Question

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

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

Answer 1

Answer: (A), (B), (D)

The correct options are
A the endothermic character of the $2^{nd}$ IP of $Cu$ is not so high
B size of ${Cu}^{2+}$ is less than ${Cu}^{+}$
D the lattice energy released for cupric compounds is much higher than ${Cu}^{+}$

Electronic configurations of cuprous $\left({Cu}^{+}\right)$ and cupric $\left({Cu}^{2+}\right)$ ions are as follows
${Cu}^{+}=\left[Ar\right]3d^{10}4s^0$
${Cu}^{2+}=\left[Ar\right]3d^{9}4s^0$

Thus, electronic configuration of ${Cu}^{+}$ is more stable but it is less stable because in ${Cu}^{2+}$ due to its small size the nuclear charge is sufficient to hold 27 electrons but in ${Cu}^{+}$ such a condition is not true.

Further, the $2^{nd}$ IP of $Cu$ is not very high as compared to its $1^{st}$ IP. Consequently, a large amount of lattice energy is released for cupric compounds as compared to ${Cu}^{+}$ compounds.

Hence, options A, B, and C are correct.