A) 4 and 2
B) 5 and 3
C) 3 and 2
D) 4 and 3
E) 3 and 1
Correct Answer: D
Solution :
Spin only magnetic moment, \[\mu =\sqrt{n(n+2)}=\sqrt{24}\] \[\Rightarrow \] \[{{n}^{2}}+2n-24=0\] \[(n+6)\,(n-4)=0\] \[\therefore \] \[n=4\] [\[\because \]\[n=-\,6\] not possible] Here, n is the number of unpaired electrons. The electronic configuration of the metal ion\[{{M}^{x+}}\]is \[Z(25)=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{5}}\] Since, four unpaired electrons are present, the oxidation state must be +3. \[\therefore \]\[{{Z}^{3+}}(25)=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{4}}\]You need to login to perform this action.
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