NEET Chemistry NEET PYQ-Coordination Chemistry

  • question_answer
    Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is [AIPMT 2014]

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

    B) \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}\]

    C) \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{2+}},\]

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

    Correct Answer: B

    Solution :

    [b] The CFSE for octahedral complex is given by
    \[CFSE=[-0.4{{t}_{2g}}{{e}^{-}}+0.6{{e}_{g}}{{e}^{-}}]\]
    For \[M{{n}^{3+}},\,[3{{d}^{4}}]\to \,t_{2g}^{3}\,e_{g}^{1}\]
    \[\therefore \]      \[CFSE=[-0.4\,\times 3+0.6\times 1]\]
     = 0.6
    For \[F{{e}^{3+}},[3{{d}^{5}}]\to t_{2g}^{3}e_{g}^{2}\]
    \[CFSE\,=[-(0.4\times 3)+(0.6\times 2)]=0\]
    For \[C{{o}^{2+}},[3{{d}^{7}}]\to t_{2g}^{5}e_{g}^{2}\]
    \[CFSE\,=[(-0.4\times 5)\,+(2\times 0.6)]\,=-0.8\]
    For \[C{{o}^{3+}},[3{{d}^{6}}]\to t_{2g}^{4}e_{g}^{2}\]
    \[CFSE=[(-0.4\times 4)\,+(2\times 0.6)]=-0.4\]


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