The noble gases and the compounds of xenon

The gases that did not react

A balloon filled with helium rises. A sign filled with neon glows orange-red. A light bulb filled with argon lasts longer. All three gases are noble gases, the elements of group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). Radon is radioactive.

Argon was found in 1894 by Rayleigh and Ramsay. They noticed that nitrogen taken from air was slightly denser than nitrogen made from ammonia, and the extra weight was a new gas. Ramsay found the others over the next few years. Their atoms have a complete outer shell, $ns^2np^6$, and helium has $1s^2$. For about sixty years chemists thought that the gases could not form compounds at all.

The trends

• Atomic radius in pm: He 120, Ne 160, Ar 190, Kr 200, Xe 220. These are van der Waals radii, because the atoms are not bonded. They look bigger than the covalent radii of the other elements in the period for this reason.
• First ionisation enthalpy in kJ/mol: He 2372, Ne 2080, Ar 1520, Kr 1351, Xe 1170, Rn 1037. It is the highest of every period and falls down the group.
• Boiling point in K: He 4.2, Ne 27, Ar 87, Kr 120, Xe 165, Rn 211. The atoms attract each other only by weak dispersion forces, and these grow with the number of electrons. Helium has the lowest boiling point of any substance.
• Abundance: argon makes up about 0.93 percent of dry air by volume, much more than the others. Helium collects in natural gas wells.

The gases come from air by the fractional distillation of liquid air. Helium comes from natural gas. The uses: helium for balloons and for cooling the magnets of scanners (liquid helium at 4.2 K), and in the breathing mixture of divers; neon in lighting signs; argon to give an inert atmosphere for welding and in bulbs; xenon in lamps and in some anaesthetics.

Bartlett and the first compound

In 1962 Neil Bartlett made a red solid from oxygen and platinum hexafluoride, $\text{O}_2^+[\text{PtF}_6]^-$. He noticed that the ionisation enthalpy of the oxygen molecule, 1175 kJ/mol, is almost the same as that of xenon, 1170 kJ/mol. So he tried xenon with the same reagent, and a solid formed that was first written $\text{Xe}^+[\text{PtF}_6]^-$. Later work showed that the product is a mixture of compounds, but the idea was right. Within months, xenon and fluorine were made to give XeF4.

Compounds of the lighter gases, helium, neon and argon, are not known under ordinary conditions. Krypton forms KrF2, and radon forms fluorides.

The fluorides of xenon

Xenon reacts directly with fluorine. NCERT gives these conditions:

• $\text{Xe}\ +\ \text{F}_{2}\ \xrightarrow{673\ \text{K},\ \ 1\ \text{bar}}\ \text{XeF}_{2}$
• $\text{Xe}\ +\ 2\text{F}_{2}\ \xrightarrow{873\ \text{K},\ \ 7\ \text{bar}}\ \text{XeF}_{4}$
• $\text{Xe}\ +\ 3\text{F}_{2}\ \xrightarrow{573\ \text{K},\ \ 60\ \text{to}\ 70\ \text{bar}}\ \text{XeF}_{6}$

Other books use a ratio to control the product: an excess of xenon for XeF2 (2 : 1), 1 : 5 of xenon to fluorine for XeF4 and 1 : 20 for XeF6. Another route to XeF6 is $\text{XeF}_{4}\ +\ \text{O}_{2}\text{F}_{2}\ \rightarrow \ \text{XeF}_{6}\ +\ \text{O}_{2}$. The three are colourless crystalline solids. They are strong fluorinating agents.

Shapes. Xenon has 8 outer electrons. Each F uses one, so the lone pairs are $\dfrac{8 - n}{2}$ for $n$ fluorine atoms.

• XeF2: 2 bonds and 3 lone pairs. Steric number 5, $sp^3d$. The lone pairs lie in the middle plane, so the atoms are linear, F-Xe-F at 180°.
• XeF4: 4 bonds and 2 lone pairs. Steric number 6, $sp^3d^2$. The lone pairs sit above and below the plane, so the atoms are square planar.
• XeF6: 6 bonds and 1 lone pair. Steric number 7, $sp^3d^3$. The shape is a distorted octahedron.

Hydrolysis and the oxides

The fluorides react with water. The oxidation state of xenon tells us what to expect:

• $2\text{XeF}_{2}\ +\ 2\text{H}_{2}\text{O}\ \rightarrow \ 2\text{Xe}\ +\ 4\text{HF}\ +\ \text{O}_{2}$
• $6\text{XeF}_{4}\ +\ 12\text{H}_{2}\text{O}\ \rightarrow \ 4\text{Xe}\ +\ 2\text{XeO}_{3}\ +\ 24\text{HF}\ +\ 3\text{O}_{2}$. This is a disproportionation of xenon($+4$): it goes to 0 and to $+6$.
• $\text{XeF}_{6}\ +\ 3\text{H}_{2}\text{O}\ \rightarrow \ \text{XeO}_{3}\ +\ 6\text{HF}$. This is complete hydrolysis, and the oxidation state stays at $+6$.
• $\text{XeF}_{6}\ +\ \text{H}_{2}\text{O}\ \rightarrow \ \text{XeOF}_{4}\ +\ 2\text{HF}$ and $\text{XeF}_{6}\ +\ 2\text{H}_{2}\text{O}\ \rightarrow \ \text{XeO}_{2}\text{F}_{2}\ +\ 4\text{HF}$ are the partial steps.

To count the pairs in an oxide, treat each Xe=O as one region of electron density that uses two electrons of xenon.

• XeO3: 3 oxygen atoms use 6 of the 8 electrons. One lone pair remains. Steric number 4, $sp^3$. The shape is a trigonal pyramid. It is explosive.
• XeOF4: one O and four F. One lone pair. Steric number 6, $sp^3d^2$. Square pyramidal.
• XeO2F2: two O and two F. One lone pair. Steric number 5, $sp^3d$. See-saw.
• XeO4: four O, no lone pair. Tetrahedral.

Xenon fluorides also give and take fluoride ions. XeF2 gives one to a strong acceptor: $\text{XeF}_{2}\ +\ \text{PF}_{5}\ \rightarrow \ [\text{XeF}]^{+}[\text{PF}_{6}]^{-}$. XeF4 and XeF6 behave in the same way: $\text{XeF}_{4}\ +\ \text{SbF}_{5}\ \rightarrow \ [\text{XeF}_{3}]^{+}[\text{SbF}_{6}]^{-}$ and $\text{XeF}_{6}\ +\ \text{BF}_{3}\ \rightarrow \ [\text{XeF}_{5}]^{+}[\text{BF}_{4}]^{-}$. XeF6 can also take a fluoride ion from sodium fluoride: $\text{XeF}_{6}\ +\ \text{NaF}\ \rightarrow \ \text{Na}^{+}[\text{XeF}_{7}]^{-}$.

Choose 1 fluorine molecule and read the bar: 2 bonded atoms and 3 lone pairs, so 5 regions and a linear shape. Choose 3 and the red bar shrinks to one lone pair. The mass of the product follows from the molar mass.

Practise with new numbers

Try it in Python

The program counts the lone pairs of seven xenon compounds and names their shapes. A double bond to oxygen counts as one region and uses 2 electrons of xenon.

Common mistakes

• Saying that all noble gases form compounds. Only xenon, krypton and radon form well known ones.
• Forgetting that a double bond to oxygen is one region. XeO3 has a steric number of 4, not 6.
• Putting the lone pairs of XeF4 in the plane. They are above and below it, and that keeps the molecule square planar.
• Using the covalent radius for the noble gases. The radii quoted are van der Waals radii.

Remember these

• The noble gases are $ns^2np^6$. Ionisation enthalpy is the highest in each period. Boiling points rise down the group, with He at 4.2 K.
• Bartlett, 1962: xenon and PtF6. Xenon fluorides: XeF2 (linear), XeF4 (square planar), XeF6 (distorted octahedral).
• Lone pairs on xenon: XeF2 3, XeF4 2, XeF6 1. Hybridisation $sp^3d$, $sp^3d^2$, $sp^3d^3$.
• Hydrolysis gives XeO3 (pyramidal), XeOF4 (square pyramidal) or XeO2F2 (see-saw).

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why do the noble gases have the highest ionisation enthalpy in their period?

2. How many lone pairs are on xenon in XeF4, and what shape does it have?

3. What happens when XeF6 is hydrolysed completely?

4. Why was Bartlett led to try xenon with platinum hexafluoride?

References

• NCERT, Chemistry Part I for Class XII, the chapter The p-Block Elements, on group 18.
• OpenStax, Chemistry 2e, Chapter 18 (the noble gases) and Chapter 7 (molecular geometry).