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The electronic configuration of \(\mathrm{Cu(II)}\) is \[ 3d^9 \] whereas that of \(\mathrm{Cu(I)}\) is \[ 3d^{10} \] Which of the following is correct? 

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Stability of oxidation states in transition metals depends on: \[ \text{ligands, hydration energy and lattice energy} \]
Updated On: May 30, 2026
  • \(\mathrm{Cu(II)}\) is more stable
  • \(\mathrm{Cu(I)}\) is less stable
  • \(\mathrm{Cu(I)}\) and \(\mathrm{Cu(II)}\) are equally stable
  • Stability of \(\mathrm{Cu(I)}\) and \(\mathrm{Cu(II)}\) depends on nature of copper salts
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Transition metal stability is governed by multiple factors beyond simple electronic configuration.
While a fully filled \( d^{10} \) configuration (as in \( Cu^+ \)) is inherently stable in an isolated gaseous state due to symmetry, the situation changes in a condensed phase like an aqueous solution or a solid salt.
The stability of an oxidation state depends on the balance between ionization energy, hydration energy (in water), and lattice energy (in solids).
Step 2: Detailed Explanation:
Let's compare the two copper ions:
1. Cu(I) [\( 3d^{10} \)]: It has a completely filled d-subshell. In the gas phase, it is very stable. However, in water, \( Cu^+(aq) \) is unstable and undergoes disproportionation into \( Cu^{2+}(aq) \) and \( Cu(s) \). This is because the hydration energy released by \( Cu^{2+} \) is much higher than that for \( Cu^+ \), making the higher oxidation state more favorable in aqueous environments.
2. Cu(II) [\( 3d^9 \)]: This is the most common and stable oxidation state of copper in aqueous solution. Despite having an unpaired electron, the thermodynamic stability provided by hydration more than compensates for the energy required to remove the second electron from copper.
However, the "stability" is not absolute. It depends heavily on the chemical environment:
- In the presence of soft ligands like cyanide (\( CN^- \)) or iodide (\( I^- \)), the Cu(I) state is highly stabilized, and Cu(II) is often reduced.
- In simple chloride or sulfate solutions, Cu(II) is much more stable.
- In the solid state, certain salts of Cu(I) are perfectly stable and insoluble, like \( CuCl \).
Therefore, whether Cu(I) or Cu(II) is "stable" is a relative term that depends on the surrounding anions, ligands, and the specific salt being formed.
Step 3: Final Answer:
The relative stability of \( Cu^{1+} \) versus \( Cu^{2+} \) is a function of the entire thermodynamic cycle including the nature of the ligands and the salt lattice.
It is scientifically most accurate to say stability depends on the nature of the copper salts.
The correct option is (D).
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