Animated Solution for Chemistry - Coordination Compounds: The one that can exhibit highest paramagnetic behaviour among the following is $\text{gly}=\text{glycinato}$; $\text{bpy}=2,2^{\prime }\text{-bipyridine}$

Paramagnetism and Unpaired Electrons

When we talk about the magnetic properties of coordination complexes, the most crucial factor is the number of unpaired electrons. Paramagnetism arises directly from these unpaired spins. The magnetic moment, denoted by $\mu$, is calculated using the spin-only formula:

where $n$ is the number of unpaired electrons. To find the complex with the highest paramagnetic behaviour, our mission is simple: determine the electronic configuration of the central metal ion in each complex and count the unpaired electrons!

Let's start with the first option: $[\text{Pd}(\text{gly}{)}_2]$.

Here, the central metal is Palladium (Pd). Since glycinato (gly) has a $-1$ charge, Palladium must be in a $+2$ oxidation state. The electronic configuration of ${\text{Pd}}^{2+}$ is a $4d^8$ system.

A golden rule of coordination chemistry: Metals from the $4d$ and $5d$ series almost exclusively form low-spin, square planar complexes when they have a $d^8$ configuration. Because it's low-spin, all 8 electrons will pair up in the lower energy orbitals.

Therefore, the number of unpaired electrons ($n$) is $0$. This complex is diamagnetic.

Moving on to the second complex: $[\text{Fe}(\text{en})(\text{bpy})({\text{NH}}_3{)}_2{]}^{2+}$.

Iron (Fe) is in a $+2$ oxidation state, making it a $3d^6$ system. Now, look at the ligands surrounding it: ethylenediamine (en), bipyridine (bpy), and ammonia (${\text{NH}}_3$). All of these are strong field ligands!

Strong field ligands cause a large crystal field splitting (${\Delta }_o>P$), forcing the electrons to pair up in the lower $t_{2g}$ orbitals rather than occupying the higher $e_g$ orbitals. The configuration becomes $t_{2g}^6{e}_g^0$.

Once again, all electrons are paired. $n=0$, and the complex is diamagnetic.

Next, we examine $[\text{Co}(\text{OX}{)}_2(\text{OH}{)}_2{]}^{-}$.

Cobalt (Co) is in a $+3$ oxidation state, which is another $3d^6$ system. The problem explicitly gives us a massive hint: $({\Delta }_o>P)$.

This condition literally defines a low-spin complex! It tells us that the crystal field splitting energy (${\Delta }_o$) is greater than the pairing energy ($P$). The electrons will take the path of least resistance and pair up in the $t_{2g}$ level.

Just like the iron complex, the configuration is $t_{2g}^6{e}_g^0$. All electrons are paired, so $n=0$. This complex is also diamagnetic.

Finally, let's look at $[\text{Ti}({\text{NH}}_3{)}_6{]}^{3+}$.

Titanium (Ti) has an atomic number of 22. In a $+3$ oxidation state, it loses three electrons (two from the $4s$ and one from the $3d$), leaving it as a $3d^1$ system.

With only one electron in the $d$-orbitals, it will occupy one of the lower $t_{2g}$ orbitals. The configuration is $t_{2g}^1{e}_g^0$.

Since there is only one electron, it is impossible for it to pair up! Therefore, we have exactly $1$ unpaired electron ($n=1$).

Comparing our findings: - $[\text{Pd}(\text{gly}{)}_2]$ has $0$ unpaired electrons. - $[\text{Fe}(\text{en})(\text{bpy})({\text{NH}}_3{)}_2{]}^{2+}$ has $0$ unpaired electrons. - $[\text{Co}(\text{OX}{)}_2(\text{OH}{)}_2{]}^{-}$ has $0$ unpaired electrons. - $[\text{Ti}({\text{NH}}_3{)}_6{]}^{3+}$ has $1$ unpaired electron.

Since $[\text{Ti}({\text{NH}}_3{)}_6{]}^{3+}$ is the only complex with an unpaired electron, it is the only paramagnetic species among the choices. Consequently, it exhibits the highest paramagnetic behaviour.