Question

What is the correct order of spin only magnetic moment (in $BM$) of $Mn ^{2+} Cr ^{2+}$ 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. $Cr ^{2+}> V ^{2+}> Mn ^{2+}$

Answer 1

Answer: (C)

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.
${ }_{25} M n=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{2}$
$M n^{2+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{0}$

Number of unpaired electrons $=5$
${ }_{24} C r=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{1}$
$C r^{2+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{4}, 4 s^{0}$

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
$M n^{2+}>C r^{2+}>V^{2+}$