Exercises — Magnetic moments of complexes

WHAT: plug $n=2$ into $\mu =\sqrt{n(n+2)}$. WHY: the formula is the definition of the spin-only moment; nothing else is needed. $\mu =\sqrt{2(2+2)}=\sqrt{2\cdot 4}=\sqrt{8}=2.83{\mu }_B.$ Answer: $2.83{\mu }_B$.

Diamagnetic means all electrons paired, so $n=0$. Check: $\mu =\sqrt{0(0+2)}=\sqrt{0}=0$. The $n=1$ case gives $\sqrt{3}=1.73{\mu }_B$ — that one is paramagnetic. Answer: the $n=0$ complex.

Step 1 — oxidation state. Water is neutral; overall charge $+3$; so Cr is $+3$, i.e. ${\text{Cr}}^{3+}$. Step 2 — d-count. Neutral Cr is $[\text{Ar}]3d^{5}4s^1$. Remove the $4s$ electron first, then two $3d$: ${\text{Cr}}^{3+}$ is $3d^3$. (See Electronic Configuration of d-block ions.) Step 3 — spin state. In an octahedral field, $d^3$ has only three electrons — they each take a separate $t_{2g}$ orbital by Hund's rule. There is no high-spin/low-spin choice for $d^3$: it is $n=3$ either way. Step 4 — formula. $\mu =\sqrt{3(3+2)}=\sqrt{15}=3.87{\mu }_B.$ Answer: $3.87{\mu }_B$.

Step 1. Overall $+2$, water neutral → ${\text{Mn}}^{2+}$. Step 2. Neutral Mn is $[\text{Ar}]3d^{5}4s^2$. Remove both $4s$ → ${\text{Mn}}^{2+}$ is $3d^5$. Step 3. ${\text{H}}_2\text{O}$ is a weak field ligand (Spectrochemical Series) → small ${\Delta }_o$ → high spin. Five $d$-electrons spread over all five $d$-orbitals, all parallel → $n=5$. Step 4. $\mu =\sqrt{5(5+2)}=\sqrt{35}=5.92{\mu }_B.$ Answer: $5.92{\mu }_B$ (the maximum possible for a 3d ion).

Step 1 — d-count. Co is $[\text{Ar}]3d^{7}4s^2$. ${\text{Co}}^{3+}$: remove two $4s$ + one $3d$ → $3d^6$. Step 2 — rank the ligands. In the Spectrochemical Series, ${\text{F}}^{-}$ is weak, ${\text{NH}}_3$ is strong. Step 3 — fill $d^6$ in each case using Crystal Field Theory splitting $t_{2g}$/$e_g$:

• Weak field $[{\text{CoF}}_6{]}^{3-}$ (high spin): fill all five orbitals singly (5 electrons), then the 6th pairs up. Config $t_{2g}^4{e}_g^2=(\uparrow \downarrow )(\uparrow )(\uparrow )(\uparrow )(\uparrow )$ → 4 unpaired, $n=4$. $\mu =\sqrt{4(4+2)}=\sqrt{24}=4.90{\mu }_B.$
• Strong field $[{\text{Co(NH}}_3{)}_6{]}^{3+}$ (low spin): large ${\Delta }_o$ forces all six into $t_{2g}$: $(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )$ → 0 unpaired, $n=0$. $\mu =0\Rightarrow \text{diamagnetic}.$ Answer: $[{\text{CoF}}_6{]}^{3-}$ = high spin, $4.90{\mu }_B$; $[{\text{Co(NH}}_3{)}_6{]}^{3+}$ = low spin, diamagnetic ($0{\mu }_B$).

d-count: Fe is $[\text{Ar}]3d^{6}4s^2$; ${\text{Fe}}^{2+}$ = $3d^6$.

• $[{\text{Fe(H}}_2\text{O}{)}_6{]}^{2+}$ — ${\text{H}}_2\text{O}$ weak → high spin $t_{2g}^4{e}_g^2$ → $n=4$ → $\mu =\sqrt{24}=4.90{\mu }_B$.
• $[{\text{Fe(CN)}}_6{]}^{4-}$ — ${\text{CN}}^{-}$ strong → low spin $t_{2g}^6$ → $n=0$ → diamagnetic. Answer: $4.90{\mu }_B$ and $0{\mu }_B$ respectively.

Step 1 — d-count. ${\text{Ni}}^{2+}$ = $[\text{Ar}]3d^8$ in both. Step 2 — two possible 4-coordinate shapes (Square Planar vs Tetrahedral Geometry):

• Tetrahedral splitting is small; the $d^8$ electrons stay maximally unpaired → 2 unpaired ($n=2$).
• Square planar pushes the single $d_{x^2-y^2}$ orbital very high; the 8 electrons fill the four lower orbitals as pairs → 0 unpaired ($n=0$). Step 3 — read the measurements backwards.
• $[{\text{NiCl}}_4{]}^{2-}$: $\mu =2.83\Rightarrow n(n+2)=8\Rightarrow n=2\Rightarrow$ tetrahedral (weak-field ${\text{Cl}}^{-}$).
• $[{\text{Ni(CN)}}_4{]}^{2-}$: $\mu =0\Rightarrow n=0\Rightarrow$ square planar (strong-field ${\text{CN}}^{-}$). Answer: paramagnetic ${\text{NiCl}}_4^{2-}$ = tetrahedral; diamagnetic ${\text{Ni(CN)}}_4^{2-}$ = square planar. The magnetic moment is direct structural evidence.

${\text{Fe}}^{2+}$ is $d^6$. To get $\mu =0$ you need $n=0$, i.e. all six electrons paired in $t_{2g}$ → low spin → you need a strong-field ligand (top of the Spectrochemical Series, e.g. ${\text{CN}}^{-}$, CO). Then $t_{2g}^6{e}_g^0$, $n=0$, and $\mu =\sqrt{0(0+2)}=0$. A weak-field ligand would give high-spin $n=4$, $\mu =4.90$, defeating the design. Answer: choose a strong-field ligand; the complex is $t_{2g}^6$, $n=0$, $\mu =0$.

(a) Solve for $n$. Square both sides: $n(n+2)=5.92^2\approx 35$. Solve $n^2+2n-35=0\Rightarrow (n+7)(n-5)=0\Rightarrow n=5$ (reject the negative root — you cannot have a negative count). (b) Configuration. Five unpaired electrons is the maximum for a 3d ion → high-spin $d^5$, e.g. ${\text{Mn}}^{2+}$ or ${\text{Fe}}^{3+}$. (c) Ligand field. To keep all five unpaired the field must be weak (small ${\Delta }_o$) — e.g. ${\text{H}}_2\text{O}$, ${\text{F}}^{-}$. A strong field would pair them up and drop $\mu$ toward $1.73$. Answer: $n=5$, high-spin $d^5$ (Mn²⁺/Fe³⁺), weak-field ligand.

Step 1 — get $n$. $1.73^2=3.00=n(n+2)$. Solve $n^2+2n-3=0\Rightarrow (n+3)(n-1)=0\Rightarrow n=1$. Step 2 — reconcile with $d^5$. Five electrons giving only one unpaired means four of them are paired: $t_{2g}^5=(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow )$. That is low spin. Step 3 — spin state → ligand. Low-spin $d^5$ demands a strong-field ligand. A textbook example is $[{\text{Fe(CN)}}_6{]}^{3-}$. Answer: $n=1$, low-spin $d^5$, e.g. $[{\text{Fe(CN)}}_6{]}^{3-}$.

Spin-only prediction: ${\mu }_{\text{so}}=\sqrt{7(7+2)}=\sqrt{63}=7.94{\mu }_B$. The measured $6.9$ (actually the true $J$-based value for Gd³⁺ is $7.94$, but for most lanthanides spin-only fails badly). What went wrong in general: for $4f$ lanthanides the orbital angular momentum is not quenched by the ligand field — the $4f$ orbitals are buried deep inside the atom, shielded from the ligands. So the spin-only formula, which assumes orbital contribution is frozen out, does not apply; you must use the full $\mu =g_J\sqrt{J(J+1)}{\mu }_B$. Answer: spin-only gives $7.94{\mu }_B$; it is unreliable for $4f$ ions because orbital angular momentum survives. (Gd³⁺ is a lucky special case where $L=0$, so spin-only happens to work.)