Question

Calculate the spin only magnetic moment of $\text{F}{{\text{e}}^{+2}}$.[At. no. = 26]

Answer 1

Hint: The spin magnetic moment can be found by writing the correct electronic configuration of $\text{F}{{\text{e}}^{+2}}$. $\text{Fe}$ is a d- block element, after removing 2 electrons $\text{F}{{\text{e}}^{+2}}$ will be formed.

Complete answer:
First, write the electronic configuration of $\text{Fe}$. (atomic number is 26):
$\text{Fe}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{2}}3{{\text{d}}^{6}}$
Now, after removing 2 electrons from $\text{Fe}$, $\text{F}{{\text{e}}^{+2}}$ will be formed. Its electronic configuration will be
$\text{F}{{\text{e}}^{+2}}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{0}}3{{\text{d}}^{6}}$
The 2 electrons will be removed from the outermost shell of $\text{Fe}$ which is the 4th shell respectively.
The spin magnetic moment is a magnetic moment caused due to the spin of particles. The formula to find the spin magnetic moment is$\sqrt{\text{n(n}+\text{2)}}$, where n is the number of unpaired electrons. The unpaired electrons of $\text{F}{{\text{e}}^{+2}}$ , by the shell configuration$\begin{matrix}
   \uparrow \downarrow & \uparrow & \uparrow & \uparrow & \uparrow \\
\end{matrix}$will be 4 as the 2 electrons are paired up. So, n= 4.
The spin magnetic moment will be$\sqrt{4(4+2)}=\sqrt{4\times 6}$ , is equal to$2\sqrt{6}$.
The spin magnetic moment of $\text{F}{{\text{e}}^{+2}}$ is 4.89 B.M. (Bohr Magneton). Bohr Magneton is the unit of spin magnetic moment.

Additional Information: -Uses of $\text{F}{{\text{e}}^{+2}}$:
(1) Iron (II) or $\text{F}{{\text{e}}^{+2}}$ is used as a pigment.
(2) It is used in cosmetics.
(3) It is used in tattoo inks.
(4) It can be used as a phosphate remover from home aquaria.

Note: While writing the electronic configuration, $\text{F}{{\text{e}}^{+2}}$ the electrons will be removed from the outer shell of $\text{Fe}$, which is 4th shell ($4\text{s}$), not from ($3\text{d}$). It will be $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{0}}3{{\text{d}}^{6}}$ not $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{2}}3{{\text{d}}^{4}}$ while writing the configuration of $\text{F}{{\text{e}}^{+2}}$.