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JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    Consider the following statements, (I) \[[Mn{{({{H}_{2}}O)}_{4}}]S{{O}_{4}}\]is paramagnetic and square planar. (II) crystal field splitting energy \[(i.e.{{\Delta }_{0}})\] in \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{+3}}\] is higher than in \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{+2}}\] (III) Wilkinson catalyst, a red - violet complex \[[RhCl{{(P{{H}_{3}}P)}_{3}}]\] is diamagnetic and square planar (IV)\[Hg[Co{{(SCN)}_{4}}]\] a deep blue complex is paramagnetic and tetrahedral and of these select the correct one from the given codes.

    A)  I and IV only                     

    B)  II, III and IV only

    C)  I, III and IV only               

    D)  All of these

    Correct Answer: D

    Solution :

    (I) High CFSE in low spin \[{{d}^{5}}\] configuration. It is reported square planar and contains one unpaired electron. (II) \[{{\Delta }_{0}}\]increases with increase in charge on central metal ion. (III) \[4{{d}^{8}}\] configuration \[\to \] high CFSE compels for pairing of electrons and thus squares planar (i.e., \[ds{{p}^{2}}\] and diamagnetic) red - violet colour may be accounted for charge transfer (of triphenyl phosphine group). (IV) \[SC{{N}^{-}}\] is weak field ligand and thus, no empty d-orbital is available for \[ds{{p}^{2}}\] hybridization. As a result of this Co (II) complex is \[s{{p}^{2}}\] hybridized.


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