A) \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},\,\,\,{{[COC{{l}_{4}}]}^{2-}}\]
B) \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}},\,\,\,{{[Co{{F}_{6}}]}^{3-}}\]
C) \[{{[Mn{{({{H}_{2}}O)}_{6}}]}^{2+}},\,\,\,{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
D) \[{{[COC{{l}_{4}}]}^{2-}},\,\,\,{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
Correct Answer: B
Solution :
| Same magnetic moment = Same number of unpaired electrons \[=\sqrt{n(n+2}\] |
| Where n= number of unpaired electrons |
| \[C{{o}^{2+}}=3{{d}^{7}},3\] unpaired electrons |
| \[C{{r}^{2+}}=3{{d}^{4}},4\] unpaired electrons |
| \[M{{n}^{2+}}=3{{d}^{+5}},5\] unpaired electrons |
| \[C{{o}^{3+}}=3{{d}^{6}},4\] unpaired electrons |
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