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 $C{u}^{2+}$ is less than $C{u}^{+}$
• C. $C{u}^{2+}$ has stabler electronic configuration as compared to $C{u}^{+}$
• D. the lattice energy released for cupric compounds is much higher than $C{u}^{+}$

Answer 1

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

Electronic configurations of cuprous $\left(C{u}^{+}\right)$and cupric $\left(C{u}^{2+}\right)$ ions are as follows
$C{u}^{+}=[Ar]3d^{10}4s^{\circ }$
$C{u}^{2+}=[Ar]3d^{9}4s^0$
Thus, electronic configuration of $C{u}^{+}$is more stable but it is less stable because in $C{u}^{2+}$ due to its small size the nuclear charge is sufficient to hold $27$ electrons but in $C{u}^{+}$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 lattice energy is released for cupric compounds as compared to $C{u}^{+}$compounds.