Question

Why is $Zn$ not a transition metal, whereas silver is a transition element? Which bivalent cation in $3d$ transition series is most paramagnetic and why?

Answer 1

Zinc is the second-most-abundant trace mineral presence in the body with atomic number $30$. And it is a d-block element occurring in ${12^{th}}$ group and the ${4^{th}}$ period. A transition element is a metallic element with variable valency and strong tendency to form coordination compounds and most of the compounds are coloured. The bivalent cation is also called divalent cation having $+ 2$ charge.

Complete answer:
We have to remember that zinc is not a transition element. Because, in the transition element, the valence electrons are occupied in d-orbital. The transition metal properties occur because of the presence of incomplete or partially filled d-orbital and their oxidation state.
In the case of silver and zinc, both have completely filled d-orbitals. The electronic configuration of $Ag$ and $Zn$ can be written as,
$Ag = \left[ {Kr} \right]4{d^{10}}5{s^1}$
$Zn = \left[ {Ar} \right]3{d^{10}}4{s^2}$
And by $+ 2$ oxidation state of silver, it can attain partially filled d-orbitals, but zinc cannot attain this state by any of the oxidation states. Hence, zinc is not considered as a transition metal.
The compound with the highest number of unpaired electrons is the most paramagnetic element. Among 3d transition series, $M{n^{2 + }}$ions have the highest number of unpaired electrons, which is equal to five. The electronic configuration of $M{n^{2 + }}$ is equal to, $\left[ {Ar} \right]3{d^5}$. Here, d-orbital contains five unpaired electrons. Hence, $M{n^{2 + }}$ is the most paramagnetic bivalent cation among $3d$ transition series.

Note:
We must know that zinc is not considered as a transition metal, because it does not have a partially filled d-orbital. But silver is a transition metal. Because, by the removal of two electrons from its outer orbitals, it attains $A{g^{2 + }}$oxidation state. And the d-orbital becomes partially filled. Among 3d transition series, $M{n^{2 + }}$is the most paramagnetic bivalent cation due to the presence of the highest number of unpaired electrons.