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

Before we count anything, the two ideas the whole page rests on:

Definition The one physical law behind every shape (VSEPR in one breath)

Electrons are negatively charged, and like charges repel. Around a central atom the electrons live in "clouds" called regions of electron density (a single bond, a double bond, or a lone pair each counts as one region — one cloud). Those clouds push each other as far apart as possible on the surface of an imaginary sphere. So:

2 regions → $180°$ apart (linear)
3 regions → $120°$ (trigonal planar)
4 regions → $109.5°$ (tetrahedral)
5 regions → trigonal bipyramid (3 equatorial at $120°$ + 2 axial)
6 regions → octahedron ( $90°$ all round)
7 regions → pentagonal bipyramid (5 equatorial + 2 axial)

A lone pair is fatter than a bonding pair (it is held by one atom, not shared), so it pushes harder and claims the roomiest seat. That single fact settles every "where do the lone pairs go?" question below.

Formula The general version of the SAME rule (double bonds AND charge)

A doubly-bonded O (Xe=O) still occupies one region of electron density, but it takes 2 of Xe's electrons instead of 1. To keep ONE unified procedure that also handles ions, we compute the lone pairs left on Xe directly, charge included: $electrons left on Xe = V - (1 \times n_{single}) - (2 \times n_{double}) - C, lp = \frac{electrons left}{2}$ $total regions = n_{single} + n_{double} + lp$ where $n_{single}$ = number of single bonds (e.g. Xe–F), $n_{double}$ = number of Xe=O double bonds, and $C$ = overall charge with sign. Consistency check (why this really is one rule): with $n_{double} = 0$ , lp $= \frac{1}{2} (V - n_{single} - C)$ and total regions $= n_{single} + \frac{1}{2} (V - n_{single} - C) = \frac{1}{2} (V + n_{single} - C)$ — exactly the master rule with $M = n_{single}$ . So the general box reproduces the single-bond box for any charge $C$ , and needs no ad-hoc fixes for ions. We use this box explicitly in Examples 5, 6 and 7.

Keep these four boxes in view — every example is "count regions → place lone pairs → read shape + hybridisation."

The scenario matrix

Every question this topic can throw is one of the cells below. Each example that follows is tagged with the cell it covers.

Cell	What makes it different	Example
A. 0 lone pairs (degenerate)	central atom uses all pairs bonding — shape is "textbook perfect"	Ex 1 (SF₆ vs XeF₆ contrast)
B. 1 lone pair, all single bonds	odd region count, distortion	Ex 2 (XeF₆)
C. 2 lone pairs (even)	lone pairs go trans	Ex 3 (XeF₄)
D. 3 lone pairs	lone pairs go equatorial in TBP	Ex 4 (XeF₂)
E. Double bonds present (Xe=O)	O uses 2e⁻ but 1 region	Ex 5 (XeO₃)
F. Mixed F + O + lone pair	count both bond types	Ex 6 (XeOF₄)
G. Charged species (ion)	charge shifts the electron count	Ex 7 (XeF₅⁻, XeF₃⁺)
H. Word / real-world problem	apply reactivity + isolation logic	Ex 8 (feed ratio design)
I. Exam twist / "spot the error"	catch a wrong reflex	Ex 9 (why not tetrahedral / octahedral)

Example 1 — Cell A: the degenerate "no lone pair" case

Step 1 — Count pairs for $SF_{6}$ . S has $V = 6$ valence electrons. $M = 6$ (six single F bonds). $C = 0$ . $pairs = \frac{6 + 6 - 0}{2} = 6.$ Why this step? The master rule tells us the total number of regions of electron density pushing around the centre.

Step 2 — Split. bp $= 6$ , lp $= 6 - 6 = 0$ . Why? Six bonds use all six pairs; nothing is left over — this is the degenerate case.

Step 3 — Place them. 6 regions, 0 lone pairs ⇒ perfect octahedron. Six regions → $s p^{3} d^{2}$ (from the hybridisation list). No lone pair means nothing distorts it.

Step 4 — Contrast with $XeF_{6}$ . Xe has $V = 8$ , so $(8 + 6) /2 = 7$ pairs ⇒ 6 bp + 1 lp. That extra lone pair is the whole difference. See the next example.

Verify: pairs $= (6 + 6) /2 = 6 =$ bp, lp $= 0$ . Octahedron ✓. Compare Interhalogen and Oxyfluoride compounds where $SF_{6}$ -like counting recurs.

Example 2 — Cell B: one lone pair (odd, distorted)

Step 1 — Count. $V = 8$ , $M = 6$ , $C = 0$ : $pairs = (8 + 6) /2 = 7$ . Why? Standard master rule.

Step 2 — Split. bp $= 6$ , lp $= 7 - 6 = 1$ .

Step 3 — Seat the electron regions (the electron-geometry). Seven regions spread out most evenly at the vertices of a pentagonal bipyramid (5 in a plane, 1 top, 1 bottom) — this is the electron-pair geometry. Because there is 1 lone pair, standard VSEPR places it in an equatorial position of that pentagonal bipyramid (the roomier belt), and the six F atoms occupy the remaining vertices. Why start with the electron-geometry? Shape is decided in two layers: first seat all regions (bonds + lone pairs) in their lowest-repulsion arrangement, then hide the lone pair to read the molecular shape from the atoms only.

Step 4 — Read the molecular shape. Hiding the equatorial lone pair leaves six F atoms that are close to, but not exactly, an octahedron: the equatorial lone pair pushes its neighbouring F atoms aside, so the F-skeleton is a distorted octahedron. (Contrast this carefully: the electron-geometry is pentagonal bipyramidal; the molecular shape — atoms only — is described as distorted octahedral.) Seven regions → hybridisation $s p^{3} d^{3}$ (see the hybridisation box).

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

Figure 1 — XeF₆. Pink Xe at the centre; six blue F atoms; the pale-yellow lobe (upper-left) is the single equatorial lone pair, and the yellow arrow shows it shoving the nearest F atoms outward. The three F atoms nearest the lobe are nudged off their ideal positions while the far three stay put — that displacement is the "distortion" that makes the F-skeleton distorted-octahedral rather than a perfect octahedron.

Verify: $(8 + 6) /2 = 7$ ; bp $= 6$ , lp $= 1$ . Electron-geometry pentagonal bipyramidal; molecular shape distorted octahedral; $s p^{3} d^{3}$ ✓.

Example 3 — Cell C: two lone pairs (trans placement)

Step 1 — Count. $V = 8$ , $M = 4$ : $pairs = (8 + 4) /2 = 6$ .

Step 2 — Split. bp $= 4$ , lp $= 6 - 4 = 2$ .

Step 3 — Base geometry. 6 regions ⇒ octahedral electron arrangement. Six regions → $s p^{3} d^{2}$ .

Step 4 — Place the 2 lone pairs. Lone pairs repel each other strongly, so they take positions $180°$ apart (trans / axial). This leaves the 4 F atoms in one plane. Why? Trans placement maximises the lone-pair–lone-pair angle (the biggest possible separation), minimising repulsion — the fatter-cloud rule from the VSEPR box.

Step 5 — Read off the shape. 4 F in a plane, $90°$ apart ⇒ square planar.

Figure 2 — XeF₄. The two yellow "lp" circles sit directly above and below the pink Xe (dashed arrows front and back) — the trans placement of Step 4. The four blue F atoms are left in a flat square, each corner $90°$ from the next, which is exactly why the molecular shape reads as square planar.

Verify: $(8 + 4) /2 = 6$ ; bp $= 4$ , lp $= 2$ trans. Square planar, $s p^{3} d^{2}$ ✓.

Example 4 — Cell D: three lone pairs (equatorial in TBP)

Step 1 — Count. $V = 8$ , $M = 2$ : $pairs = (8 + 2) /2 = 5$ .

Step 2 — Split. bp $= 2$ , lp $= 5 - 2 = 3$ .

Step 3 — Base geometry. 5 regions ⇒ trigonal bipyramidal (TBP). Five regions → $s p^{3} d$ . A TBP has 3 equatorial positions (spread $120°$ in a plane) and 2 axial positions ( $180°$ , top & bottom).

Step 4 — Place the 3 lone pairs. All three go equatorial. Why? An equatorial position has only 2 close ( $90°$ ) neighbours; an axial position has 3. Fat lone pairs want the roomiest seats, so they claim all three equatorial spots.

Step 5 — Where do the F's land? The 2 F atoms are forced into the axial positions, $180°$ apart ⇒ linear.

Figure 3 — XeF₂. Three yellow lone-pair arrows fan out $120°$ apart in the horizontal (equatorial) belt around the pink Xe — Step 4. The two blue F atoms are pushed to the top and bottom (axial), and the dotted pink line marks the $180°$ F–Xe–F angle that makes the molecule linear.

Verify: $(8 + 2) /2 = 5$ ; bp $= 2$ , lp $= 3$ equatorial. Linear ( $180°$ ), $s p^{3} d$ ✓.

Example 5 — Cell E: double bonds (Xe=O)

Step 1 — Use the general rule (double bonds present). From the general box: $n_{single} = 0$ , $n_{double} = 3$ , $C = 0$ . $electrons left on Xe = 8 - (1 \times 0) - (2 \times 3) - 0 = 8 - 6 = 2 \Rightarrow lp = \frac{2}{2} = 1.$ Why this box and not the plain master rule? The plain rule assumes every terminal atom donates one electron (a single bond). Oxygen here forms a double bond and takes two of Xe's electrons, so we use the general formula — which, remember, is the same rule specialised to $n_{double} \neq = 0$ .

Step 2 — Total regions. $regions = n_{single} + n_{double} + lp = 0 + 3 + 1 = 4$ . Why? A double bond still occupies only one region of electron density (VSEPR counts clouds, not individual bonds), so three Xe=O groups plus one lone pair make four regions.

Step 3 — Base geometry. 4 regions ⇒ tetrahedral electron arrangement. Four regions → $s p^{3}$ . Why? Four clouds push apart to the roomiest symmetric arrangement, the tetrahedron ( $109.5°$ ) — the 4-region result from the VSEPR box.

Step 4 — Read off molecular shape. 4 regions with 1 lone pair ⇒ trigonal pyramidal — exactly like $NH_{3}$ . Why? Shape is named from the atoms only; hide the invisible lone pair and the three O atoms sit as a tripod under Xe. The bulky lone pair also squeezes the O–Xe–O angle slightly below $109.5°$ .

Figure 4 — XeO₃. The pink double lines are the three Xe=O bonds forming a tripod base (yellow O atoms); the single yellow lobe pointing up from Xe is the lone pair. It sits opposite the tripod and presses the three O's slightly together — the " $< 109.5°$ " label.

Verify: electrons left $= 8 - 6 = 2 = 1$ lp; regions $= 4$ . Trigonal pyramidal, $s p^{3}$ ✓. Balance the equation: O $= 3$ (from 3 H₂O) $= 3$ (in XeO₃) ✓; H: $6 = 6$ (HF) ✓; F: $6 = 6$ ✓.

Example 6 — Cell F: mixed F + O + lone pair

Step 1 — Use the general rule (mixed bonds). Now both bond types appear: $n_{single} = 4$ (four Xe–F), $n_{double} = 1$ (one Xe=O), $C = 0$ . $electrons left on Xe = 8 - (1 \times 4) - (2 \times 1) - 0 = 8 - 6 = 2 \Rightarrow lp = 1.$ Why this step? This is exactly the unified formula from the general box — built to handle a mix of single and double bonds (and charge) in one line, so we never invent ad-hoc bookkeeping. Each F drains 1 electron, the O drains 2.

Step 2 — Total regions. $regions = n_{single} + n_{double} + lp = 4 + 1 + 1 = 6$ . Why? Again each bond (single or double) is one region, plus the one lone pair — six clouds in all.

Step 3 — Base geometry. 6 regions ⇒ octahedral electron arrangement. Six regions → $s p^{3} d^{2}$ . Why? Six clouds repel to the octahedron ( $90°$ neighbours all round) — the 6-region result, same base shape we met for XeF₄.

Step 4 — Place the lone pair, then read the shape. Put the single lone pair at one octahedral vertex; the five bonding groups (one O + four F) occupy the other five. The result is a square pyramid: the four F's form the square base and the O sits at the apex opposite the lone pair. Why place it there? With only one lone pair every octahedral vertex is equivalent, so there is no cis/trans choice — wherever it goes, five bonds remain in a square-pyramidal skeleton. The lone pair simply pushes the four basal F's slightly down, tilting the pyramid.

Verify: electrons left $= 8 - 6 = 2 = 1$ lp; regions $= 6$ . Square pyramidal, $s p^{3} d^{2}$ ✓. Equation check — F: $6$ left $= 4$ (XeOF₄) $+ 2$ (HF) $= 6$ ✓; O: $1 = 1$ ✓; H: $2 = 2$ ✓.

Example 7 — Cell G: charged species (ions)

(a) $XeF_{5}^{-}$ : $C = - 1$ .

Step 1 — Count with the general rule (charge included). $n_{single} = 5$ , $n_{double} = 0$ , $C = - 1$ : $electrons left on Xe = 8 - (1 \times 5) - (2 \times 0) - (- 1) = 8 - 5 + 1 = 4 \Rightarrow lp = \frac{4}{2} = 2.$ Why $- (- 1)$ ? The ion gained an electron; subtracting $C = - 1$ adds it back. This is precisely why the general box carries a $- C$ term — no ad-hoc fix needed. (Cross-check with the plain master rule: $pairs = (V + M - C) /2 = (8 + 5 + 1) /2 = 7$ , bp $= 5$ , lp $= 7 - 5 = 2$ ✓ — same answer.)

Step 2 — Total regions. $regions = 5 + 0 + 2 = 7$ . Why? Five Xe–F bonds plus two lone pairs — seven clouds.

Step 3 — Shape. 7 regions ⇒ base pentagonal bipyramid; the 5 F's occupy the pentagonal (equatorial) plane and the 2 lone pairs take the two axial (top & bottom) positions. Molecular shape $=$ pentagonal planar. Seven regions → $s p^{3} d^{3}$ . Why put the lone pairs axial? In a pentagonal bipyramid the two axial sites are $180°$ from each other — the maximum possible separation for two fat lone pairs, so they minimise repulsion there and leave the five F atoms flat in the equatorial plane.

(b) $XeF_{3}^{+}$ : $C = + 1$ .

Step 1 — Count with the general rule. $n_{single} = 3$ , $n_{double} = 0$ , $C = + 1$ : $electrons left on Xe = 8 - (1 \times 3) - 0 - (+ 1) = 8 - 3 - 1 = 4 \Rightarrow lp = \frac{4}{2} = 2.$ Why $- (+ 1)$ ? The cation lost an electron — subtracting $C = + 1$ removes it. (Cross-check: $pairs = (8 + 3 - 1) /2 = 5$ , bp $= 3$ , lp $= 5 - 3 = 2$ ✓.)

Step 2 — Total regions. $regions = 3 + 0 + 2 = 5$ .

Step 3 — Shape. 5 regions ⇒ trigonal-bipyramidal base. Five regions → $s p^{3} d$ . Both lone pairs go equatorial (roomiest seats, same argument as XeF₂), leaving three positions — two axial F and one equatorial F ⇒ T-shaped (like $ClF_{3}$ ). Why equatorial lone pairs? Equatorial sites have only two close $90°$ neighbours versus three for axial sites, so fat lone pairs prefer them — bending the remaining F atoms into the letter "T".

Verify: (a) electrons left $= 8 - 5 + 1 = 4$ , lp $= 2$ , regions $= 7$ , pentagonal planar, $s p^{3} d^{3}$ ✓. (b) electrons left $= 8 - 3 - 1 = 4$ , lp $= 2$ , regions $= 5$ , T-shaped, $s p^{3} d$ ✓. See Interhalogen and Oxyfluoride compounds for the ClF₃ analogy.

Example 8 — Cell H: real-world / word problem

Step 1 — Choose feed ratio. Use a large excess of F₂: $Xe : F_{2} = 1 : 20$ . Why? More F₂ per Xe raises the chance that each Xe grabs the maximum (6) fluorines. A low ratio (Xe in excess) would starve the reaction of F and give XeF₂ instead.

Step 2 — Choose pressure & temperature. High pressure $60 - 70 bar$ , moderate heat $573 K$ . Why? High pressure (Le Chatelier) drives the gas-phase reaction toward the side with fewer gas moles and stabilises the highest fluoride; moderate heat gives enough energy to react without decomposing the product.

Step 3 — Write the balanced reaction. $Xe + 3 F_{2} 1 : 20 573 K, 60 - 70 bar XeF_{6} .$ Why check the balance? To confirm three F₂ molecules supply exactly the six F atoms XeF₆ needs: left $3 \times 2 = 6$ F atoms; right XeF₆ has $6$ ✓.

Step 4 — Why Ar fails. Fluorination needs F₂ to rip electrons off the noble gas. Ar has a much higher ionization enthalpy than Xe (see Ionization Enthalpy trends) — its electrons are held too tightly for even F₂ to remove. Why can't we just crank pressure higher? Pressure only shifts an equilibrium that already runs; it cannot overcome a fundamental energy barrier to removing Ar's tightly-held electrons. No noble-gas fluoride forms if the first electron won't budge.

Verify: stoichiometry $1 Xe + 3 F_{2}$ : F atoms $6 = 6$ ✓. Reactivity order Xe > Kr > Ar follows falling IE down the group ✓.

Example 9 — Cell I: exam twist ("spot the wrong reflex")

Step 1 — Fix the XeF₄ error. $CH_{4}$ : C has $V = 4$ , so $(4 + 4) /2 = 4$ pairs $=$ 4 bp, 0 lp ⇒ tetrahedral. But $XeF_{4}$ : $(8 + 4) /2 = 6$ pairs $=$ 4 bp + 2 lp. The two lone pairs make it square planar, not tetrahedral. Why did the student go wrong? They counted bonds but never counted the lone pairs. Xe brings $V = 8$ valence electrons versus carbon's $4$ ; those extra four electrons become two lone pairs that carbon simply does not have, changing the whole geometry.

Step 2 — Fix the XeF₆ error. $SF_{6}$ : S has $V = 6$ , $(6 + 6) /2 = 6$ pairs $=$ 6 bp, 0 lp ⇒ perfect octahedron. $XeF_{6}$ : $(8 + 6) /2 = 7$ pairs $=$ 6 bp + 1 lp ⇒ electron-geometry pentagonal bipyramidal, molecular shape distorted octahedral. Why did the student go wrong here? Same blind spot: they matched Xe to S by counting six bonds, but Xe's two extra valence electrons ( $V = 8$ vs $6$ ) create one stereochemically active lone pair. That lone pair dents the octahedron — so "perfect octahedron" is wrong.

Step 3 — State the general resolution. The number of terminal atoms alone never fixes the shape. Always run the counting rule to get the lone pairs (which come from the central atom's valence-electron count $V$ ), then place them. Two molecules with the same number of bonds but different $V$ almost always have different shapes — as $CH_{4}$ vs $XeF_{4}$ and $SF_{6}$ vs $XeF_{6}$ both prove. Why this is the takeaway? It converts a memorised list of shapes into one reliable procedure that works for any new species the exam invents.

Verify: CH₄: $(4 + 4) /2 = 4$ , lp $= 0$ , tetrahedral ✓. XeF₄: $(8 + 4) /2 = 6$ , lp $= 2$ , square planar ✓. SF₆: $(6 + 6) /2 = 6$ , lp $= 0$ , octahedral ✓. XeF₆: $(8 + 6) /2 = 7$ , lp $= 1$ , distorted octahedral ✓.

Recall Self-test: match each species to its cell

XeF₂ → Cell D (3 lp) ::: Linear, sp³d XeF₄ → Cell C (2 lp trans) ::: Square planar, sp³d² XeF₆ → Cell B (1 lp) ::: Electron-geometry pentagonal bipyramidal; molecular shape distorted octahedral, sp³d³ XeO₃ → Cell E (double bonds) ::: Trigonal pyramidal, sp³ XeOF₄ → Cell F (mixed) ::: Square pyramidal, sp³d² XeF₅⁻ → Cell G (anion) ::: Pentagonal planar, sp³d³ (7 regions, 2 lp) XeF₃⁺ → Cell G (cation) ::: T-shaped, sp³d (5 regions, 2 lp)

Recall The one rule to remember

General rule (handles single bonds, double bonds AND charge): $lp = \frac{1}{2} [V - n_{single} - 2 n_{double} - C]$ , then $regions = n_{single} + n_{double} + lp$ . Hybridisation is read from the region count (4→sp³, 5→sp³d, 6→sp³d², 7→sp³d³). Then place lone pairs in the roomiest seats (VSEPR Theory).

Connections

Parent topic (Hinglish)
VSEPR Theory — the shape-prediction engine used in every example
Hybridisation (sp3d, sp3d2, sp3d3) — sp³d / sp³d² / sp³d³ frameworks
Ionization Enthalpy trends — why Xe reacts but Ar/Ne don't (Ex 8)
Interhalogen and Oxyfluoride compounds — XeOF₄, XeF₃⁺ ↔ ClF₃ analogies
Fractional Distillation of Liquid Air — how Xe is isolated in the first place
p-Block general trends — reactivity down the group