe with oxygen).
Bartlett made the first noble gas compound in 1962 ( $\mathrm{XePtF_6}$ ). $\mathrm{XeF_2}$ , $\mathrm{XeF_4}$ and $\mathrm{XeF_6}$ form directly from the elements; they are strong oxidising agents.
VSEPR shapes: $\mathrm{XeF_2}$ linear, $\mathrm{XeF_4}$ square planar, $\mathrm{XeF_6}$ distorted octahedral, $\mathrm{XeO_3}$ trigonal pyramidal.

Sources and credits

Image: HD.6B.090 (10578112165).jpg by the U.S. Department of Energy, via Wikimedia Commons, public domain.
Text adapted from Noble gas, Xenon hexafluoroplatinate, Xenon tetrafluoride and Xenon hexafluoride; ionisation energies from Molar ionization energies of the elements; all by Wikipedia contributors, licensed under CC BY-SA 4.0, adapted by Ganzaa.
Video: Xenon Bonds - Periodic Table of Videos, Periodic Videos, embedded from YouTube.
The text, tables, charts, diagrams, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Seven o'clock on a Friday evening

2Why the noble gases are so unreactive

3Why xenon reacts but helium does not

4The xenon fluorides

5Shapes from VSEPR

Seven o'clock on a Friday evening

Why the noble gases are so unreactive

Why xenon reacts but helium does not

The xenon fluorides

Shapes from VSEPR