VBT applied to complexes — inner vs outer orbital, hybridization, magnetism: Question bank

Symbols and units used on this page (define before you use)

The arrow-in-box picture and how orbitals split are the visual backbone of everything below — study the figures first, then attack the questions.

Arrow-in-box: what "paired" vs "unpaired" looks like.

Why octahedral splitting is bigger than tetrahedral (ligand approach).

The Spectrochemical Series as an energy ladder (defined right here).

Three words to have anchored (from the parent note), now with pictures above:

Unpaired electron = a box with a single arrow $↑$ and no opposite-spin partner (figure s01, left).
Inner orbital complex = uses the metal's own $(n - 1) d$ orbitals → hybridization written $d^{2} s p^{3}$ (the $d$ comes first).
Outer orbital complex = uses the higher $n d$ orbitals → $s p^{3} d^{2}$ (the $d$ tags on at the end).

True or false — justify

TF — "A diamagnetic complex has zero unpaired electrons."

True. Diamagnetic means the ion is not pulled into a magnetic field, i.e.

μ = 0

; since every lone arrow contributes a spin-magnet,

μ = 0

can only happen when no box holds a single arrow — every electron has an opposite-spin partner whose magnet cancels it (figure s01, right).

TF — "Every octahedral complex uses six

d

-orbitals."

False. Octahedral hybridization mixes only two

d

-orbitals with one

s

and three

p

(figure s04): the set has six lobes pointing to six ligands, but only two of the six ingredients are

d

-orbitals — hence

d^{2} s p^{3}

s p^{3} d^{2}

TF — "A weak-field ligand can never give a low-spin complex."

True for the ordinary cases here. On the energy ladder (figure s03) a weak ligand makes only a tiny splitting, so it is cheaper for electrons to spread out singly (high spin) than to pay the pairing cost; only strong ligands make the gap big enough to force pairing (low spin).

TF — "

s p^{3} d^{2}

and

d^{2} s p^{3}

describe the same shape."

True on shape, false on identity. Both give the octahedron of figure s04, but

d^{2} s p^{3}

borrows inner

(n - 1) d

orbitals (electrons had to pair up first → low spin) while

s p^{3} d^{2}

borrows outer

n d

orbitals (electrons stayed spread → high spin); the measured

μ

separates them.

TF — "Pairing electrons up always lowers the number of unpaired electrons."

True by definition. Pairing drops a second arrow into an already-occupied box (figure s01), converting two lonely arrows into one cancelling couple, so the count

n

of single arrows can only go down.

TF — "VBT can predict whether a complex is blue or red."

False. Colour comes from the size of the

d

-orbital energy gap

Δ

(figure s02/s03) that an electron jumps across; VBT never computes

Δ

, so it is blind to colour — that needs Crystal Field Theory.

TF — "A tetrahedral

d^{8}

complex and a square-planar

d^{8}

complex have the same magnetism."

False. Square-planar

d^{8}

(strong ligands) pairs down to

n = 0

, diamagnetic; tetrahedral

d^{8}

keeps two lonely arrows (

n = 2

μ = 2.83

BM) — same electron count, different shape (figure s04), different magnet.

TF — "The spin-only formula ignores orbital contribution to the magnetic moment."

True.

μ = n (n + 2)

counts only the spin magnets (the arrows); electrons also circulate, adding an orbital magnet the formula deliberately drops, which is why measured moments sometimes differ slightly.

Spot the error

Error — "

Fe^{2 +}

[Ar] 3 d^{4} 4 s^{2}

because

4 s

fills before

3 d

The mistake is emptying

3 d

before

4 s

. For cations the $4 s$ electrons leave first, so

Fe^{2 +} = [Ar] 3 d^{6}

— Aufbau governs filling, not the order of removal.

Error — "

[FeF_{6}]^{3 -}

d^{2} s p^{3}

because it is octahedral."

Octahedral shape alone doesn't fix the hybridization.

F^{-}

sits low on the ladder (figure s03, weak), so

d^{5}

stays fully unpaired with no empty inner

3 d

; it must use outer orbitals →

s p^{3} d^{2}

, not

d^{2} s p^{3}

Error — "

[NiCl_{4}]^{2 -}

is square planar because Ni

^{2 +}

d^{8}

The

d^{8}

count doesn't guarantee square planar.

Cl^{-}

is weak-field and won't force pairing, so it stays high-spin tetrahedral (

s p^{3}

); only a strong ligand like

CN^{-}

gives the square-planar

d s p^{2}

Error — "Coordination number 6 means six unpaired electrons."

Coordination number counts ligands/bonds, unrelated to unpaired electrons. It equals the number of hybrid orbitals needed (six lobes for octahedral, figure s04), regardless of

n

Error — "A strong-field ligand always makes an inner-orbital complex."

Strong-field forces pairing only if there are electrons to pair into a smaller set. For

d^{1}

–

d^{3}

octahedral there is nothing to pair, so strong vs weak makes no difference to

n

Error — "

[Ni (CN)_{4}]^{2 -}

uses

s p^{3} d^{2}

That's a six-orbital octahedral set, but this complex has coordination number 4. Square planar needs exactly four hybrids:

d s p^{2}

(figure s04).

Error — "Because CN⁻ is a strong ligand,

[Fe (CN)_{6}]^{4 -}

is strongly paramagnetic."

Strong ligand → pairing → fewer lonely arrows, not more. Here

d^{6}

pairs completely to

n = 0

, so it is diamagnetic — the opposite of strongly paramagnetic.

Why questions

Why — "Why does VBT need the spectrochemical order at all?"

Because VBT can't compute the splitting gap itself; the Spectrochemical Series (figure s03) is fed in as an external fact telling us strong-field (pairs → inner) from weak-field (no pairing → outer).

Why — "Why do we pair electrons to make an inner-orbital complex?"

Pairing squeezes the

d

-electrons into fewer boxes (figure s01), freeing empty $(n - 1) d$ orbitals so they can join the hybrid set and give the lower-energy

d^{2} s p^{3}

octahedron.

Why — "Why can we run VBT backwards from a measured moment?"

A measured

μ

pins down

n

via

μ = n (n + 2)

; once

n

is known we infer whether the metal paired (inner) or not (outer), making Magnetic Properties of Complexes a diagnostic tool.

Why — "Why is

4 s

removed before

3 d

when forming a cation, even though it fills first?"

Once

3 d

is occupied it sinks below

4 s

in energy, so the outermost (highest-energy) electrons to strip off become the

4 s

pair — filling order and removal order are simply different questions.

Why — "Why does the factor of

\frac{1}{2}

disappear in the spin-only formula?"

The spin magnitude carries a

\frac{1}{2} n (n + 2)

, but the electron's gyromagnetic factor

g \approx 2

multiplies it, and

2 \times \frac{1}{2} = 1

cancels to leave

μ = n (n + 2)

BM.

Why — "Why are tetrahedral complexes almost always high spin?"

In a tetrahedron the four ligands approach between the axes, missing the

d

-orbitals' lobes, so the splitting

Δ_{tet}

is small (figure s02) — smaller than the pairing cost — so electrons stay spread out and unpaired.

Why — "Why is

Δ_{tet} < Δ_{oct}

in terms of orbital orientation?"

In an octahedron six ligands point straight at two of the

d

-orbitals, repelling their electrons hard (big split); in a tetrahedron only four ligands, and they slot between the lobes, so the repulsion — and the gap — is much weaker (figure s02). Roughly

Δ_{tet} \approx \frac{4}{9} Δ_{oct}

Edge cases

Edge — "What does VBT predict for a

d^{0}

ion like

Sc^{3 +}

in an octahedral complex?"

With no

d

-electrons there's nothing to pair or leave unpaired, so

n = 0

μ = 0

(diamagnetic); inner vs outer is moot since empty

3 d

orbitals are available either way.

Edge — "For

d^{10}

ions such as

Zn^{2 +}

, does ligand field strength change the magnetism?"

No. All ten

d

-electrons are already paired, so

n = 0

regardless of ligand; the complex is diamagnetic and strong/weak-field is irrelevant to spin.

Edge — "Where in the

d^{n}

range does strong-vs-weak field actually change

n

for octahedral complexes?"

Only for

d^{4}

–

d^{7}

. Below

d^{4}

there are too few electrons to force pairing, and

d^{8}

–

d^{10}

fill to the same unpaired count either way.

Edge — "Is there ever an intermediate-spin state for

d^{4}

octahedral complexes?"

Yes — at ligand fields near the crossover, a

d^{4}

ion can adopt an intermediate arrangement between high-spin (

n = 4

) and low-spin (

n = 2

), so

n = 3

is possible. VBT's clean strong/weak split can't capture this borderline case; it needs the continuous

Δ

of Crystal Field Theory.

Edge — "If a ligand sits mid-series (like

H_{2} O

), how does VBT decide inner or outer?"

VBT can't decide alone; it needs the experimental

μ

or the Spectrochemical Series placement, since borderline ligands can give either spin state depending on the metal ion.

Edge — "For a linear two-coordinate complex, why is 'inner vs outer' not even a question?"

Linear geometry uses

s p

hybridization with no $d$ -orbitals (figure s04), so there is neither an inner nor outer

d

choice to make.

Edge — "

d^{5}

with a weak-field ligand gives the maximum

μ

for a first-row metal — why?"

Five orbitals each take one arrow before any pairing is forced, so

n = 5

is the largest unpaired count achievable, giving

μ = 35 \approx 5.92

BM.

Edge — "

[Cu (H_{2} O)_{6}]^{2 +}

(

d^{9}

) is octahedral but distorted — can VBT predict that?"

No. VBT gives a symmetric octahedron and

n = 1

, but an unevenly filled

d

-set drives a Jahn–Teller distortion (two bonds stretch) to lower energy. VBT is blind to such distortions; they follow from orbital-energy arguments in Crystal Field Theory.

Recall One-line self-test before you close

Given only " $μ = 0$ , coordination number 6, first-row $d^{6}$ ion," name the hybridization and spin state. Zero unpaired ⟹ fully paired $d^{6}$ ⟹ two empty inner $3 d$ freed ⟹ $d^{2} s p^{3}$ , octahedral, low-spin, inner orbital (as in $[Co (NH_{3})_{6}]^{3 +}$ ).