Question

Which of the following are diamagnetic?
(I)${{K}_{4}}[Fe{{(CN)}_{6}}]$
(II)${{K}_{3}}[Cr{{(CN)}_{6}}]$
(III)${{K}_{3}}[Co{{(CN)}_{6}}]$
(IV)${{K}_{2}}[Ni{{(CN)}_{4}}]$

Answer 1

The answer here includes the calculation of total number of unpaired electrons in the given complex which gives the information about whether the complex is diamagnetic or paramagnetic.

Complete answer:
The concepts of diamagnetism and paramagnetism and also about ferromagnetism are familiar to us from the lower classes of chemistry.
Now we shall see how to calculate the total unpaired electrons in the given complex which helps us to deduce which among them is diamagnetic.
- In the first complex ${{K}_{4}}[Fe{{(CN)}_{6}}]$, the iron atom which is central metal atom ahs the outer electronic configuration as$3{{d}^{6}}4{{s}^{2}}$ and the calculation of charge on this atom is,
$4+x-6=0$
$\Rightarrow x=+2$
Here, the cyanide ligand is strong field ligand and thus the electrons in atomic orbital pair up where, the hybridisation will be ${{d}^{2}}s{{p}^{3}}$ and thus there are no unpaired electrons present in this and hence this complex is diamagnetic.
Similarly if we calculate for the complex ${{K}_{3}}[Cr{{(CN)}_{6}}]$ the complex will have unpaired electrons where this complex is paramagnetic in nature.
In the complex ${{K}_{3}}[Co{{(CN)}_{6}}]$ all the electrons are paired where the complex therefore will be diamagnetic.
In case of the complex ${{K}_{2}}[Ni{{(CN)}_{4}}]$, there is a pairing of electrons here too and thus the complex is diamagnetic in nature.
Thus the correct answer is, the diamagnetic complexes among these will be ${{K}_{4}}[Fe{{(CN)}_{6}}]$,${{K}_{3}}[Co{{(CN)}_{6}}]$ and ${{K}_{2}}[Ni{{(CN)}_{4}}]$

Note:
Note that in the orbital of the central atom where the pairing of electrons takes place or not will depend on the ligands to which the central atom is attached. If the ligands attached are strong field ligands according to the spectrochemical series, then there is pairing up of electrons in the orbital otherwise there will be no pairing for the weak field ligands.