Question

Why is $ Co(CN)_5^{2 - } $ paramagnetic?

Answer 1

Hint: In order to find the magnetic nature (paramagnetic or diamagnetic) of the given compound, we will first write the electronic configuration of the central metal atom or ion. Then we will observe this electronic configuration. If the central metal ion contains at least one unpaired electron then, the compound will be paramagnetic and if there are no unpaired electrons, then the compound will be diamagnetic.

Complete Step By Step Answer:
We should know that a compound can be paramagnetic or diamagnetic based on the electronic configuration of the central metal ion in the coordination complex. The given coordination complex is $ Co(CN)_5^{2 - } $ .
Now, we will first find the oxidation state of Cobalt (Co) in this complex by the substitution method. Let us assume that the oxidation state of Co be x and that of cyanide ion is $ - 1. $ We will now find the oxidation state of cobalt ion in the given coordination complex. So, now we will use the substitution method as follows:
 $ x + 5( - 1) = - 2 $
 $ x = + 3 $
$Co(CN)_5^{2 - } $
Therefore, the oxidation state of Cobalt (Co) in this complex is $ + 3 $ and now, we will write the electronic configuration of the cobalt ion.
Now, the electronic configuration of $ C{o^{3 + }} $ is given as:
 $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6} $
So, according to this electronic configuration, $ C{o^{3 + }} $ will be paramagnetic.
Hence, the given complex is paramagnetic.

Note:
We should remember that the paramagnetic materials or substances are weakly magnetized under the action of an external magnetic field while the diamagnetic materials or substances are opposed to magnetism under an external magnetic field in the same direction as the applied magnetic field.