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Question #44782

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The Question

The standard reduction potentials for the $+2/+1$ couple of Group 11 elements show a stark contrast: $E^\circ(\ce{Cu^{2+}/Cu^+}) = +0.16\text{ V}$ , whereas $E^\circ(\ce{Ag^{2+}/Ag^+}) = +1.98\text{ V}$ . Consequently, $\ce{Ag^{2+}(aq)}$ is a powerful oxidizing agent that rapidly oxidizes water, while $\ce{Cu^{2+}(aq)}$ is thermodynamically stable in the absence of disproportionation drivers. Despite the extreme instability of the aqueous $\ce{Ag^{2+}}$ ion, the binary solid Silver(II) fluoride, $\ce{AgF_2(s)}$ , is a stable compound at room temperature.

Given that the second ionization enthalpies ( $IE_2$ ) are $1958\text{ kJ mol}^{-1}$ for $\ce{Cu}$ and $2070\text{ kJ mol}^{-1}$ for $\ce{Ag}$ , which of the following statements correctly identifies the thermodynamic principles governing these observations?

Options

A.
The instability of $\ce{Ag^{2+}(aq)}$ is primarily due to its significantly less exothermic hydration enthalpy compared to $\ce{Cu^{2+}(aq)}$ , which fails to compensate for the high $IE_2$ of Silver. However, the high lattice enthalpy of $\ce{AgF_2(s)}$ provides sufficient exothermic stabilization to overcome the $IE_2$ .
B.
The high $E^\circ(\ce{Ag^{2+}/Ag^+})$ reflects the extreme instability of the $4d^9$ configuration. $\ce{AgF_2(s)}$ bypasses this instability by crystallizing as a diamagnetic mixed-valence compound, $\ce{Ag^I[Ag^{III}F_4]}$ , whereas $\ce{CuF_2(s)}$ remains a true paramagnetic $\ce{Cu^{II}}$ species.
C.
The fluoride ion acts as a strong-field ligand in the solid state, forcing $\ce{Ag^{2+}}$ into a low-spin configuration with a massive Crystal Field Stabilization Energy (CFSE) that offsets the high $IE_2$ . This CFSE stabilization is absent in the weak-field $\ce{[Ag(H_2O)_6]^{2+}}$ complex.
D.
The higher $E^\circ$ of Silver is a consequence of the inert pair effect stabilizing the $+1$ oxidation state. $\ce{AgF_2(s)}$ is only stable because the extreme electronegativity of fluorine reduces the effective nuclear charge ( $Z_{eff}$ ) on Silver, lowering the actual energy required to remove the second electron.

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Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Electrode Potential Dependencies|Stability of High O.S. Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

This question tests the 'Chemical Exception' where students conflate aqueous stability (driven by hydration enthalpy) with solid-state stability (driven by lattice enthalpy).

A common trap is assuming that because an ion is a powerful oxidizing agent in water (like Ag2+ or Co3+), its solid compounds must also be unstable.

Furthermore, it introduces the 'Hidden Variable' of the mixed-valence trap (AgO vs AgF2), forcing students to differentiate between true high oxidation state stabilization via hard-hard lattice interactions and disproportionation phenomena.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Electrochemistry - Class 12
Concept	Electrode Potential Dependencies
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free