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

Before we begin, one shared toolbox — every symbol here was built in the parent note, but let us restate it in plain words so nothing is assumed.

Definition Two units you will meet on this page

cm⁻¹ ("wavenumbers" or "reciprocal centimetres"). Chemists measure orbital-splitting energies not in joules but in how many wave-crests fit in one centimetre for the light that carries that much energy. Because energy $E = h c / λ$ is directly proportional to $1/ λ$ , the quantity $1/ λ$ (units cm⁻¹) is an energy scale — bigger number = more energetic. So " $Δ_{o} = 22000 cm^{- 1}$ " just means "the splitting equals the energy of light with 22000 crests per cm". Use it as a plain energy; you may add and subtract cm⁻¹ values freely. Every energy on this page — $Δ_{o}$ , $Δ_{t}$ , and the pairing energy $P$ — is measured in these same cm⁻¹, so they can be compared and combined directly.
$μ_{B}$ (the Bohr magneton). The natural "unit of magnetism" for a single electron — the magnetic strength one lone spinning electron produces. When we write a magnetic moment as " $4.90 μ_{B}$ ", we mean "as magnetic as about five unpaired electrons' worth of spin". See Magnetic Properties of Coordination Compounds (spin-only formula).

Formula The toolbox for every problem below

For an octahedral complex, each of the three lower orbitals ( $t_{2 g}$ ) sits $0.4 Δ_{o}$ below the average line (the barycentre), and each of the two upper orbitals ( $e_{g}$ ) sits $0.6 Δ_{o}$ above it. So: $CFSE = [- 0.4 n_{t_{2 g}} + 0.6 n_{e_{g}}] Δ_{o} + m P$

$n_{t_{2 g}}$ = how many electrons sit in the lower set,
$n_{e_{g}}$ = how many sit in the upper set,
$P$ = pairing energy (the "annoyance" cost of forcing two electrons to share one orbital), measured in cm⁻¹ like $Δ_{o}$ ,
$m$ = number of extra pairs the ligand field forces, counted beyond the maximum-unpaired (high-spin) baseline.

"Extra pairs" is subtle, so read this rule: count how many pairs the low-spin picture has that the high-spin picture does NOT. That difference is $m$ . (In the high-spin picture $m = 0$ by our convention — it is the baseline.)

Look at the reference diagram below whenever you place electrons: lower shelf = $t_{2 g}$ , upper shelf = $e_{g}$ , the dashed line is the barycentre.

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

Level 1 — Recognition

Problem L1.1

State which real $d$ -electron counts (in an octahedral field) can ever be high-spin or low-spin, and which are forced into a single arrangement. Explain in one line why.

Recall Solution

WHAT we do: list where the "fork in the road" exists. The fork only appears once the lower $t_{2 g}$ shelf is exactly full-of-singles and the next electron must choose: climb to $e_{g}$ (cost $Δ_{o}$ ) or pair down in $t_{2 g}$ (cost $P$ ).

$d^{0}$ : no $d$ -electrons at all — nothing to place, no choice (and CFSE $= 0$ ).
$d^{1}, d^{2}, d^{3}$ : only 1–3 electrons — they each take a separate $t_{2 g}$ orbital, no crowding, no choice.
$d^{4}, d^{5}, d^{6}, d^{7}$ : choice exists — high-spin vs low-spin.
$d^{8}, d^{9}, d^{10}$ : the $t_{2 g}$ is full (6) and electrons must go into $e_{g}$ ; the remaining filling is forced, no choice.

WHY (the one line): ==a genuine choice exists only when the $t_{2 g}$ shelf already holds one electron in each of its three orbitals (needs $\geq 3$ electrons) and there are still electrons left over that could either pair down or climb up (needs $\leq 7$ )== — that window is exactly $d^{4}$ through $d^{7}$ . Below $d^{4}$ nothing has to decide (empty $e_{g}$ , no crowding); above $d^{7}$ the choice is already made for you (the $e_{g}$ must start filling once $t_{2 g}$ is full at 6).

Answer: ==only $d^{4}$ – $d^{7}$ are ambiguous==; $d^{0}$ – $d^{3}$ and $d^{8}$ – $d^{10}$ are forced.

Problem L1.2

For $[Cr(H_{2} O)_{6}]^{3 +}$ , identify the $d$ -electron count and write its ground configuration.

Recall Solution

WHAT: find the oxidation state, subtract from the group. Cr is group 6, so the neutral atom sketch gives Cr $^{3 +}$ = $d^{3}$ (chromium(III) always). Water is a mid-field ligand, but $d^{3}$ is forced anyway. By Hund's rule (one electron per orbital before pairing, see Pairing Energy and Hund's Rule), the three electrons occupy the three $t_{2 g}$ orbitals singly: $t_{2 g}^{3} e_{g}^{0}$ Answer: == $d^{3}$ , configuration $t_{2 g}^{3} e_{g}^{0}$ ==.

Level 2 — Application

Problem L2.1

Compute the CFSE (in units of $Δ_{o}$ , ignore $P$ for now) for a high-spin $d^{5}$ octahedral ion, e.g. $[Mn(H_{2} O)_{6}]^{2 +}$ .

Recall Solution

WHAT: place 5 electrons in the maximum-unpaired way, then plug into the formula. High-spin $d^{5}$ = one electron in every orbital: $t_{2 g}^{3} e_{g}^{2}$ . $CFSE = [- 0.4 \times 3 + 0.6 \times 2] Δ_{o} = [- 1.2 + 1.2] Δ_{o} = 0$ WHY it's zero: each of the 3 lower electrons earns $- 0.4 Δ_{o}$ of stabilization, but each of the 2 upper electrons pays $+ 0.6 Δ_{o}$ . The books balance: $3 \times (- 0.4) = - 1.2$ saved, $2 \times (+ 0.6) = + 1.2$ spent. A perfectly symmetric spread means the barycentre is untouched — no net stabilization from the field. Answer: == $CFSE = 0 Δ_{o}$ == (this is why high-spin $d^{5}$ complexes like Mn $^{2 +}$ are pale/colourless and unusually labile).

Problem L2.2

Compute the CFSE (in $Δ_{o}$ , ignore $P$ ) for low-spin $d^{5}$ , e.g. $[Fe(CN)_{6}]^{3 -}$ .

Recall Solution

WHAT: strong field CN⁻ forces all 5 electrons into the lower shelf. Config: $t_{2 g}^{5} e_{g}^{0}$ . $CFSE = [- 0.4 \times 5 + 0.6 \times 0] Δ_{o} = - 2.0 Δ_{o}$ WHY this number: now all five electrons sit on the deep $t_{2 g}$ shelf, each earning $- 0.4 Δ_{o}$ — that is $5 \times (- 0.4) = - 2.0 Δ_{o}$ of pure stabilization, with nobody paying the $+ 0.6 Δ_{o}$ upper-shelf cost because $e_{g}$ is empty. Every electron is doing stabilizing work; that is the essence of why strong-field low-spin has such a deep CFSE. WHAT IT LOOKS LIKE: all five electrons crammed onto the lower shelf of the figure, two orbitals doubly filled and one singly. Compared to L2.1's $0$ , this is a huge $- 2.0 Δ_{o}$ — that is exactly why CN⁻ complexes are so stable. Answer: == $- 2.0 Δ_{o}$ == (before the pairing penalty).

Problem L2.3

Compute CFSE for high-spin $d^{7}$ ( $t_{2 g}^{5} e_{g}^{2}$ ).

Recall Solution

WHAT: place 7 electrons the weak-field way: fill $t_{2 g}$ (holding 6) then the last two spread singly, but high-spin means the $e_{g}$ pair up as late as possible — for $d^{7}$ that is $t_{2 g}^{5} e_{g}^{2}$ . $CFSE = [- 0.4 \times 5 + 0.6 \times 2] Δ_{o} = [- 2.0 + 1.2] Δ_{o} = - 0.8 Δ_{o}$ WHY the two terms: the 5 lower electrons save $5 \times (- 0.4) = - 2.0 Δ_{o}$ , but 2 electrons were forced up into $e_{g}$ and each pays back $+ 0.6 Δ_{o}$ , returning $+ 1.2 Δ_{o}$ . The net $- 0.8 Δ_{o}$ is the tug-of-war between deep-shelf savings and upper-shelf cost — smaller than the $d^{5}$ low-spin case precisely because two electrons are "wasted" up high. Answer: == $- 0.8 Δ_{o}$ ==.

Level 3 — Analysis

Problem L3.1

For octahedral $d^{6}$ , write the CFSE (including $P$ ) for both spin states, then find the exact condition on $Δ_{o}$ and $P$ for the low-spin state to be more stable.

Recall Solution

WHAT: build both, compare. High-spin $d^{6} = t_{2 g}^{4} e_{g}^{2}$ . Orbital part $= [- 0.4 \times 4 + 0.6 \times 2] Δ_{o} = - 0.4 Δ_{o}$ . Extra pairs $m = 0$ (baseline). $CFSE_{H S} = - 0.4 Δ_{o}$ Low-spin $d^{6} = t_{2 g}^{6} e_{g}^{0}$ . Orbital part $= - 0.4 \times 6 = - 2.4 Δ_{o}$ . Now count extra pairs: LS has 3 doubly-filled $t_{2 g}$ orbitals (3 pairs); HS ( $t_{2 g}^{4}$ ) has 1 pair. So extra $m = 3 - 1 = 2$ (the two amber brackets in the $m$ -figure above). $CFSE_{L S} = - 2.4 Δ_{o} + 2 P$ WHY the $2 P$ , not $3 P$ : we count pairs beyond the high-spin baseline. HS already carries one forced pair; the field only adds two more. (The parent note's $d^{6}$ example uses a different, free-ion baseline giving $+ P$ vs $+ 3 P$ — the difference $2 P$ is identical, and that difference is all that matters for comparison. Both bookkeepings agree.) Condition: low-spin wins when $- 2.4 Δ_{o} + 2 P < - 0.4 Δ_{o} \Rightarrow 2 P < 2 Δ_{o} \Rightarrow Δ_{o} > P$ Answer: low-spin favoured iff == $Δ_{o} > P$ ==.

Problem L3.2

A certain octahedral $d^{6}$ complex has $Δ_{o} = 22000 cm^{- 1}$ and $P = 19000 cm^{- 1}$ . Which spin state is the ground state? By how much (in cm⁻¹) is it more stable than the other?

Recall Solution

WHAT: plug numbers into both CFSE expressions from L3.1. (Recall cm⁻¹ is just an energy scale — see the definition callout up top — so we add and subtract these numbers directly.) $CFSE_{H S} = - 0.4 (22000) = - 8800 cm^{- 1}$ $CFSE_{L S} = - 2.4 (22000) + 2 (19000) = - 52800 + 38000 = - 14800 cm^{- 1}$ Since $Δ_{o} = 22000 > P = 19000$ , low-spin should win — and indeed $- 14800 < - 8800$ , low-spin is lower (more stable). Difference $= ∣ - 14800 - (- 8800) ∣ = 6000 cm^{- 1}$ . Answer: ==low-spin, more stable by $6000 cm^{- 1}$ ==.

Level 4 — Synthesis

Problem L4.1

$[Fe(H_{2} O)_{6}]^{2 +}$ and $[Fe(CN)_{6}]^{4 -}$ are both iron(II) ( $d^{6}$ ). Predict, for each: (a) spin state, (b) number of unpaired electrons, (c) spin-only magnetic moment $μ = n (n + 2) μ_{B}$ , and (d) which absorbs longer-wavelength light.

Recall Solution

WHAT: use the spectrochemical series to fix the field strength, then read off everything. H₂O sits in the middle of the series (weak-to-moderate); CN⁻ is at the strong end (π-acceptor, see Ligand Field Theory & π-backbonding). (a) $[Fe(H_{2} O)_{6}]^{2 +}$ : weak field ⇒ high-spin $t_{2 g}^{4} e_{g}^{2}$ . $[Fe(CN)_{6}]^{4 -}$ : strong field ⇒ low-spin $t_{2 g}^{6} e_{g}^{0}$ . (b) Count singly-occupied orbitals. HS $d^{6}$ : draw $t_{2 g}^{4} e_{g}^{2}$ → one $t_{2 g}$ orbital is paired, the other two $t_{2 g}$ + both $e_{g}$ are single ⇒ 4 unpaired. LS $d^{6}$ : $t_{2 g}^{6}$ all paired ⇒ 0 unpaired. (c) Using the spin-only formula $μ = n (n + 2)$ (answer comes out in Bohr magnetons $μ_{B}$ , the per-electron magnetism unit defined at the top): $μ_{H S} = 4 (6) = 24 \approx 4.90 μ_{B}, μ_{L S} = 0 = 0 μ_{B} (diamagnetic) .$ (d) The colour comes from a $d$ – $d$ jump of size $Δ_{o}$ (see Colour of Transition-Metal Complexes (d-d transitions, $\Delta_o$ and λ)). Photon energy $E = h c / λ$ , so larger $Δ_{o}$ ⇒ shorter $λ$ absorbed. CN⁻ gives the larger $Δ_{o}$ , so it absorbs the shorter wavelength; the aqua complex (smaller $Δ_{o}$ ) absorbs the longer wavelength. Answer: aqua = high-spin, 4 unpaired, $μ \approx 4.90 μ_{B}$ , absorbs longer $λ$ ; cyano = low-spin, 0 unpaired, diamagnetic, absorbs shorter $λ$ .

Problem L4.2

Explain, using both barycentre logic and the ratio $Δ_{t} \approx \frac{4}{9} Δ_{o}$ , why the tetrahedral complex $[FeCl_{4}]^{2 -}$ ( $d^{6}$ ) is essentially forced high-spin — and give its CFSE in terms of $Δ_{t}$ .

Recall Solution

First, what is $Δ_{t}$ ? Just as $Δ_{o}$ is the energy gap between the split $d$ -orbital sets in an octahedral field, == $Δ_{t}$ is the same idea for a tetrahedral field==: the energy gap between its two split sets. It is measured in cm⁻¹, exactly like $Δ_{o}$ and $P$ .

The splitting is inverted (barycentre logic). In tetrahedral geometry there are only 4 ligands, and — crucially — none of them points straight down a Cartesian axis; they sit in the "between-axis" directions. So now the orbitals that point between axes ( $d_{x y}, d_{y z}, d_{x z}$ , the $t_{2}$ set) are the ones closest to ligands and get raised, while the axis-pointing orbitals ( $d_{z^{2}}, d_{x^{2} - y^{2}}$ , the $e$ set) are lowered. The order flips versus octahedral (see Crystal Field Theory — Octahedral vs Tetrahedral Splitting). Now apply barycentre logic (the "average line" from the definition above): the lower $e$ set has 2 orbitals and the upper $t_{2}$ set has 3. For the total down-push to equal the total up-push, let each $e$ drop by $a$ and each $t_{2}$ rise by $b$ : $2 a = 3 b (balance), a + b = Δ_{t} (gap) .$ From the balance $a = \frac{3}{2} b$ ; substitute: $\frac{3}{2} b + b = Δ_{t} \Rightarrow \frac{5}{2} b = Δ_{t} \Rightarrow b = 0.4 Δ_{t}, a = 0.6 Δ_{t}$ . So each $e$ sits at $- 0.6 Δ_{t}$ and each $t_{2}$ at $+ 0.4 Δ_{t}$ — the $0.6$ and $0.4$ swap places versus octahedral, precisely because the orbital counts (2 vs 3) swapped.

WHY the $\frac{4}{9}$ ratio matters. Two geometric facts shrink the tetrahedral gap: (i) there are only 4 ligands instead of 6 (less total repulsion), and (ii) no ligand points directly at any $d$ -orbital (weaker, glancing repulsion). A full point-charge calculation combines these into $Δ_{t} \approx \frac{4}{9} Δ_{o} \approx 0.44 Δ_{o}$ — under half the octahedral gap. Since $Δ_{o}$ is itself modest for a weak-field ligand like Cl⁻, $Δ_{t}$ comes out far smaller than the pairing energy $P$ . With $Δ_{t} ≪ P$ , climbing to the upper set is always cheaper than pairing, so the electrons never pair ⇒ tetrahedral is essentially always high-spin.

CFSE. High-spin $d^{6}$ tetrahedral fills the lower $e$ first (max 4), then $t_{2}$ : $e^{3} t_{2}^{3}$ . $CFSE = [- 0.6 \times 3 + 0.4 \times 3] Δ_{t} = [- 1.8 + 1.2] Δ_{t} = - 0.6 Δ_{t}$ Answer: essentially always high-spin (because $Δ_{t} \approx \frac{4}{9} Δ_{o} ≪ P$ ); == $CFSE = - 0.6 Δ_{t}$ ==.

Level 5 — Mastery

Problem L5.1

Two isomeric $d^{4}$ octahedral complexes have identical ligands and $Δ_{o}$ , but one shows a Jahn–Teller distortion and the other does not. (a) Which spin state distorts and why? (b) Compute the un-distorted CFSE (in $Δ_{o}$ , plus $P$ -count) for both spin states, and state the $Δ_{o}$ – $P$ condition selecting low-spin.

Recall Solution

WHAT: build both $d^{4}$ pictures, spot the unevenly-filled degenerate set, connect to distortion. High-spin $d^{4} = t_{2 g}^{3} e_{g}^{1}$ . Orbital part $= [- 0.4 \times 3 + 0.6 \times 1] Δ_{o} = [- 1.2 + 0.6] Δ_{o} = - 0.6 Δ_{o}$ ; extra pairs $m = 0$ . $CFSE_{H S} = - 0.6 Δ_{o}$ Low-spin $d^{4} = t_{2 g}^{4} e_{g}^{0}$ . Orbital part $= - 0.4 \times 4 = - 1.6 Δ_{o}$ ; the 4th electron pairs in $t_{2 g}$ , so extra pairs $m = 1$ . $CFSE_{L S} = - 1.6 Δ_{o} + P$ (a) Jahn–Teller: the effect (see Jahn–Teller Distortion) strikes when a degenerate set is unevenly filled. HS $d^{4}$ has $e_{g}^{1}$ — one electron shared unevenly across the two degenerate $e_{g}$ orbitals ( $d_{z^{2}}$ vs $d_{x^{2} - y^{2}}$ ), which point straight at ligands ⇒ a strong (large) Jahn–Teller distortion. So the high-spin form distorts. (b) Condition: LS wins when $- 1.6 Δ_{o} + P < - 0.6 Δ_{o} \Rightarrow P < Δ_{o}$ , i.e. == $Δ_{o} > P$ == (same fork as always). Answer: high-spin $d^{4}$ distorts (uneven $e_{g}^{1}$ ); $CFSE_{H S} = - 0.6 Δ_{o}$ , $CFSE_{L S} = - 1.6 Δ_{o} + P$ ; low-spin selected iff $Δ_{o} > P$ .

Problem L5.2

Rank these three $d^{6}$ octahedral complexes by increasing $Δ_{o}$ and justify each move up the ladder using the three rules that raise $Δ_{o}$ : $[CoF_{6}]^{3 -}$ , $[Co(NH_{3})_{6}]^{3 +}$ , $[Rh(NH_{3})_{6}]^{3 +}$ .

Recall Solution

WHAT: apply the three rules that enlarge $Δ_{o}$ — (i) ligand position in the spectrochemical series, (ii) higher oxidation state, (iii) going down a group (3d < 4d < 5d) — comparing the complexes two at a time.

$[CoF_{6}]^{3 -}$ (smallest $Δ_{o}$ ): Co(III), a 3d metal, bound by F⁻, which sits near the weak-field end of the spectrochemical series. Weak ligand ⇒ smallest splitting. (Consistent with it being high-spin and paramagnetic.)
$[Co(NH_{3})_{6}]^{3 +}$ (middle): same metal and same oxidation state as the fluoride, so rules (ii) and (iii) are unchanged — the only thing that moves is the ligand. Since NH₃ lies higher in the series than F⁻ (rule i), it produces a larger $Δ_{o}$ . (Consistent with it being low-spin and diamagnetic — the magnetic clue from the parent note.)
$[Rh(NH_{3})_{6}]^{3 +}$ (largest): same ligand (NH₃) and same oxidation state (+3) as the cobalt ammine, so rules (i) and (ii) are unchanged — the only thing that moves is the metal row. Rh is 4d, and descending a group sharply raises $Δ_{o}$ (rule iii) ⇒ the largest splitting of the three; strongly low-spin and kinetically inert, cf. Stability & Kinetic Inertness (Cr³⁺, Co³⁺).

Answer (increasing $Δ_{o}$ ): $[CoF_{6}]^{3 -} < [Co(NH_{3})_{6}]^{3 +} < [Rh(NH_{3})_{6}]^{3 +}$

Recall Self-test checklist

Can you place electrons for any $d^{n}$ in both spin states? ::: Fill $t_{2 g}$ first singly (Hund), then decide the 4th–7th electron by comparing $Δ_{o}$ vs $P$ ; low-spin only when $Δ_{o} > P$ and only for $d^{4}$ – $d^{7}$ . Do you keep $Δ_{o}$ -terms and $P$ -terms separate until given their ratio? ::: Yes — CFSE = orbital part (in $Δ_{o}$ ) + pairing part (in $P$ ); combine only with numbers. Do you swap to $- 0.6/ + 0.4$ and count pairs from a consistent baseline for tetrahedral? ::: Yes — tetrahedral inverts to $e$ below $t_{2}$ ; $Δ_{t} \approx \frac{4}{9} Δ_{o}$ forces high-spin.