A) \[{{\left[ Cr{{({{H}_{2}}O)}_{6}} \right]}^{3+}}\]
B) \[{{\left[ Fe{{({{H}_{2}}O)}_{6}} \right]}^{2+}}\]
C) \[{{\left[ Cu{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\]
D) \[{{\left[ Zn{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{2+}}\]
Correct Answer: B
Solution :
\[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\] has four unpaired electrons, \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{3+}},\]\[{{[Cu{{({{H}_{2}}O)}_{6}}]}^{2+}}\] and \[{{[Zn{{({{H}_{2}}O)}_{6}}]}^{2+}}\]have 3, 1, 0 unpaired electrons respectively.
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