What is the correct order of spin only magnetic moment (in BM) of $\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}$ and $\mathrm{V}^{2+}$ ?

• A

  $\mathrm{Mn}^{2+}>\mathrm{V}^{2+}>\mathrm{Cr}^{2+}$
• B

  $\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
• Correct answer
  C

  $\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}$
• D

  $\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Mn}^{2+}$

Solution

Spin only magnetic moments depend upon the number of unpaired electrons, more the number of unpaired electron, greater will be the spin only magnetic moment. ${ }_{25} \mathrm{Mn}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5, 4 s^2$ $\mathrm{Mn}^{2+}=1 s^2, 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5, 4 s^0$
Number of unpaired electrons $=5$ ${ }_{24} \mathrm{Cr}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5, 4 s^1$ $\mathrm{Cr}^{2+}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^4, 4 s^0$
Number of unpaired electrons $=4$ ${ }_{23} \mathrm{~V}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^3, 4 s^2$ $\mathrm{V}^{2+}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^3, 4 s^0$
Number of unpaired electrons $=3$ So, the correct order of spin only magnetic moment is $\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}$

Hence, the correct answer is option: 'C'.