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