BCECE Medical BCECE Medical Solved Papers-2002

  • question_answer
    Coordination number and oxidation number of Cr in \[{{K}_{3}}Cr{{({{C}_{2}}{{O}_{4}})}_{3}}\] are respectively :

    A)  6 and +3    

    B)  4 and-2

    C)  3 and 0       

    D)  3 and +3

    Correct Answer: A

    Solution :

    Key Idea: (i) Coordination of central atom is number of lone pair of electrons donated by ligands (ii) The sum of oxidation number of all elements in coordination compound is always zero.                 \[{{K}_{3}}[Cr\,{{({{C}_{2}}{{O}_{4}})}_{3}}]\] \[\because \] Oxalate is bidentate ligand so coordination number of Cr = 6 Let oxidation number of Cr in                 \[{{K}_{3}}[Cr\,{{({{C}_{2}}{{O}_{4}})}_{3}}]=x\] or            \[(+1\times 3)+x+(-2\times 3)=0\] or            \[3+x-6=0\] \[\therefore \] \[x=+3\] \[\therefore \] Coordination number of Cr in given compound is 6 and oxidation number + 3


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