question_answer

Which of the following option is having maximum number of unpaired electrons -

A)  A tetrahedral \[{{d}^{6}}\] ion

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

C)  A square planar \[{{d}^{7}}\] ion

D)  A co-ordination compound with magnetic moment of \[5.92\text{ }B.M.\]

Correct Answer: D

Solution :

                            [a] For tetrahedral \[{{d}^{6}}\]ion,   4 unpaired electrons [b] For \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{3+}},\] 0 unpaired electrons [c] For square planar \[{{d}^{7}}\] ion,    1 unpaired electrons [d] B.M. \[=\sqrt{n(n+2)},n=\] unpaired electrons \[5.92B.M.=\sqrt{n(n+2)}\] \[n=5\] unpaired electrons