Question

${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$ is an diamagnetic complex because?
A. ${C_2}{O_4}^{2 - }$ is a strong field ligand hence causes pairing of electrons
B. ${C_2}{O_4}^{2 - }$ is a bidentate ligand hence causes pairing of electrons
C. $Co$ is a strong central atom hence in all complexes of Co electrons are paired
D. ${C_2}{O_4}^{2 - }$ is a strong field ligand hence causing splitting of d-orbitals

Answer 1

We can predict the magnetic nature of the complex by knowing the nature of the ligand. The nature of the ligand decides the pairing of electrons and the type of magnetic behaviour could be predicted.

Complete step by step answer:
We can say a strong ligand or a strong field ligand is a ligand that could result in a higher crystal field splitting.
We can say a weak ligand or a weak field ligand is a ligand that could result in a lower crystal field splitting.
The magnetic nature of a complex is dependent on the number of unpaired electrons in the complex.
We have to know the strong field ligand leads to pairing of electrons and diamagnetic behaviour is favored and a weak field ligand does not lead to pairing of electrons and paramagnetic behaviour is favored.
We know that in ${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$, the central metal atom is cobalt. The electronic configuration of cobalt 3+ is $\left[ {Ar} \right]3{d^6}$ and ${\left( {{C_2}{O_4}} \right)^{2 - }}$ is the ligand.
In the complex ${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$, the ligand ${C_2}{O_4}$ leads to pairing of unpaired electrons in d-orbitals and so, there is no unpaired electron.
We know that ${\left( {{C_2}{O_4}} \right)^{2 - }}$ is a strong field ligand and it results in pairing of electrons. Therefore, ${\left[ {Co{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}$ is a diamagnetic complex.

Therefore, the option A is correct.

Note:
The binding of a strong field ligand causes a higher difference between the higher and lower energy level orbitals. Examples: $C{N^-}$ (cyanide ligands), $N{O_2}^-$ (nitro ligand) and $CO$(carbonyl ligands)
The binding of a weak field ligand leads a lower difference between the higher and lower energy level orbitals because the low difference between the two orbital levels leads to repulsions between electrons present in those energy levels, the higher energy orbitals could be easily filled with electrons when compared to electrons present in low energy orbitals. Examples: ${I^-}$ (iodide ligand), $B{r^-}$ (bromide ligand), etc.