1. We need the arrangement that is not correct.
   So we check each option against the stated property.

2. Option A:
   $Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}$ Property: ionic size

Across the first transition series, for ions of the same charge $M^{2+}$, ionic radius generally decreases from left to right because effective nuclear charge increases.

Thus: $Mn^{2+} > Fe^{2+} > Co^{2+} > Ni^{2+}$ So in increasing order: $Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}$ This is correct.

3. Option B:
   $Co^{3+} < Fe^{3+} < Cr^{3+} < Sc^{3+}$ Property: stability in aqueous solution

In aqueous solution, $M^{3+}$ ions are stabilized when they have favorable $d$-electron configurations and high hydration enthalpy.
Among these:

• $Sc^{3+}$ is $3d^0$ and quite stable
• $Cr^{3+}$ is $3d^3$ and very stable
• $Fe^{3+}$ is $3d^5$ and stable
• $Co^{3+}$ is less stable in aqueous solution and tends to get reduced to $Co^{2+}$

Hence increasing stability in aqueous solution is: $Co^{3+} < Fe^{3+} < Cr^{3+} < Sc^{3+}$ So this is correct.

4. Option C:
   $Sc < Ti < Cr < Mn$ Property: number of oxidation states

Check the common oxidation states:

• $Sc$: mainly $+3$ → very few
• $Ti$: $+2,+3,+4$
• $Cr$: $+2,+3,+6$ (and others possible)
• $Mn$: shows maximum variety, from $+2$ to $+7$

So the number of oxidation states increases as: $Sc < Ti < Cr < Mn$ This is correct.

5. Option D:
   $V^{2+} < Cr^{2+} < Mn^{2+} < Fe^{2+}$ Property: paramagnetic behaviour

Paramagnetic behaviour depends on the number of unpaired electrons.

Now find configurations:

• $V^{2+}$: $[Ar]3d^3$ → $3$ unpaired electrons
• $Cr^{2+}$: $[Ar]3d^4$ → $4$ unpaired electrons
• $Mn^{2+}$: $[Ar]3d^5$ → $5$ unpaired electrons
• $Fe^{2+}$: $[Ar]3d^6$

For free ion/high spin $3d^6$, $Fe^{2+}$ has 4 unpaired electrons, not 6. So paramagnetism order based on unpaired electrons should be: $V^{2+}(3) < Cr^{2+}(4) < Fe^{2+}(4) < Mn^{2+}(5)$ More accurately, $Cr^{2+}$ and $Fe^{2+}$ have comparable spin-only magnetic moments based on 4 unpaired electrons, and $Mn^{2+}$ should be the highest.

Therefore the given order $V^{2+} < Cr^{2+} < Mn^{2+} < Fe^{2+}$ is incorrect because $Fe^{2+}$ should not be more paramagnetic than $Mn^{2+}$.

6. Conclusion

The arrangement that does not represent the correct order is: $\boxed{D}$