JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    Identify the correct trend given below: (Atomic No.: Ti = 22, Cr = 24 and Mo = 42)   JEE Main Online Paper (Held On 09 April 2016)

    A) \[{{\Delta }_{o}}\]of\[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}<{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}\]and\[{{\Delta }_{o}}\]of\[{{[Ti{{({{H}_{2}}O)}_{6}}]}^{3+}}>{{[Ti{{({{H}_{2}}O)}_{6}}]}^{2+}}\]

    B) \[{{\Delta }_{o}}\]of\[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}>{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}\]and\[{{\Delta }_{o}}\]of\[{{[Ti{{({{H}_{2}}O)}_{6}}]}^{3+}}<{{[Ti{{({{H}_{2}}O)}_{6}}]}^{2+}}\]

    C) \[{{\Delta }_{o}}\]of\[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}>{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}\]and \[{{\Delta }_{o}}\]of\[{{[Ti{{({{H}_{2}}O)}_{6}}]}^{3+}}<{{[Ti{{({{H}_{2}}O)}_{6}}]}^{2+}}\]

    D) \[{{\Delta }_{o}}\]of\[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}<{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}\]and \[{{\Delta }_{o}}\]of\[{{[Ti{{({{H}_{2}}O)}_{6}}]}^{3+}}>{{[Ti{{({{H}_{2}}O)}_{6}}]}^{2+}}\]

    Correct Answer: A

    Solution :

                    \[{{\Delta }_{o}}\propto CFSE\](Crystal field stabilization energy) \[{{\Delta }_{o}}\]of\[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}<{{\Delta }_{o}}\]of\[{{[Mo{{({{H}_{2}}O)}_{6}}]}^{2+}}\]Because here \[{{\Delta }_{o}}\]depends on \[{{Z}_{eff}}\And {{Z}_{eff}}\]of 4d series is more than 3d series. But \[{{\Delta }_{o}}\]of\[{{[Ti{{({{H}_{2}}O)}_{6}}]}^{3+}}<\]\[{{\Delta }_{o}}\]of\[{{[Ti{{({{H}_{2}}O)}_{6}}]}^{2+}}\]


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