Question

Which among the following cations will form lowest stability complex if the ligand remains the same?

• A. $\mathrm{Cu^{2+}}$
• B. $\mathrm{Fe^{2+}}$
• C. $\mathrm{Cd^{2+}}$
• D. $\mathrm{Ni^{2+}}$

Hint:

Remember the Irving-Williams series for $3d$ metals, and keep in mind that larger down-group elements like $\mathrm{Cd^{2+}}$ have a lower charge-to-size ratio, making their complexes inherently less stable than their smaller $3d$ counterparts.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to identify the divalent metal cation that forms the least stable coordination complex when bonded to the same ligand.

Step 2: Key Formula or Approach:
For divalent transition metal ions of the first row, the stability of complexes follows the Irving-Williams series: $$\mathrm{Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}}$$ This stability order is primarily governed by the decrease in ionic radius and the increase in crystal field stabilization energy (CFSE). When comparing first-row transition metals with post-transition or second-row elements like $\mathrm{Cd^{2+}}$, we look at the charge-to-size ratio (ionic potential).

Step 3: Detailed Explanation:
Among the given options, $\mathrm{Cu^{2+}}$, $\mathrm{Ni^{2+}}$, and $\mathrm{Fe^{2+}}$ belong to the $3d$ transition series, whereas $\mathrm{Cd^{2+}}$ belongs to the $4d$ series.
The ionic radius of $\mathrm{Cd^{2+}}$ is significantly larger than the radii of the $3d$ divalent cations due to the presence of an extra electron shell.
Since the charge on all the given ions is $+2$, the larger size of $\mathrm{Cd^{2+}}$ results in a much lower charge-to-size ratio (ionic potential).
A lower ionic potential means weaker electrostatic attraction between the metal cation and the lone pairs of the ligand, leading to a significantly less stable complex.
Therefore, the stability order is: $\mathrm{Cu^{2+} > Ni^{2+} > Fe^{2+} > Cd^{2+}}$.

Step 4: Final Answer:
The cation forming the lowest stability complex is $\mathrm{Cd^{2+}}$, which corresponds to option (C).