Question

What are the correct orders of stability for the following compounds?

• A. $\text{VF}_5 > \text{VCl}_5$ ; $\text{CuCl}_2 > \text{CuI}_2$
• B. $\text{VCl}_5 > \text{VF}_5$ ; $\text{CuCl}_2 > \text{CuI}_2$
• C. $\text{VCl}_5 > \text{VF}_5$ ; $\text{CuI}_2 > \text{CuCl}_2$
• D. $\text{VF}_5 > \text{VCl}_5$ ; $\text{CuI}_2 > \text{CuCl}_2$

Hint:

Fluoride stabilizes the highest oxidation state of transition metals due to strong metal-ligand bonding.
Iodide fails to stabilize high oxidation states because of its strong reducing nature, making $\text{CuI}_2$ decompose instantly into stable white insoluble $\text{CuI}$.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question asks us to determine the relative stability orders for two pairs of transition metal halides: vanadium pentahalides ($\text{VF}_5$ vs $\text{VCl}_5$) and copper dihalides ($\text{CuCl}_2$ vs $\text{CuI}_2$).
Step 2 : Key Formulas and Approach:
The stability of transition metal halides in high oxidation states depends on the size and electronegativity of the halogen.
Higher oxidation states are stabilized by smaller, highly electronegative ligands (like fluorine).
The stability of halides in lower oxidation states is affected by the reducing ability of the halide ion.
Step 3 : Detailed Explanation:

Vanadium halides ($\text{VF}_5$ vs $\text{VCl}_5$): Vanadium is in its maximum $+5$ oxidation state.
Fluorine, being extremely small and the most electronegative element, can stabilize this high $+5$ oxidation state due to high lattice energy and steric feasibility.
In contrast, chlorine is larger and less electronegative, making $\text{VCl}_5$ highly unstable; it decomposes readily into $\text{VCl}_4$ and $\text{Cl}_2$.
Thus, $\text{VF}_5>\text{VCl}_5$.

Copper halides ($\text{CuCl}_2$ vs $\text{CuI}_2$): Copper is in the $+2$ oxidation state.
Iodide ion ($\text{I}^-$) is a strong reducing agent. It readily reduces $\text{Cu}^{2+}$ to $\text{Cu}^+$, oxidizing itself to iodine ($\text{I}_2$):
\[ 2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} \, (\text{s}) + \text{I}_2 \]
Because of this spontaneous redox reaction, $\text{CuI}_2$ does not exist in stable form.
Chloride ion ($\text{Cl}^-$) is not a strong enough reducing agent to reduce $\text{Cu}^{2+}$, so $\text{CuCl}_2$ is stable.
Thus, $\text{CuCl}_2>\text{CuI}_2$.

Step 4 : Final Answer:
The correct stability orders are $\text{VF}_5>\text{VCl}_5$ and $\text{CuCl}_2>\text{CuI}_2$.
This matches Option (A).