NEET AIPMT SOLVED PAPER SCREENING 2008

  • question_answer
    In which of the following coordination entities the magnitude of \[{{\Delta }_{o}}\](CCFSE in octahedral field) will be maximum? (Atomic number Co = 27)

    A) \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{3+}}\]      

    B) \[{{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}\]

    C) \[{{[Co{{(CN)}_{6}}]}^{3-}}\]      

    D) \[{{[Co{{({{C}_{2}}{{O}_{4}})}_{3}}]}^{3-}}\]

    Correct Answer: C

    Solution :

                    Key Idea:   The magnitude of \[{{\Delta }_{oct}},\](the orbital splitting energy) is decided by the nature of ligand. Strong field ligand has highest \[{{\Delta }_{oct}}.\] The increasing field strength is as \[{{I}^{-}}<B{{r}^{-}}<C{{l}^{-}}<{{F}^{-}}<O{{H}^{-}}<{{H}_{2}}O<{{C}_{2}}O_{4}^{2-}\] \[<N{{H}_{3}}<en<NO_{2}^{-}<C{{N}^{-}}\] The \[C{{N}^{-}}\] is the strongest ligand among these, hence the magnitude of \[{{\Delta }_{oct}}\] will be maximum in\[{{[Co{{(CN)}_{6}}]}^{3-}}.\] Note: The magnitude of \[{{\Delta }_{oct}}\] is also decided by oxidation state of the metal ion. Thus, greater the ionic charge on the central metal ion, the greater the value of \[{{\Delta }_{oct}}\]. But here the oxidatinon state of Co metal ion is same in all complexes.


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