Question

Write the condensed ground-state electron configurations of these transition metal ions and the state which is paramagnetic:
(c) Mn2+\mathrm{Mn}^{2+}

Solution

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Let's review the configurations of transition elements and the process of ionization of their atoms, in order to write the condensed configurations of the given ions and determine whether they are paramagnetic or not.

Paramagnetic species\textbf{Paramagnetic species} have unpaired electrons\textit{unpaired electrons}. Therefore, to determine if the given ions are paramagnetic, we should look at their configurations to see if there are any unpaired electrons.

The best way to determine the number of unpaired electrons is to draw the partial valence-level\textit{partial valence-level} orbital diagram of both atoms and ions that will form. For the transition elements, valence electrons\textit{valence electrons} are those in both the outer s\textit{s} and in the d\textit{d} sublevels. The fraction of configuration that is represented by the symbol of the previous noble gas in brackets will always contain only paired\textit{paired} electrons.

To draw a correct orbital diagram and fill the sublevels we have to respect all the rules that apply when filling orbitals, making sure that n\textit{n}s orbital is filled up before (n-1)\textit{(n-1)}d orbitals. Each orbital may contain a maximum of two electrons, the consequence of the exclusion principle. We must also follow Hund's rule, meaning that d orbitals of the same energy\textit{the same energy} are filled each with one electron of parallel spins first, before pairing the electrons up. Be careful because there are some exceptions in the configuration of elements in Groups 6B and 1B.

Now let's remember that when the transition elements form ions, ns\textit{ns} electrons will be lost before (n-1)d\textit{(n-1)d} electrons. Therefore, when we write the condensed configuration of Mn, we will see that in order to form Mn2+^{2+} ion Mn atom has to lose both electrons from 4s orbital, marked in red\color{#c34632}red.

This leaves Mn2+^{2+} ion with 5 unpaired electrons\textit{5 unpaired electrons}, meaning it will be paramagnetic\textbf{paramagnetic}. The condensed configuration of the given ion as well as a partial orbital diagram showing the number of unpaired electrons are represented in the picture below.

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