Every part of this question reduces to one balance: the sum of the oxidation states and ligand charges inside the square brackets must equal the charge carried by the complex ion, which is fixed by whatever ions sit outside it.For a formula (counter-ion)\(\displaystyle _a\)[M(ligand)\(\displaystyle _n\)](counter-ion)\(\displaystyle _b\), the whole compound is neutral, so the complex ion's charge is exactly opposite to the total charge of the counter-ions. Once you know the metal's oxidation state, remove that many electrons from the free-atom configuration — the \(\displaystyle ns\) electrons first, then \(\displaystyle (n-1)d\) — to get the \(\displaystyle d\)-electron count. The number of unpaired electrons in that count then gives the magnetic moment through the spin-only formula
\[\mu = \sqrt{n(n+2)} \ \text{BM} \]
where \(\displaystyle n\) is the number of unpaired electrons.
The step people get wrong: whether those \(\displaystyle d\)-electrons pair up depends on the ligand's field strength. Strong-field ligands (\(\displaystyle NH_3\), \(\displaystyle CN^{-}\)) force pairing into the lower \(\displaystyle t_{2g}\) set before any electron reaches \(\displaystyle e_g\) (low spin); weak-to-medium field ligands (\(\displaystyle H_2O\), \(\displaystyle \mathrm{Cl^{-}}\), oxalato, pyridine) generally don't (high spin). For \(\displaystyle d^3\) it makes no difference — three electrons singly fill three \(\displaystyle t_{2g}\) orbitals either way — but for \(\displaystyle d^5\) and \(\displaystyle d^6\) it changes the magnetic moment completely.
(i) \(\displaystyle K[Cr(H_2O)_2(C_2O_4)_2]\cdot 3H_2O\)The complex ion \(\displaystyle [Cr(H_2O)_2(C_2O_4)_2]^{-}\) must carry charge \(\displaystyle -1\) to balance the one \(\displaystyle K^+\) outside it (the water of crystallisation carries no charge). Aqua is neutral; oxalato (\(\displaystyle C_2O_4^{2-}\)) is bidentate and contributes \(\displaystyle -2\) each:
\[x + 2(0) + 2(-2) = -1 \ \Rightarrow\ x = +3 \]
Chromium is in the
+$\displaystyle 3$ oxidation state.
Cr (Z = $\displaystyle 24$) has ground-state configuration \(\displaystyle [Ar]3d^54s^1\). Removing $\displaystyle 3$ electrons for \(\displaystyle \mathrm{Cr^{3+}}\) (the \(\displaystyle 4s\) electron first, then two from \(\displaystyle 3d\)) gives \(\displaystyle [Ar]3d^3\), i.e. \(\displaystyle t_{2g}^3e_g^0\).
Coordination number: $\displaystyle 2$ aqua ligands (monodentate, $\displaystyle 1$ donor atom each) + $\displaystyle 2$ oxalato ligands (bidentate, $\displaystyle 2$ donor atoms each) \(\displaystyle = 2+4 = 6\). Stereochemistry:
octahedral.
Three electrons occupy three separate \(\displaystyle t_{2g}\) orbitals (Hund's rule) whatever the field strength, so \(\displaystyle n=3\):
\[\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ \text{BM} \]
— paramagnetic.
IUPAC name:
potassium diaquadioxalatochromate(III) trihydrate.
(ii) \(\displaystyle [Co(NH_3)_5Cl]Cl_2\)The complex ion \(\displaystyle [Co(NH_3)_5Cl]^{2+}\) must be \(\displaystyle +2\) to balance the two chloride ions outside:
\[x + 5(0) + (-1) = +2 \ \Rightarrow\ x = +3 \]
Cobalt is in the
+$\displaystyle 3$ oxidation state.
Co (Z = $\displaystyle 27$): \(\displaystyle [Ar]3d^74s^2\). Removing $\displaystyle 3$ electrons for \(\displaystyle \mathrm{Co^{3+}}\) gives \(\displaystyle [Ar]3d^6\).
Coordination number: \(\displaystyle 5(NH_3) + 1(Cl) = 6\),
octahedral.
\(\displaystyle NH_3\) is a strong-field ligand, so all six \(\displaystyle d\) electrons pack into the lower set before touching \(\displaystyle e_g\): \(\displaystyle t_{2g}^6e_g^0\), leaving \(\displaystyle n=0\) unpaired electrons.
\[\mu = \sqrt{0(0+2)} = 0 \ \text{BM} \]
— diamagnetic. (This is exactly the case where field strength changes the outcome: the same \(\displaystyle d^6\) count with weak-field ligands would give $\displaystyle 4$ unpaired electrons instead of 0.)
IUPAC name:
pentaamminechloridocobalt(III) chloride.
(iii) \(\displaystyle [CrCl_3(py)_3]\)This complex carries no ions outside it, so it is neutral overall:
\[x + 3(-1) + 3(0) = 0 \ \Rightarrow\ x = +3 \]
Chromium is in the
+$\displaystyle 3$ oxidation state — the same as part (i), so the electronic configuration is again \(\displaystyle [Ar]3d^3\) (\(\displaystyle t_{2g}^3e_g^0\)).
Coordination number: \(\displaystyle 3(Cl) + 3(py) = 6\),
octahedral.
As in (i), \(\displaystyle d^3\) gives $\displaystyle 3$ unpaired electrons regardless of field strength:
\[\mu = \sqrt{3\times5} = \sqrt{15} \approx 3.87 \ \text{BM} \]
— paramagnetic.
IUPAC name:
trichloridotripyridinechromium(III).
(iv) \(\displaystyle Cs[FeCl_4]\)The complex ion \(\displaystyle [FeCl_4]^{-}\) balances the single \(\displaystyle Cs^+\):
\[x + 4(-1) = -1 \ \Rightarrow\ x = +3 \]
Iron is in the
+$\displaystyle 3$ oxidation state.
Fe (Z = $\displaystyle 26$): \(\displaystyle [Ar]3d^64s^2\). Removing $\displaystyle 3$ electrons for \(\displaystyle \mathrm{Fe^{3+}}\) gives \(\displaystyle [Ar]3d^5\).
Coordination number:
$\displaystyle 4$ — only four chloride ions are bound. With a coordination number this low around a first-row \(\displaystyle M^{3+}\) ion, the geometry is
tetrahedral, not octahedral (the ligand-field splitting in a tetrahedral field is too small to favour anything else with a weak-field halide).
Tetrahedral splitting is small, so \(\displaystyle \mathrm{Cl^{-}}\) (already weak-field) gives high spin: all five \(\displaystyle d\) orbitals singly occupied, \(\displaystyle n=5\):
\[\mu = \sqrt{5\times7} = \sqrt{35} \approx 5.92 \ \text{BM} \]
— strongly paramagnetic.
IUPAC name:
caesium tetrachloridoferrate(III).
(v) \(\displaystyle K_4[Mn(CN)_6]\)The complex ion \(\displaystyle [Mn(CN)_6]^{4-}\) balances four \(\displaystyle K^+\) ions:
\[x + 6(-1) = -4 \ \Rightarrow\ x = +2 \]
Manganese is in the
+$\displaystyle 2$ oxidation state.
Mn (Z = $\displaystyle 25$): \(\displaystyle [Ar]3d^54s^2\). Removing $\displaystyle 2$ electrons for \(\displaystyle \mathrm{Mn^{2+}}\) (both from \(\displaystyle 4s\)) gives \(\displaystyle [Ar]3d^5\) — the same \(\displaystyle d^5\) count as iron in part (iv), but here the ligand is different.
Coordination number: $\displaystyle 6$,
octahedral.
\(\displaystyle \mathrm{CN^{-}}\) is a strong-field ligand, so all five electrons pack into the three \(\displaystyle t_{2g}\) orbitals before any reach \(\displaystyle e_g\): \(\displaystyle t_{2g}^5e_g^0\), leaving only \(\displaystyle n=1\) unpaired electron. (Compare this directly with the same \(\displaystyle d^5\) count in part (iv), which was high-spin with \(\displaystyle n=5\) because \(\displaystyle Cl^-\) is weak-field and the geometry was tetrahedral — same electron count, opposite spin state.)
\[\mu = \sqrt{1\times3} = \sqrt{3} \approx 1.73 \ \text{BM} \]
— weakly paramagnetic.
IUPAC name:
potassium hexacyanidomanganate(II).
Answer: (i) potassium diaquadioxalatochromate(III) trihydrate — Cr(+$\displaystyle 3$), \(\displaystyle 3d^3\), CN $\displaystyle 6$, octahedral, \(\displaystyle \mu\approx3.87\) BM; (ii) pentaamminechloridocobalt(III) chloride — Co(+$\displaystyle 3$), \(\displaystyle 3d^6\) low spin, CN $\displaystyle 6$, octahedral, \(\displaystyle \mu=0\) BM; (iii) trichloridotripyridinechromium(III) — Cr(+$\displaystyle 3$), \(\displaystyle 3d^3\), CN $\displaystyle 6$, octahedral, \(\displaystyle \mu\approx3.87\) BM; (iv) caesium tetrachloridoferrate(III) — Fe(+$\displaystyle 3$), \(\displaystyle 3d^5\) high spin, CN $\displaystyle 4$, tetrahedral, \(\displaystyle \mu\approx5.92\) BM; (v) potassium hexacyanidomanganate(II) — Mn(+$\displaystyle 2$), \(\displaystyle 3d^5\) low spin, CN $\displaystyle 6$, octahedral, \(\displaystyle \mu\approx1.73\) BM.