Crystal Field Stabilization Energy (CFSE) — high-spin vs low-spin; spectrochemical series: Worked examples

This is the practice arena for the parent CFSE topic. We will not introduce one new idea — we will drill the machine you already built: the $- 0.4/ + 0.6$ rule and the pairing-energy tax. But we will do it for every kind of input the topic can hand you, so that when an exam question lands you already know which cell of the table it lives in.

Recall The three tools we reuse (each earned in the parent note)

$Δ_{o}$ (read "delta-oh") ::: the energy gap between the lower $t_{2 g}$ orbitals and the upper $e_{g}$ orbitals in an octahedral complex.
$E (t_{2 g}) = - 0.4 Δ_{o}$ and $E (e_{g}) = + 0.6 Δ_{o}$ ::: the shift of each orbital away from the barycentre (the "average" level of the unsplit orbitals).
CFSE $= [- 0.4 n_{t_{2 g}} + 0.6 n_{e_{g}}] Δ_{o} + m P$ ::: sum the electrons' orbital energies, then add $P$ (pairing energy) once for each extra pair the field forces.

The scenario matrix

Every CFSE question is one of these cells. The worked examples below are labelled with the cell they hit, and together they touch all of them.

Cell	What makes it special	Covered by
A. Forced fill, low count ( $d^{1}$ – $d^{3}$ )	No high/low-spin choice; only $t_{2 g}$ used	Ex 1
B. The fork, weak field ( $d^{4}$ – $d^{7}$ , small $Δ_{o}$ )	High-spin branch; $e_{g}$ populated	Ex 2
C. The fork, strong field ( $d^{4}$ – $d^{7}$ , large $Δ_{o}$ )	Low-spin branch; extra pairs taxed	Ex 3
D. Forced fill, high count ( $d^{8}$ – $d^{10}$ )	No choice again; $e_{g}$ partly/fully filled	Ex 4
E. Degenerate / zero cases ( $d^{0}$ , $d^{10}$ )	CFSE $= 0$ ; sanity anchors	Ex 5
F. The crossover / limiting value ( $Δ_{o} = P$ )	The exact tipping point between HS and LS	Ex 6
G. Tetrahedral (inverted, always HS)	$e$ below $t_{2}$ , $Δ_{t} = \frac{4}{9} Δ_{o}$	Ex 7
H. Reverse problem (magnetism → config)	Given unpaired-electron count, deduce spin state	Ex 8
I. Real-world word problem	Colour / spectrum ties to $Δ_{o}$	Ex 9
J. Exam twist (compare two complexes)	Decide which is more stable with $P$ included	Ex 10

The whole matrix is really just two questions asked in order — how many $d$ -electrons, then (if it's a fork case) weak or strong field. The map below draws exactly that flow: read it left-to-right along the $d$ -count line, and notice the dashed magenta box marking the only region ( $d^{4}$ – $d^{7}$ ) where a fork exists.

Figure — Crystal Field Stabilization Energy (CFSE) — high-spin vs low-spin; spectrochemical series

Cell A — forced fill, low electron count

Worked example Ex 1 · Cell A ·

$d^{3}$ : $[Cr(H_{2} O)_{6}]^{3 +}$ Find the CFSE.

Forecast: guess before reading — will any electron go into $e_{g}$ ? How many pairs form?

Identify the $d$ -count. Cr is group 6; Cr $^{3 +}$ has lost 3 electrons → $d^{3}$ . Why this step? CFSE depends only on how many $d$ -electrons there are and where they sit — nothing else.
Fill by Hund's rule. Three electrons, three $t_{2 g}$ orbitals → one each, all unpaired: $t_{2 g}^{3} e_{g}^{0}$ . Why this step? Electrons avoid sharing an orbital until forced. With exactly 3 lower orbitals available, none is forced up into $e_{g}$ and none pairs. So high-spin = low-spin — there is no fork here.
Plug into the formula. $CFSE = (- 0.4 \times 3 + 0.6 \times 0) Δ_{o} + 0 \cdot P = - 1.2 Δ_{o}$ Why this step? $n_{t_{2 g}} = 3, n_{e_{g}} = 0$ ; $m = 0$ because the spread-out arrangement already had 0 pairs and so does the complex.

Verify: magnitude $1.2 Δ_{o}$ is the largest CFSE any first-row config can reach without paying $P$ — this is exactly why Cr $^{3 +}$ is so stable and inert (see Stability & Kinetic Inertness (Cr³⁺, Co³⁺)). Sign is negative = stabilizing. ✓

Cells B & C — the fork, both branches

The picture below shows both branches of the $d^{6}$ fork side by side. On the left (high-spin) two electrons ride up into $e_{g}$ and only one pair forms in $t_{2 g}$ ; on the right (low-spin) all six pack into the lower $t_{2 g}$ set, forming three pairs and leaving $e_{g}$ empty. Compare the arrow counts: the extra pairs on the right are exactly the electrons paying the $P$ tax.

Worked example Ex 2 · Cell B ·

$d^{5}$ high-spin: $[Mn(H_{2} O)_{6}]^{2 +}$ Find the CFSE. Water is a weak-field ligand, so $Δ_{o} < P$ .

Forecast: a famously "boring" number is coming. Guess it.

$d$ -count: Mn is group 7; Mn $^{2 +}$ → $d^{5}$ .
Weak field ⇒ maximise unpaired. Fill all five orbitals singly: $t_{2 g}^{3} e_{g}^{2}$ , five unpaired electrons. Why this step? Since $Δ_{o} < P$ , climbing to $e_{g}$ is cheaper than pairing, so every orbital gets one electron before any doubles.
Formula: $CFSE = (- 0.4 \times 3 + 0.6 \times 2) Δ_{o} + 0 \cdot P = (- 1.2 + 1.2) Δ_{o} = 0$ Why this step? $m = 0$ — no pairs beyond the spread-out baseline.

Verify: the up-shift of two $e_{g}$ electrons exactly cancels the down-shift of three $t_{2 g}$ electrons — this is the barycentre working itself out. High-spin $d^{5}$ has zero CFSE, which is why such ions form weak, labile complexes and are nearly colourless. ✓

Worked example Ex 3 · Cell C ·

$d^{6}$ low-spin: $[Fe(CN)_{6}]^{4 -}$ Find the CFSE. CN $^{-}$ is a strong-field ligand ( $Δ_{o} > P$ ).

Forecast: will there be electrons in $e_{g}$ ? How many extra pairs?

$d$ -count: Fe $^{2 +}$ → $d^{6}$ .
Choose ONE baseline and state it. We adopt the fully-spread baseline: imagine all six electrons occupying five orbitals as spread out as possible with zero pairs (a hypothetical reference), so $p_{free} = 0$ . Then $m = p_{complex}$ directly, and — crucially — we will use this same baseline for the high-spin branch too, so the comparison is fair. Why this step? Recall the warning in the " $m$ = extra pairs" box: the absolute value of $m$ depends on the baseline, but a comparison between two branches is only valid if both use the same one. Choosing the fully-spread reference makes $m$ equal to the plain pair-count, which is the convention the parent note's $d^{6}$ example uses ( $+ 3 P$ for LS, $+ 1 P$ for HS).
Count pairs in each branch against this baseline.
- HS $d^{6}$ : $t_{2 g}^{4} e_{g}^{2}$ has 1 pair → $m_{HS} = 1$ .
- LS $d^{6}$ : $t_{2 g}^{6} e_{g}^{0}$ has 3 pairs → $m_{LS} = 3$ . Why this step? With $p_{free} = 0$ , $m$ is simply "how many doubly-occupied orbitals does this configuration have."
Formula (LS branch): $CFSE_{LS} = (- 0.4 \times 6) Δ_{o} + 3 P = - 2.4 Δ_{o} + 3 P$ and the HS branch is $CFSE_{HS} = - 0.4 Δ_{o} + 1 P$ .

Verify: low-spin wins when $- 2.4 Δ_{o} + 3 P < - 0.4 Δ_{o} + 1 P \Rightarrow - 2 Δ_{o} < - 2 P \Rightarrow Δ_{o} > P$ — exactly the condition for a strong field. Note the difference in pair count, $3 - 1 = 2$ , is what actually drives the fork; the shared baseline cancels out. The algebra reproduces the fork rule, so our pair-counting is self-consistent. $[Fe(CN)_{6}]^{4 -}$ is diamagnetic (0 unpaired) ✓, matching $t_{2 g}^{6}$ .

Cell D — forced fill, high electron count

Worked example Ex 4 · Cell D ·

$d^{8}$ : $[Ni(H_{2} O)_{6}]^{2 +}$ Find the CFSE.

Forecast: with 8 electrons, is there still a fork?

$d$ -count: Ni $^{2 +}$ → $d^{8}$ .
Fill. Fill $t_{2 g}$ fully (6 electrons, 3 pairs), then $e_{g}$ gets 2, one each: $t_{2 g}^{6} e_{g}^{2}$ , two unpaired. Why this step? For $d^{8}$ there is only one way to arrange electrons regardless of $Δ_{o}$ vs $P$ — the last two must enter $e_{g}$ and, being in separate orbitals, don't add a pair beyond what filling $t_{2 g}$ already required. No fork.
Baseline pairs. Fully-spread $d^{8}$ : fill all 5 singly (5 used), remaining 3 pair up → 3 pairs. Complex also has 3 pairs. So $m = 0$ .
Formula: $CFSE = (- 0.4 \times 6 + 0.6 \times 2) Δ_{o} = (- 2.4 + 1.2) Δ_{o} = - 1.2 Δ_{o}$

Verify: two unpaired electrons ⇒ paramagnetic. Spin-only moment $μ = 2 (2 + 2) = 8 \approx 2.83 μ_{B}$ (see Magnetic Properties of Coordination Compounds (spin-only formula)). Both HS and LS give the same $- 1.2 Δ_{o}$ , confirming "no fork." ✓

Cell E — degenerate / zero cases

Cell F — the crossover (limiting value)

Worked example Ex 6 · Cell F ·

$d^{6}$ exactly at $Δ_{o} = P$ A $d^{6}$ ion sits at the tipping point where $Δ_{o} = P$ . What is the CFSE on each branch, and which does nature pick?

Forecast: guess whether the two branches give equal or different CFSE at the crossover.

Write both branches using the SAME (fully-spread) baseline.
- HS: $- 0.4 Δ_{o} + 1 P$
- LS: $- 2.4 Δ_{o} + 3 P$ Why this step? We must reuse the identical baseline chosen in Ex 3 (fully-spread, $p_{free} = 0$ ). A crossover is a comparison of the two branches, and — per the warning in the " $m$ = extra pairs" box — a comparison is only meaningful when both sides count pairs from the same reference. Mixing baselines here would shift one branch relative to the other and put the tie in the wrong place.
Set $P = Δ_{o}$ and evaluate.
- HS: $- 0.4 Δ_{o} + Δ_{o} = + 0.6 Δ_{o}$
- LS: $- 2.4 Δ_{o} + 3 Δ_{o} = + 0.6 Δ_{o}$ Why this step? Substituting the crossover condition tests whether the fork is a true tie there.
Compare: both equal $+ 0.6 Δ_{o}$ → exact tie.

Verify: the two straight lines CFSE $_{HS} (P)$ and CFSE $_{LS} (P)$ (as functions of $P$ for fixed $Δ_{o}$ ) must cross precisely where $Δ_{o} = P$ — that is the definition of the crossover. Getting identical values confirms the lines intersect exactly there, not merely near it. ✓

Cell G — tetrahedral (inverted, always high-spin)

Worked example Ex 7 · Cell G ·

$d^{7}$ tetrahedral: $[CoCl_{4}]^{2 -}$ Find the CFSE in units of $Δ_{t}$ , then in units of $Δ_{o}$ .

Forecast: the labels flip — which set is lower now?

Recall the inverted diagram. In a tetrahedral field the lower set is $e$ (2 orbitals) and the upper set is $t_{2}$ (3 orbitals) — opposite to octahedral (see Crystal Field Theory — Octahedral vs Tetrahedral Splitting). Why this step? Same barycentre logic, but now 2 orbitals go down and 3 go up.
Get the energies by barycentre. Let $e$ drop by $x$ , $t_{2}$ rise by $y$ , gap $x + y = Δ_{t}$ , and the balance condition $2 x = 3 y$ . Why " $2 x = 3 y$ "? The barycentre — the average energy of all five orbitals — cannot move when we merely redistribute energy (splitting neither creates nor destroys energy). "Average unchanged" means the total downward push must equal the total upward push. Two $e$ orbitals each drop by $x$ (total down $= 2 x$ ); three $t_{2}$ orbitals each rise by $y$ (total up $= 3 y$ ). Setting total-down $=$ total-up gives $2 x = 3 y$ . Look at the figure: the two orange levels sit deeper below the dotted barycentre than the three magenta levels sit above it, precisely because 2 must balance 3. Solving with $x + y = Δ_{t}$ : $y = \frac{2}{5} Δ_{t} = 0.4 Δ_{t}$ (up), $x = \frac{3}{5} Δ_{t} = 0.6 Δ_{t}$ (down). So $E (e) = - 0.6 Δ_{t}, E (t_{2}) = + 0.4 Δ_{t}$ .
Fill $d^{7}$ . $Δ_{t}$ is small (always $< P$ ), so always high-spin: fill singly first — $e^{2} t_{2}^{3}$ (5 unpaired) — then the 6th and 7th pair up in the lower $e$ set: $e^{4} t_{2}^{3}$ . $CFSE = (- 0.6 \times 4 + 0.4 \times 3) Δ_{t} = (- 2.4 + 1.2) Δ_{t} = - 1.2 Δ_{t}$
Convert to $Δ_{o}$ . Since $Δ_{t} \approx \frac{4}{9} Δ_{o}$ : $CFSE \approx - 1.2 \times \frac{4}{9} Δ_{o} = - \frac{4.8}{9} Δ_{o} \approx - 0.533 Δ_{o}$

Verify: because $Δ_{t}$ is only ~44% of $Δ_{o}$ , a tetrahedral complex can never afford to pay $P$ — hence "essentially always high-spin," exactly as the parent note warns. The magnitude in $Δ_{o}$ units ( $\approx 0.53$ ) is modest, matching tetrahedral complexes' generally low stabilization. ✓

Cell H — reverse problem (magnetism → configuration)

Worked example Ex 8 · Cell H · Deduce spin state from

$μ$ An octahedral Co $^{3 +}$ ( $d^{6}$ ) complex has measured magnetic moment $μ \approx 0 μ_{B}$ (diamagnetic). A different Co $^{3 +}$ complex has $μ \approx 4.9 μ_{B}$ . Assign each as high- or low-spin.

Forecast: which one is CN $^{-}$ -like, which is F $^{-}$ -like?

Invert the spin-only formula. $μ = n (n + 2)$ where $n$ = unpaired electrons. Solve for $n$ : $μ = 0 \Rightarrow n = 0$ ; $μ = 4.9 \Rightarrow n (n + 2) = 24 \Rightarrow n = 4$ . Why this step? Magnetism counts unpaired electrons — it's our experimental window into the configuration.
Match $n$ to a $d^{6}$ config.
- $n = 0$ : all paired → $t_{2 g}^{6} e_{g}^{0}$ → low-spin (strong field, large $Δ_{o}$ ).
- $n = 4$ : four unpaired → $t_{2 g}^{4} e_{g}^{2}$ → high-spin (weak field, small $Δ_{o}$ ).
Name the ligands. Low-spin diamagnetic Co $^{3 +}$ is e.g. $[Co(NH_{3})_{6}]^{3 +}$ ; high-spin is e.g. $[CoF_{6}]^{3 -}$ .

Verify: the parent note's magnetic-clue example says $[Co(NH_{3})_{6}]^{3 +}$ is diamagnetic ⇒ low-spin — matches our $n = 0$ branch. And $4 (4 + 2) = 24 = 4.90 μ_{B}$ reproduces the given moment. ✓

Cell I — real-world word problem (colour)

Worked example Ex 9 · Cell I · Colour tells you

$Δ_{o}$ $[Ti(H_{2} O)_{6}]^{3 +}$ is a $d^{1}$ ion that absorbs light at $λ \approx 500 nm$ (green light) and therefore looks purple. Estimate $Δ_{o}$ in kJ/mol and its CFSE.

Forecast: the lone electron jumps from $t_{2 g}$ to $e_{g}$ — the photon energy is $Δ_{o}$ .

Energy of one photon. $E = \frac{h c}{λ}$ with $h = 6.626 \times 1 0^{- 34} J\cdots$ , $c = 3.0 \times 1 0^{8} m/s$ , $λ = 500 \times 1 0^{- 9} m$ . $E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{500 \times 1 0 ^{- 9}} \approx 3.98 \times 1 0^{- 19} J$ Why this step? The $d$ - $d$ transition absorbs exactly the gap $Δ_{o}$ , so the photon energy = $Δ_{o}$ per molecule (see Colour of Transition-Metal Complexes (d-d transitions, $\Delta_o$ and λ)).
Scale to a mole. A CFSE is normally quoted per mole, so multiply the per-molecule energy by Avogadro's number $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ and divide by 1000 to reach kJ: $Δ_{o} \approx \frac{( 3.98 \times 1 0 ^{- 19} J ) \times ( 6.022 \times 1 0 ^{23} mol ^{- 1} )}{1000} \approx 240 kJ/mol$ Why this step? One photon promotes one electron in one molecule; a mole of molecules absorbs a mole of photons, so the molar gap is just $N_{A}$ times bigger.
CFSE of $d^{1}$ . One electron sits in $t_{2 g}$ , so $n_{t_{2 g}} = 1, n_{e_{g}} = 0, m = 0$ : $CFSE = - 0.4 Δ_{o} \approx - 0.4 \times 240 = - 96 kJ/mol$ Why this step? Direct application of the CFSE formula for a single lower-set electron.

Verify: measured $Δ_{o}$ for $[Ti(H_{2} O)_{6}]^{3 +}$ is ~243 kJ/mol (20,300 cm $^{- 1}$ ) — our 240 kJ/mol lands right on it. Units track: (J/photon) × (mol $^{- 1}$ ) ÷ 1000 = kJ/mol ✓. Absorbing green ⇒ transmitting red+blue ⇒ purple, matching the observed colour. ✓

Cell J — exam twist (compare two complexes)

Worked example Ex 10 · Cell J · Which is more stable — and is low-spin

always the winner? For a $d^{6}$ metal, take a weak-field case with $Δ_{o} = 0.8 P$ and a strong-field case with $Δ_{o} = 1.5 P$ . Compute the net CFSE (including the pairing tax) for the spin state nature actually chooses in each, expressing answers in units of $P$ .

Forecast: does the big raw $- 2.4 Δ_{o}$ of low-spin guarantee more stability? Steel-man the trap first.

Weak field, $Δ_{o} = 0.8 P < P$ ⇒ high-spin. Use the HS branch $- 0.4 Δ_{o} + 1 P$ (same fully-spread baseline as Ex 3): $CFSE = - 0.4 (0.8 P) + 1 P = - 0.32 P + P = + 0.68 P$ Why this step? Since $Δ_{o} < P$ , the fork rule sends us to high-spin; we substitute $Δ_{o}$ in terms of $P$ to get a single number.
Strong field, $Δ_{o} = 1.5 P > P$ ⇒ low-spin. Use the LS branch $- 2.4 Δ_{o} + 3 P$ : $CFSE = - 2.4 (1.5 P) + 3 P = - 3.6 P + 3 P = - 0.6 P$ Why this step? $Δ_{o} > P$ sends us to low-spin; same substitution.
Compare net stabilization. Weak-field HS gives $+ 0.68 P$ (net destabilized once pairing is included); strong-field LS gives $- 0.6 P$ (net stabilized). The strong-field complex is genuinely more stable — but only because we honoured the $+ m P$ tax. Why this step? This directly defuses Mistake #1 from the parent note: the raw $- 2.4 Δ_{o}$ looks huge, but $+ 3 P$ eats most of it.

Verify: if we had forgotten $P$ , the LS "CFSE" would read $- 2.4 (1.5 P) = - 3.6 P$ — wildly overstating stability and "predicting" low-spin even for weak fields. Including $P$ shrinks it to $- 0.6 P$ , and the sign of the net values ( $+ 0.68 P$ vs $- 0.6 P$ ) correctly ranks them. ✓

Recall Quick self-test across the matrix

A $d^{4}$ octahedral ion has $μ = 4.9 μ_{B}$ . High- or low-spin, and its CFSE? ::: $n = 4$ unpaired ⇒ high-spin $t_{2 g}^{3} e_{g}^{1}$ ; CFSE $= (- 0.4 \times 3 + 0.6 \times 1) Δ_{o} = - 0.6 Δ_{o}$ , $m = 0$ . A $d^{4}$ low-spin ion is $t_{2 g}^{4} e_{g}^{0}$ — how many unpaired, what CFSE? ::: 2 unpaired; CFSE $= - 0.4 \times 4 Δ_{o} + 1 P = - 1.6 Δ_{o} + P$ . Why is tetrahedral $d^{7}$ never low-spin? ::: $Δ_{t} \approx \frac{4}{9} Δ_{o}$ is too small to ever exceed $P$ , so pairing is never worth it.

See also: Jahn–Teller Distortion · Pairing Energy and Hund's Rule · Ligand Field Theory & π-backbonding.