question_answer

What is the correct order of spin only magnetic moment (in BM) of \[M{{n}^{2+}},C{{r}^{2+}}\]and \[{{V}^{2+}}\]?

A) \[M{{n}^{2+}}>{{V}^{2+}}>C{{r}^{2+}}\]

B)  \[{{V}^{2+}}>C{{r}^{2+}}>M{{n}^{2+}}\]

C)  \[M{{n}^{2+}}>C{{r}^{2+}}>{{V}^{2+}}\]

D)  \[C{{r}^{2+}}>{{V}^{2+}}>M{{n}^{2+}}\]

Correct Answer: C

Solution :

 Spin only magnetic moment depends upon die  number of unpaired electrons, more the number of unpaired electrons, greater will be the spin only magnetic moment. \[_{25}Mn=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{2}}\] \[M{{n}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{o}}\] Number of unpaired electrons =5 \[{{\,}_{24}}Cr=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{d}^{5}},4{{s}^{1}}\] \[C{{r}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{4}},4{{s}^{0}}\] Number of unpaired electrons =4 Number of unpaired electrons =3 So, the correct order of spin only magnetic moment is \[M{{n}^{2+}}>C{{r}^{2+}}>{{V}^{2+}}\]