Question

Which among the given ions form complex compounds that are the most stable?
(a) $C{u^{ + + }}$
(b) $N{i^{ + + }}$
(c) $F{e^{ + + }}$
(d) $M{n^{ + + }}$

Answer 1

Hint : Complex compounds or coordination complexes are the chemical compositions which have a central atom that would be a metal, along with non-metallic or neutral atoms bound to this central atom. Remember that the atoms that are attached to this central atom are called ligands.

Complete Step By Step Answer:
A coordination or complex compound is actually a chemical compound where the central atom or ion is attached to a certain number of ions, molecules, or atoms. The central ions or atoms, within any coordination compound, are usually a Lewis Acid which serve as the acceptors of pairs of electrons.
Here the point we need to note is that stability of a complex compound comes from the charge density of the central ions. When this charge density goes higher it means that the metallic lattice is of higher strength. This charge density should be high for the central metal ion in order to end up with the most stable complex or coordination compound.
We see from the options that we are given four divalent (two is the valency) metal ions. They all belong to the same transition series which is the first transition series. Here if we consider in general, that the ligands (surrounding atoms of the central atom) will be nitrogen or oxygen then, the stability magnitude increases in this order:
$M{n^{ + + }} < F{e^{ + + }} < N{i^{ + + }} < C{u^{ + + }}$
Since the stability increases as charge density increases, the order of charge densities of the given metallic atoms are the same as given above.
Look at each option;
Option (a) $C{u^{ + + }}$ is correct; since in the increasing order of charge densities, $C{u^{ + + }}$ takes the highest value. So as this charge density increases, it is able to form the most stable complex compound.
Option (b) $N{i^{ + + }}$ is incorrect; since it has a charge density less than $C{u^{ + + }}$ from the given options, so it will form a comparatively stable compound but it will not be the most stable.
Option (c) $F{e^{ + + }}$ is incorrect; since it is second in the increasing order of charge densities, it will not be the most stable among the given options.
Option (d) $M{n^{ + + }}$ is incorrect; since it has the least charge density among the given options, it cannot form the most stable complex compound.
Therefore the right option of the ion that forms the most stable compound will be option (a) $C{u^{ + + }}$ .

Note :
Transition metals are those which have electrons in their outer shell which are unpaired. The existence of unpaired electrons in such transitions metals can absorb light while reacting with other atoms within complex compounds, giving color to such compounds.