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, \[\mu =\sqrt{n(n+2)}=\sqrt{24}\] \[\Rightarrow \]               \[{{n}^{2}}+2n-24=0\]                 \[(n+6)(n-4)=0\]                 \[n=4\] \[\therefore \]  [\[\because \]\[n=-6\]not possible) Here, n is the number of unpaired electrons. The electronic configuration of the metal ion \[{{M}^{x+}}\]is \[Z(25)=1{{s}^{2}},\text{ }2{{s}^{2}},\text{ }2{{p}^{6}},\text{ }3{{s}^{2}},\text{ }3{{p}^{6}},\text{ }4{{s}^{2}},\text{ }3{{d}^{5}}\] Since, four unpaired electrons are present, the oxidation state must be + 3. \[\therefore \]\[{{Z}^{3+}}(25)=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{4}}\]