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Step-by-Step Solution
Step 1: Identify Each Complex and Its Metal Center
1) $[Co(\text{ox})_{2}(\text{OH})_{2}]^{-}$
2) $[Ti(NH_{3})_{6}]^{3+}$
3) $[V(\text{gly})_{2}(\text{OH})_{2}(NH_{3})_{2}]^{+}$
4) $[Fe(\text{en})(\text{bpy})(NH_{3})_{2}]^{2+}$
Step 2: Determine the Oxidation State and d-Electron Configuration
$[Co(\text{ox})_{2}(\text{OH})_{2}]^{-}$
- Oxalate ($\text{ox}$) is a bidentate ligand with charge $-2$, and there are two such ligands: total charge $-4$.
- Each hydroxide ($\text{OH}^{-}$) has charge $-1$, and there are two of them: total charge $-2$.
- Overall complex has a charge of $-1$.
Let the oxidation state of cobalt be $x$. Then,
$\displaystyle x + (-4) + (-2) = -1 \quad \Rightarrow \quad x = +5.$
Cobalt in +5 oxidation state has an electronic configuration: $[Ar]\,3d^{4}$.
$[Ti(NH_{3})_{6}]^{3+}$
- Titanium in +3 oxidation state gives: $[Ar]\,3d^{1}$.
$[V(\text{gly})_{2}(\text{OH})_{2}(NH_{3})_{2}]^{+}$
- Glycine (gly) is neutral in its usual coordination form; each OH is $-1$.
- Overall charge is $+1$.
If vanadium is in +5 state, its configuration becomes $[Ar]\,3d^{0}$.
$[Fe(\text{en})(\text{bpy})(NH_{3})_{2}]^{2+}$
- Iron in +2 oxidation state: $[Ar]\,3d^{6}$.
Step 3: Assess the Ligand Field Strength
- Oxalate ($\text{ox}^{2-}$) and hydroxide ($\text{OH}^{-}$) are usually considered weak-field ligands, so they do not force electron pairing strongly.
- Ammonia ($NH_{3}$), ethylenediamine (en), and bipyridyl (bpy) are relatively stronger-field ligands, promoting electron pairing in many cases.
Step 4: Determine the Number of Unpaired Electrons
$[Co(\text{ox})_{2}(\text{OH})_{2}]^{-}$
For $\text{Co}^{5+}$ ($3d^{4}$) in a weak-field environment, there are typically 4 unpaired electrons.
$[Ti(NH_{3})_{6}]^{3+}$
$\text{Ti}^{3+}$ has $3d^{1}$, so it has 1 unpaired electron.
$[V(\text{gly})_{2}(\text{OH})_{2}(NH_{3})_{2}]^{+}$
$\text{V}^{5+}$ is $3d^{0}$, so it has 0 unpaired electrons.
$[Fe(\text{en})(\text{bpy})(NH_{3})_{2}]^{2+}$
$\text{Fe}^{2+}$ ($3d^{6}$) in the presence of strong-field ligands (en, bpy, $NH_{3}$) can lead to a low-spin complex. This low-spin configuration usually results in either no unpaired electrons or fewer unpaired electrons than a high-spin case. The given information suggests it has 0 unpaired electrons.
Step 5: Compare and Conclude
Among the complexes, $[Co(\text{ox})_{2}(\text{OH})_{2}]^{-}$ with 4 unpaired electrons has the highest paramagnetism. Hence, it exhibits the maximum paramagnetic behavior.
