Group 18 Elements: The Noble Gases

Q: Why are group 18 elements so unreactive?

Group 18 elements have a complete octet, ns²np⁶ (helium is 1s²). They therefore have the highest ionisation enthalpies in their periods, so they do not lose electrons, and positive electron gain enthalpies, so they do not gain electrons either. With no tendency to share, they stay monatomic.

Q: What was the first noble gas compound?

Xenon hexafluoroplatinate, written [Xe]⁺[PtF₆]⁻, a red crystalline solid made by Neil Bartlett in 1962. He made it after finding that PtF₆ oxidises dioxygen to [O₂]⁺[PtF₆]⁻, and noting that xenon and O₂ have almost the same ionisation enthalpy, 1170 and 1175 kJ per mol. Xenon fluorides and oxides followed within months.

Q: What are the shapes of the xenon fluorides?

XeF₂ is linear (sp³d, three lone pairs), XeF₄ is square planar (sp³d², two lone pairs) and XeF₆ is a distorted octahedron (sp³d³, one lone pair). In each case the lone pairs take the least crowded positions, which is what decides the shape.

ChemistryThe p-Block Elements: Group 15, 16, 17 And 18For JEE aspirants

Group 18 elements are the noble gases helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radioactive radon (Rn). All are monatomic, colourless gases with a complete octet ( $n s^{2} n p^{6}$ ; helium is $1 s^{2}$ ), the highest ionisation enthalpies in their periods and positive electron gain enthalpies, which is why the group 18 elements are so unreactive. This page covers their trends, isolation, the xenon fluorides, oxides and oxyfluorides with their shapes, and their uses. JEE Main and NEET test the trends; JEE Advanced also asks about xenon compounds.

On this page1At a glance2Trends3Isolation4Bartlett5Xenon fluorides6Oxides and shapes7Uses8Solved examples

Key Formulas - Quick Reference

Configuration $n s^{2} n p^{6}$ (He: $1 s^{2}$ ); ionisation enthalpy is the highest in each period and electron gain enthalpy is positive.
Boiling points are very low but rise down the group (He 4.2 K to Rn 211 K) as dispersion forces grow.
★ Must learn First noble gas compound (Bartlett, 1962): $Xe + Pt F_{6} \to [Xe]^{+} [Pt F_{6}]^{-}$ , a red solid.
★ Must learn $Xe + F_{2} 673 K, 1 bar Xe F_{2}$ (2 : 1); $Xe + 2 F_{2} 873 K, 7 bar Xe F_{4}$ (1 : 5); $Xe + 3 F_{2} 573 K, 60 - 70 bar Xe F_{6}$ (1 : 20).
★ Must learn Complete hydrolysis: $Xe F_{6} + 3 H_{2} O \to Xe O_{3} + 6HF$ ; partial: $Xe F_{6} + H_{2} O \to XeO F_{4} + 2HF$ .
$2Xe F_{2} + 2 H_{2} O \to 2Xe + 4HF + O_{2}$ ; $6Xe F_{4} + 12 H_{2} O \to 4Xe + 2Xe O_{3} + 24HF + 3 O_{2}$ .
★ Must learn Shapes: $Xe F_{2}$ linear ( $s p^{3} d$ , 3 lp), $Xe F_{4}$ square planar ( $s p^{3} d^{2}$ , 2 lp), $Xe F_{6}$ distorted octahedral ( $s p^{3} d^{3}$ , 1 lp).
★ Must learn $Xe O_{3}$ pyramidal ( $s p^{3}$ , 1 lp); $XeO F_{4}$ square pyramidal ( $s p^{3} d^{2}$ , 1 lp); $Xe O_{2} F_{2}$ see-saw ( $s p^{3} d$ , 1 lp).
Xenon fluorides attack glass, $2Xe F_{6} + 3Si O_{2} \to 3Si F_{4} + 2Xe O_{3}$ , so they are kept in nickel or Monel vessels.
They are strong fluorinating agents and act as $F^{-}$ donors or acceptors: $Xe F_{6} + NaF \to Na [Xe F_{7}]$ .

1. The Noble Gases at a Glance

Group 18 contains helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn); the synthetic element oganesson (Og) also belongs here. They are called the noble gases (once the inert or rare gases) because of their very low chemical reactivity. All of them exist as single atoms, not molecules.

Occurrence. Except radon, all are present in the atmosphere: argon is the most abundant (0.93% of dry air by volume) and xenon the least. Helium is rare in air but is found in natural gas (up to about 2%) in the USA, Poland and Russia, and in minerals such as cleveite, where it comes from alpha decay. Radon is formed by the radioactive decay of radium.

Figure 1: Group 18 at a glance. Every electron gain enthalpy is positive, ionisation enthalpies are the highest in each period, and boiling points are very low but rise down the group as the atoms grow more polarisable.

1.1 How They Were Found

Gas	Discovery
Argon	Rayleigh and Ramsay (1894). Nitrogen from air was about 0.5% denser than nitrogen made chemically. They removed oxygen with heated copper and nitrogen with heated magnesium; the leftover gas had a vapour density of 20 and an atomic mass of 40
Helium	Janssen (1868) saw its lines in the Sun's spectrum during an eclipse, before it was found on Earth
Neon, krypton, xenon	Ramsay and Travers (1898), by the fractional distillation of liquid argon under reduced pressure
Radon	Dorn (1900), as a decay product of radium: $_{88}^{226} Ra$ gives $_{86}^{222} Rn$ + $_{2}^{4} He$

2Cu + O_{2} removing O_{2} 2CuO

3Mg + N_{2} removing N_{2} M g_{3} N_{2}

2. Trends in Physical Properties

Property	He	Ne	Ar	Kr	Xe	Rn
Configuration	$1 s^{2}$	[He] $2 s^{2} 2 p^{6}$	[Ne] $3 s^{2} 3 p^{6}$	[Ar] $3 d^{10} 4 s^{2} 4 p^{6}$	[Kr] $4 d^{10} 5 s^{2} 5 p^{6}$	[Xe] $4 f^{14} 5 d^{10} 6 s^{2} 6 p^{6}$
Atomic (van der Waals) radius / pm	120	160	190	200	220	-
1st ionisation enthalpy / kJ mol $^{- 1}$	2372	2080	1520	1351	1170	1037
Electron gain enthalpy / kJ mol $^{- 1}$	+48	+116	+96	+96	+77	+68
Melting point / K	-	24.6	83.8	115.9	161.3	202
Boiling point / K	4.2	27.1	87.2	119.7	165.0	211
In dry air / % by volume	$5.2 \times 1 0^{- 4}$	$1.8 \times 1 0^{- 3}$	0.934	$1.1 \times 1 0^{- 4}$	$8.7 \times 1 0^{- 6}$	trace

All are monatomic, colourless, odourless and tasteless gases, and only sparingly soluble in water; solubility increases down the group.
Melting and boiling points are extremely low, because the only forces between the atoms are weak London (dispersion) forces. These grow as the atoms become larger and more polarisable, so both rise from He to Rn. Helium has the lowest boiling point of any substance, 4.2 K, and does not solidify at all under ordinary pressure.
Ionisation enthalpies are the highest in each period, because the octet is complete; they fall down the group as the atom grows.
Electron gain enthalpies are positive: energy has to be supplied to add an electron, since the next electron would have to enter a new shell.
Radon is radioactive, formed from radium; it is the only member with no stable isotope.

Exam Trick

Everything about noble gases follows from 'complete octet, weak forces'. Complete octet gives the high ionisation enthalpy, the positive electron gain enthalpy and the inertness. Weak dispersion forces give the very low boiling points, the difficulty of liquefying them, and the steady rise in boiling point down the group (Solved Examples 1 and 3).

Key idea

Noble gases are unreactive because of a complete octet, very high ionisation enthalpy and positive electron gain enthalpy: they neither lose nor gain electrons easily.

3. Isolation of the Noble Gases

Except helium and radon, the noble gases are obtained from air. Carbon dioxide and water vapour are removed, the air is liquefied, and the liquid is distilled in Claude's apparatus into three fractions. Chemical methods clean up each fraction: calcium carbide takes out nitrogen, and heated copper takes out oxygen.

Ca C_{2} + N_{2} CaC N_{2} + C

2Cu + O_{2} 2CuO

Figure 2: Isolating the noble gases. Liquid air is distilled into three fractions; nitrogen is removed with calcium carbide and oxygen with hot copper, leaving He and Ne, then Ar, then Kr and Xe.

Helium is obtained more cheaply from natural gas by liquefying the other components, and radon is collected from the decay of radium salts.

Quick Recall: tap to check

Why do boiling points rise from He to Rn?

Larger atoms are more polarisable, so the van der Waals forces between them get stronger.

Which noble gas is not obtained from air in the usual way?

Radon: it comes from the radioactive decay of radium.

Why is the electron gain enthalpy of noble gases positive?

An added electron must enter the next, higher shell, which is energetically unfavourable.

4. Chemical Properties

The noble gases are almost completely unreactive. There are three reasons:

they have a stable, complete octet ( $1 s^{2}$ for helium);
their ionisation enthalpies are very high, so they do not lose electrons;
their electron gain enthalpies are positive, so they do not gain electrons.

4.1 Bartlett's Discovery

In 1962 Neil Bartlett found that the powerful oxidant $Pt F_{6}$ removes an electron from dioxygen to give the orange solid $[O_{2}]^{+} [Pt F_{6}]^{-}$ . He then noticed that the ionisation enthalpy of xenon (1170 kJ mol $^{- 1}$ ) is almost the same as that of $O_{2}$ (1175 kJ mol $^{- 1}$ ), so xenon should react in the same way. Mixing the two gases gave a red crystalline solid, the first noble gas compound:

O_{2} + Pt F_{6} [O_{2}]^{+} [Pt F_{6}]^{-}

Xe + Pt F_{6} red crystalline solid [Xe]^{+} [Pt F_{6}]^{-}

Figure 3: Why xenon was the first noble gas to react. Its ionisation enthalpy (1170 kJ mol

^{- 1}

) is almost the same as that of

O_{2}

(1175), so Bartlett reasoned that if

Pt F_{6}

can pull an electron out of

O_{2}

, it can also do so with xenon.

Only krypton, xenon and radon form true compounds, and almost all of them are with fluorine and oxygen, the two most electronegative elements. Xenon chemistry is by far the richest; krypton gives $Kr F_{2}$ , and radon gives $Rn F_{2}$ but is hard to study because it is radioactive. Helium, neon and argon form no true compounds; they only form clathrates, in which atoms are trapped in cavities of a solid such as quinol or ice, held by physical forces alone.

Key idea

Bartlett's logic: if $Pt F_{6}$ can ionise $O_{2}$ (1175 kJ mol $^{- 1}$ ), it can ionise Xe (1170 kJ mol $^{- 1}$ ). Only the heavier noble gases, with the lowest ionisation enthalpies, react.

5. Xenon Fluorides

5.1 Preparation

Xenon and fluorine combine directly in a sealed nickel vessel; the ratio of the two gases and the conditions decide the product. $Xe F_{6}$ is also made from $Xe F_{4}$ and dioxygen difluoride at 143 K.

Xe + F_{2} 673 K, 1 bar Xe : F_{2} = 2 : 1 Xe F_{2}

Xe + 2 F_{2} 873 K, 7 bar Xe : F_{2} = 1 : 5 Xe F_{4}

Xe + 3 F_{2} 573 K, 60-70 bar Xe : F_{2} = 1 : 20 Xe F_{6}

Xe F_{4} + O_{2} F_{2} 143 K Xe F_{6} + O_{2}

Figure 4: Xenon chemistry in one picture. The ratio of xenon to fluorine and the conditions decide which fluoride forms; water then converts the fluorides into xenon,

Xe O_{3}

XeO F_{4}

Xe O_{2} F_{2}

, depending on how much water is used.

5.2 Properties

All three are colourless crystalline solids that sublime readily: $Xe F_{2}$ melts at 413 K, $Xe F_{4}$ at 390 K and $Xe F_{6}$ at 322.6 K.
They are powerful fluorinating agents and strong oxidising agents.
They attack glass, so they are stored in nickel or Monel metal vessels.

(a) With hydrogen they are reduced to xenon:

Xe F_{2} + H_{2} Xe + 2HF

Xe F_{4} + 2 H_{2} Xe + 4HF

Xe F_{6} + 3 H_{2} Xe + 6HF

(b) Hydrolysis. $Xe F_{2}$ hydrolyses slowly to xenon and oxygen, while $Xe F_{4}$ and $Xe F_{6}$ react with even traces of moisture. How much water is used decides the product from $Xe F_{6}$ (Figure 4):

2Xe F_{2} + 2 H_{2} O 2Xe + 4HF + O_{2}

2Xe F_{2} + 4O H^{-} alkaline 2Xe + 4 F^{-} + 2 H_{2} O + O_{2}

6Xe F_{4} + 12 H_{2} O 4Xe + 2Xe O_{3} + 24HF + 3 O_{2}

Xe F_{6} + 3 H_{2} O complete Xe O_{3} + 6HF

Xe F_{6} + H_{2} O partial XeO F_{4} + 2HF

Xe F_{6} + 2 H_{2} O partial Xe O_{2} F_{2} + 4HF

2Xe F_{6} + 3Si O_{2} why glass cannot be used 3Si F_{4} + 2Xe O_{3}

(c) As fluorinating agents they hand fluorine to metals and to other molecules:

Xe F_{4} + Pt Xe + Pt F_{4}

Xe F_{4} + 2Hg Xe + 2Hg F_{2}

2NO + Xe F_{2} 2NOF + Xe

2N O_{2} + Xe F_{2} 2N O_{2} F + Xe

4NO + Xe F_{4} Xe + 4NOF

3Xe F_{4} + 4BC l_{3} 4B F_{3} + 3Xe + 6C l_{2}

Xe F_{2} + I_{2} 2IF + Xe

Xe F_{2} + 2HCl Xe + 2HF + C l_{2}

Xe F_{6} + 6HCl Xe + 6HF + 3C l_{2}

(d) With fluoride ion donors and acceptors. The fluorides behave both as fluoride acceptors (Lewis acids) with strong fluoride donors, and as fluoride donors with strong acceptors:

Xe F_{2} + 2Sb F_{5} [XeF]^{+} [S b_{2} F_{11}]^{-}

Xe F_{2} + I F_{5} [XeF]^{+} [I F_{6}]^{-}

Xe F_{4} + Sb F_{5} [Xe F_{3}]^{+} [Sb F_{6}]^{-}

Xe F_{6} + As F_{5} [Xe F_{5}]^{+} [As F_{6}]^{-}

Xe F_{6} + B F_{3} [Xe F_{5}]^{+} [B F_{4}]^{-}

Xe F_{6} + NaF Na [Xe F_{7}]

Xe F_{6} + HF [Xe F_{5}]^{+} [H F_{2}]^{-}

(e) Other reactions. $Xe F_{4}$ heated with excess xenon gives $Xe F_{2}$ , and $Xe F_{6}$ decomposes on strong heating:

Xe F_{4} + Xe excess Xe 2Xe F_{2}

2Xe F_{6} Δ Xe F_{2} + Xe F_{4} + 3 F_{2}

Complete hydrolysis

Xe F_{6} + 3 H_{2} O \to Xe O_{3} + 6HF

: all six F replaced;

Xe O_{3}

is a colourless, explosive solid.

Partial hydrolysis

Xe F_{6} + H_{2} O \to XeO F_{4} + 2HF

and

Xe F_{6} + 2 H_{2} O \to Xe O_{2} F_{2} + 4HF

: only some F replaced.

6. Xenon Oxides, Oxyfluorides and Their Shapes

Xenon does not react with oxygen directly. Its oxides come from the hydrolysis of the fluorides. $Xe O_{3}$ is a colourless, explosive solid; $XeO F_{4}$ is a colourless volatile liquid; $Xe O_{2} F_{2}$ and $XeO F_{2}$ are also known. The shapes follow from VSEPR theory: count the bond pairs and lone pairs around xenon, remembering that a lone pair always takes the position where it is least crowded.

Compound	Oxidation state of Xe	Bond pairs + lone pairs	Hybridisation	Shape
$Xe F_{2}$	+2	2 + 3	$s p^{3} d$	linear
$Xe F_{4}$	+4	4 + 2	$s p^{3} d^{2}$	square planar
$Xe F_{6}$	+6	6 + 1	$s p^{3} d^{3}$	distorted octahedral
$Xe O_{3}$	+6	3 + 1	$s p^{3}$	pyramidal
$XeO F_{2}$	+4	3 + 2	$s p^{3} d$	T-shaped
$Xe O_{2} F_{2}$	+6	4 + 1	$s p^{3} d$	see-saw (distorted trigonal bipyramidal)
$XeO F_{4}$	+6	5 + 1	$s p^{3} d^{2}$	square pyramidal

In counting, each Xe=O double bond is treated as one electron pair position, so $Xe O_{3}$ behaves like three bond pairs and one lone pair.

Figure 5: Promotion explains the even-numbered fluorides. Each 5p to 5d promotion unpairs two electrons, so xenon forms

Xe F_{2}

Xe F_{4}

and

Xe F_{6}

, never an odd fluoride.

Figure 6: Shapes of the xenon fluorides. Lone pairs (shaded lobes) take the roomiest positions: three equatorial ones make

Xe F_{2}

linear, two opposite ones make

Xe F_{4}

square planar, and the single lone pair of

Xe F_{6}

distorts its octahedron.

Figure 7: Shapes of

Xe O_{3}

XeO F_{4}

and

Xe O_{2} F_{2}

. Count only the

σ

bonds (one per O or F) plus the lone pairs; each Xe=O adds a

π

bond but no extra electron pair.

Exam Trick

Three lone pairs, two lone pairs, one lone pair. Going $Xe F_{2} \to Xe F_{4} \to Xe F_{6}$ , each extra $F_{2}$ uses up one lone pair: 3, 2, 1. Lone pairs sit in the equatorial plane, so $Xe F_{2}$ ends up linear and $Xe F_{4}$ square planar; the single lone pair in $Xe F_{6}$ only distorts the octahedron.

Figure 8: One formula for every xenon compound: lone pairs

= \frac{8 - b - 2 a}{2}

for

Xe O_{a} F_{b}

; add the

σ

bonds and read the shape family.

JEE Advanced

Why is $Xe F_{6}$ not a regular octahedron? Xenon carries seven electron pairs (six bond pairs and one lone pair). The lone pair is stereochemically active, so the gas-phase molecule is a distorted (capped) octahedron that keeps flipping between equivalent forms. In the solid, $Xe F_{6}$ exists as $[Xe F_{5}]^{+}$ and $F^{-}$ units linked into tetramers and hexamers. The same fluoride-donor habit shows in $Xe F_{6} + MF \to M^{+} [Xe F_{7}]^{-}$ (M = Na, K, Rb, Cs), while $Xe F_{2} + P F_{5} \to [XeF]^{+} [P F_{6}]^{-}$ shows it as a fluoride donor.

Quick Recall: tap to check

What is the shape of

XeO F_{4}

Square pyramidal: 5

σ

bond pairs + 1 lone pair (octahedral family).

Which xenon fluoride is linear, and why?

Xe F_{2}

: 2 bond pairs + 3 lone pairs, all lone pairs equatorial in a trigonal bipyramid.

What forms when

Xe F_{6}

is completely hydrolysed?

Xe O_{3}

and HF.

7. Uses of the Noble Gases

Figure 9: Uses of the noble gases. Every use follows from one property: they are chemically inert, they glow in a discharge tube, or (for helium) they are light, poorly soluble and extremely cold when liquid.

Helium: non-inflammable and light, so it fills airships, observation balloons and weather balloons; a helium-oxygen mixture is used by deep-sea divers, because helium is much less soluble in blood than nitrogen and does not cause the bends; it is used to treat asthma, to inflate aeroplane tyres, to give an inert atmosphere for welding and melting easily oxidised metals, and (as liquid helium, b.p. 4.2 K) as a cryogenic coolant for superconducting magnets in NMR and MRI.
Neon: discharge tubes and neon signs, each gas giving its own colour; beacon lights for air navigation, because the light penetrates fog and mist; lamps in botanical gardens and greenhouses.
Argon: fills electric bulbs and fluorescent tubes (it is more inert than nitrogen and does not attack the filament); provides an inert atmosphere for arc welding and for high-temperature metallurgy.
Krypton and xenon: high-efficiency and flash lamps; radon: radiotherapy for cancer.

Key idea

Every xenon shape comes from one count: lone pairs = $(8 - n_{F} - 2 n_{O}) /2$ , then add the

σ

bonds and read the VSEPR family.

Figure 10: Group 18 on one page. Inertness comes from the complete octet; xenon breaks the rule only with the most electronegative partners, F and O, and every xenon shape follows from counting pairs.

8. Solved Examples

Solved Example 1

Why is the liquefaction of noble gases difficult?

Solution:

Their atoms are held together only by weak van der Waals (dispersion) forces, so very low temperatures are needed to condense them. Because these forces grow with atomic size, liquefaction becomes easier down the group; helium, the smallest, is the hardest of all (b.p. 4.2 K).

Solved Example 2

Which noble gas is radioactive?

Solution:

Radon, formed by the alpha decay of radium; it has no stable isotope.

Solved Example 3

Why do the boiling points of the noble gases increase from helium to radon?

Solution:

As the atoms get larger, their electron clouds are more easily distorted (more polarisable), so the dispersion forces between atoms become stronger and more energy is needed to separate them.

Solved Example 4

The molecular shapes of

S F_{4}

C F_{4}

and

Xe F_{4}

are
(A) the same, with 2, 0 and 1 lone pairs
(B) the same, with 1, 1 and 1 lone pairs
(C) different, with 0, 1 and 2 lone pairs
(D) different, with 1, 0 and 2 lone pairs

Solution:

Answer: (D). $S F_{4}$ has 4 bond pairs + 1 lone pair (see-saw), $C F_{4}$ has 4 bond pairs and no lone pair (tetrahedral), and $Xe F_{4}$ has 4 bond pairs + 2 lone pairs (square planar). All three shapes are different.

Solved Example 5

Which of the following is planar?
(A)

Xe O_{4}

(B)

Xe O_{3}

(C)

Xe O_{2} F_{2}

(D)

Xe F_{4}

Solution:

Answer: (D). $Xe F_{4}$ is square planar: its two lone pairs sit above and below the plane of the four fluorines. $Xe O_{4}$ is tetrahedral, $Xe O_{3}$ pyramidal and $Xe O_{2} F_{2}$ see-saw shaped.

Solved Example 6

Xe F_{6}

is hydrolysed step by step. Identify A, B and C:

Xe F_{6}

+ 1

H_{2} O

gives A; A +

H_{2} O

gives B; B +

H_{2} O

gives C.

Solution:

A = $XeO F_{4}$ , B = $Xe O_{2} F_{2}$ , C = $Xe O_{3}$ .

Xe F_{6} + H_{2} O XeO F_{4} + 2HF

XeO F_{4} + H_{2} O Xe O_{2} F_{2} + 2HF

Xe O_{2} F_{2} + H_{2} O Xe O_{3} + 2HF

Each mole of water replaces two F atoms with one O atom, so complete hydrolysis with three moles of water gives $Xe O_{3}$ directly.

Solved Example 7

2.45 g of

Xe F_{6}

is completely hydrolysed. Find the mass of

Xe O_{3}

formed and the moles of HF produced. (

Xe F_{6}

= 245,

Xe O_{3}

= 179 g mol

^{- 1}

)

Solution:

Xe F_{6} + 3 H_{2} O Xe O_{3} + 6HF

Moles of $Xe F_{6} = 2.45/245 = 0.0100$ mol.

Moles of $Xe O_{3} = 0.0100$ mol, so its mass $= 0.0100 \times 179 =$ 1.79 g.

Moles of HF $= 6 \times 0.0100 =$ 0.0600 mol (1.2 g).

Solved Example 8

Why do the noble gases form compounds almost only with fluorine and oxygen?

Solution:

A noble gas atom gives up an electron only under great compulsion. Fluorine and oxygen are the two most electronegative elements, so only they can pull electron density from xenon strongly enough to form bonds. The heavier noble gases react because their ionisation enthalpies are lower and their outer electrons are far from the nucleus.

Solved Example 9

Xe F_{6}

has seven electron pairs around xenon. What is its hybridisation and why is the octahedron distorted?

Solution:

Six bond pairs and one lone pair give seven pairs, so xenon is $s p^{3} d^{3}$ hybridised. The lone pair is stereochemically active: it occupies one of the seven positions and pushes the six Xe-F bonds away from it, so the shape is a distorted octahedron, not a regular one.

Solved Example 10

Why is a helium-oxygen mixture, rather than air, used by deep-sea divers?

Solution:

Under high pressure, nitrogen from air dissolves in the blood; when the diver comes up, it bubbles out and causes the painful and dangerous 'bends'. Helium is far less soluble in blood, so a mixture of about 80% helium and 20% oxygen avoids the problem; it is also non-inflammable.

Practice Questions

(i) Give the formula of the first compound of a noble gas. (ii) Give the hybridisation and structure of $Xe O_{3}$ .Answer: (i) $[Xe]^{+} [Pt F_{6}]^{-}$ (Bartlett, 1962). (ii) $s p^{3}$ ; pyramidal, with one lone pair on xenon.
Which compound led to the discovery of noble gas compounds?Answer: $[O_{2}]^{+} [Pt F_{6}]^{-}$ : because $Pt F_{6}$ could oxidise $O_{2}$ , whose ionisation enthalpy is close to that of xenon, Bartlett tried xenon next.
The total number of lone pairs of electrons on the central atom in $XeO F_{4}$ is (A) 0 (B) 1 (C) 2 (D) 3Answer: (B) 1; with five bond pairs it gives a square pyramidal shape ( $s p^{3} d^{2}$ ).
The shape of $O_{2} F_{2}$ is similar to that of (A) $C_{2} F_{2}$ (B) $H_{2} O_{2}$ (C) $H_{2} F_{2}$ (D) $C_{2} H_{2}$ Answer: (B) $H_{2} O_{2}$ : both have the non-planar 'open book' shape.
In $Br F_{3}$ the lone pairs occupy equatorial positions to minimise (A) lone pair-bond pair repulsion only (B) bond pair-bond pair repulsion only (C) lone pair-lone pair and lone pair-bond pair repulsions (D) lone pair-lone pair repulsion onlyAnswer: (C): equatorial lone pairs are at 120° to each other and have only two bond pairs at 90°, so both repulsions are smallest.
Which of the following has the maximum number of lone pairs on the central atom? (A) $Cl O_{3}^{-}$ (B) $Xe F_{4}$ (C) $S F_{4}$ (D) $I_{3}^{-}$ Answer: (D) $I_{3}^{-}$ , with three lone pairs on the central iodine; $Xe F_{4}$ has two, $Cl O_{3}^{-}$ and $S F_{4}$ one each.
Which noble gas has the highest ionisation enthalpy, and why?Answer: Helium (2372 kJ mol $^{- 1}$ ): its two electrons are in the $1 s$ shell, closest to the nucleus and with no shielding.
Why can $Xe F_{6}$ not be stored in glass vessels?Answer: It reacts with the silica of glass: $2Xe F_{6} + 3Si O_{2} \to 3Si F_{4} + 2Xe O_{3}$ . Nickel or Monel metal vessels are used.

Common Mistakes to Avoid

Watch out

Saying no noble gas forms compounds. Krypton, xenon and radon do; He, Ne and Ar form only clathrates.
Writing helium's configuration as $n s^{2} n p^{6}$ . Helium is $1 s^{2}$ , with only two electrons.
Calling the electron gain enthalpy of noble gases zero. It is positive: energy must be supplied.
Drawing $Xe F_{4}$ as tetrahedral. It is square planar, because two lone pairs sit above and below the plane.
Calling $Xe F_{6}$ a regular octahedron. Its seventh electron pair (a lone pair) distorts it.
Forgetting the lone pair in $Xe O_{3}$ and calling it planar; it is pyramidal.
Storing xenon fluorides in glass; they attack silica and are kept in nickel or Monel vessels.
Naming $Xe F_{2}$ as the first noble gas compound. The first was Bartlett's $[Xe]^{+} [Pt F_{6}]^{-}$ .

Frequently Asked Questions

Why are group 18 elements so unreactive?

Group 18 elements have a complete octet, $n s^{2} n p^{6}$ (helium is $1 s^{2}$ ). They therefore have the highest ionisation enthalpies in their periods, so they do not lose electrons, and positive electron gain enthalpies, so they do not gain electrons either. With no tendency to share, they stay monatomic.

Why does xenon form compounds when helium and neon do not?

Ionisation enthalpy falls down the group, from 2372 kJ per mol for helium to 1170 for xenon, because the outer electrons are farther from the nucleus and well shielded. Xenon's value is close to that of the oxygen molecule, so strong oxidants such as $Pt F_{6}$ and fluorine can pull electrons from it.

What was the first noble gas compound?

Xenon hexafluoroplatinate, written $[Xe]^{+} [Pt F_{6}]^{-}$ , a red crystalline solid made by Neil Bartlett in 1962. He made it after finding that $Pt F_{6}$ oxidises dioxygen to $[O_{2}]^{+} [Pt F_{6}]^{-}$ , and noting that xenon and $O_{2}$ have almost the same ionisation enthalpy, 1170 and 1175 kJ per mol. Xenon fluorides and oxides followed within months.

What are the shapes of the xenon fluorides?

$Xe F_{2}$ is linear ( $s p^{3} d$ , three lone pairs), $Xe F_{4}$ is square planar ( $s p^{3} d^{2}$ , two lone pairs) and $Xe F_{6}$ is a distorted octahedron ( $s p^{3} d^{3}$ , one lone pair). In each case the lone pairs take the least crowded positions, which is what decides the shape.

What happens when xenon hexafluoride is hydrolysed?

It depends on how much water is used. One mole of water gives $XeO F_{4}$ and HF, two moles give $Xe O_{2} F_{2}$ , and complete hydrolysis with three moles of water gives $Xe O_{3}$ and 6HF. $Xe O_{3}$ is a colourless, explosive, pyramidal solid that is a powerful oxidising agent.

How are the noble gases obtained?

Dry air is liquefied and fractionally distilled in Claude's apparatus into three fractions: helium with neon, argon with oxygen, and krypton with xenon. Nitrogen is removed by calcium carbide and oxygen by heated copper. Helium also comes from natural gas, and radon from the decay of radium.

What does NEET ask from group 18 elements?

NEET covers the general trends of group 18 elements: electronic configuration, atomic radii, ionisation enthalpy, positive electron gain enthalpy, boiling points and the reasons for their inertness. NCERT facts on xenon fluorides, their hydrolysis and shapes, and the uses of helium and argon are also asked.

Which group 18 topics matter most for JEE Advanced?

JEE Advanced focuses on xenon fluorides: how each is prepared, hydrolysis products, their behaviour as fluorinating agents and as fluoride donors or acceptors, and above all the shapes and hybridisation of $Xe F_{2}$ , $Xe F_{4}$ , $Xe F_{6}$ , $Xe O_{3}$ , $XeO F_{4}$ and $Xe O_{2} F_{2}$ from VSEPR theory.

Previous year questions on Group 18 Elements: The Noble Gases

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