Question

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

• A. \( V^{2+}<Cr^{2+}<Fe^{2+}<Mn^{2+} \): paramagnetic behaviour
• B. \( Ni^{2+}<Co^{2+}<Fe^{2+}<Mn^{2+} \): ionic size
• C. \( Co^{3+}<Fe^{3+}<Cr^{3+}<Sc^{3+} \): stability in aqueous solution
• D. \( Sc<Ti<Cr<Mn \): number of oxidation states

Hint:

The paramagnetic nature of transition metal ions is determined by the number of unpaired electrons. The correct order for paramagnetic behavior should be \( V^{2+}<Cr^{2+}<Fe^{2+}<Mn^{2+} \), but Fe\(^{2+} \) has fewer unpaired electrons than Mn\(^{2+}\), making the given order incorrect.

Answer 1

The Correct Option is A

Let's analyze the given options:Paramagnetic behaviour (Option a): 
The number of unpaired electrons determines paramagnetism. The electronic configurations are: \[ V^{2+} = 3d^3, \quad Cr^{2+} = 3d^4, \quad Fe^{2+} = 3d^6, \quad Mn^{2+} = 3d^5 \] 
Since Mn\(^{2+}\) has the maximum unpaired electrons (5), it should be the most paramagnetic. However, Fe\(^{2+}\) has only 4 unpaired electrons, meaning Mn\(^{2+}\) should be more paramagnetic than Fe\(^{2+}\). The given order \( V^{2+}<Cr^{2+}<Fe^{2+}<Mn^{2+} \) is incorrect.  Ionic size (Option b): 
Across a period, ionic size decreases due to increasing nuclear charge. The given order \( Ni^{2+}<Co^{2+}<Fe^{2+}<Mn^{2+} \) is correct because Mn\(^{2+}\) is the largest and Ni\(^{2+}\) is the smallest.  Stability in aqueous solution (Option c): 
The stability of transition metal ions in aqueous solution generally increases as the oxidation state increases. The given order \( Co^{3+}<Fe^{3+}<Cr^{3+}<Sc^{3+} \) is correct.  Oxidation states (Option d): 
The correct order of oxidation states is: \[ Sc (1)<Ti (2,3,4)<Cr (2,3,4,5,6)<Mn (2,3,4,5,6,7) \] 
Since Mn shows the highest number of oxidation states, the given order \( Sc<Ti<Cr<Mn \) is correct. 

Final Answer: Option (A) does not represent the correct order.