A) Fe
B) Mn
C) Ni
D) Cu
Correct Answer: D
Solution :
Cu \[_{26}Fe = 3{{d}^{6}} 4{{s}^{2}}~~~~~~\left( 4 unpaired electrons \right)\] \[_{25}Mn = 3{{d}^{5}} 4{{s}^{2}}~~~~~\left( 5 unpaired electrons \right)\] \[_{28}Ni = 3{{d}^{8}} 4{{s}^{2}}~~~~~~\left( 2 unpaired electrons \right)\] \[_{29}Cu = 3{{d}^{10}} 4{{s}^{1}}~~~~~\left( 1 unpaired electrons \right)\] Thus Cu contains least number of unpaired electrons.
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