The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.
The electronic configuration shows the number of electrons available for donation or the number of empty orbitals to accept electrons.
Option (i) has a 3s1 configuration which allows the donation of 1 electron is possible to gain +1 oxidation number. Also it can accept on electron and gain ON = -1, but being an s block element, alkali earth metals, are highly resistive to reduction by accepting electron. Also the have very high negative electrode potential suggesting that it can only donate one electron having only one oxidation state.
In Option (ii) the given configuration is for element Scandium. Now we consider the outer s orbital which can give 2 electrons and exhibit an oxidation state of +2, the following lone d electron can be further donated in order to make the d-orbital empty thus attaining a lower energy state. This removal of this electron gives an oxidation state of +3. Thus Sc exhibits +2, +3 oxidation state.
Option (iii) similar to the said example the subsequent valence orbital of 4s can lose two electrons giving a +2 ON, while the d-orbital electron is given away in the presence of a strong oxidizing agent. Hence Ti can exhibit three Oxidation states based on configuration i.e. +2, +3, +4.
Option (iv) this configuration suggests that, there are 3 empty p orbitals that can accommodate 3 foreign electrons thus implying a -3 ON, whereas it can also donate those 3 orbital to acquire lower energy giving 3+ ON. The inner 3s orbital can loose electrons to further give an oxidation state of +5.
The element with 3s2 3p3configuration is phosphorous, which is seen to exhibit many compounds of hydrides and oxides, which have large application in fertilizer industry. So study of oxidation number plays an important role.