Group 18 (Noble gases) — discovery, isolation, compounds of Xe (XeF₂, XeF₄, XeF₆, XeO₃) — structure and bonding: Exercises

Definition Three words we will keep straight (define once, use forever)

These three phrases sound alike; fix them now so counting never gets muddled.

Electron pair = two electrons sitting together. There are two kinds: bonding pairs (shared between Xe and a partner) and lone pairs (sitting only on Xe). A double bond holds two bonding pairs (one σ, one π).
Electron domain (= bonding direction) = one region of space pointing out from Xe that VSEPR must give a spot to. A single bond, a double bond, or a lone pair each count as exactly one domain. A double bond is one domain even though it holds two pairs, because both pairs point the same way.
Bond pair / bonding domain = a domain that points at a bonded atom. Number of bonding domains = number of atoms bonded to Xe.

Rule of thumb used everywhere below: shape is decided by the number of domains (bonds + lone pairs), while hybridisation labels count total pairs — so we always compute both, and we are careful that a double bond adds pairs but not domains.

Formula The master recipe (single AND double bonds)

Step 1 — total electron pairs (this sets the hybridisation label): $Total pairs = \frac{( valence e ^{-} of Xe ) + ( shared e ^{-} from partners ) + ( charge e ^{-} )}{2}$ Fill "shared $e^{-}$ from partners":

Each single-bonded atom (like F) contributes 1 electron.
Each doubly-bonded atom (like O in Xe=O) contributes 2 electrons (it brings two shared pairs into the count).
For an anion, add 1 electron per unit of negative charge.

Step 2 — split the pairs, treating π-pairs explicitly:

$π -pairs =$ number of double bonds (each Xe=O hides one extra π-pair inside its bonding direction).
$σ -bonding pairs =$ number of atoms bonded to Xe (this equals the number of bonding domains).
$Lone pairs = Total pairs - (σ -bonding pairs) - (π -pairs)$ In words: take out one pair for each bonding direction, then take out the extra π-pairs that ride along double bonds; whatever is left is genuinely non-bonding.
For an all-single-bond molecule $π -pairs = 0$ , so this reduces to the familiar $Lone pairs = Total pairs - bonds$ .

Step 3 — count domains for the shape: $Electron domains = (bonding domains) + (lone pairs)$ Arrange the domains for maximum spacing, park lone pairs in the roomiest spots, and read off the shape.

Xe always brings 8 valence electrons (its full $n s^{2} n p^{6}$ shell).

Level 1 — Recognition

These test recall. No calculation, just knowing the facts cold.

L1.1 Name the Group 18 elements in order. State which valence configuration is the exception, and why chemistry courses usually list only the first six.

L1.2 Who made the first noble-gas compound, in what year, and what was its formula?

L1.3 Give the shape and hybridisation of each: XeF₂, XeF₄, XeF₆, XeO₃.

Recall Solution L1

L1.1 The full column is ==He, Ne, Ar, Kr, Xe, Rn, and Og (Oganesson, $Z = 118$ )==. Each has a full valence shell $n s^{2} n p^{6}$ ; the exception is Helium, which is $1 s^{2}$ (no $p$ subshell exists in the first shell, so it simply fills the $1 s$ ). Courses usually discuss only the first six (He→Rn) because Og is a synthetic, super-heavy element made a few atoms at a time, existing for only milliseconds — it has no practical chemistry to teach and is predicted (from relativistic effects) not even to behave like a normal noble gas. So "the six noble gases" is a convenience of ordinary chemistry, not the complete group.

L1.2 Neil Bartlett, 1962, formula $Xe^{+} [PtF_{6}]^{-}$ . His clue: $PtF_{6}$ had already ripped an electron off $O_{2}$ , and the ionization enthalpy of Xe ( $1170$ ) is almost identical to that of $O_{2}$ ( $1175 kJ/mol$ ). See Ionization Enthalpy trends.

L1.3

Molecule	Shape	Hybridisation label
XeF₂	Linear	$s p^{3} d$
XeF₄	Square planar	$s p^{3} d^{2}$
XeF₆	Distorted octahedral	$s p^{3} d^{3}$
XeO₃	Trigonal pyramidal	$s p^{3}$

(On the hybridisation labels: see the note at the end of L2. These labels are the traditional bookkeeping tags used in exams; modern calculations show the shapes come mainly from s and p orbitals plus electron-pair repulsion, not from heavy real d-orbital use. Learn the labels for marks, but read the caveat.)

Level 2 — Application

Now use the master recipe. Show the pair count each time.

L2.1 Using the recipe, compute the total electron pairs, bond pairs, and lone pairs around Xe in XeF₄. Deduce the parent geometry and the final shape.

L2.2 Do the same for XeF₆.

L2.3 XeOF₄ has 4 Xe–F single bonds, 1 Xe=O double bond, and Xe as centre. Work out its electron-domain count and shape, using the double-bond rule from the master recipe.

Recall Solution L2

L2.1 — XeF₄ (all single bonds, so $π -pairs = 0$ )

Total pairs $= (8 + 4) /2 = 6$ .
$σ$ -bonding pairs = bonding domains = $4$ ; $π$ -pairs $= 0$ .
Lone pairs $= 6 - 4 - 0 = 2$ .
Domains $= 4 + 2 = 6$ ⇒ parent geometry octahedral. The 2 lone pairs sit trans (opposite, $180°$ apart) to minimise lone-pair–lone-pair repulsion. Removing those two axial positions from the octahedron leaves 4 F atoms in a plane ⇒ ==square planar==. Hybridisation label $s p^{3} d^{2}$ . See Hybridisation (sp3d, sp3d2, sp3d3).

Figure s01 — XeF₄, square planar. Alt text: a central blue Xe atom with four orange F atoms bonded in one flat plane at $90°$ to each other, and two faded red lone-pair clouds sitting directly above and below Xe (trans, $180°$ apart). The picture makes the key insight visible: the two lone pairs took the top and bottom vertices of the octahedron (they are the most out-of-the-way spots for two mutually-repelling clouds), which forces all four F atoms into the single horizontal plane you see — that flatness is "square planar." Notice the four F–Xe–F angles inside the plane are each $90°$ (green label).

Figure — Group 18 (Noble gases) — discovery, isolation, compounds of Xe (XeF₂, XeF₄, XeF₆, XeO₃) — structure and bonding

L2.2 — XeF₆ (all single bonds, so $π -pairs = 0$ )

Total pairs $= (8 + 6) /2 = 7$ .
$σ$ -bonding pairs = bonding domains = $6$ ; $π$ -pairs $= 0$ .
Lone pairs $= 7 - 6 - 0 = 1$ .
Domains $= 6 + 1 = 7$ . The correct parent electron-domain geometry for 7 domains is the pentagonal bipyramid (five domains in an equatorial pentagon, one straight up, one straight down). The single lone pair is stereochemically active: it occupies one of these seven positions and squeezes the six Xe–F bonds together, so the molecular framework (the F atoms only) is a distorted / non-regular octahedron. The "distorted octahedral" name describes the shape of the six F atoms; the "pentagonal bipyramid" describes the arrangement of all seven domains including the lone pair — they are two views of the same molecule, not a contradiction. Shape label = ==distorted octahedral==, hybridisation label $s p^{3} d^{3}$ .

Figure s02 — XeF₆, pentagonal-bipyramid domains → distorted octahedron. Alt text: a central blue Xe; five orange F atoms in a horizontal pentagon around it, one orange F straight up, and a faded red lone-pair cloud in the sixth (straight-down) position, pushing the surrounding F atoms upward. Reading the figure: seven domains genuinely start as a pentagonal bipyramid, but because one of those seven is the lone pair (bottom, red), the six F atoms form a lopsided, non-regular octahedron — the lone pair's push is exactly the "distortion."

L2.3 — XeOF₄ (one double bond, so $π -pairs = 1$ )

Step 1 — total pairs: single-bonded F atoms contribute $4 \times 1 = 4$ ; the doubly-bonded O contributes $2$ . Total pairs $= (8 + 4 + 2) /2 = 14/2 = 7$ .
Step 2 — split with the π-rule: $σ$ -bonding pairs = atoms bonded to Xe $= 4 (F) + 1 (O) = 5$ ; $π$ -pairs $= 1$ (the one Xe=O double bond). Lone pairs $= 7 - 5 - 1 = 1$ . The subtraction of the π-pair is no longer ad hoc — it is Step 2 of the master recipe.
Step 3 — domains: bonding domains $= 5$ , lone pairs $= 1$ , so electron domains $= 5 + 1 = 6$ ⇒ octahedral parent. One lone pair takes one octahedral vertex; the remaining 5 bonds form a ==square pyramidal== shape (4 F in a base square, 1 O at the apex). Hybridisation label $s p^{3} d^{2}$ .

Level 3 — Analysis

Explain why, don't just state.

L3.1 In XeF₂ the three lone pairs go equatorial, not axial, in the trigonal-bipyramidal arrangement. Explain why, and why the result is a linear molecule.

L3.2 Both SF₆ and XeF₆ have six Xe/S–F bonds, yet SF₆ is a perfect octahedron while XeF₆ is distorted. Explain the difference from domain counting.

L3.3 XeF₄ is square planar (all F–Xe–F cis angles $90°$ ). Compute the total number of F–Xe–F angles and how many are $90°$ vs $180°$ .

Recall Solution L3

L3.1 In a trigonal bipyramid (TBP) there are two axial positions ( $180°$ apart, each with three near-neighbours at $90°$ ) and three equatorial positions ( $120°$ apart, each with only two near-neighbours at $90°$ ). A lone pair is a fat, un-anchored cloud that repels harder than a bond pair. Repulsion is worst at $90°$ . An equatorial lone pair has only two $90°$ neighbours; an axial one has three. So each lone pair minimises strain by sitting equatorial. With all three lone pairs equatorial, the only remaining positions are the two axial ones — occupied by the two F atoms directly opposite each other. F–Xe–F is therefore $180°$ ⇒ ==linear==.

Figure s03 — XeF₂, why linear. Alt text: central blue Xe with two orange F atoms bonded straight up and straight down (the axial positions, F–Xe–F $= 180°$ ), and three faded red lone-pair clouds spread around the middle at $120°$ (the equatorial positions). Read the picture as the answer to "where do three fat clouds go to stay farthest apart?": they claim the equatorial belt (only two $90°$ neighbours each), which leaves the two F atoms no choice but the axial top and bottom — and two atoms on a straight line through Xe is exactly a linear molecule.

L3.2 SF₆: valence of S = $6$ , six single bonds add $6$ , total pairs $= (6 + 6) /2 = 6$ = exactly 6 bonding pairs, 0 lone pairs ⇒ 6 domains, perfect octahedron. XeF₆: total pairs $= 7$ = 6 bonding pairs + 1 lone pair ⇒ 7 domains. That single lone pair is stereochemically active (it takes a real spot and changes the angles — see the definition in L2.2), so it pushes the bonds away, bending the F–Xe–F angles off $90°$ ⇒ distorted octahedron. The presence or absence of that one lone pair (and hence 6 vs 7 domains) is the whole difference.

L3.3 With 4 F atoms, the number of F–Xe–F pairs $= (2 4) = \frac{4 \times 3}{2} = 6$ . In a square, each corner has two adjacent corners (at $90°$ ) and one diagonal (at $180°$ ). Adjacent (cis, $90°$ ) pairs = the 4 sides of the square = $4$ . Diagonal (trans, $180°$ ) pairs = the 2 diagonals = $2$ . Check: $4 + 2 = 6$ ✓.

Level 4 — Synthesis

Combine reactions and reasoning.

L4.1 Starting from Xe and F₂ only, write the three balanced fluoride-forming reactions with their conditions, and explain in one sentence what physically controls which fluoride forms.

L4.2 Write the complete and partial hydrolysis reactions of XeF₆, balanced, and name the Xe-containing product of each with its shape.

L4.3 A student mixes Xe and F₂ in a $1 : 20$ molar ratio at $573 K$ and $65 bar$ . Predict the product, then predict the geometry and hybridisation of that product without recomputing from scratch — reason from the ratio.

Recall Solution L4

L4.1 $Xe + F_{2} Xe in excess 673 K, 1 bar XeF_{2}$ $Xe + 2 F_{2} 1 : 5 873 K, 7 bar XeF_{4}$ $Xe + 3 F_{2} 1 : 20 573 K, 60 - 70 bar XeF_{6}$ The F₂-to-Xe ratio in the feed (plus pressure) controls how many F atoms stick: more F₂ and higher pressure push Xe to bond to more fluorines. See Interhalogen and Oxyfluoride compounds.

L4.2 Complete hydrolysis (excess water): $XeF_{6} + 3 H_{2} O \to XeO_{3} + 6 HF$ Product: XeO₃, trigonal pyramidal (3 Xe=O bonding domains + 1 lone pair = 4 domains, $s p^{3}$ ), an explosive solid.

Partial hydrolysis (limited water): $XeF_{6} + H_{2} O \to XeOF_{4} + 2 HF$ Product: XeOF₄, square pyramidal (from L2.3: 5 bonding domains + 1 lone pair = 6 domains, $s p^{3} d^{2}$ label).

L4.3 A $1 : 20$ Xe:F₂ feed at $573 K$ , $60 - 70 bar$ is exactly the recipe for XeF₆ (compare the third reaction in L4.1). Since we already established XeF₆ has 6 bonding domains + 1 lone pair = 7 domains (L2.2), its geometry is distorted octahedral with hybridisation label $s p^{3} d^{3}$ . No recomputation needed — the ratio is the fingerprint of the product.

Level 5 — Mastery

Full reasoning, multi-step, no scaffolding.

L5.1 Predict the shape of the species XeF₅⁻ (5 Xe–F bonds + the extra electron from the negative charge). Show the domain count and name the geometry. (This anion really exists — pentagonal planar.)

L5.2 Bartlett noticed IE(O₂) $= 1175$ and IE(Xe) $= 1170 kJ/mol$ . Kr has IE $= 1351$ and Ar has IE $= 1521 kJ/mol$ . Explain quantitatively why PtF₆ oxidises Xe readily but not Ar, using the difference in IE as your argument.

L5.3 Rank the boiling points of He, Ne, Ar, Kr, Xe from lowest to highest and justify with the correct intermolecular force. Then state which single element has the highest ionization enthalpy of all elements and why.

Recall Solution L5

L5.1 — XeF₅⁻ (all single bonds, so $π -pairs = 0$ ; anion adds 1 electron)

Total pairs $= (8 + 5 + 1) /2 = 14/2 = 7$ .
$σ$ -bonding pairs = bonding domains = $5$ ; $π$ -pairs $= 0$ . Lone pairs $= 7 - 5 - 0 = 2$ .
Domains $= 5 + 2 = 7$ ⇒ pentagonal bipyramidal parent. The 2 lone pairs take the 2 axial positions (opposite each other, $180°$ apart, minimising their mutual repulsion). The 5 F atoms fill the equatorial pentagon. Result = ==pentagonal planar==, hybridisation label $s p^{3} d^{3}$ .

Figure s04 — XeF₅⁻, pentagonal planar. Alt text: central blue Xe with five orange F atoms arranged in a flat five-pointed ring at $72°$ spacing, and two faded red lone-pair clouds directly above and below Xe. The figure shows why the shape is flat: the two lone pairs grab the top and bottom axial spots (farthest apart at $180°$ ), leaving all five F atoms in a single equatorial plane — a regular pentagon with $72°$ between neighbours (green label), which is "pentagonal planar."

L5.2 PtF₆ is an extremely strong oxidiser — it is one of the few reagents powerful enough to pull an electron off $O_{2}$ (its oxidising strength sits high enough to ionise anything whose IE is around $1175 kJ/mol$ or lower; that observed reaction with $O_{2}$ is the yardstick we calibrate against, since no simple redox-potential number is given). So the practical rule Bartlett used is: PtF₆ can oxidise a species whose IE is at or below ≈ IE(O₂) = 1175. Now compare:

Xe: IE $= 1170$ , which is $1175 - 1170 = 5 kJ/mol$ below the yardstick — easily oxidised. ✓
Ar: IE $= 1521$ , which is $1521 - 1175 = 346 kJ/mol$ above the yardstick (and $1521 - 1170 = 351 kJ/mol$ above Xe) — far beyond PtF₆'s reach. ✗
Kr: IE $= 1351$ , i.e. $1351 - 1170 = 181 kJ/mol$ harder than Xe — still too costly for easy reaction.

The argument is quantitative because it rests on the measured benchmark reaction $O_{2} + PtF_{6} \to O_{2}^{+} [PtF_{6}]^{-}$ , which fixes how much ionization cost PtF₆ can pay. Down the group, size ↑ ⇒ outer electrons farther and screened ⇒ IE ↓ ⇒ reactivity ↑.

L5.3 Boiling points (lowest → highest): ==He < Ne < Ar < Kr < Xe==. All noble gases are monatomic and non-polar, so the only attraction between atoms is London dispersion — momentary dipoles from electron-cloud fluctuations. Bigger, more electron-rich atoms have more polarizable clouds ⇒ stronger dispersion ⇒ higher boiling point. Xe (biggest of the stable ones) boils highest; He (smallest, 2 electrons) boils lowest. The element with the highest ionization enthalpy of all is ==Helium== — a tiny atom whose two $1 s$ electrons sit closest to the nucleus and feel the least shielding, so they are the hardest of all to remove. See p-Block general trends.

Recall Feynman check — say it in one breath

Xenon holds its outer toys loosely, so greedy F and O can pull them out and make XeF₂ (linear), XeF₄ (square, thanks to two invisible elbow-pairs), XeF₆ (a squashed octahedron because of one elbow), and XeO₃ (a pyramid). Count domains = bonds + invisible lone pairs, spread them out as far as possible, and the shape falls right out.

Connections

VSEPR Theory — the pair-counting engine used in every problem
Hybridisation (sp3d, sp3d2, sp3d3) — labels the orbital sets $s p^{3} d$ , $s p^{3} d^{2}$ , $s p^{3} d^{3}$
Ionization Enthalpy trends — the numbers behind L1.2 and L5.2
London Dispersion Forces — the boiling-point argument in L5.3
Interhalogen and Oxyfluoride compounds — XeOF₄ and the fluoride syntheses
p-Block general trends — down-group size/IE reasoning