This question tests your understanding of periodic trends and properties of transition metal ions. Let's analyze each option step by step to identify which arrangement does NOT correctly represent the stated property.

Option 1: Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺ : Ionic Size

Ionic size generally increases down a group and decreases across a period due to increasing effective nuclear charge. For isoelectronic ions (same number of electrons), size decreases with increasing atomic number. All these ions have the electronic configuration [Ar] 3dⁿ.

Atomic numbers: Ni (28), Co (27), Fe (26), Mn (25). Since all are M²⁺ ions, ionic radius should increase as atomic number decreases. The correct order should be: Ni²⁺ (69 pm) < Co²⁺ (74 pm) < Fe²⁺ (78 pm) < Mn²⁺ (83 pm). This arrangement is correct.

Option 2: V¹⁺ < Cr²⁺ < Mn²⁺ < Fe²⁺ : Paramagnetic Behaviour

Paramagnetic behaviour depends on the number of unpaired electrons. More unpaired electrons means greater paramagnetism.

Let's calculate unpaired electrons:

• V¹⁺ (Vanadium +1): [Ar] 3d⁴ → 4 unpaired electrons
• Cr²⁺ (Chromium +2): [Ar] 3d⁴ → 4 unpaired electrons
• Mn²⁺ (Manganese +2): [Ar] 3d⁵ → 5 unpaired electrons
• Fe²⁺ (Iron +2): [Ar] 3d⁶ → 4 unpaired electrons

So the number of unpaired electrons: V¹⁺ = Cr²⁺ = 4 < Mn²⁺ = 5 > Fe²⁺ = 4. Therefore, the order of paramagnetic behaviour should be: V¹⁺ ≈ Cr²⁺ < Mn²⁺ > Fe²⁺. The given order V¹⁺ < Cr²⁺ < Mn²⁺ < Fe²⁺ is incorrect because Fe²⁺ has fewer unpaired electrons than Mn²⁺.

Option 3: Sc < Ti < Cr < Mn : Number of Oxidation States

Transition metals exhibit various oxidation states. Generally, the number of oxidation states increases across the period until the middle, then decreases.

Common oxidation states:

• Sc: +3
• Ti: +2, +3, +4
• Cr: +2, +3, +6
• Mn: +2, +3, +4, +6, +7

So the number of oxidation states: Sc (1) < Ti (3) < Cr (3) < Mn (5). This order is correct.

Option 4: Co³⁺ < Fe³⁺ < Cr³⁺ < Sc³⁺ : Stability in Aqueous Solution

Stability in aqueous solution for M³⁺ ions is related to their hydration energy and lattice energy. Generally, smaller ions with higher charge are more stable due to greater hydration energy. The order of stability is often: Sc³⁺ < Cr³⁺ < Fe³⁺ < Co³⁺? Actually, Co³⁺ is unstable in water and oxidizes water, while Cr³⁺ is very stable. The correct order of stability is: Sc³⁺ < Fe³⁺ < Cr³⁺ (Co³⁺ is least stable). So the given order Co³⁺ < Fe³⁺ < Cr³⁺ < Sc³⁺ is incorrect, but we already found Option 2 is incorrect. However, Option 2 is clearly wrong based on unpaired electrons, making it the primary incorrect arrangement.

Final Answer

The arrangement that does not represent the correct order is Option 2: V¹⁺ < Cr²⁺ < Mn²⁺ < Fe²⁺ for paramagnetic behaviour, because Fe²⁺ has fewer unpaired electrons (4) than Mn²⁺ (5), so it should be less paramagnetic.

Related Topics & Formulae

Paramagnetism: Substances with unpaired electrons are attracted to a magnetic field. The magnetic moment is given by $\mu =\sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons.

Ionic Size: For isoelectronic ions, size decreases with increasing nuclear charge: $r_{\text{ionic}}\propto \frac{1}{Z}$.

Oxidation States: Depend on the number of valence electrons and stability of half-filled or fully-filled orbitals.

Stability in Solution: Influenced by hydration energy, charge density, and tendency to oxidize/reduce.