Idea used

In transition elements, the element showing the maximum number of oxidation states is generally the one with a partially filled $d$-subshell and favorable electronic configuration.

Among the first transition series, manganese is known to show the largest number of oxidation states.

Check each option

• Ti: common oxidation states are $+2, +3, +4$
• Fe: common oxidation states are $+2, +3$; higher states like $+6$ exist in ferrate, but the number is still less than Mn
• Co: common oxidation states are $+2, +3$; some higher states exist, but not as many as Mn
• Mn: shows oxidation states from $+2$ to $+7$ commonly, such as $+2, +3, +4, +5, +6, +7$ Hence it has the maximum variety of oxidation states among the given metals.

Why Mn shows many oxidation states

Manganese has electronic configuration: $[Ar] \, 3d^5 4s^2$ Since both $4s$ and $3d$ electrons can participate in bonding, Mn can lose different numbers of electrons and attain many oxidation states.

Conclusion

Therefore, the metal that shows the highest and maximum number of oxidation states is: $\boxed{\text{Mn}}$

Comparison with stored answer

Stored correct answer is B, which corresponds to Mn. This matches our derived answer.