A) \[M{{n}^{2+}}\]
B) \[F{{e}^{2+}}\]
C) \[T{{i}^{2+}}\]
D) \[C{{r}^{2+}}\]
Correct Answer: A
Solution :
Magnetic moment depends upon the number of unpaired electron. Higher the number of unpaired electron in the given ion, higher will be its magnetic moment. \[\mu =\sqrt{n(n+2)}\] where \[n=\] number of unpaired electron \[_{25}Mn=[Ar]4{{s}^{2}},3{{d}^{5}}\] \[M{{n}^{2+}}[Ar]3{{d}^{5}},4{{s}^{o}}\]
Number of unpaired electrons \[=5\] \[_{26}Fe=[Ar]4{{s}^{2}},3{{d}^{6}}\] \[F{{e}^{2+}}=[Ar]3{{d}^{6}},4{{s}^{o}}\] 
Number of unpaired electrons \[=4\] \[_{22}Ti=[Ar]4{{s}^{2}},\,3{{d}^{2}}\] \[T{{i}^{2+}}=[Ar]3{{d}^{2}},4{{s}^{0}}\] 
Number of unpaired electrons \[=2\] \[_{24}Cr=[Ar]3{{d}^{5}},4{{s}^{1}}\] \[C{{r}^{2+}}=[Ar]3{{d}^{4}},4{{s}^{0}}\] 
\[\therefore \,M{{n}^{2+}}\] has five unpaired electron and hence the dipole moment \[(\mu =n\sqrt{n+2})\] will be maximum for manganese.
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