Question

Highest (+7) oxidation state is shown by:
A) Co
B) Cr
C) V
D) Mn

Answer 1

Transition metal elements which make the group III to group XII are known as the Transition metal elements. As the name suggests, these elements have a wide range of transitions in their oxidation states. The transition metals reside in the d-block of the periodic table.

Complete answer:
Most of the transition elements exhibit multiple oxidation states, since it is easy to lose electrons for transition compounds compared to the alkali metals or alkaline earth metals. Transition metals are more complex and exhibit a wide range of oxidation states due to the removal of electrons from the d-orbitals.
Out of all the transition metal elements, Manganese which is present in the middle of the transition series exhibits the highest number of oxidation states since it has five unpaired electrons in its d-orbital. Other elements like scandium, chromium show one or two variable oxidation states only. Manganese exhibits a variable oxidation states of: $+ 2, + 3, + 4, + 6, + 7$
Mn atom in the permanganate ion shows +7 oxidation state. The oxygen in it shows -2 oxidation state. Since there are 4 oxygen atoms the overall charge on oxygen becomes -8. In addition, the overall charge of the ion is -1. To compensate for all the charges Mn must have a O.S of +7, which gives us $M{n^{ + 7}}\& 4{O^{2 - }}$ giving us the resulting $MnO_4^ -$ . Mn forms multiple $p\pi - d\pi$ bonds using the 2p orbital of oxygen.

Hence the correct answer is Option (D).

Note:
Mn shows an oxidation state of +4 generally, because of single bond formation, which is due to the unavailability of 2p orbitals. For example: In Mn compounds with F, Fluorine doesn’t have p-orbitals. Hence Mn shows +4 Oxidation state there.