A) \[T{{i}^{3+}}\]
B) \[N{{i}^{2+}}\]
C) \[C{{r}^{3+}}\]
D) \[M{{n}^{2+}}\]
Correct Answer: B
Solution :
Magnetic moment is given by \[\mu =\sqrt{n(n+2)}\] Here, n = number of unpaired electrons \[\Rightarrow \] \[2.83=\sqrt{n(n+2)}\] \[\Rightarrow \] \[{{(2.83)}^{2}}=n(n+2)\] \[\Rightarrow \] \[8.00={{n}^{2}}+2n\] \[\Rightarrow \] \[{{n}^{2}}+2n-8=0\] \[\Rightarrow \] \[{{n}^{2}}+4n-2n-8=0\] \[\Rightarrow \] \[(n+4)(n-2)=0\] \[\therefore \] \[n=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\because \,n\ne -4]\] Among the given ions: Hence, \[N{{i}^{2+}}\] possesses a magnetic moment of 2.83 BM.You need to login to perform this action.
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