WB JEE Medical WB JEE Medical Solved Paper-2014

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

    A) The endothermic character of the 2nd IP c 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 \[C{{u}^{+}}\]

    Correct Answer: A

    Solution :

    Electronic configurations of cuprous \[(C{{u}^{+}})\]  and cupric \[(C{{u}^{2+}})\] ions are as follows \[C{{u}^{+}}=[Ar]3{{d}^{10}}4{{s}^{0}}\] \[C{{u}^{+}}=[Ar]3{{d}^{9}}4{{s}^{0}}\] 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 1st IP. Consequently a large amount of lattice energy is released for cupric compounds as compared to \[C{{u}^{+}}\]compounds.


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