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 electronsYou need to login to perform this action.
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