Visual walkthrough — Coordination number and geometry — 2 (linear), 4 (tetrahedral - square planar), 6 (octahedral)

We start from the most primitive idea possible.

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Step 1 — What is a ligand, really? Points that push apart

WHAT: We replace the whole complicated molecule by a single dot of negative charge at a fixed distance $r$ from the metal.

WHY: Because the only thing that decides geometry at this level is that these negative dots repel each other (like charges push apart). Their internal chemistry does not matter yet — so we throw it away and keep the physics that matters.

PICTURE: the metal at the centre, ligands as cyan dots at distance $r$, red arrows showing the repulsion between two of them.

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Step 2 — CN = 2: two points on a sphere → 180°

WHAT: Put two dots on a sphere and slide them until $d_{12}$ is largest.

WHY: With only one pair, minimising $E_{\text{rep}}=k/d_{12}$ means maximising the single distance $d_{12}$. The farthest two points on a sphere can be is diametrically opposite — the two poles.

PICTURE: two ligands at the north and south poles; the angle measured at the metal between the two bonds is the straight angle.

Who actually does this? $d^{10}$ ions (Ag⁺, Au⁺, Cu⁺). A filled $d^{10}$ shell is spherical and gains nothing from extra ligands, so it holds the minimum useful number — two — as far apart as possible.

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Step 3 — CN = 4, first guess: four points → tetrahedron (109.5°)

WHAT: Add two more dots, so four dots on the sphere, and again spread them for minimum repulsion.

WHY: You might guess "put them in a flat square (90° apart)." But a square is cramped — the dots are all in one plane. Lifting them into 3D lets each dot back away from all three others at once. The unique arrangement where all four are equidistant from each other is the tetrahedron.

PICTURE: the four ligands at alternating corners of a cube — that is a tetrahedron — with the equal edges highlighted.

Tetrahedral is chosen when the metal has no special electronic preference — $d^0,d^5,d^{10}$ — or when ligands are bulky (a tetrahedron gives more elbow room) or weak field. Examples: $[{\text{NiCl}}_4{]}^{2-}$, $[{\text{CoCl}}_4{]}^{2-}$, $[{\text{Zn(NH}}_3{)}_4{]}^{2+}$.

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Step 4 — CN = 6: six points → octahedron (90°)

WHAT: Six dots on the sphere, spread out.

WHY: The best you can do is put one dot on each of the six directions $\pm x,\pm y,\pm z$ — the corners of an octahedron. Every dot then has four neighbours at $90^{\circ }$ and one opposite at $180^{\circ }$, and this is the minimum-repulsion layout for six.

PICTURE: four ligands forming a square in the plane, one straight up, one straight down.

Why most common? Six ligands satisfy the metal's charge/bonding needs while $90^{\circ }$ keeps repulsion tolerable — a sweet spot. Most ${\text{Co}}^{3+},{\text{Fe}}^{3+},{\text{Cr}}^{3+}$ complexes are octahedral.

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Step 5 — The twist: why $d^8$ + strong field breaks the tetrahedral guess

Steps 1–4 used only ligand–ligand repulsion. For transition metals we must now let the metal's own d-electrons vote. This is Crystal Field Theory.

WHAT: Watch what a square-planar arrangement of the four ligands does to the metal's five d-orbitals.

WHY: In a square plane the ligands lie along $\pm x,\pm y$. The orbital $d_{x^2-y^2}$ points straight at those four ligands, so its electrons are shoved very high in energy. The other four d-orbitals point between ligands and stay low.

PICTURE: the square-planar energy ladder — four low orbitals, then a big gap, then a lonely high $d_{x^2-y^2}$.

Same metal ${\text{Ni}}^{2+}$ ($d^8$), two answers: $[{\text{Ni(CN)}}_4{]}^{2-}$ is square planar, $[{\text{NiCl}}_4{]}^{2-}$ is tetrahedral.

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Step 6 — Degenerate & edge cases (don't get ambushed)

PICTURE: one ligand with two hands (bidentate, like en) filling two sphere positions from a single molecule.

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The one-picture summary

One sphere, one rule (dots spread out), one twist ($d^8$ + strong field re-shuffles CN = 4):

dots on sphere ($\mathrm{CN}$)	spread-out answer	key angle	electronic override
2	linear	$180^{\circ }$	(loved by $d^{10}$)
4	tetrahedral	$109.5^{\circ }$	→ square planar if $d^8$ + strong field
6	octahedral	$90^{\circ }/180^{\circ }$	most ions

Recall Feynman retelling (say it to a 12-year-old)

Glue negative dots onto a ball; the dots hate each other and slide as far apart as they can. Two dots → opposite ends → a straight line. Four dots → they climb into a little 3D pyramid (a tetrahedron) because a flat square would squeeze them. Six dots → a square belt plus a top and bottom cap → an octahedron. That's the whole shape story from pushing alone. One special exception: when the metal happens to have exactly eight d-electrons and the dots are the "strong, pinching" kind, it becomes worth it to flatten the four dots into a square — because doing so leaves one very expensive electron-seat empty and saves a lot of energy. Also remember: some dots come with two hands, so one molecule can fill two spots on the ball — that's why you count hands (donor atoms), not molecules.

Recall

Where does 109.5° come from? ::: $\arccos (-1/3)$ = the angle between two tetrahedron vertices seen from the centre. Why linear for CN = 2? ::: Two repelling points on a sphere are farthest apart at opposite poles → 180°. What flips $d^8$ from tetrahedral to square planar? ::: A strong-field ligand makes the gap big enough to leave $d_{x^2-y^2}$ empty, which is strongly stabilising. CN of $[\text{Cr(EDTA)}{]}^{-}$? ::: 6 (EDTA hexadentate, $1\times 6$).

See also: Isomerism in coordination compounds — square-planar vs tetrahedral CN = 4 give different isomer counts, a direct consequence of the geometry derived here.