The element that can exhibit the highest number of oxidation states among the following is:

Mn

The correct answer is option 1. Mn.

Let us delve into why manganese (Mn) can exhibit the highest number of oxidation states among the given elements: Mn, Co, V, and Ni.

Electronic Configurations and Oxidation States

Manganese (Mn):

Atomic Number: 25

Ground State Electronic Configuration: \([Ar] 3d^5 4s^2\)

Oxidation States: \(+2, +3, +4, +5, +6, +7\)

Manganese has a partially filled 3d subshell with five unpaired electrons and two electrons in the 4s orbital. This allows manganese to lose various numbers of electrons, leading to multiple stable oxidation states. Each oxidation state corresponds to different chemical species:

+2: \(Mn^{2+}\) (common in many manganese compounds)

+3: \(Mn^{3+}\) (in some oxides and fluorides)

+4: \(MnO_2\) (manganese dioxide)

+5: \(MnO_4^{3-}\)  (in some oxoanions)

+6: \(MnO_4^{2-}\) (in some oxoanions)

+7: \(MnO_4^-\) (permanganate ion, a strong oxidizing agent)

Cobalt (Co):

Atomic Number: 27

Ground State Electronic Configuration: \([Ar] 3d^7 4s^2\)

Oxidation States: +2, +3

Cobalt typically exhibits oxidation states of +2 \((Co^{2+})\) and +3 \((Co^{3+})\). The +2 state is the most stable and common, while the +3 state is less common but can be found in some complex compounds.

Vanadium (V):

Atomic Number: 23

Ground State Electronic Configuration: \([Ar] 3d^3 4s^2\)

Oxidation States: \(+2, +3, +4, +5\)

Vanadium can lose electrons from both the \(3d\) and \(4s\) orbitals, leading to multiple oxidation states:

+2: \(V^{2+}\) (in some vanadium salts)

+3: \(V^{3+}\) (in some oxides and complexes)

+4: \(VO^{2+}\) (vanadyl ion)

+5: \(VO_4^{3-}\) (vanadate ion)

Nickel (Ni):

Atomic Number: 28

Ground State Electronic Configuration: \([Ar] 3d^8 4s^2\)

Oxidation States: \(+2, +3\) (less common)

Nickel primarily exhibits the \(+2\) oxidation state \((Ni^{2+})\), which is the most stable and common. The \(+3\) state \((Ni^{3+})\) is rare and usually found in specific coordination complexes or oxides.

Summary:

The variety of oxidation states for each element depends on its electronic configuration and the ability to lose electrons from both the \(3d\) and \(4s\) orbitals. Manganese, with its half-filled \(d^5\) configuration, has the greatest number of stable oxidation states (from +2 to +7) due to the relative ease of losing different numbers of electrons from the \(3d\) and \(4s\) orbitals. This makes it capable of exhibiting a wider range of chemical behaviors and compounds. Therefore, the element that can exhibit the highest number of oxidation states is \(Mn\) (Manganese).