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

Definition What "

$n$ ", " $(n - 1) d$ " and " $n d$ " mean — inner vs outer orbitals Here $n$ is the principal quantum number of the metal's valence shell — the "floor number" of the highest occupied shell. For a first-row transition metal like Fe, Co, Ni the valence shell is the 4th shell, so $n = 4$ .

The valence shell ( $n = 4$ ) holds the $4 s$ and $4 p$ orbitals, and (empty, higher up) the $4 d$ .
The shell one below it is $(n - 1) = 3$ , holding the $3 d$ orbitals where the $d$ -electrons actually live.

So for a 3d metal:

$== (n - 1) d = 3 d ==$ = the inner $d$ -orbitals (lower in energy, close to the nucleus). Using these gives an inner-orbital complex ( $d^{2} s p^{3}$ or $d s p^{2}$ ).
$== n d = 4 d ==$ = the outer $d$ -orbitals (higher, further out). Using these gives an outer-orbital complex ( $s p^{3} d^{2}$ ).

"Inner" and "outer" literally refer to these two different shells of $d$ -orbitals. For a second-row metal (e.g. Pd, $n = 5$ ) the same words mean $4 d$ inner and $5 d$ outer — just shift $n$ .

Intuition Why

$4 s$ and $4 p$ are the orbitals that get hybridized (the "why" behind step 6) Hybridization mixes orbitals that are close in energy — you cannot blend a low orbital with one far above it and expect a useful bonding set. After the $3 d$ shell, the next orbitals climbing in energy are the valence $4 s$ , then the three $4 p$ , then (much higher, empty) the $4 d$ . So the metal takes them in energy order:

It always grabs $4 s$ and the needed $4 p$ first (they are the cheapest empty valence orbitals).
The remaining slots come from the cheapest $d$ -orbitals still empty: if pairing freed inner $3 d$ , it uses those (inner, $d^{2} s p^{3}$ ); if not, it is forced up to the costlier outer $4 d$ (outer, $s p^{3} d^{2}$ ).

This is exactly why weak-field complexes end up "outer": with no empty inner $3 d$ , the metal must climb to higher-energy $4 d$ orbitals to complete its hybrid set. (Later shells $5 s /5 p$ play the same role for heavier metals — just replace $4$ by $n$ .)

This page tests everything from the parent: Oxidation State Determination, Coordination Number and Geometry, Spectrochemical Series, Hybridization, and Magnetic Properties of Complexes.

Definition The two

$d$ -orbital sets: $t_{2 g}$ (lower) and $e_{g}$ (upper) When six ligands approach a metal along the $x, y, z$ axes (octahedron), they push harder on the two $d$ -orbitals that point straight at them and less on the three that point between the axes. This splits the five $d$ -orbitals into two groups:

$== e_{g} ==$ = the upper, higher-energy pair ( $d_{z^{2}}$ , $d_{x^{2} - y^{2}}$ ) — they aim at the ligands.
$== t_{2 g} ==$ = the lower, lower-energy trio ( $d_{x y}$ , $d_{y z}$ , $d_{x z}$ ) — they aim between ligands.

The Figure s02 splitting diagram below makes this concrete: the single left-hand level is the free ion (all five $d$ -orbitals at one energy); on the right the trio drops to become $t_{2 g}$ (magenta) and the pair rises to become $e_{g}$ (violet), with the vertical axis showing increasing energy upward. When this note says "the lower set of three orbitals," it means $t_{2 g}$ ; "the upper set" means $e_{g}$ . Filling the $t_{2 g}$ trio first (before $e_{g}$ ) is exactly what "pairing to free inner $d$ " does.

Intuition The real world is a spectrum: intermediate spin and spin-crossover

VBT deliberately gives you a clean two-way switch (strong ⟹ pair, weak ⟹ don't). Nature is more continuous. A ligand of intermediate field strength (near $H_{2} O$ ) sits close to the tipping point where the pairing-energy cost and the splitting energy nearly cancel. Then:

Intermediate-spin states can appear (for $d^{4}$ – $d^{6}$ , a partial pairing that is neither fully high- nor fully low-spin).
Spin-crossover: the same complex can flip between high-spin and low-spin when you change temperature or pressure — a whole "grey zone" between the two columns of the spectrochemical rule.

On this page every ligand is chosen to be clearly strong or clearly weak so the switch is unambiguous — but be aware this borderline quadrant exists, and that predicting it properly needs Crystal Field Theory, not VBT alone.

Level 1 — Recognition

(Read the config, name the hybridization, count the electrons — no traps yet.)

Recall Solution Q1.1

Rule: neutral atom config, then remove electrons — $4 s$ FIRST, then $3 d$ .

$Fe$ is $[Ar] 3 d^{6} 4 s^{2}$ → $Fe^{3 +}$ : remove $4 s^{2}$ and one $3 d$ → $== 3 d^{5} ==$ .
$Co$ is $[Ar] 3 d^{7} 4 s^{2}$ → $Co^{3 +}$ : remove $4 s^{2} + 13 d$ → $== 3 d^{6} ==$ .
$Ni$ is $[Ar] 3 d^{8} 4 s^{2}$ → $Ni^{2 +}$ : remove $4 s^{2}$ → $== 3 d^{8} ==$ .
$Mn$ is $[Ar] 3 d^{5} 4 s^{2}$ → $Mn^{2 +}$ : remove $4 s^{2}$ → $== 3 d^{5} ==$ .
$Cr$ is $[Ar] 3 d^{5} 4 s^{1}$ → $Cr^{3 +}$ : remove $4 s^{1} + 23 d$ → $== 3 d^{3} ==$ .

Recall Solution Q1.2

Hybridization	C.N.	Geometry	Inner / Outer
$s p$	2	linear	— (no $d$ )
$s p^{3}$	4	tetrahedral	outer (no inner $d$ used)
$d s p^{2}$	4	square planar	inner ( $(n - 1) d$ )
$s p^{3} d$	5	trigonal bipyramidal	outer ( $n d$ )
$s p^{3} d^{2}$	6	octahedral	outer ( $n d$ )
$d^{2} s p^{3}$	6	octahedral	inner ( $(n - 1) d$ )

Reading trick: if the letter $d$ sits in front, an inner $(n - 1) d$ orbital was used → inner complex. If $d$ trails at the end, an outer $n d$ orbital was borrowed → outer complex. (The five-coordinate $s p^{3} d$ set, geometry trigonal bipyramidal, appears in complexes such as $[Fe (CO)_{5}]$ .)

Recall Solution Q1.3

$μ = n_{up} (n_{up} + 2) = 3 \cdot 5 = 15 \approx== 3.87 ==$ BM. $μ > 0 \Rightarrow$ paramagnetic (it has unpaired electrons, so a magnet attracts it).

Level 2 — Application

(Run the full recipe on standard complexes.)

Recall Solution Q2.1

Oxidation: $NH_{3}$ is neutral $\Rightarrow x + 0 = + 3 \Rightarrow Cr^{3 +}$ . Config: $Cr^{3 +} = 3 d^{3}$ . Ligand: $NH_{3}$ is strong field (right of $H_{2} O$ in the series) — but $d^{3}$ has only three electrons, each already alone in a separate $t_{2 g}$ orbital. There is nothing to pair; strong vs weak makes no difference here. Empty $3 d$ : $d^{3} =↑ ↑ ↑__$ → two empty inner $3 d$ orbitals sit ready. Hybridize: two empty inner $3 d$ are available, so the metal builds $d^{2} s p^{3}$ (two $3 d$ + $4 s$ + three $4 p$ ) → octahedral, inner. Magnetism: $n_{up} = 3 \Rightarrow μ = 15 = 3.87$ BM, paramagnetic.

Recall Solution Q2.2

Oxidation: $H_{2} O$ neutral $\Rightarrow Mn^{2 +}$ . Config: $Mn^{2 +} = 3 d^{5}$ . Ligand: $H_{2} O$ is weak field (it sits at the weak end of the series) → no pairing. Distribution: $d^{5} =↑ ↑ ↑ ↑ ↑$ — all five orbitals singly filled, no empty inner $3 d$ . Hybridize: with no inner $3 d$ free, the metal must climb to the next empty orbitals in energy order — the valence $4 s$ and $4 p$ fill first, then it is forced up to the outer $4 d$ orbitals to finish the six-orbital set: $s p^{3} d^{2}$ → octahedral, outer, high spin. These $4 d$ are high-energy (virtual) orbitals — normally empty and well above the $3 d$ ; the metal only "borrows" them because no cheaper inner $3 d$ is available. This is the general reason weak-field complexes are outer: they are forced to use higher-lying orbitals. Magnetism: $n_{up} = 5 \Rightarrow μ = 5 \cdot 7 = 35 \approx 5.92$ BM. Strongly paramagnetic.

Recall Solution Q2.3

Oxidation: $x + 6 (- 1) = - 3 \Rightarrow Co^{3 +}$ . Config: $Co^{3 +} = 3 d^{6}$ . Ligand: $CN^{-}$ strong (top of the series) → forces pairing. Pairing: $d^{6} =↑↓ ↑↓ ↑↓__$ (fills the $t_{2 g}$ trio) → two empty inner $3 d$ . Hybridize: two empty inner $3 d$ available → $d^{2} s p^{3}$ → octahedral, inner, low spin. Magnetism: $n_{up} = 0 \Rightarrow μ = 0$ → diamagnetic.

Level 3 — Analysis

(Same metal, different ligand — the geometry/magnetism flip.)

The Figure s01 below shows the two outcomes side by side: the left panel is $[Ni (CN)_{4}]^{2 -}$ (strong CN⁻ pairs the $d^{8}$ , freeing one inner $3 d$ → flat square-planar $d s p^{2}$ , all electrons paired, $μ = 0$ ); the right panel is $[NiCl_{4}]^{2 -}$ (weak Cl⁻ leaves two electrons unpaired, no inner $3 d$ free → three-dimensional tetrahedral $s p^{3}$ , $μ = 2.83$ BM). Read the caption arrow: same metal, same coordination number 4, but the ligand alone decides both the shape and the magnetism.

Recall Solution Q3.1

Both are $Ni^{2 +} = 3 d^{8}$ , coordination number 4. The ligand strength decides the outcome — see Figure s01 above.

$[Ni (CN)_{4}]^{2 -}$ — CN⁻ strong: $d^{8}$ pairs into $↑↓ ↑↓ ↑↓ ↑↓_$ → one empty inner $3 d$ . With one inner $3 d$ free plus $4 s$ and two $4 p$ , the only 4-orbital set available is $d s p^{2}$ → square planar. $n_{up} = 0 \Rightarrow μ = 0$ → diamagnetic.

$[NiCl_{4}]^{2 -}$ — Cl⁻ weak: No pairing. $d^{8} =↑↓ ↑↓ ↑↓ ↑ ↑$ → no inner $3 d$ free. With no inner $d$ available, the metal uses $4 s$ + three $4 p$ = $s p^{3}$ → tetrahedral. $n_{up} = 2 \Rightarrow μ = 2 \cdot 4 = 8 \approx 2.83$ BM → paramagnetic.

Difference: one metal, one coordination number — but strong CN⁻ pairs electrons and opens an inner orbital (square planar, diamagnetic), while weak Cl⁻ leaves electrons unpaired and forces the outer $s p^{3}$ set (tetrahedral, paramagnetic).

Recall Solution Q3.2

$[Co (NH_{3})_{6}]^{3 +}$ — NH₃ strong: pairs to $↑↓ ↑↓ ↑↓__$ (fills $t_{2 g}$ ), two empty inner $3 d$ → $d^{2} s p^{3}$ , inner, $n_{up} = 0$ , $μ = 0$ → diamagnetic. $[CoF_{6}]^{3 -}$ — F⁻ weak: no pairing, $d^{6} =↑↓ ↑ ↑ ↑ ↑$ , $n_{up} = 4$ → no inner $3 d$ free so $s p^{3} d^{2}$ , outer, $μ = 4 \cdot 6 = 24 \approx 4.90$ BM → paramagnetic. Lesson: identical ion, identical geometry (octahedral), yet the ligand alone drives $n_{up}$ from $4$ down to $0$ .

Level 4 — Synthesis

(Reverse the logic: from measured $μ$ back to structure.)

Recall Solution Q4.1

$Co^{3 +} = 3 d^{6}$ . Diamagnetic $\Rightarrow μ = 0 \Rightarrow n_{up} = 0 \Rightarrow$ all six $d$ -electrons paired into the $t_{2 g}$ trio → two inner $3 d$ freed. Two empty inner $3 d \Rightarrow$ hybridization $d^{2} s p^{3}$ → inner octahedral, low spin. Full pairing of $d^{6}$ requires a strong-field ligand (e.g. CN⁻, NH₃, right of H₂O in the series).

Recall Solution Q4.2

Invert the formula. Square both sides first (so we do algebra, not guessing): $μ = n_{up} (n_{up} + 2) = 5.92 \Rightarrow n_{up} (n_{up} + 2) = 35 \Rightarrow n_{up}^{2} + 2 n_{up} - 35 = 0.$ Factor: $(n_{up} - 5) (n_{up} + 7) = 0 \Rightarrow n_{up} = 5$ or $n_{up} = - 7$ . The number of unpaired electrons cannot be negative, so we discard $n_{up} = - 7$ and keep $n_{up} = 5$ . Five unpaired electrons needs five singly-occupied $d$ -orbitals $\Rightarrow$ the ion must be $d^{5} \Rightarrow Fe^{3 +}$ (a $d^{6}$ ion maxes out at $n_{up} = 4$ ). $n_{up} = 5$ = maximum for $d^{5} \Rightarrow$ no pairing $\Rightarrow$ weak-field ligand, no inner $3 d$ free $\Rightarrow$ outer $s p^{3} d^{2}$ , high spin.

Recall Solution Q4.3

$Ni^{2 +} = 3 d^{8}$ . Diamagnetic $\Rightarrow n_{up} = 0 \Rightarrow$ all electrons paired: $↑↓ ↑↓ ↑↓ ↑↓_$ → one empty inner $3 d$ → $d s p^{2}$ → square planar. Not tetrahedral: $s p^{3}$ (tetrahedral) uses no inner $d$ , so the $3 d^{8}$ electrons keep their two unpaired spins → that would be paramagnetic ( $n_{up} = 2$ ). Diamagnetism forces the square-planar $d s p^{2}$ path.

Level 5 — Mastery

(Full multi-step reasoning and edge/limiting cases.)

Recall Solution Q5.1

Maximum unpaired = fill all five $d$ -orbitals singly first (Hund), then pair.

$d^{n}$	max unpaired $n_{up,max}$	$μ_{m a x}$ (BM)
$d^{1}$	1	1.73
$d^{2}$	2	2.83
$d^{3}$	3	3.87
$d^{4}$	4	4.90
$d^{5}$	5	5.92
$d^{6}$	4	4.90
$d^{7}$	3	3.87
$d^{8}$	2	2.83
$d^{9}$	1	1.73
$d^{10}$	0	0

Strong-field can force $μ = 0$ only for even octahedral $d$ -counts that can fully pair into the lower $t_{2 g}$ set of three orbitals: $d^{6}$ (low-spin $t_{2 g}^{6}$ , $n_{up} = 0$ ). $d^{10}$ is diamagnetic regardless of field. $d^{4}$ (→ $n_{up} = 2$ ) and $d^{5}$ (→ $n_{up} = 1$ ) drop but do not reach zero even at strong field.

Recall Solution Q5.2

The lower $t_{2 g}$ set of an octahedral field holds three orbitals.

$d^{1}$ – $d^{3}$ : electrons fit singly into those three lower orbitals no matter what — strong and weak give the same $n_{up}$ . ( $d^{3} =↑ ↑ ↑$ , $n_{up} = 3$ always.)
$d^{4}$ – $d^{7}$ : the 4th (and further) electron faces a choice — pair up in the $t_{2 g}$ set (strong field, low spin) or go to the upper $e_{g}$ set unpaired (weak field, high spin). This is where inner vs outer truly diverges — and, at borderline field strength, exactly where intermediate-spin and spin-crossover live.
$d^{8}$ – $d^{10}$ : the upper $e_{g}$ set is now forced to take electrons regardless; $n_{up}$ is fixed again. ( $d^{8} = n_{up} = 2$ octahedral, same either way.) So the "decision zone" is exactly $d^{4}$ – $d^{7}$ .

Recall Solution Q5.3

Oxidation: $x + 4 (- 1) = - 2 \Rightarrow Mn^{2 +}$ , so $3 d^{5}$ . Geometry: tetrahedral, $s p^{3}$ — uses no inner $d$ . Distribution: $d^{5} =↑ ↑ ↑ ↑ ↑$ → $n_{up} = 5$ . $μ = 5 \cdot 7 = 35 \approx 5.92$ BM. Why no strong/weak debate: tetrahedral field splitting is small (about $\frac{4}{9}$ of the octahedral value), so it can never overcome the pairing cost. Tetrahedral complexes are essentially always high spin — you just apply Hund's rule and count.

Recall Solution Q5.4

$Zn^{2 +} = 3 d^{10}$ → all orbitals full, $n_{up} = 0$ → $μ = 0$ → diamagnetic (a full $d$ -shell edge case).
$Sc^{3 +} = 3 d^{0}$ → no $d$ -electrons at all, $n_{up} = 0$ → $μ = 0$ → diamagnetic (an empty $d$ -shell edge case). Lesson: both extremes ( $d^{0}$ and $d^{10}$ ) give $μ = 0$ no matter the ligand — magnetism can't distinguish ligand strength when there are no unpaired electrons to redistribute.