question_answer

A compound in which a metal ion\[{{M}^{x+}}(Z=25)\] has a spin only magnetic moment of \[\sqrt{24}\] BM. The number of unpaired electrons in the compound and the oxidation state of the metal ion are respectively

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. \[\therefore \]\[{{Z}^{3+}}(25)=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{4}}\]