Isomerism — structural (linkage, ionization, coordination, hydrate) and stereo (geometrical, optical): Visual walkthrough

We link back to the parent Isomerism topic and use ideas from Coordination Number and Geometry, Werner's Theory and Coordination Sphere, and Chirality in Organic Chemistry.

Step 1 — Draw the empty stage: what "octahedral" actually is

WHAT: We start with one metal atom (call it $M$ ) at the centre and six empty positions where ligands can attach. A ligand is just an atom or group that donates a lone pair of electrons to $M$ (see Werner's Theory and Coordination Sphere). Six positions arranged so that each is at $9 0^{\circ}$ to its neighbours is called octahedral.

WHY this shape and not a hexagon: In three dimensions, six points that spread out as far from each other as possible sit at the tips of an octahedron — imagine two square-based pyramids glued base to base. This is the natural low-energy arrangement, so it is the geometry we must reason about (details in Coordination Number and Geometry).

PICTURE: Look at the figure. Four positions form a square around the equator, numbered $1, 2, 3, 4$ going around the square in order (so $1$ and $3$ are across from each other, and $2$ and $4$ are across from each other). Two more positions sit straight up and down the vertical axis (top $= 5$ , bottom $= 6$ ).

Step 2 — The simplest case that shows nothing: $M A_{6}$ and $M A_{5} B$

WHAT: Fill all six positions with the same ligand $A$ . That is $M A_{6}$ . Now swap exactly one $A$ for a different ligand $B$ : that is $M A_{5} B$ .

WHY start here: Before we hunt for isomers we must see when there are none — otherwise we can't tell what actually causes an isomer. Isomers need a choice of arrangement. With all-identical ligands there is no choice.

PICTURE: In $M A_{6}$ every position is identical, so there is nothing to rearrange. In $M A_{5} B$ , wherever you place the single $B$ , you can rotate the whole octahedron to bring that $B$ to the top. So every $M A_{5} B$ is the same molecule — one isomer only.

Step 3 — First real fork: $M A_{4} B_{2}$ gives cis and trans

WHAT: Now use two $B$ ligands and four $A$ ligands: $M A_{4} B_{2}$ . Place the two $B$ 's. There are exactly two genuinely different placements.

WHY exactly two: Fix one $B$ at the top (position 5) — we are allowed to, by the rotation trick of Step 2. The second $B$ can now go in only two kinds of spot relative to the first:

next to it (any equatorial position, $9 0^{\circ}$ away) → the two $B$ 's are cis,
opposite it (the bottom, position 6, $18 0^{\circ}$ away) → the two $B$ 's are trans.

All four equatorial spots are equivalent by rotation about the vertical axis, so "next to it" is a single case, not four.

PICTURE: Left, the two orange $B$ 's share an edge — cis. Right, they are on the top–bottom axis — trans. No rotation turns one into the other, so these are two distinct isomers.

$metal centre M four identical ligands A_{4} two identical ligands whose relative angle is the choice: 9 0^{\circ} or 18 0^{\circ} B_{2}$

Each symbol: $M$ is the atom holding everything; $A_{4}$ fills the four positions the $B$ 's don't; $B_{2}$ is the pair whose relative angle — $9 0^{\circ}$ (cis) or $18 0^{\circ}$ (trans) — is the only thing that differs between the two isomers.

Step 4 — The other geometrical fork: $M A_{3} B_{3}$ gives fac and mer

WHAT: Split the six positions evenly: three $A$ 's and three $B$ 's, $M A_{3} B_{3}$ . Place the three $B$ 's. Again there are exactly two genuinely different placements — but they have different names.

WHY exactly two: Three positions on an octahedron can group in only two ways:

All three $B$ 's meet at one triangular face — every pair of them is $9 0^{\circ}$ (cis) to each other. This is facial, or fac.
The three $B$ 's lie in a single plane through the metal (a "meridian," like a line of longitude on a globe) — so one pair is $18 0^{\circ}$ (trans) and the other two are $9 0^{\circ}$ . This is meridional, or mer.

There is no third option: once two $B$ 's are placed, the last $B$ is either on the face they share (fac) or across from one of them (mer).

PICTURE: Left, the three orange $B$ 's cap one triangular face — all mutually $9 0^{\circ}$ (fac). Right, they string along a meridian, so the top and bottom $B$ are $18 0^{\circ}$ apart (mer). No rotation turns one into the other.

$three of each ligand M A_{3} B_{3} ⟹ 3 B’s on one face all pairs 9 0^{\circ} fac and 3 B’s on a meridian one pair 18 0^{\circ} mer$

Each label reads: fac = the three like ligands share a face (all cis); mer = they share a meridional plane (one trans pair). We will run the mirror test on both in Step 8 — spoiler: both turn out achiral because each keeps a mirror plane, but you must check, never assume.

Step 5 — What a mirror image is (built from zero, with axes pinned to the octahedron)

WHAT: Take any isomer and reflect it in a flat mirror. The mirror swaps left and right but keeps up/down and front/back. The result is a new drawing called the mirror image (or enantiomer if it turns out to be different).

WHY we need this: Geometrical isomerism (Steps 3–4) came from angles. The other family, optical isomerism, comes from handedness — and handedness only shows up when you compare a shape to its reflection. So we must first know precisely what "reflect" does.

Pinning the axes to our numbered picture (so the coordinates mean something): put the metal $M$ at the origin. Then match each straight line joining opposite positions (the $18 0^{\circ}$ pairs we listed in Step 1) to one axis:

the vertical axis is $y$ — it joins position 5 (top) ↔ position 6 (bottom);
the left–right axis is $x$ — it joins position 1 ↔ position 3 (the pair across the square);
the front–back axis is $z$ — it joins position 2 ↔ position 4 (the other pair across the square).

So the numbering from Step 1 fixes the axes with no ambiguity: $1 \leftrightarrow 3$ is $x$ , $2 \leftrightarrow 4$ is $z$ , $5 \leftrightarrow 6$ is $y$ .

PICTURE: A single point at $(x, y, z)$ reflected in the vertical $y z$ -mirror plane (the plane containing the up–down axis and the front–back axis) becomes $(- x, y, z)$ — the $x$ (left–right) sign flips, the others stay. Watch the blue point and its red reflection.

$(x, y, z) mirror in y z -plane (- x, y, z)$

Term by term: $x$ is how far left/right a ligand sits (along the position-1↔3 line); the minus sign in front of it is the entire action of the mirror — it negates left–right only. $y$ (up–down, along 5↔6) and $z$ (front–back, along 2↔4) pass through unchanged, so this mirror never moves anything vertically.

Step 6 — Test trans- $[M (e n)_{2} X_{2}]$ : the horizontal equatorial plane maps X onto X ⇒ achiral

WHAT: Introduce a bidentate ligand — one ligand with two donor atoms that grabs two adjacent positions of the octahedron at once (it must span a $9 0^{\circ}$ edge; it cannot reach across $18 0^{\circ}$ ). The standard example is en = ethylenediamine, drawn as a curved bridge $⌢$ linking two neighbouring spots (see Ambidentate vs Polydentate Ligands). Take $[M (e n)_{2} X_{2}]$ with the two $X$ 's trans (opposite, on the vertical $y$ -axis, i.e. positions 5 and 6).

WHY this case first: It is the "boring" one — and finding why it is boring reveals the symmetry test in action.

PICTURE: With both $X$ 's on the vertical axis, the two en bridges lie in the horizontal equatorial belt. The relevant mirror is the horizontal equatorial plane through the metal centre — the flat plane at $y = 0$ containing the four equatorial positions. Reflecting through it flips top↔bottom: it sends the top $X$ (position 5) exactly onto the bottom $X$ (position 6) and back, and leaves the equatorial en bridges in place. Because reflection lands the molecule perfectly on itself, this equatorial plane is a genuine internal mirror plane ⇒ achiral.

Step 7 — Test cis- $[M (e n)_{2} X_{2}]$ : no plane survives ⇒ two enantiomers

WHAT: Same complex, but now the two $X$ 's are cis (adjacent, $9 0^{\circ}$ ). The two en bridges are forced to wrap around like the threads of a screw.

WHY it flips to chiral: Put the two $X$ 's on two adjacent positions. Now hunt for any internal mirror plane, exactly as we did for $t r an s$ :

the horizontal equatorial plane no longer works — one $X$ is on the axis, one is in the belt, so reflection sends an $X$ where an en donor sits;
every other candidate plane cuts an en bridge the wrong way.

No plane survives the test. The two en bridges spiral either clockwise ( $Δ$ , "delta") or anticlockwise ( $Λ$ , "lambda"), like left-handed and right-handed screws.

PICTURE: Left = the $Δ$ (right-handed) spiral of the two en bridges; right = its mirror image, the $Λ$ (left-handed) spiral. Try to rotate one onto the other — a right-handed screw never becomes a left-handed one by turning. So these are two distinct molecules: a pair of enantiomers.

$cis- [M (en)_{2} X_{2}] ⟹ right-handed twist Δ and left-handed twist Λ (non-superimposable)$

The symbols $Δ$ and $Λ$ are names for the two handednesses of the spiral — they are not numbers; they simply label which way the chelate rings screw around the vertical axis.

Step 8 — Run the same mirror test on the rest: fac, mer, and the general mixed-ligand case

WHAT & WHY: We now push every remaining arrangement through the identical narrative — empty geometry → angle choice → mirror test — so no reader ever meets an unshown scenario. Crucially, for fac and mer we construct the specific mirror plane by naming which positions it fixes and which it swaps, so "plane present" is shown, not asserted.

Back to $M A_{3} B_{3}$ (fac and mer from Step 4). Constructing each mirror plane:

fac: put the three $B$ 's on the face made of positions ${5, 1, 2}$ (top, and two adjacent equatorial); the three $A$ 's then cap the opposite face ${6, 3, 4}$ . Take the plane that contains the $M$ , the $5 - 6$ axis, and bisects the angle between positions $1$ and $2$ . This plane fixes the $5$ - $B$ and $6$ - $A$ (they lie in it) and swaps $1 \leftrightarrow 2$ (two $B$ 's) and $3 \leftrightarrow 4$ (two $A$ 's) — like reflecting onto like. Every atom lands on an identical atom ⇒ genuine mirror plane ⇒ achiral.
mer: put the three $B$ 's on the meridian ${5, 6, 1}$ (top, bottom, one equatorial). Take the plane that contains all three of these $B$ positions ( $5$ , $6$ , $1$ ) plus $M$ and position $3$ (which is opposite $1$ ). This plane fixes $5, 6, 1, 3$ (all lie in it) and swaps the two off-plane $A$ 's, $2 \leftrightarrow 4$ . Again like maps to like ⇒ genuine mirror plane ⇒ achiral.

So $M A_{3} B_{3}$ gives 2 geometrical isomers, both achiral ⇒ 2 stereoisomers total — and we built the plane in each case, we did not assume it.

The general mixed-ligand case $M A_{2} B_{2} C_{2}$ (the recipe finally on a hard target): three pairs of identical ligands. Run the identical three-step recipe:

Angle step. Each pair can be trans or cis. Enumerate: all-trans (every pair opposite); one pair trans, two pairs cis; all-cis. Sorting out the rotational duplicates leaves exactly 5 geometrical arrangements.
Mirror step, on each. Four of them keep an internal mirror plane (built exactly as for fac/mer above) ⇒ achiral. One arrangement — the all-cis one — has no plane and no centre ⇒ chiral ⇒ counts as a $Δ/Λ$ pair.
Count. $4 \times 1 + 1 \times 2 = 6$ stereoisomers.

This is the promised payoff: the same "geometry → angles → mirror" loop that gave $M A_{4} B_{2}$ its 2 and $[M (e n)_{2} X_{2}]$ its 3 now cracks a formula the parent note never mentioned.

The other degenerate octahedral cases, same narrative:

$M A_{6}$ / $M A_{5} B$ (Step 2): only one angle arrangement, and it keeps full symmetry ⇒ one isomer, achiral.
$M (e n)_{3}$ : three bidentate bridges, all spiralling; run the mirror test and — just like cis- $[M (e n)_{2} X_{2}]$ — no plane survives, so it is a chiral $Δ/Λ$ pair, even with no " $X$ " ligand at all.

Stepping off the octahedron (to see the boundary of our rules): the very same recipe on a tetrahedron shows why the octahedral results are special. A tetrahedron has no opposite ( $18 0^{\circ}$ ) positions — every vertex is $109. 5^{\circ}$ from every other — so the angle step produces no cis/trans fork at all. Then the mirror step: $M A_{2} B_{2}$ keeps a mirror plane (achiral, 1 isomer), but four different ligands $M ab c d$ leave no plane ⇒ chiral, exactly the handedness of a carbon centre in Chirality in Organic Chemistry. Same recipe, different stage.

PICTURE: each remaining case tagged with the verdict its three-step test produced.

Recall Check yourself on the edges

How many stereoisomers does $M A_{3} B_{3}$ have, and are they chiral? ::: Two — fac and mer — and both are achiral because each keeps an internal mirror plane (built explicitly in Step 8). How many stereoisomers does $M A_{2} B_{2} C_{2}$ have? ::: Six — 5 geometrical arrangements, of which the all-cis one is chiral and so counts twice ( $4 \times 1 + 1 \times 2 = 6$ ). How many stereoisomers does tetrahedral $M A_{2} B_{2}$ have, and why? ::: One — every vertex is equivalent (no opposite position, and it has a mirror plane), so no cis/trans and no chirality. Is $[C o (e n)_{3}]^{3 +}$ chiral? ::: Yes — three spiralling chelate rings give a $Δ$ / $Λ$ pair with no internal mirror plane.

The one-picture summary

The whole derivation on one canvas: start from the empty octahedron → add ligands → the angle choice ( $9 0^{\circ}$ vs $18 0^{\circ}$ ; face vs meridian) splits geometrical isomers (cis/trans, fac/mer) → for each, the mirror test splits optical isomers (achiral vs $Δ$ / $Λ$ ). Counting rule: chiral geometrical isomers count twice.

Recall Feynman: tell the whole walk to a 12-year-old

Picture a ball with six sticks — four around the middle like a compass, one up, one down. That's our stage. If all six ends are the same, or only one is different, there's nothing to argue about: only one way to build it. But hang two special ends and you get a real choice — put them next to each other (cis) or across from each other (trans). Hang three special ends and you get a different choice — bunch them on one face (fac) or spread them along a line of longitude (mer). Those are the "angle" isomers.

Now hold each toy up to a mirror. Sometimes the reflection is just the same toy again — that means the toy already had a "fold line" of symmetry, and we call it achiral (fac, mer, and trans all do this). But sometimes the reflection is a left-handed version you can never turn into the right-handed one — like your two hands (the cis one with two curly en straps does this). Those come in pairs, so we count them twice. That's the entire secret: angles make the cis/trans and fac/mer pairs, and the mirror makes the left-hand/right-hand pairs. Add them up and you have every stereoisomer — even for a monster like $M A_{2} B_{2} C_{2}$ , which comes out to six.

Recall Links to widen the picture

The colour/field-strength side of the story: Crystal Field Theory.
Why $c i s$ - $[P t (N H_{3})_{2} C l_{2}]$ matters medically: Cisplatin and Bioinorganic Chemistry.
The organic mirror-image analogy: Chirality in Organic Chemistry.