Question

What is the spin only magnetic moment of ${\text{Mn}}^{2+}$(aq) ion?

Answer 1

Step 1: The electronic configuration of $\text{Mn}$:

• First, write the electronic configuration of $\text{Mn}$ (the atomic number is $25$):
  $\text{Mn}$: $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{5}4{\text{s}}^2$

Step 2: The electronic configuration of ${\text{Mn}}^{2+}$

• Now, after removing 2 electrons from. ${\text{Mn}}^{2+}$ will be formed. Its electronic configuration will be
  ${\text{Mn}}^{2+}$: $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{0}3{\text{d}}^5$
• The 2 electrons will be removed from the outermost shell of $\text{Mn}$ which is the 4th shell respectively.

Step 3: Calculate the spin-only magnetic moment

• The spin magnetic moment is a magnetic moment caused due to the spin of particles.
• The formula to calculate spin only magnetic moment is
• $\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}$

Where n= number of unpaired electrons.

The unpaired electrons of ${\text{Mn}}^{2+}$ , by the shell configuration $\begin{array}{ccccc}\uparrow & \uparrow & \uparrow & \uparrow & \uparrow \end{array}$will be $5$.

So,$\mathrm{n}=5$.
The spin magnetic moment will be:

$$\begin{gathered} \mathrm{Spin}\mathrm{magnetic}\mathrm{moment}=\sqrt{5(5+2)} \\ =\sqrt{5\times 7} \\ =5.916\text{B.M} \end{gathered}$$
Thus, the spin magnetic moment of ${\text{Mn}}^{2+}$ is 5.916 B.M. (Bohr Magneton). Bohr Magneton is the unit of the spin magnetic moment.