Exercises — Crystal Field Theory (CFT) — Δ_oct, Δ_tet, splitting diagrams

What we do: find the neutral atom's configuration, remove electrons for the charge (remove $4s$ before $3d$), count the leftover $3d$ electrons.

• Ti = [Ar]$3d^{2}4s^2$; Ti${}^{3+}$ removes 3 electrons ($4s^2$ then one $3d$) → $3d^1$ → $d^1$.
• Fe = [Ar]$3d^{6}4s^2$; Fe${}^{3+}$ removes $4s^2$ + one $3d$ → $3d^5$ → $d^5$.
• Co = [Ar]$3d^{7}4s^2$; Co${}^{3+}$ removes $4s^2$ + one $3d$ → $3d^6$ → $d^6$.
• Ni = [Ar]$3d^{8}4s^2$; Ni${}^{2+}$ removes $4s^2$ → $3d^8$ → $d^8$.
• Zn = [Ar]$3d^{10}4s^2$; Zn${}^{2+}$ removes $4s^2$ → $3d^{10}$ → $d^{10}$.

Why the question matters: a "choice" only exists when, after putting one electron in each of the three $t_{2g}$ orbitals, the next electron faces the fork "climb to $e_g$ (cost $\Delta$) or pair (cost $P$)." That happens only for $d^4$–$d^7$.

• $d^1,d^3,d^8$ → filling is forced, no choice.
• $d^4,d^6,d^7$ → two possibilities (high- or low-spin). ✔

Answer: $d^4,d^6,d^7$.

What we do: high-spin = spread out singly first (Hund's rule), because the small gap makes climbing cheaper than pairing. Put one electron in each of $t_{2g}(3)$ then one in $e_g$: $t_{2g}^3{e}_g^1$. Every electron is alone → 4 unpaired.

Configuration: $d^7$ high-spin fills all five orbitals once (5 electrons), then two more must pair in $t_{2g}$: $t_{2g}^5{e}_g^2$. Apply the recipe (recall $\Delta \equiv {\Delta }_{oct}$): $\text{CFSE}=5(-0.4{\Delta }_{oct})+2(+0.6{\Delta }_{oct})=-2.0{\Delta }_{oct}+1.2{\Delta }_{oct}=-0.8{\Delta }_{oct}.$ (Unpaired electrons: $t_{2g}$ has one pair + one single, $e_g$ has two singles → 3 unpaired.)

Configuration: strong field → fill $t_{2g}$ completely (6 electrons, 3 pairs), then the 7th must go to $e_g$: $t_{2g}^6{e}_g^1$. $\text{CFSE}=6(-0.4{\Delta }_{oct})+1(+0.6{\Delta }_{oct})=-2.4{\Delta }_{oct}+0.6{\Delta }_{oct}=-1.8{\Delta }_{oct}.$ 1 unpaired electron.

Why this formula: the absorbed photon's energy exactly equals the gap it jumps (${\Delta }_{oct}=h\nu =hc/\lambda$), from Colour & d–d Transitions. ${\Delta }_{oct}=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3.00\times 10^8)}{500\times 10^{-9}}.$ Numerator $=6.626\times 10^{-34}\times 3.00\times 10^8=1.9878\times 10^{-25}$. Divide by $\lambda =5.00\times 10^{-7}\text{m}$: ${\Delta }_{oct}=\frac{1.9878\times 10^{-25}}{5.00\times 10^{-7}}=3.9756\times 10^{-19}\text{J}\approx 3.98\times 10^{-19}\text{J per ion}.$

Why the $\frac{4}{9}$ factor (brief justification, not just quoted): two independent weakenings multiply together.

1. Fewer ligands. A tetrahedron has only 4 point charges vs an octahedron's 6, so the repulsion budget is scaled roughly by $\frac{4}{6}=\frac{2}{3}$.
2. Worse aim. In an octahedron the $e_g$ orbitals point straight at ligands; in a tetrahedron no orbital points directly at a ligand — the closest ($t_2$) only points toward the ligand region, so the geometric "hit efficiency" is again reduced by a factor of about $\frac{2}{3}$.

Multiplying the two effects: $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}$. Hence ${\Delta }_{tet}=\frac{4}{9}{\Delta }_{oct}=\frac{4}{9}(18,000)=8,000{\text{cm}}^{-1}.$

Ligand strength (from Spectrochemical Series & Ligand Strength): F${}^{-}$ is a weak field ligand; NH${}_3$ is strong.

• $[{\text{CoF}}_6{]}^{3-}$: weak field → ${\Delta }_{oct}<P$ → high-spin $d^6$: $t_{2g}^4{e}_g^2$ → $t_{2g}$ has one pair + two singles, $e_g$ two singles → 4 unpaired. Paramagnetic.
• $[{\text{Co(NH}}_3{\text{)}}_6{]}^{3+}$: strong field → ${\Delta }_{oct}>P$ → low-spin $d^6$: $t_{2g}^6{e}_g^0$ → all paired → 0 unpaired. Diamagnetic.

Use $\mu =\sqrt{n(n+2)}$.

• $[{\text{CoF}}_6{]}^{3-}$, $n=4$: $\mu =\sqrt{4\cdot 6}=\sqrt{24}=4.90\text{BM}$.
• $[{\text{Co(NH}}_3{\text{)}}_6{]}^{3+}$, $n=0$: $\mu =\sqrt{0}=0\text{BM}$.

Configuration: $t_{2g}^3{e}_g^2$ (five singly-filled orbitals). $\text{CFSE}=3(-0.4{\Delta }_{oct})+2(+0.6{\Delta }_{oct})=-1.2{\Delta }_{oct}+1.2{\Delta }_{oct}=0.$ Why zero: the three stabilised $t_{2g}$ electrons exactly cancel the two destabilised $e_g$ electrons. Consequence: ions like Mn${}^{2+}$, Fe${}^{3+}$ gain no crystal-field stability, so they are relatively labile (ligands swap easily) and their spin-forbidden $d$–$d$ transitions are weak → pale colours (see Stability of Complexes & CFSE).

Count pairs.

• Free ion $d^6$: 5 orbitals get one electron each, the 6th pairs → 1 pair.
• Low-spin $d^6$: $t_{2g}^6$ = three doubly-occupied orbitals → 3 pairs.
• Extra pairs created by the field $=3-1=2$.

Orbital CFSE: $6(-0.4{\Delta }_{oct})+0(+0.6{\Delta }_{oct})=-2.4{\Delta }_{oct}$. Total: ${\text{CFSE}}_{\text{total}}=-2.4{\Delta }_{oct}+2P.$ The big $-2.4{\Delta }_{oct}$ (plus needing ${\Delta }_{oct}>P$) is exactly why strong-field $d^6$ complexes are so stable.

Key fact: ${\Delta }_{tet}=\frac{4}{9}{\Delta }_{oct}$ is so small it is essentially always $<P$ → tetrahedral is always high-spin. Tetrahedral ordering: lower $e$ (2 orbitals), upper $t_2$ (3 orbitals). Fill 6 electrons high-spin:

• $e$: 2 orbitals, put 1 each → then the pairing begins: $e^3{t}_2^3$.
• Occupation: $e^3{t}_2^3$ → $e$ has one pair + one single, $t_2$ has three singles → 4 unpaired. $\mu =\sqrt{4\cdot 6}=\sqrt{24}=4.90\text{BM}.$

Configuration $e^3{t}_2^3$: $\text{CFSE}=3(-0.6{\Delta }_{tet})+3(+0.4{\Delta }_{tet})=-1.8{\Delta }_{tet}+1.2{\Delta }_{tet}=-0.6{\Delta }_{tet}.$

(a) High-spin $d^4$: $t_{2g}^3{e}_g^1$ → 3 + 1 = 4 unpaired. (b) The $e_g$ set holds 1 electron in 2 degenerate orbitals — an uneven (asymmetric) occupation. The single $e_g$ electron can sit in $d_{z^2}$ or $d_{x^2-y^2}$, which point differently in space. The molecule lowers its energy by distorting (e.g. elongating along $z$) so those two orbitals stop being equal in energy and the electron settles into the now-lower one. Uneven filling of the $e_g$ set (which points at ligands) causes a strong Jahn–Teller distortion — this is why Cr${}^{2+}$ and Mn${}^{3+}$ ($d^4$) complexes are famously distorted.

From L2.1 and L2.2:

• High-spin ($t_{2g}^5{e}_g^2$): CFSE $=-0.8{\Delta }_{oct}$.
• Low-spin ($t_{2g}^6{e}_g^1$): CFSE $=-1.8{\Delta }_{oct}$. Difference $=(-1.8)-(-0.8)=-1.0{\Delta }_{oct}$. The low-spin form is more stabilised by $1.0{\Delta }_{oct}$ (before subtracting the extra pairing cost $P$ that low-spin must pay). Whether low-spin actually wins depends on whether ${\Delta }_{oct}>P$.

Set-up. Let $t_{2g}$ drop by $a$ (so its energy is $-a$) and $e_g$ rise by $b$ (energy $+b$). Two conditions:

1. The gap is ${\Delta }_{oct}$: $b-(-a)=a+b={\Delta }_{oct}$.
2. Barycentre = 0 (energy just redistributes): the summed displacement weighted by orbital count is zero. Three $t_{2g}$ orbitals each at $-a$, two $e_g$ each at $+b$: $3(-a)+2(+b)=0⟹3a=2b.$ Solve. From $3a=2b$, $a=\frac{2}{3}b$. Substitute into $a+b={\Delta }_{oct}$: $\frac{2}{3}b+b={\Delta }_{oct}⟹\frac{5}{3}b={\Delta }_{oct}⟹b=0.6{\Delta }_{oct},a=0.4{\Delta }_{oct}.$ So $E(e_g)=+0.6{\Delta }_{oct}$, $E(t_{2g})=-0.4{\Delta }_{oct}$. ∎

High-spin $d^4$ unpaired count ::: 4 CFSE of high-spin $d^7$ (octahedral, ignore $P$) ::: $-0.8{\Delta }_{oct}$ CFSE of low-spin $d^7$ (octahedral, ignore $P$) ::: $-1.8{\Delta }_{oct}$ ${\Delta }_{tet}$ in terms of ${\Delta }_{oct}$ ::: $\frac{4}{9}{\Delta }_{oct}$ Why high-spin $d^5$ has zero CFSE ::: the three $t_{2g}$ ($-1.2{\Delta }_{oct}$) exactly cancel the two $e_g$ ($+1.2{\Delta }_{oct}$) Spin-only $\mu$ for $n=4$ ::: $\sqrt{24}=4.90$ BM