A) A tetrahedral d6 ion
B) \[{{[Co{{({{H}_{2}}O)}_{6}}]}^{3+}}\]
C) A square planar d7 ion
D) A co-ordination compound with magnetic moment of 5.92 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] \[B.M.=\sqrt{n(n+2)},=\]unpaired electrons \[5.92\,B.M.=\sqrt{n(n+2)}\] n = 5 unpaired electrons.You need to login to perform this action.
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