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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