VBT applied to complexes — inner vs outer orbital, hybridization, magnetism: Worked examples

You have met the VBT recipe on the parent note. This page is the drill ground. We enumerate every kind of situation VBT can throw at you, then work an example for each — so you never meet a case in an exam that you have not already seen solved here.

Before we start, one reminder of the two counting tools we lean on every single time:

Recall The only two numbers you ever need

Unpaired electrons $n$ ::: gives magnetism via $μ = n (n + 2)$ Bohr magnetons. $d$ -electron count $d^{n}$ ::: comes from oxidation state; decides whether pairing is even possible.

The scenario matrix

Every VBT question is really one of these case classes. Think of the matrix as the full "menu" — each worked example below is tagged with the cell(s) it fills.

Cell	What varies	The tricky part	Example
A — C.N. 6, strong ligand	forces pairing	inner, $d^{2} s p^{3}$ , low spin	Ex 1
B — C.N. 6, weak ligand	no pairing	outer, $s p^{3} d^{2}$ , high spin	Ex 2
C — same ion, ligand flips spin	strong vs weak on one metal	inner ⇄ outer	Ex 3
D — C.N. 4, geometry decided by ligand	$d s p^{2}$ vs $s p^{3}$	square planar vs tetrahedral	Ex 4
E — C.N. 2, degenerate low count	too few $d$ -electrons to matter	$s p$ , linear	Ex 5
F — $d^{0}$ / $d^{10}$ edge cases	zero or full $d$	ligand strength irrelevant	Ex 6
G — run it backwards from measured $μ$	find $n$ from experiment	deduce inner/outer	Ex 7
H — real-world / biology word problem	strip the story to a formula	oxygen carrier	Ex 8
I — exam twist: strong ligand, still high spin	when strong ≠ low spin	$d^{1}$ – $d^{3}$ or tetrahedral	Ex 9

The signs/quadrants of trigonometry have an analogue here: the "quadrants" of VBT are the $d$ -count ranges $d^{0}$ , $d^{1}$ – $d^{3}$ , $d^{4}$ – $d^{7}$ , $d^{8}$ – $d^{10}$ . Strong-vs-weak ligand only changes the answer inside $d^{4}$ – $d^{7}$ . Outside that band, the ligand can shout all it likes and nothing changes — Cells F and I exist precisely to make you feel where the boundary is.

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

Example 1 — Cell A: octahedral, strong ligand, inner

Worked example

$[Fe (CN)_{6}]^{4 -}$ Forecast: Strong ligand, six of them. Will Fe pair up? How many unpaired electrons remain — and is it magnetic? Guess before reading.

Step 1 — Oxidation state. $CN^{-}$ is $- 1$ each; six give $- 6$ ; overall charge $- 4$ . So $x + (- 6) = - 4 \Rightarrow x = + 2$ . Why this step? We cannot know the $d$ -count until we know the ion's charge.

Step 2 — $d$ -count. $Fe = [Ar] 3 d^{6} 4 s^{2}$ . Remove the two $4 s$ electrons first (cations lose $4 s$ before $3 d$ ) ⟹ $Fe^{2 +} = 3 d^{6}$ . Why this step? $d^{6}$ sits in the $d^{4}$ – $d^{7}$ band — the band where ligand strength actually decides the answer.

Step 3 — Ligand class. $CN^{-}$ is near the top of the Spectrochemical Series ⟹ strong field ⟹ forces pairing. Why this step? Only a strong ligand can compress $d^{6}$ into three orbitals.

Step 4 — Pair up. $d^{6} \to↑↓ ↑↓ ↑↓__$ . Two inner $3 d$ orbitals are now empty. Why this step? Two empty inner $d$ are exactly what $d^{2} s p^{3}$ needs.

Step 5 — Hybridize. $2 \times (3 d) + 1 \times (4 s) + 3 \times (4 p) = d^{2} s p^{3}$ ⟹ octahedral, inner orbital, low spin.

Step 6 — Magnetism. $n = 0 \Rightarrow μ = 0 (0 + 2) = 0$ ⟹ diamagnetic.

Verify: Charge check $+ 2 - 6 = - 4$ . ✓ Electron count: $6 = 6$ paired, $0$ unpaired. ✓ $μ = 0$ matches the "all paired" picture.

Example 2 — Cell B: octahedral, weak ligand, outer

Worked example

$[FeF_{6}]^{3 -}$ Forecast: Same metal family, but weak ligand. More or fewer unpaired electrons than Ex 1? Bigger or smaller $μ$ ?

Step 1 — Oxidation state. $F^{-} = - 1$ ; $x + 6 (- 1) = - 3 \Rightarrow x = + 3$ . Why this step? Different charge ⟹ different $d$ -count ⟹ possibly different story.

Step 2 — $d$ -count. $Fe^{3 +} = 3 d^{5}$ . Why this step? $d^{5}$ is still inside the $d^{4}$ – $d^{7}$ band, so the ligand matters.

Step 3 — Ligand class. $F^{-}$ is low in the Spectrochemical Series ⟹ weak field ⟹ no pairing.

Step 4 — Distribute. $d^{5} \to↑ ↑ ↑ ↑ ↑$ . All five inner $3 d$ are half-filled and occupied, so none is empty. Why this step? With no empty inner $d$ , inner hybridization is impossible — the metal must reach outward.

Step 5 — Hybridize outward. $4 s + 4 p_{x} + 4 p_{y} + 4 p_{z} + 4 d_{x^{2} - y^{2}} + 4 d_{z^{2}} = s p^{3} d^{2}$ ⟹ octahedral, outer orbital, high spin.

Step 6 — Magnetism. $n = 5 \Rightarrow μ = 5 \cdot 7 = 35 \approx 5.92$ BM. Strongly paramagnetic.

Verify: Charge $+ 3 - 6 = - 3$ . ✓ $n = 5$ is the maximum for the $d$ shell — the "most magnetic" a first-row ion gets.

Example 3 — Cell C: same ion, ligand flips the spin

Worked example

$[Co (NH_{3})_{6}]^{3 +}$ vs $[CoF_{6}]^{3 -}$ Forecast: Both are $Co^{3 +} = 3 d^{6}$ . One ligand is strong ( $NH_{3}$ ), one weak ( $F^{-}$ ). Will one be diamagnetic and the other paramagnetic? Predict both $μ$ values.

Step 1 — Common ground. In each, Co is $+ 3$ (check $[Co (NH_{3})_{6}]^{3 +}$ : $x + 0 = + 3$ ; $[CoF_{6}]^{3 -}$ : $x - 6 = - 3$ ). Both give $3 d^{6}$ . Why this step? Isolating the only difference (the ligand) is the whole point of this cell.

Step 2 — Strong branch ( $NH_{3}$ ). Pair up: $↑↓ ↑↓ ↑↓__$ ⟹ $d^{2} s p^{3}$ , inner, $n = 0$ , $μ = 0$ ⟹ diamagnetic.

Step 3 — Weak branch ( $F^{-}$ ). No pairing: $↑↓ ↑ ↑ ↑ ↑$ ⟹ $n = 4$ ⟹ $s p^{3} d^{2}$ , outer ⟹ $μ = 4 \cdot 6 = 24 \approx 4.90$ BM ⟹ paramagnetic. Why this step? Same $d$ -count, opposite outcome — this is the sharpest demonstration that the ligand decides everything in the $d^{4}$ – $d^{7}$ band.

Verify: $NH_{3}$ (strong): $n = 0$ , $μ = 0$ . $F^{-}$ (weak): $n = 4$ , $μ = 4.90$ . Two different answers from one ion — exactly the contrast we forecast.

Example 4 — Cell D: C.N. 4, ligand picks the shape

Worked example

$[Ni (CN)_{4}]^{2 -}$ vs $[NiCl_{4}]^{2 -}$ Forecast: Both are $Ni^{2 +} = 3 d^{8}$ , coordination number 4. Same shape? Same magnetism? Guess the geometries.

Step 1 — Oxidation & count. $[Ni (CN)_{4}]^{2 -}$ : $x + 4 (- 1) = - 2 \Rightarrow x = + 2$ ; $[NiCl_{4}]^{2 -}$ : same, $x = + 2$ . So $Ni^{2 +} = 3 d^{8}$ in both.

Step 2 — Strong branch ( $CN^{-}$ ). Pair the two lone $d$ electrons: $↑↓ ↑↓ ↑↓ ↑↓_$ frees one inner $3 d$ . Mix $1 \times (3 d) + 1 \times (4 s) + 2 \times (4 p) = d s p^{2}$ ⟹ square planar, $n = 0$ , $μ = 0$ , diamagnetic. Why this step? Four orbitals with one inner $d$ ⟹ the $d s p^{2}$ set points to four corners of a square.

Step 3 — Weak branch ( $Cl^{-}$ ). No pairing: $↑↓ ↑↓ ↑↓ ↑ ↑$ leaves $n = 2$ . Use $s p^{3}$ (no $d$ ) ⟹ tetrahedral, $μ = 2 \cdot 4 = 8 \approx 2.83$ BM, paramagnetic. Why this step? No spare inner $d$ ⟹ the four orbitals must be the four $s p^{3}$ lobes ⟹ tetrahedron.

Verify (see figure): $CN^{-}$ → square planar, $μ = 0$ ; $Cl^{-}$ → tetrahedral, $μ = 2.83$ . Same metal, same C.N., but geometry AND magnetism both flip with the ligand.

Compare geometries with Coordination Number and Geometry and the hybrid sets with Hybridization.

Example 5 — Cell E: C.N. 2, degenerate low count

Example 6 — Cell F: $d^{0}$ and $d^{10}$ edges (ligand strength irrelevant)

Worked example

$[Sc (H_{2} O)_{6}]^{3 +}$ ( $d^{0}$ ) and $[Zn (NH_{3})_{6}]^{2 +}$ ( $d^{10}$ ) Forecast: One has an empty $d$ shell, one a full $d$ shell. Does calling the ligand "strong" or "weak" change anything here?

Step 1 — $Sc^{3 +}$ . $Sc = [Ar] 3 d^{1} 4 s^{2}$ ; lose 3 electrons ⟹ $3 d^{0}$ . No $d$ electrons at all ⟹ nothing to pair, nothing unpaired. $n = 0$ , $μ = 0$ . Why this step? With $d^{0}$ , "pairing to free inner $d$ " is a non-question — the inner $d$ are already empty. Uses $d^{2} s p^{3}$ trivially (empty inner $d$ available), octahedral.

Step 2 — $Zn^{2 +}$ . $Zn = [Ar] 3 d^{10} 4 s^{2}$ ; lose the $4 s$ ⟹ $3 d^{10}$ . Full shell ⟹ no empty inner $d$ , no unpaired electrons ⟹ $s p^{3} d^{2}$ outer, octahedral, $n = 0$ , $μ = 0$ . Why this step? $d^{10}$ is the mirror edge: no empty inner $d$ , so it is forced outer regardless of ligand.

Step 3 — The lesson. In both edges $μ = 0$ whatever the ligand. Strong-vs-weak only matters in the $d^{4}$ – $d^{7}$ band.

Verify: $Sc^{3 +}$ : $n = 0$ , $μ = 0$ . $Zn^{2 +}$ : $n = 0$ , $μ = 0$ . Both diamagnetic — the ligand label was a red herring.

Example 7 — Cell G: run it backwards from measured $μ$

Worked example A

$Mn^{2 +}$ octahedral complex is measured at $μ \approx 5.92$ BM. Is it inner or outer? Forecast: From the number alone, how many unpaired electrons? Then which hybridization?

Step 1 — Solve $μ = n (n + 2)$ for $n$ . $5.9 2^{2} = 35.05 \approx n (n + 2)$ . Try $n = 5$ : $5 \cdot 7 = 35$ . ✓ So $n = 5$ . Why this step? $μ$ is what the lab measures; $n$ is what we want. Inverting the formula is the whole trick.

Step 2 — Match to $d$ -count. $Mn^{2 +} = 3 d^{5}$ . Five unpaired means no pairing at all — every $d$ orbital singly occupied. Why this step? $n = 5$ out of $d^{5}$ ⟹ high spin ⟹ weak-field ligand ⟹ outer.

Step 3 — Conclude. No empty inner $3 d$ ⟹ $s p^{3} d^{2}$ ⟹ outer orbital, high spin.

Verify: $5 \cdot 7 = 35 = 5.916 \approx 5.92$ BM. ✓ The measured moment reproduces exactly — we ran VBT backwards, as promised on the parent note. See Magnetic Properties of Complexes.

Example 8 — Cell H: real-world word problem

Worked example Oxyhemoglobin. In hemoglobin the iron is

$Fe^{2 +}$ , six-coordinate; when $O_{2}$ binds (a strong-field ligand set), the complex becomes diamagnetic. How many unpaired electrons must $O_{2}$ binding remove, and what hybridization does the iron adopt? Forecast: Deoxy-iron ( $H_{2} O$ -like weak field) has some unpaired electrons; oxy-iron has zero. What was the drop in $n$ ?

Step 1 — Strip the story. "Diamagnetic" ⟹ $μ = 0 \Rightarrow n = 0$ . $Fe^{2 +} = 3 d^{6}$ . Why this step? Biology words ("carries oxygen", "diamagnetic") are just VBT inputs in disguise.

Step 2 — Oxy state. $d^{6}$ fully paired: $↑↓ ↑↓ ↑↓__$ ⟹ two empty inner $3 d$ ⟹ $d^{2} s p^{3}$ , octahedral, inner, $n = 0$ , $μ = 0$ .

Step 3 — Deoxy comparison. Weak-field (high-spin) $d^{6}$ would be $↑↓ ↑ ↑ ↑ ↑$ , $n = 4$ . So $O_{2}$ binding drops $n$ from $4$ to $0$ — a fall of 4 unpaired electrons, and $μ$ from $4.90$ to $0$ BM. Why this step? The measurable change in magnetism is what confirmed low-spin oxy-heme historically.

Verify: Oxy: $n = 0$ , $μ = 0$ . Deoxy (high spin): $n = 4$ , $μ = 24 = 4.90$ . Drop in $n = 4 - 0 = 4$ . ✓

Example 9 — Cell I: exam twist — strong ligand, still high spin

Worked example

$[Cr (H_{2} O)_{6}]^{3 +}$ — does field strength change $n$ here? Forecast: $Cr^{3 +} = 3 d^{3}$ . Trap: is this the $d^{4}$ – $d^{7}$ band where strong-vs-weak matters? Predict $n$ for a strong ligand and a weak ligand.

Step 1 — Oxidation & count. $x + 6 (0) = + 3 \Rightarrow Cr^{3 +} = 3 d^{3}$ .

Step 2 — Where does $d^{3}$ sit? In the $d^{1}$ – $d^{3}$ range. Three electrons drop one-per-orbital into three separate $3 d$ orbitals ( $↑ ↑ ↑$ ) with no pairing choice to make — there is no second electron in any orbital yet. Why this step? Pairing only becomes a decision from $d^{4}$ onward. Below that, strong and weak give the same configuration.

Step 3 — Hybridize. Two empty inner $3 d$ remain either way ⟹ $d^{2} s p^{3}$ , octahedral, inner-type. $n = 3$ ⟹ $μ = 3 \cdot 5 = 15 \approx 3.87$ BM.

Step 4 — The twist resolved. Even though $H_{2} O$ (and even $CN^{-}$ ) is not weak, $n = 3$ regardless of ligand strength. Strong ≠ low spin outside the $d^{4}$ – $d^{7}$ band.

Verify: $3 \cdot 5 = 15 = 3.873 \approx 3.87$ BM. ✓ Same $n$ for any ligand — the exam trap defused.

Recall Rapid self-test across the matrix

$[FeF_{6}]^{3 -}$ ( $d^{5}$ , weak): $n$ and $μ$ ? ::: $n = 5$ , $μ = 5.92$ BM. $[Ni (CN)_{4}]^{2 -}$ : geometry and $μ$ ? ::: square planar ( $d s p^{2}$ ), $μ = 0$ . $[Cr (H_{2} O)_{6}]^{3 +}$ : $μ$ ? ::: $3.87$ BM (ligand-independent, $d^{3}$ ). A measured $μ = 2.83$ BM means how many unpaired electrons? ::: $n = 2$ .

Prerequisites & neighbours: Oxidation State Determination · Spectrochemical Series · Hybridization · Coordination Number and Geometry · Magnetic Properties of Complexes · Crystal Field Theory.