3.2.11p-Block

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

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1. Discovery — WHY they were missed for so long

  • Argon (1894, Rayleigh & Ramsay): "Atmospheric N₂" was denser than chemically-made N₂. WHY? Atmospheric "nitrogen" hid a heavier unreactive gas → Argon (Greek argos = lazy).
  • Helium: First seen as a yellow line in the Sun's spectrum (1868) before being found on Earth (helios = sun). Found terrestrially in uranium minerals (it's emitted α\alpha-particles = He nuclei).
  • Ne, Kr, Xe: Found by fractional distillation of liquid air (Ramsay & Travers). Names: neos=new, kryptos=hidden, xenos=stranger.
  • Radon: Radioactive, from decay of radium.

2. Isolation

WHY fractional distillation works: each noble gas has a distinct, very low boiling point; warming the liquid slowly boils them off one at a time.


3. The first compound — HOW Bartlett reasoned (Forecast-then-Verify)


4. Xenon fluorides — preparation (all from Xe + F₂, vary the ratio)


5. Structure & bonding — derive with VSEPR

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

XeF₂

  • Xe valence = 8; 2 bonds to F (each F shares 1 e⁻).
  • Pairs =(8+2)/2=5= (8+2)/2 = 52 bond pairs + 3 lone pairs.
  • 5 pairs ⇒ trigonal bipyramidal arrangement; 3 lone pairs go equatorial (least repulsion).
  • Shape = linear (F–Xe–F, 180°180°); hybridisation sp3dsp^3d.
  • Why this step? Lone pairs need maximum space → equatorial positions of TBP → axial F's left → linear.

XeF₄

  • Pairs =(8+4)/2=6= (8+4)/2 = 64 bond pairs + 2 lone pairs.
  • 6 pairs ⇒ octahedral; 2 lone pairs go opposite (trans) to each other.
  • Shape = square planar; hybridisation sp3d2sp^3d^2.
  • Why? trans lone pairs minimise lone-pair–lone-pair repulsion (max 180°180° apart).

XeF₆

  • Pairs =(8+6)/2=7= (8+6)/2 = 76 bond pairs + 1 lone pair.
  • 7 pairs ⇒ based on pentagonal bipyramid / distorted octahedron.
  • Shape = distorted octahedral (the lone pair distorts it); hybridisation sp3d3sp^3d^3. Not a perfect octahedron because of the stereochemically active lone pair.

XeO₃ (with O, double bonds Xe=O)

  • XeF6+3H2OXeO3+6HF\text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF} (complete hydrolysis; explosive solid).
  • Xe: 3 Xe=O bonds + 1 lone pair → 4 electron domains.
  • Shape = pyramidal (trigonal pyramidal), like NH₃; sp3sp^3.

6. Quick property logic


Who discovered the first noble gas compound and in what year?
Neil Bartlett, 1962 — Xe⁺[PtF₆]⁻
What clue made Bartlett try Xe?
IE of Xe (1170) ≈ IE of O₂ (1175 kJ/mol), and PtF₆ had oxidised O₂
Geometry of XeF₂ and its hybridisation?
Linear; sp³d (2 bp + 3 lp, TBP base)
Geometry of XeF₄ and hybridisation?
Square planar; sp³d² (4 bp + 2 lp trans, octahedral base)
Geometry of XeF₆ and hybridisation?
Distorted octahedral; sp³d³ (6 bp + 1 lp)
Geometry of XeO₃?
Trigonal pyramidal, sp³ (3 Xe=O + 1 lp)
How is XeF₂ made?
Xe + F₂ (Xe in excess), 673 K, 1 bar
How is XeF₆ made?
Xe + 3F₂, ratio 1:20, 573 K, 60–70 bar
Product of complete hydrolysis of XeF₆?
XeO₃ + HF (explosive XeO₃)
Product of partial hydrolysis of XeF₆?
XeOF₄ + 2HF (square pyramidal)
How was argon first noticed?
Atmospheric N₂ denser than chemical N₂ (Rayleigh–Ramsay)
Where was helium first detected?
In the Sun's spectrum (yellow line) before Earth
Why are Ne and Ar essentially unreactive while Xe reacts?
Higher IE / electrons more tightly held; Xe has low IE
Number of electron domains in XeF₄?
6 (4 bond pairs + 2 lone pairs)
Recall Feynman: explain to a 12-year-old

Imagine some kids who already have all the toys they want — they don't trade with anyone. Those are the noble gases. For ages people thought they'd never trade. But one big kid, Xenon, holds his outer toys loosely. So if a super-greedy kid named Fluorine (or Oxygen) comes, he can snatch toys and force Xenon to make friends — that's XeF₂, XeF₄, XeF₆, XeO₃. The shapes of these friend-groups depend on how many fluorine arms Xenon has PLUS how many "invisible" lone pairs push them around, like extra elbows making everyone stand a certain way.

Connections

  • VSEPR Theory — predicts every shape above
  • Hybridisation (sp3d, sp3d2, sp3d3) — bonding framework for expanded octet
  • Ionization Enthalpy trends — explains why only Xe reacts
  • Fractional Distillation of Liquid Air — isolation method
  • Interhalogen and Oxyfluoride compounds — XeOF₄ analogy
  • London Dispersion Forces — why noble gases have low boiling points
  • p-Block general trends

Concept Map

means

discovered as

Ar denser N2

isolates

Xe low IE 1170

IE close to O2 1175

makes

reacts with F2

vary ratio

reacts with O

renamed

proves

Noble gases full shell

Low reactivity inert

Left-over gas of air

Fractional distillation of air

He Ne Ar Kr Xe

Xenon reactive

Bartlett 1962

Xe+ PtF6- first compound

XeF2 XeF4 XeF6

Different stoichiometry

XeO3

Noble not inert

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Noble gases (He, Ne, Ar, Kr, Xe, Rn) ka outer shell full hota hai, isliye inhe "inert" kehte the — yeh kisi se react nahi karte. Discovery ka mazza yeh hai ki yeh air mein "bacha hua gas" tha: jab N₂, O₂, CO₂, water sab nikaal do, tab bhi kuch reh jaata tha — wahi Argon nikla. Helium toh pehle Sun ke spectrum mein dikha tha, baad mein Earth pe mila. Isolation ke liye liquid air ka fractional distillation karte hain, kyunki har gas ka boiling point alag hota hai.

Ab twist: Xenon group mein neeche hai, iske electrons nucleus se door aur dheele bandhe hote hain (low ionization enthalpy, O₂ jaisa). 1962 mein Bartlett ne dekha ki PtF₆ ne O₂ ka electron kheench liya — aur socha "Xe ka IE bhi waisa hi hai, toh Xe bhi react karega!" Forecast verify ho gaya — Xe⁺[PtF₆]⁻ bana, pehla noble gas compound. Isliye sirf bahut greedy partners — F aur O — hi Xe ko react kara paate hain.

Shapes ke liye VSEPR use karo: Xe ke around bond pairs + lone pairs count karo. XeF₂ = 2 bond + 3 lone = linear. XeF₄ = 4 bond + 2 lone = square planar. XeF₆ = 6 bond + 1 lone = distorted octahedral (lone pair thoda push karta hai). XeO₃ = 3 double bonds + 1 lone = pyramidal (NH₃ jaisa). Yaad rakho: lone pair hamesha shape ko bigaadta hai, isliye XeF₄ tetrahedral nahi aur XeF₆ perfect octahedron nahi.

Go deeper — visual, from zero

Test yourself — p-Block

Connections