A) \[N{{i}^{2+}}\]
B) \[T{{i}^{3+}}\]
C) \[C{{r}^{3+}}\]
D) \[C{{O}^{2+}}\]
Correct Answer: A
Solution :
Magnetic moment, \[\mu =\sqrt{n(n+2)}\]BM where, n = number of unpaired electrons \[\mu =2.84(given)\] \[\therefore \] \[284=\sqrt{n(n+2)}B.M\] \[{{(2.84)}^{2}}=n(n+2)\] \[8={{n}^{2}}+2n\] \[{{n}^{2}}+2n-8=0\] \[{{n}^{2}}+4n-2n-8=0\] \[n(n+4)-2(n+4)=0\] \[n=2\] \[N{{i}^{2+}}=[Ar]3{{d}^{8}}4{{s}^{0}}\](two unpaired electrons) \[T{{i}^{3+}}=[Ar]3{{d}^{1}}4{{s}^{0}}\] (one unpaired electrons) \[C{{r}^{3+}}=[Ar]3{{d}^{3}}\], (three unpaired electrons) \[C{{o}^{2+}}=[Ar],3{{d}^{7}},4{{s}^{0}}\](three unpaired electrons) So, only \[N{{i}^{2+}}\] has 2 unpaired electrons.
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