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Step-by-Step Solution
Step 1: Identify the Oxidation States and d-Electron Count
(i) For the complex $[FeCl_{4}]^{2-}$, iron appears as $Fe^{2+}$. Hence, the electronic configuration is $[Ar]\,3d^{6}$.
(ii) For the complex $[Co(C_{2}O_{4})_{3}]^{3-}$, cobalt appears as $Co^{3+}$. Thus, the electronic configuration is $[Ar]\,3d^{6}$.
(iii) For $MnO_{4}^{2-}$, manganese is in the $+6$ oxidation state, giving $Mn^{6+}$ with an electronic configuration $[Ar]\,3d^{1}$.
Step 2: Determine the Nature of the Ligands (Strong or Weak Field) and Unpaired Electrons
(i) $Cl^{-}$ is a weak field ligand. Hence, no pairing of electrons takes place for $Fe^{2+}$ ($3d^{6}$). The number of unpaired electrons $n = 4$.
(ii) $C_{2}O_{4}^{2-}$ (oxalate) is generally considered a strong field ligand for $Co^{3+}$ complexes, causing pairing of all electrons. Therefore, the number of unpaired electrons $n = 0$.
(iii) For $Mn^{6+}$ ($3d^{1}$), there is only one d-electron. Hence, the number of unpaired electrons $n = 1$.
Step 3: Use the Spin-Only Magnetic Moment Formula
The spin-only magnetic moment $ \mu $ (in Bohr magnetons, BM) is given by:
$$
\mu = \sqrt{n(n + 2)} \, \text{BM},
$$
where $n$ is the number of unpaired electrons.
Step 4: Calculate the Magnetic Moments
(i) For $[FeCl_{4}]^{2-}$: $n = 4$
$$
\mu = \sqrt{4 \times (4 + 2)} \, \text{BM}
= \sqrt{24} \, \text{BM}
\approx 4.90 \, \text{BM}.
$$
(ii) For $[Co(C_{2}O_{4})_{3}]^{3-}$: $n = 0$
All electrons are paired, hence:
$$
\mu = 0 \, \text{BM}.
$$
(iii) For $MnO_{4}^{2-}$: $n = 1$
$$
\mu = \sqrt{1 \times (1 + 2)} \, \text{BM}
= \sqrt{3} \, \text{BM}
\approx 1.73 \, \text{BM}.
$$
Step 5: Final Answer
Therefore, the magnetic moments (spin-only) for
$[FeCl_{4}]^{2-}$,
$[Co(C_{2}O_{4})_{3}]^{3-}$,
and $MnO_{4}^{2-}$
are
$4.90\,\text{BM}$,
$0\,\text{BM}$,
and
$1.73\,\text{BM}$
respectively.
