The variety of oxidation states in transition elements arises from the participation of both the $(n-1)d$ and $ns$ electrons in bonding. The element with the maximum number of unpaired electrons in the d-subshell plus electrons in the s-subshell will exhibit the largest number of oxidation states.

1. Analyze Configurations: $3d^3 4s^2$ (Vanadium): $3+2=5$ valence electrons. Max O.S. is +5. $3d^5 4s^1$ (Chromium): $5+1=6$ valence electrons. Max O.S. is +6. $3d^5 4s^2$ (Manganese): $5+2=7$ valence electrons. Max O.S. is +7. $3d^2 4s^2$ (Titanium): $2+2=4$ valence electrons. Max O.S. is +4.

2. Conclusion: Manganese ($3d^5 4s^2$) can utilize all 7 valence electrons, showing oxidation states from +2 to +7, which is the maximum number in the 3d series.