The one that can exhibit highest paramagnetic behaviour among the following is: gly = glycinato; bpy = 2, 2'-bipyridine

First we recall that the magnitude of paramagnetism of a coordination compound is directly related to the number of unpaired electrons present in the metal ion. More unpaired electrons → larger magnetic moment → stronger paramagnetic behaviour.

Therefore, for every complex given in the options we will find

   • the oxidation state of the central metal,

   • the d-electron count,

   • whether the complex is high-spin or low-spin,

   • the exact number of unpaired electrons.

After that we will compare the numbers of unpaired electrons to decide which complex shows the highest paramagnetism.

----------------------------------------------------------

Option A : $$[Pd(gly)_2]$$

Each glycinato ligand is bidentate with charge $$-1$$. Two such ligands give a total ligand charge of $$-2$$. The complex is neutral, so let the oxidation state of Pd be $$x$$. We write

$$x + (-2) = 0 \;\;\Rightarrow\;\; x = +2$$

Hence the metal centre is $$\text{Pd}^{2+}$$. Palladium has atomic number $$46$$, electronic configuration $$[Kr]\,4d^{10}\,5s^{0}$$. Removing two electrons gives

$$\text{Pd}^{2+} : [Kr]\,4d^{8}$$

A $$d^8$$ platinum-group metal ion almost always forms a square-planar complex. In a square-planar field all eight electrons pair up in the lower energy orbitals, giving

number of unpaired electrons $$= 0$$.

So $$[Pd(gly)_2]$$ is diamagnetic.

----------------------------------------------------------

Option B : $$[Fe(en)(bpy)(NH_3)_2]^{2+}$$

The ligands ethylenediamine (en), 2,2'-bipyridine (bpy) and ammonia ($$NH_3$$) are all neutral. The overall charge is $$+2$$, therefore

$$x + 0 = +2 \;\;\Rightarrow\;\; x = +2$$

so the metal ion is $$\text{Fe}^{2+}$$. Iron has atomic number $$26$$, ground-state configuration $$[Ar]\,3d^{6}\,4s^{2}$$. Removing two electrons (from 4s first) gives

$$\text{Fe}^{2+} : [Ar]\,3d^{6}$$

bpy is a strong π-acceptor, en is a reasonably strong field ligand, and even $$NH_3$$ is of moderate field strength. Together they create a large octahedral crystal field splitting $$\Delta_0$$. For $$d^{6}$$, when $$\Delta_0 > P$$ (pairing energy), a low-spin configuration is adopted:

$$t_{2g}^6\,e_g^0$$

All six electrons are paired, so

number of unpaired electrons $$= 0$$.

Thus this complex is also diamagnetic.

----------------------------------------------------------

Option C : $$[Co(OX)_2(OH)_2]^- \quad (\Delta_0 > P)$$

Here $$OX^{2-}$$ is the oxalato ligand and $$OH^-$$ is hydroxide.

Total ligand charge $$= 2(-2) + 2(-1) = -6$$.

The overall charge of the complex is $$-1$$, so

$$x + (-6) = -1 \;\;\Rightarrow\;\; x = +5$$

Therefore the metal ion is $$\text{Co}^{5+}$$, which is a $$d^{4}$$ system because cobalt (atomic number $$27$$) loses five electrons:

$$\text{Co} : [Ar]\,3d^{7}\,4s^{2}$$

After removing five electrons → $$[Ar]\,3d^{4}$$.

The statement $$\Delta_0 > P$$ explicitly tells us that the complex is low-spin. For an octahedral low-spin $$d^{4}$$ ion, the electron filling is

$$t_{2g}^4\,e_g^0$$

We put the first three electrons singly into the three $$t_{2g}$$ orbitals, and the fourth electron pairs up in one of them (because pairing is preferred to promotion when $$\Delta_0 > P$$). Hence the occupation looks like

  • one $$t_{2g}$$ orbital has ↑↓ (paired)
  • the other two $$t_{2g}$$ orbitals have ↑ (single unpaired each)

So the total number of unpaired electrons is

$$n = 2$$

and the complex is paramagnetic.

----------------------------------------------------------

Option D : $$[Ti(NH_3)_6]^{3+}$$

All six $$NH_3$$ ligands are neutral, so

$$x + 0 = +3 \;\;\Rightarrow\;\; x = +3$$

Thus the metal ion is $$\text{Ti}^{3+}$$. The ground-state configuration of Ti is $$[Ar]\,3d^{2}\,4s^{2}$$; removing three electrons gives

$$\text{Ti}^{3+} : [Ar]\,3d^{1}$$

This is a $$d^{1}$$ octahedral ion. No matter whether the ligand field is weak or strong, the single electron lives in a $$t_{2g}$$ orbital, so

number of unpaired electrons $$= 1$$.

----------------------------------------------------------

Now we summarise the unpaired electrons in each option:

Option A : 0    Option B : 0    Option C : 2    Option D : 1

The largest value is $$2$$, occurring in Option C.

Hence, the correct answer is Option C.