Exercises — Magnetic properties — paramagnetism via spin-only formula μ = √(n(n+2)) BM

Step — WHAT: find the ion configuration. Ti is $[\text{Ar}]3d^{2}4s^2$. To make ${\text{Ti}}^{3+}$ we remove 3 electrons, and the rule (Electronic Configuration of Ions) is remove 4s before 3d. Two 4s electrons leave, then one 3d electron → ${\text{Ti}}^{3+}=3d^1$. WHY: in ions the 4s level sits above 3d, so it empties first. Step — count singles. One electron sits alone in one box (look at the s01 picture, top row) → $n=1$. $\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73\text{BM}.$

Rule: paramagnetic needs $n\ge 1$; diamagnetic means $n=0$ (all paired).

• ${\text{Sc}}^{3+}$, $d^0$: no d-electrons at all → $n=0$ → diamagnetic.
• ${\text{Zn}}^{2+}$, $d^{10}$: all five boxes hold a full pair ↑↓ → every magnet cancels → $n=0$ → diamagnetic.
• ${\text{Cu}}^{2+}$, $d^9$: four boxes paired, one box has a lone ↑ → $n=1$ → paramagnetic, $\mu =1.73$ BM.

${\text{Mn}}^{2+}$, $d^5$: by Hund's rule (Hund's Rule & Electron Configuration) each of the five boxes gets exactly one ↑ before any pairing → all 5 unpaired → $n=5$. $\mu =\sqrt{5\cdot 7}=\sqrt{35}=5.92\text{BM}.$ This is the maximum for a d-block ion — the half-filled $d^5$ peak. ${\text{Ni}}^{2+}$, $d^8$: fill five boxes singly (5 electrons), then the remaining 3 electrons pair up in three boxes. That leaves 2 boxes with a lone ↑ → $n=2$. $\mu =\sqrt{2\cdot 4}=\sqrt{8}=2.83\text{BM}.$

Remove 3 electrons, 4s first: $4s^2$ gone (2), then one 3d gone (1) → ${\text{Fe}}^{3+}=3d^5$. Five boxes, one ↑ each → $n=5$. $\mu =\sqrt{35}=5.92\text{BM}.$ Note ${\text{Fe}}^{3+}$ and ${\text{Mn}}^{2+}$ are both $d^5$ → identical $\mu$. Same electron count, same magnet.

WHY reverse-solve: experiment gives $\mu$; we want the hidden electron count $n$. Use $n=-1+\sqrt{1+{\mu }^2}$. $n=-1+\sqrt{1+3.87^2}=-1+\sqrt{1+14.98}=-1+\sqrt{15.98}=-1+3.998\approx 3.$ So $n=3$. A candidate: ${\text{Cr}}^{3+}$ ($d^3$, three lone electrons) or ${\text{V}}^{2+}$ ($d^3$). Check forward: $\sqrt{3\cdot 5}=\sqrt{15}=3.87$ ✓.

In an octahedral field (Crystal Field Theory) the five d-orbitals split into a lower set $t_{2g}$ (3 boxes) and an upper set $e_g$ (2 boxes), separated by a gap $\Delta$. See the s02 figure. Weak field (H₂O), high spin: $\Delta$ is small, so it's cheaper to put electrons in the upper $e_g$ boxes than to pair them. Filling: $t_{2g}^4{e}_g^2$. Counting singles: $t_{2g}$ has boxes (↑↓, ↑, ↑) = 2 lone, $e_g$ has (↑, ↑) = 2 lone → $n=4$. $\mu =\sqrt{4\cdot 6}=\sqrt{24}=4.90\text{BM}.$ Strong field (CN⁻), low spin: $\Delta$ is large, so pairing in the lower set is cheaper than climbing to $e_g$. Filling: $t_{2g}^6{e}_g^0$ — all three lower boxes are ↑↓ pairs → $n=0$. $\mu =\sqrt{0}=0\text{BM (diamagnetic)}.$ Lesson: magnetism is a probe of ligand-field strength — same ion, the magnet reveals the ligand.

We place 4 electrons into the split $t_{2g}$ (3 boxes) / $e_g$ (2 boxes) set. The s03 figure shows both fillings side by side. High spin (weak field, small $\Delta$): spread out to avoid pairing. First 3 electrons go one-each into $t_{2g}$ (↑, ↑, ↑). The 4th electron faces a choice: pair in $t_{2g}$ (cost = pairing energy $P$) or climb to $e_g$ (cost = $\Delta$). Small $\Delta <P$ → it climbs. Config $t_{2g}^3{e}_g^1$ → all 4 electrons are lone ↑ → $n=4$. ${\mu }_{\text{HS}}=\sqrt{4\cdot 6}=\sqrt{24}=4.90\text{BM}.$ Low spin (strong field, large $\Delta$): now $\Delta >P$, so the 4th electron prefers to pair in $t_{2g}$ rather than climb. Config $t_{2g}^4{e}_g^0$: boxes (↑↓, ↑, ↑) → 2 lone electrons → $n=2$. ${\mu }_{\text{LS}}=\sqrt{2\cdot 4}=\sqrt{8}=2.83\text{BM}.$ So a single $d^4$ ion can read 4.90 BM or 2.83 BM depending on the ligand.

Find each $n$, then $\mu$:

• ${\text{Ti}}^{3+}$, $d^1$: $n=1$, $\mu =1.73$.
• ${\text{Ni}}^{2+}$, $d^8$: $n=2$, $\mu =2.83$.
• HS ${\text{Cr}}^{2+}$, $d^4$: $n=4$, $\mu =4.90$.
• ${\text{Mn}}^{2+}$, $d^5$: $n=5$, $\mu =5.92$. Increasing order: ${\text{Ti}}^{3+}(1.73)<{\text{Ni}}^{2+}(2.83)<{\text{Cr}}^{2+}(4.90)<{\text{Mn}}^{2+}(5.92)$.

(a) Reverse-solve: $n=-1+\sqrt{1+2.83^2}=-1+\sqrt{1+8.0}=-1+3=2$. So $n=2$. From L4.1, a $d^4$ ion has $n=2$ only in the low-spin configuration $t_{2g}^4{e}_g^0$. (b) Low spin means pairing won over climbing: $\Delta >P$. The field is strong. (c) Strong-field ligands high on the spectrochemical series — e.g. ${\text{CN}}^{-}$, $\text{CO}$, or ${\text{NO}}_2^{-}$ (Crystal Field Theory).

Reverse solve: $n=-1+\sqrt{1+1.8^2}=-1+\sqrt{1+3.24}=-1+\sqrt{4.24}=-1+2.06\approx 1$. So $n=1$. A $d^5$ ion normally has $n=5$ (high spin, $t_{2g}^3{e}_g^2$). To reach $n=1$ the electrons must have paired up: low-spin $d^5$ is $t_{2g}^5{e}_g^0$ = boxes (↑↓, ↑↓, ↑) → 1 lone electron → $n=1$. $\mu =\sqrt{1\cdot 3}=\sqrt{3}=1.73\text{BM}(\approx 1.8).$ This is a very strong field forcing maximal pairing — e.g. $[{\text{Fe(CN)}}_6{]}^{3-}$ (${\text{Fe}}^{3+}$, $d^5$, low spin).

The spin-only formula drops the orbital contribution $L$, keeping only spin. In first-row transition complexes the ligand field quenches the orbital motion ($L\to 0$), so $\mu =\sqrt{n(n+2)}$ is accurate. In lanthanides the magnetic $4f$ orbitals are buried deep inside the atom, shielded from ligands, so $L$ is not quenched. You must use the full expression $\mu =\sqrt{4S(S+1)+L(L+1)}$ (really the total-$J$ formula). See Lanthanide Magnetism. Trusting spin-only for, say, ${\text{Nd}}^{3+}$ gives badly wrong numbers.

Give μ for high-spin $d^4$. ::: $\sqrt{24}=4.90$ BM ($n=4$). Give μ for low-spin $d^4$. ::: $\sqrt{8}=2.83$ BM ($n=2$). μ = 3.87 BM means how many unpaired electrons? ::: $n=3$. μ = 1.73 BM for a $d^5$ ion means what config? ::: Low-spin $t_{2g}^5{e}_g^0$, $n=1$. Why doesn't spin-only work for lanthanides? ::: Their $4f$ orbital contribution is not quenched.