Step 1: Understanding the Concept:
The thermodynamic stability of coordination complexes formed by transition metal ions with a given common ligand depends on several factors, including the charge density of the metal ion, ligand field stabilization energy (CFSE), and the nature of the metal-ligand bond (covalency). For divalent ions of the 3d transition series, the Irving-Williams series provides a very reliable general trend.
Step 2: Key Formula or Approach:
Approach: Apply the Irving-Williams series to firmly establish the high stability of the $\text{Cu}^{2+}$ complex, and then compare the remaining ions ($\text{Mn}^{2+}$ vs $\text{Cd}^{2+}$) based on general principles of coordination chemistry, specifically size, charge, and degree of covalency.
Step 3: Detailed Explanation:
Let's critically analyze the chemical properties of each metal ion provided:
- $\text{Cu^{2+}$ ($3d^9$):} According to the well-established empirical Irving-Williams series ($\text{Ba}^{2+}<\text{Sr}^{2+}<\text{Ca}^{2+}<\text{Mg}^{2+}<\text{Mn}^{2+}<\text{Fe}^{2+}<\text{Co}^{2+}<\text{Ni}^{2+}<Cu^{2+}>\text{Zn}^{2+}$), Copper(II) routinely forms the most stable complexes among all the divalent 3d transition metals with most ligands. This exceptional stability is largely attributed to additional stabilization derived from the Jahn-Teller effect.
- $\text{Mn^{2+}$ ($3d^5$):} This is typically a high-spin $d^5$ system, meaning its Crystal Field Stabilization Energy (CFSE) is exactly zero in weak or moderate ligand fields. Because it completely lacks this extra stabilizing energy and possesses a relatively large ionic radius for its row, it generally forms the least stable complexes among the later 3d metals.
- $\text{Cd^{2+}$ ($4d^{10}$):} Cadmium is a group 12 element, situated immediately below Zinc. Like $\text{Mn}^{2+}$, it possesses a spherically symmetric electron cloud and zero CFSE. However, being a 4d transition metal, it is larger and significantly more polarizable (often termed a "softer" acid) than $\text{Mn}^{2+}$. Because of its higher effective nuclear charge ($Z_{\text{eff}}$) compared to early 3d metals, it tends to form much more covalent, and therefore generally stronger, bonds with many common ligands (especially nitrogen or softer donors) than $\text{Mn}^{2+}$. As a comparative example, the formation constant ($\log K_f$) for the standard EDTA complex of $\text{Cd}^{2+}$ is $\approx 16.5$, which is substantially higher than the corresponding value for $\text{Mn}^{2+}$ ($\approx 13.9$).
Therefore, comparing the three ions, $\text{Cu}^{2+}$ is unequivocally the most stable due to its position in the Irving-Williams series and Jahn-Teller distortion. Between the two ions with zero CFSE, the heavier, more polarizable $\text{Cd}^{2+}$ forms more stable complexes than the lighter, harder $\text{Mn}^{2+}$.
The resulting sequence representing decreasing stability is: $\text{Cu}^{2+}>\text{Cd}^{2+}>\text{Mn}^{2+}$.
Step 4: Final Answer:
The correct decreasing order of stability is $\text{Cu}^{2+}>\text{Cd}^{2+}>\text{Mn}^{2+}$.