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: Restate the goal.
We must pick the cation that gives the least stable complex when paired with one fixed ligand. Complex stability tracks how strongly the central ion grips a donor lone pair.
Step 2: Set the comparison on a single principle.
For ions of the same charge, the controlling quantity is the ionic potential (polarising power) defined as $\phi = \dfrac{q}{r}$, where $q$ is the charge and $r$ the ionic radius. Every option here is $+2$, so $q$ is constant and only $r$ decides.
Step 3: Separate the ions by series.
$\mathrm{Cu^{2+}}$, $\mathrm{Ni^{2+}}$ and $\mathrm{Fe^{2+}}$ are $3d$ ions, while $\mathrm{Cd^{2+}}$ sits in the $4d$ series with one extra electron shell. An added shell makes $\mathrm{Cd^{2+}}$ noticeably larger.
Step 4: Compare the ionic potentials.
Since $r_{\mathrm{Cd^{2+}}} > r_{\mathrm{Fe^{2+}}} > r_{\mathrm{Ni^{2+}}} > r_{\mathrm{Cu^{2+}}}$ at fixed charge, the polarising power runs the opposite way: $\phi_{\mathrm{Cd^{2+}}}$ is the smallest. \[ \phi = \frac{q}{r} \Rightarrow \text{largest } r \Rightarrow \text{smallest } \phi \]
Step 5: Translate small $\phi$ into low stability.
A weaker pull on the ligand lone pairs means a weaker metal-ligand bond and a smaller formation constant $K_f$. So $\mathrm{Cd^{2+}}$ forms the most loosely held, least stable complex.
Step 6: Conclude.
The stability order is $\mathrm{Cu^{2+}} > \mathrm{Ni^{2+}} > \mathrm{Fe^{2+}} > \mathrm{Cd^{2+}}$, so the lowest-stability complex is from $\mathrm{Cd^{2+}}$, option (C).
\[ \boxed{\mathrm{Cd^{2+}}} \]