question_answer

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

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 the number of unpaired electrons, more the number of unpaired electrons, greater will be the spin only magnetic moment. \[_{25}Mn=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{2}}\] \[M{{n}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{0}}\] Number of unpaired electrons = 5 \[_{24}Cr=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{1}}\] \[C{{r}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{4}},4{{s}^{0}}\] Number of unpaired electrons = 4  ­Number of unpaired electrons = 4 \[_{23}V=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{3}},4{{s}^{2}}\]     \[{{V}^{2+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{3}},4{{s}^{0}}\] Number of unpaired electrons =3 So, the correct order of spin only magnetic moment is\[M{{n}^{2+}}>C{{r}^{2+}}>{{V}^{2+}}\]