Q59 / 200 · WBJEE 2014 · Chemistry

Cupric compounds are more stable than their cuprous counterparts in solid state. This is because

  • Correct answer
    A
    the endothermic character of the 2nd IP of Cu is not so high
  • Correct answer
    B
    size of $\mathrm{Cu}^{2+}$ is less than $\mathrm{Cu}^{+}$
  • C
    $\mathrm{Cu}^{2+}$ has stabler electronic contiguration as compared to $\mathrm{Cu}^{+}$
  • Correct answer
    D
    the lattice energy released for cupric compounds is much higher than $\mathrm{Cu}^{+}$

Solution

Electronic configurations of cuprous $\left(\mathrm{Cu}^{*}\right)$ and cupric $\left(\mathrm{Cu}^{2}{ }^{+}\right)$ ions are as follows $$ \begin{aligned} \mathrm{Cu}^{+} &=[\mathrm{Ar}] 3 d^{10} 4 s^{\circ} \\ \mathrm{Cu}^{2+} &=[\mathrm{Ar}] 3 d^{9} 4 \mathrm{s}^{\circ} \end{aligned} $$ Thus, electronic configuration of $\mathrm{Cu}^{+}$ is more stable but it is less stable because in $\mathrm{Cu}^{2+}$ due to its small siag the nuclear charge is sufficient to hold 27 electrons but in $Cu^+$ such a condition is not true. Further, the IInd IP of Cu is not very high as compared to its Ist IP. Consequently a large amount of latsicy energy is released for cupric compounds as compared to $Cu^+$ compounds.

Hence, the correct answer is options: 'A', 'B', 'D'.

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