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

We will build up the whole story in numbered steps. Read them in order — each figure carries the reasoning, so look at the arrows as you read.

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Step 1 — What is an "electron group" and why do we count it?

WHAT. Around the xenon atom (call it $\text{Xe}$, the central atom) there are two kinds of electron clumps:

• a bond pair — a pair of electrons shared in a bond to a fluorine or oxygen atom (this is a "leash" holding a partner atom);
• a lone pair — a pair of electrons that belongs to Xe alone, pointing into empty space (an "invisible arm").

Together we call each clump an electron group (or domain).

WHY count them? Because electron groups are all negatively charged, they repel each other. Whether a group is a bond or a lone pair, it still takes up a direction in space and shoves the others away. So the number of groups — not the number of atoms — decides the skeleton geometry. That is the whole engine of VSEPR Theory.

PICTURE. Balloons tied at one knot spread out to fill space. Two balloons → opposite ends. Three → a flat triangle. Look at how adding a balloon changes the pattern.

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Step 2 — The counting formula (earn every symbol)

WHAT. We need a way to get $n$ from the formula of the molecule alone. First fix what $b$ means in this electron-donation picture:

Now the rule for xenon fluorides, built piece by piece. Read it as "add $V$ and $b$ FIRST, then halve the whole thing":

$n=\frac{\stackrel{\text{add these, then}}{\overbrace{V+b}}}{\underset{\text{halve}}{\underbrace{2}}}$

Reading it term by term, right where each symbol sits:

• $V$ = the valence electrons of Xe = $8$. (Xe is in group 18; its outer shell is $5s^{2}5p^6$, which is $8$ electrons. These are the electrons Xe brings to the party.)
• $b$ = one electron contributed by each bonded F atom (just defined above). So we add $b$ (one per F) to Xe's $8$.
• The whole numerator $V+b$ is the total number of electrons that end up around Xe. We divide that entire sum by $2$ because electrons live in pairs, and a "group" is a pair.

WHY this formula and not just "count the atoms"? Because the lone pairs are invisible in the chemical formula — nothing in "XeF₄" tells you there are two hidden pairs. This arithmetic reveals them: once we know $n$ and we know $b$ (the number of F's), the leftover is the lone pairs, $l=n-b$.

PICTURE. A ledger: Xe puts $8$ on the table, each F adds $1$, we total the pile FIRST, then sweep it into pairs.

Let us now feed $b=2,4,6$ into this machine, one step each.

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Step 3 — XeF₂: two bonds, and where do the lone pairs go?

WHAT. Put $b=2$:

$n=\frac{8+2}{2}=5,l=5-2=3.$

So Xe has $5$ groups: $2$ bond pairs (to the two F's) and $3$ lone pairs. Five groups always arrange as a trigonal bipyramid (TBP) — three positions in a flat "equator", two positions at the "poles" (axial).

WHY the lone pairs choose the equator. In a TBP the two positions are not equal:

• an axial position (pole) has $3$ close neighbours at $90^{\circ }$;
• an equatorial position has only $2$ close neighbours at $90^{\circ }$.

A lone pair is fat and pushy (no second nucleus to pin it down), so it is the biggest source of repulsion. To minimise repulsion, lone pairs take the roomy equatorial seats. All $3$ lone pairs go equatorial; the $2$ bonds are forced into the two axial seats.

PICTURE. Watch the three amber lone pairs claim the equator and squeeze the two F's onto the vertical axis — leaving them exactly opposite: a straight line.

Result. $\text{F}\text{–}\text{Xe}\text{–}\text{F}$ is a straight line: shape = linear, $180^{\circ }$. Since $n=5$, the conventional hybridisation label is $s{p}^{3}d$.

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Step 4 — XeF₄: six groups, and why the lone pairs sit opposite

WHAT. Put $b=4$:

$n=\frac{8+4}{2}=6,l=6-4=2.$

Six groups: $4$ bond pairs + $2$ lone pairs. Six groups always start from an octahedron — six positions all $90^{\circ }$ apart, like the $6$ faces of a die pointing out.

WHY the two lone pairs go trans (opposite). In an octahedron every position is equivalent, but two lone pairs must decide how far apart to sit:

• cis (adjacent) → they are $90^{\circ }$ apart → strong lone-pair–lone-pair repulsion;
• trans (opposite) → they are $180^{\circ }$ apart → weakest possible repulsion.

Nature picks the smaller repulsion: the two lone pairs go trans, one straight up, one straight down. That leaves the $4$ fluorines in the horizontal plane.

PICTURE. The two amber lone pairs take the top and bottom poles; the four cyan Xe–F bonds flatten into one plane at $90^{\circ }$ to each other.

Result. Four F's in a flat plane at $90^{\circ }$: shape = square planar. With $n=6$, hybridisation label is $s{p}^3{d}^2$.

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Step 5 — XeF₆: seven groups on a pentagonal bipyramid

WHAT. Put $b=6$:

$n=\frac{8+6}{2}=7,l=7-6=1.$

Seven groups: $6$ bond pairs + one lone pair. In VSEPR, seven electron groups arrange as a pentagonal bipyramid (PBP) — a flat pentagon of five equatorial positions, plus two axial positions (one straight up, one straight down). This is the standard 7-group parent geometry; it is not an octahedron of any kind.

WHY the lone pair takes an equatorial (pentagon) seat. Just like the TBP in Step 3, the seats of a PBP are not equal:

• an axial seat (pole) points at all five equatorial neighbours at only $90^{\circ }$ — very crowded;
• an equatorial seat sits in the roomy pentagon, with its nearest neighbours a wider $72^{\circ }$ around the ring and the two poles at $90^{\circ }$ — more elbow room overall.

The fat, pushy lone pair wants the least-crowded seat, so the lone pair goes equatorial (into the pentagon). That leaves the six F's on the remaining seats — and because one pentagon seat is now filled by an invisible pair, the six bonds cannot sit symmetrically.

WHY the observed molecule looks like a "dented octahedron." Take those six F positions and let them relax: they end up close to octahedral, but the single equatorial lone pair keeps pushing the nearby bonds away, so the real molecule is a fluxional, distorted octahedral (non-rigid) shape rather than a perfect one. So: parent geometry = pentagonal bipyramid; observed shape = distorted octahedron.

PICTURE. Six F bonds relaxed toward octahedral positions; the amber lone pair sits in the pentagon plane and dents the cage — the nearby bonds bow away from it.

Result. Shape = distorted octahedral (parent 7-group geometry = pentagonal bipyramid).

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Step 6 — Degenerate case XeO₃: when the partner is oxygen (double bonds)

WHAT. Oxygen is different from fluorine. A single Xe–F bond needs one shared pair; but Xe=O is a double bond, and VSEPR treats a whole multiple bond as ONE electron group (it points in a single direction). So the "$+b$" trick has to be adjusted, because each O now shares two electrons with Xe, not one.

Let us do the electron arithmetic explicitly, the same "total electrons ÷ 2" spirit as before:

• $V=8$ valence electrons on Xe (as always).
• Each Xe=O double bond has Xe contribute $2$ electrons; across $3$ oxygens that uses $3\times 2=6$ of Xe's electrons in bonding.
• Electrons left on Xe as non-bonding: $8-6=2$ electrons $=1$ pair.

$\text{lone pairs }l=\frac{8-(2\times 3)}{2}=\frac{8-6}{2}=1.$

• Bonded groups (double bonds, each = one group): $3$.
• Total electron groups: $n=3+1=4$.

So the single lone pair is not asserted — it is exactly the $8-6=2$ leftover electrons.

WHY four groups → tetrahedral base → pyramid. Four groups spread to the corners of a tetrahedron (the maximum-spread shape for four directions). One corner is the lone pair; the other three are the O's. Hide the lone pair and you see three O's below Xe — a tripod.

PICTURE. Same skeleton as ammonia ${\text{NH}}_3$: one amber lone pair on top, three cyan Xe=O bonds fanning down into a pyramid.

Result. Shape = trigonal pyramidal (like ${\text{NH}}_3$); hybridisation label $s{p}^3$ ($n=4$). It comes from complete hydrolysis, ${\text{XeF}}_6+3{\text{H}}_2\text{O}\to {\text{XeO}}_3+6\text{HF}$.

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Step 7 — Reading the whole pattern in one line

WHAT. Line up the four molecules by their lone-pair count $l$. Here $b$ always means number of bonded atoms/groups around Xe (for XeO₃ that is the $3$ oxygens, each a single double-bond group):

Molecule	$b$ (bonded groups)	$n$ (electron groups)	$l=$ lone pairs	base geometry	shape
${\text{XeF}}_2$	2	5	3	trigonal bipyramid	linear
${\text{XeF}}_4$	4	6	2	octahedron	square planar
${\text{XeF}}_6$	6	7	1	pentagonal bipyramid	distorted octahedral
${\text{XeO}}_3$	3	4	1	tetrahedron	pyramidal

(For XeF₂/₄/₆, $n=(8+b)/2$. For XeO₃ each O is a double bond, so $n$ is found by the double-bond arithmetic of Step 6, not the single-bond formula.)

WHY it's satisfying. As $b$ climbs $2\to 4\to 6$ for the fluorides, the lone pairs fall $3\to 2\to 1$ (the mnemonic "3-2-1"), and each drop reshapes the molecule. Every shape is just "start from the $n$-group skeleton, then put lone pairs in the roomiest seats."

PICTURE. The four shapes side by side, colour-coded by lone-pair count.

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The one-picture summary

One figure, the whole derivation: start with the count $n=\frac{8+b}{2}$ (fluorides) or the double-bond arithmetic (XeO₃), split into bonds and lone pairs, place lone pairs in the least-crowded seats, and read off the shape.

Recall Feynman retelling — the walkthrough in plain words

Picture Xenon holding balloons. Every balloon is a group of electrons — some are bonds (a balloon with a fluorine or oxygen tied to the end), some are lone pairs (a balloon with nothing on it, just floating). Balloons hate each other and spread out as far as they can.

First I count how many balloons there are: Xenon brings $8$ electrons, each fluorine adds $1$, and — after totalling the whole pile — I bundle them into pairs; that gives the number $n$. With two fluorines that's five balloons; three of them are empty lone pairs, and they grab the roomy middle seats, squeezing the two fluorines to opposite poles → a straight line. With four fluorines it's six balloons; the two empty ones sit dead opposite each other, flattening the four fluorines into a square. With six fluorines it's seven balloons arranged as a pentagon-with-two-poles; the single empty one takes a roomy pentagon seat and dents the cage → a squashed octahedron. And with oxygen, each oxygen ties on with a double string that counts as one balloon and uses two of Xenon's electrons, so after three oxygens only two electrons are left — one empty balloon — giving four balloons total → a pyramid, just like ammonia. Every shape is the same trick: count the balloons, let the empty ones take the best seats, and look at where the atoms ended up.

Connections

• Parent topic (Hinglish)
• VSEPR Theory — the balloon-repulsion rule this whole page runs on
• Hybridisation (sp3d, sp3d2, sp3d3) — the $s{p}^{3}d$, $s{p}^3{d}^2$, $s{p}^3{d}^3$ shape labels for $n=5,6,7$
• Ionization Enthalpy trends — why Xe (not Ne/Ar) can be forced to bond at all
• Interhalogen and Oxyfluoride compounds — XeOF₄ and cousins by the same counting