Visual walkthrough — VBT applied to complexes — inner vs outer orbital, hybridization, magnetism

We follow one worker example the whole way: the iron ion inside $[\text{Fe}(\text{CN}{)}_6{]}^{4-}$ and its twin $[{\text{FeF}}_6{]}^{3-}$. By the end you will predict shape, magnetism, and the inner/outer label without looking anything up.

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Step 1 — Draw the empty rooms (what an orbital box is)

WHAT. Before any chemistry, meet the drawing we will use for everything: an orbital box. One box = one orbital = a room that can hold at most two electrons, and those two must have opposite spin (one arrow up $\uparrow$, one arrow down $\downarrow$). That "opposite-spin" rule is the whole reason magnetism will exist.

WHY a box-and-arrow? Because VBT is a bookkeeping theory: it never solves a wave equation, it just asks which rooms are empty so a ligand's lone pair can move in. A picture of labelled boxes is exactly the right level of detail — no more, no less.

PICTURE. The relevant rooms for a first-row transition metal like iron are drawn below: the five inner $3d$ boxes (written $(n-1)d$, meaning "the $d$ shell one level below the outermost"), then the outer $4s$, then the three $4p$, then five outer $4d$ boxes.

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Step 2 — Find how many electrons iron actually has

WHAT. We must know the metal's $d$-electron count before drawing arrows. This needs the ion's charge, so first we do oxidation-state bookkeeping.

For $[\text{Fe}(\text{CN}{)}_6{]}^{4-}$: each ${\text{CN}}^{-}$ carries $-1$, there are six of them, and the whole ion is $-4$. Let the iron charge be $x$:

$\underset{\text{Fe charge}}{\underbrace{x}}+\underset{{\text{six CN}}^{-}}{\underbrace{6\times (-1)}}=\underset{\text{total ion charge}}{\underbrace{-4}}\Rightarrow x=+2$

So the ion is ${\text{Fe}}^{2+}$.

WHY remove electrons in that special order? Neutral iron is $[\text{Ar}]3d^{6}4s^2$. To make ${\text{Fe}}^{2+}$ we strip two electrons — and the rule is $4s$ leaves first, giving ${\text{Fe}}^{2+}=[\text{Ar}]3d^6$ (not $3d^{4}4s^2$). Once $3d$ is occupied it sinks below $4s$ in energy, so $4s$ becomes the exit door.

PICTURE. Below: the neutral atom's arrows, then the two $4s$ electrons flying out, leaving six electrons sitting in the $3d$ boxes.

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Step 3 — Spread the six electrons using Hund's rule (the "before" picture)

WHAT. Place the six $3d$ electrons into the five $3d$ boxes. Hund's rule: fill every box with one up-arrow first, then start pairing. Six electrons into five boxes gives $\uparrow \downarrow \uparrow \uparrow \uparrow \uparrow$ — one box doubled, four singly filled.

WHY do this first? This is the metal's natural, unforced arrangement — what it looks like before any ligand pushes on it. Count the unpaired arrows: four. Hold that number; a strong ligand is about to change it.

PICTURE. Five boxes, four lonely up-arrows in pink, one paired box.

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Step 4 — Let the ligand decide: strong field pairs the electrons

WHAT. Now the six ${\text{CN}}^{-}$ arrive. On the spectrochemical series, ${\text{CN}}^{-}$ sits near the top — it is a strong-field ligand. Strong ligands push hard enough that it becomes cheaper for the metal to pair up its own $3d$ electrons than to leave boxes half-full.

Re-draw $d^6$ crammed into the lowest three boxes: $\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \_\_$. Now count unpaired: zero. And notice — two $3d$ boxes are now completely empty.

WHY does pairing matter? Those two freshly-emptied inner $3d$ boxes are the prize. They will let the metal build its bonding rooms out of inner orbitals in Step 5. A weak-field ligand (next step's twin) would not pay this pairing cost, and those inner boxes would stay blocked.

PICTURE. Left: weak-field, electrons stay spread (from Step 3). Right: strong-field ${\text{CN}}^{-}$ forces pairing, two inner boxes go empty (drawn dashed).

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Step 5 — Mix the empty rooms into six identical bonds (hybridization)

WHAT. Coordination number is 6 (six ${\text{CN}}^{-}$), so we need six empty rooms of identical shape pointing at the six corners of an octahedron. Gather them: the two empty inner $3d$ (from Step 4) $+$ the $4s$ $+$ the three $4p$. That is $2+1+3=6$ orbitals. Blending them gives the hybrid set named

$\underset{\text{two inner }3d}{\underbrace{d^2}}\underset{4s}{\underbrace{s}}\underset{\text{three }4p}{\underbrace{p^3}}=d^{2}s{p}^3$

WHY is the $d$ written first? Because it is the inner $(n-1)d$ that got used — the notation literally records which shell. $d^{2}s{p}^3$ = inner. If we had been forced to use the outer $4d$ instead, we would write $s{p}^3{d}^2$ = outer. Same octahedron shape, different rooms.

PICTURE. The six empty boxes merge into six identical fat lobes aimed at octahedron corners; each accepts one ${\text{CN}}^{-}$ lone pair (blue).

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Step 6 — Read the magnetism, and see where $\sqrt{n(n+2)}$ comes from

WHAT. Magnetism is caused only by unpaired electron spins — each lonely arrow is a tiny bar magnet. In our strong-field iron, Step 4 left $n=0$ unpaired. No lonely arrows ⇒ no net magnet ⇒ diamagnetic.

To turn a spin-count $n$ into a measurable number, use the spin-only formula. Building it from zero:

• Each electron has spin $s=\frac{1}{2}$. For $n$ aligned unpaired electrons the total spin is $S=n\cdot \frac{1}{2}=\frac{n}{2}$.
• Quantum mechanics says the size of a spin's angular momentum is $\sqrt{S(S+1)}$, and the magnet strength tracks it. Substitute $S=\frac{n}{2}$:

$S(S+1)=\frac{n}{2}\left(\frac{n}{2}+1\right)=\frac{n}{2}\cdot \frac{n+2}{2}=\frac{n(n+2)}{4}$

• Take the square root and let the gyromagnetic factor $g\approx 2$ cancel the $\frac{1}{2}$:

$\boxed{{\mu }_{\text{spin}}=\sqrt{n(n+2)}\text{BM}}$

Here $n$ = unpaired electrons, and $\text{BM}$ = Bohr magneton, the natural unit of atomic magnetism.

WHY this tool and not just "count arrows"? Counting arrows gives $n$, but experiments measure a field strength $\mu$. This formula is the dictionary between the two — so we can run it backwards: measure $\mu$, solve for $n$, deduce inner vs outer.

PICTURE. A little dial: $n$ on the left, arrows lighting up, needle swinging to $\mu$ on the right, with our two cases marked.

For $[\text{Fe}(\text{CN}{)}_6{]}^{4-}$: $n=0\Rightarrow \mu =\sqrt{0}=0$ BM → diamagnetic. ✔

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Step 7 — The weak-field twin (the case that would break a naive rule)

WHAT. Now the other branch, so no scenario surprises you. Take $[{\text{FeF}}_6{]}^{3-}$. Charge: $x+6(-1)=-3\Rightarrow {\text{Fe}}^{3+}=3d^5$. ${\text{F}}^{-}$ is weak-field (bottom of the series) → no pairing.

Five electrons in five boxes, Hund's rule: $\uparrow \uparrow \uparrow \uparrow \uparrow$ — all five unpaired, zero empty inner boxes.

WHY must it use outer orbitals? The inner $3d$ is completely blocked (one arrow in every box). The six bonding rooms cannot come from $3d$, so the metal reaches up to the outer $4d$: $4s+4p_x+4p_y+4p_z+4d_{z^2}+4d_{x^2-y^2}$ = $s{p}^3{d}^2$ → still octahedral, but outer/high-spin.

Magnetism: $n=5\Rightarrow \mu =\sqrt{5\cdot 7}=\sqrt{35}=5.92$ BM → strongly paramagnetic.

PICTURE. Side by side: strong-field Fe${}^{2+}$ (inner $d^{2}s{p}^3$, $n=0$) vs weak-field Fe${}^{3+}$ (outer $s{p}^3{d}^2$, $n=5$) — same octahedron, opposite magnetism.

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Step 8 — Degenerate & edge cases (so nothing is left unshown)

WHAT. Three boundary situations the walkthrough would otherwise skip:

• $d^0$, $d^{10}$ (e.g. ${\text{Zn}}^{2+}=3d^{10}$): no half-empty boxes to argue about; $n=0$ always → diamagnetic, and inner/outer is irrelevant because the $3d$ is either empty or completely full.
• $d^1$–$d^3$: Hund already leaves the electrons spread with empty inner boxes, so strong and weak ligands give the same arrangement — inner ($d^{2}s{p}^3$) either way, same $n$. No fork here.
• The geometry flip at C.N. 4 (${\text{Ni}}^{2+}=3d^8$): with strong ${\text{CN}}^{-}$, $d^8$ pairs to free one inner box → $ds{p}^2$ → square planar, $n=0$, diamagnetic. With weak ${\text{Cl}}^{-}$, no pairing → $s{p}^3$ → tetrahedral, $n=2$, $\mu =\sqrt{2\cdot 4}=2.83$ BM. Same metal, ligand decides the shape itself.

PICTURE. Three mini-panels: $d^{10}$ full boxes; $d^3$ unchanged by ligand strength; the Ni${}^{2+}$ square-planar-vs-tetrahedral flip.

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The one-picture summary

Everything on this page in one flow: charge → $d^n$ → ligand strength → pairing decision → hybrid name → geometry → $n$ → $\mu$. Follow the arrows; the two branches are our iron twins.

Recall Feynman retelling — the whole walkthrough in plain words

The metal ion is a hotel with empty rooms. First I check the guest bill to see how much the hotel owes — that's the charge, and it tells me how many resident electrons live in the inner $d$ rooms. I spread those residents out one-per-room (Hund), because people like their own space. Then the guests (ligands) show up. Pushy guests (CN⁻) force residents to double up, which empties out two fancy inner rooms — the hotel builds its six bonding suites from those inner rooms ($d^{2}s{p}^3$), and since everyone's paired, the building has no magnetism. Polite guests (F⁻) don't force sharing, so the inner rooms stay occupied; the hotel builds bonding suites from cheaper outer rooms instead ($s{p}^3{d}^2$), leaving many residents lonely and un-paired, so the building is magnetic. Finally I count the lonely residents $n$ and plug into $\sqrt{n(n+2)}$ to get the magnet strength $\mu$. Read it forwards to predict; read it backwards (measure $\mu$, get $n$) to deduce whether the hotel went inner or outer. That's VBT — and for why CN⁻ is pushier than F⁻, you need Crystal Field Theory.

Recall Self-check

$[\text{Fe}(\text{CN}{)}_6{]}^{4-}$: hybridization? ::: $d^{2}s{p}^3$ inner, octahedral, $n=0$, $\mu =0$ (diamagnetic). $[{\text{FeF}}_6{]}^{3-}$: hybridization and $\mu$? ::: $s{p}^3{d}^2$ outer, octahedral, $n=5$, $\mu =5.92$ BM. $[{\text{NiCl}}_4{]}^{2-}$: shape and $\mu$? ::: tetrahedral $s{p}^3$, $n=2$, $\mu =2.83$ BM. Why write $d$ first in $d^{2}s{p}^3$? ::: because the inner $(n-1)d$ orbitals were used; the position of $d$ records the shell.

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See also: Magnetic Properties of Complexes · Coordination Number and Geometry · Hybridization · Spectrochemical Series