Which of the following arrangements does not represent the correct order of the property stated against it?

We shall examine every arrangement one by one, always keeping in mind the accepted principles of transition-metal chemistry, and then decide which sequence fails to follow the principle stated.

First, recall that whenever we compare any property across the first transition series (Sc to Zn) we must look at the factors that actually control that property, for example

$$\bullet$$ stability of the hexaaqua ions $$[M(H_2O)_6]^{3+}$$ depends largely on hydration energy, crystal-field stabilisation energy (CFSE) and the tendency of the metal ion to undergo hydrolysis.

$$\bullet$$ the number of oxidation states shown by an element is mainly governed by the number of 3d and 4s electrons that can take part in bonding.

$$\bullet$$ paramagnetic behaviour depends purely on the number $$n$$ of unpaired d-electrons, the spin-only magnetic moment being $$\mu = \sqrt{n(n+2)}\; \text{BM}.$$

$$\bullet$$ ionic size decreases from left to right in a period because the effective nuclear charge increases while the principal quantum number remains the same.

Now we test every option.

Option A : $$Co^{3+} < Fe^{3+} < Cr^{3+} < Sc^{3+}$$ - stability of the hexaaqua ions in water

$$[Co(H_2O)_6]^{3+}$$ readily oxidises water and is therefore least stable, while $$[Sc(H_2O)_6]^{3+}$$ has no 3d electrons and cannot undergo $$d \rightarrow d$$ transitions, giving it very high CFSE in the form of pure electrostatic attraction. Thus the stability rises in the direction shown. So this order is correct.

Option B : $$Sc < Ti < Cr < Mn$$ - number of oxidation states

Write explicitly the common oxidation states:

$$\begin{aligned} Sc &: +3 \quad (1\ \text{state})\\[2pt] Ti &: +2,\; +3,\; +4 \quad (3\ \text{states})\\[2pt] Cr &: +2,\; +3,\; +6,\; (+1,\; +4) \quad (\ge 4\ \text{states})\\[2pt] Mn &: +2,\; +3,\; +4,\; +5,\; +6,\; +7 \quad (6\ \text{states}) \end{aligned}$$

Clearly the counts rise as $$1 < 3 < 4 < 6$$, matching the given order. Hence this arrangement is also correct.

Option C : $$V^{2+} < Cr^{2+} < Mn^{2+} < Fe^{2+}$$ - paramagnetic behaviour

We must determine the spin‐only magnetic moment of each divalent ion. First write the d-electron configuration after losing two electrons (4s and possibly 3d):

$$\begin{aligned} V^{2+} &: [Ar]\;3d^3 \quad (n = 3\ \text{unpaired})\\[2pt] Cr^{2+} &: [Ar]\;3d^4 \quad (n = 4\ \text{unpaired})\\[2pt] Mn^{2+} &: [Ar]\;3d^5 \quad (n = 5\ \text{unpaired})\\[2pt] Fe^{2+} &: [Ar]\;3d^6 \quad (n = 4\ \text{unpaired, high-spin}) \end{aligned}$$

Using the spin-only formula $$\mu = \sqrt{n(n+2)}$$ we obtain

$$\begin{aligned} V^{2+} &: \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}\\[2pt] Cr^{2+} &: \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM}\\[2pt] Mn^{2+} &: \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}\\[2pt] Fe^{2+} &: \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM} \end{aligned}$$

So the true order of increasing paramagnetism should be

$$V^{2+} < Cr^{2+} \approx Fe^{2+} < Mn^{2+}$$

but the option given places $$Fe^{2+}$$ after $$Mn^{2+}$$, implying it has the highest magnetic moment, which is wrong because $$Fe^{2+}$$ (4 unpaired) is less paramagnetic than $$Mn^{2+}$$ (5 unpaired). Therefore this arrangement is incorrect.

Option D : $$Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}$$ - ionic size

Across the period, the effective nuclear charge increases, pulling the 3d electrons closer and shrinking the ionic radius. Thus the radius drops from left to right: $$Mn^{2+} > Fe^{2+} > Co^{2+} > Ni^{2+}$$. Re-writing this in ascending order we get exactly the sequence shown. Hence this option is also correct.

We have found that Options A, B and D follow the accepted chemical principles, whereas Option C contradicts the actual magnetic moments of the ions.

Hence, the correct answer is Option C.