Question

Identify the order in which the spin only magnetic moment (in BM) increases for the following four ions:
(I) $\text{F}{{\text{e}}^{+2}}$
(II) $\text{T}{{\text{i}}^{+2}}$
(III) $\text{C}{{\text{u}}^{+2}}$
(IV) ${{\text{V}}^{+2}}$
A. I, II, IV, III
B. IV, I, II, III
C. III, IV, I, II
D. III, II, IV, I

Answer 1

The spin magnetic moment can be found out by writing the correct electronic configuration of elements $\text{F}{{\text{e}}^{+2}}$, $\text{T}{{\text{i}}^{+2}}$, $\text{C}{{\text{u}}^{+2}}$, ${{\text{V}}^{+2}}$, $\text{Fe}$,$\text{Ti}$, $\text{Cu}$, V are d- block elements, after removing 2 electrons $\text{F}{{\text{e}}^{+2}}$, $\text{T}{{\text{i}}^{+2}}$, $\text{C}{{\text{u}}^{+2}}$, ${{\text{V}}^{+2}}$ will be formed. The spin magnetic moment is represented by $\sqrt{\text{n(n}+\text{2)}}$, where n is the number of unpaired electrons.

Complete answer:
(I) 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, remove 2 electrons from$\text{Fe}$by which$\text{F}{{\text{e}}^{+2}}$ will be formed. Its electronic configuration will be
$\text{F}{{\text{e}}^{+2}}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{0}}3{{\text{d}}^{6}}$; 2 electrons will be removed from the outermost shell of $\text{Fe}$which is the 4th shell respectively.
The d- orbital configuration of $\text{F}{{\text{e}}^{+2}}$can be represented as: $\begin{matrix} \uparrow \downarrow & \uparrow & \uparrow & \uparrow & \uparrow \\\ \end{matrix}$. Thus, the number of unpaired electrons are 4. So, the spin magnetic moment will be$\sqrt{4(4+2)}=\sqrt{4\times 6}$ , which is equal to$2\sqrt{6}$. (4.89 B.M.)

(II) First, write the electronic configuration of $\text{Ti}$. (Atomic number is 22):
$\text{Ti}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{2}}3{{\text{d}}^{2}}$
Now, remove 2 electrons from$\text{Ti}$ by which $\text{T}{{\text{i}}^{+2}}$ will be formed. Its electronic configuration will be
$\text{T}{{\text{i}}^{+2}}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{0}}3{{\text{d}}^{2}}$; 2 electrons will be removed from the outermost shell of $\text{Ti}$which is the 4th shell respectively.
The d- orbital configuration of $\text{T}{{\text{i}}^{+2}}$can be represented as: $\begin{matrix} \uparrow & \uparrow & {} & {} & {} \\\ \end{matrix}$. Thus, the number of unpaired electrons are 2. So, the spin magnetic moment will be$\sqrt{2(2+2)}=\sqrt{4\times 2}$, is equal to $2\sqrt{2}$(2.82 B.M).

(III) First, write the electronic configuration of $\text{Cu}$. (Atomic number is 29):
$\text{Cu}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{1}}3{{\text{d}}^{10}}$
Now, after removing 2 electrons from $\text{Cu}$,$\text{C}{{\text{u}}^{+2}}$ will be formed. Its electronic configuration will be
$\text{C}{{\text{u}}^{+2}}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{0}}3{{\text{d}}^{9}}$; 2 electrons will be removed from the outermost shell of $\text{Cu}$which is the 4th shell respectively.
The d- orbital configuration of $\text{C}{{\text{u}}^{+2}}$can be represented as: $\begin{matrix} \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \downarrow & \uparrow \\\ \end{matrix}$. Thus, the number of unpaired electrons is 1. So, the spin magnetic moment will be$\sqrt{1(1+2)}=\sqrt{1\times 3}$, is equal to $\sqrt{3}$(1.73 B.M).

(IV) First, write the electronic configuration of $\text{V}$. (Atomic number is 23):
$\text{V}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{2}}3{{\text{d}}^{3}}$
Now, after removing 2 electrons from$\text{V}$,${{\text{V}}^{+2}}$ will be formed. Its electronic configuration will be
${{\text{V}}^{+2}}$: $1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}4{{\text{s}}^{0}}3{{\text{d}}^{3}}$; 2 electrons will be removed from the outermost shell of $\text{V}$which is the 4th shell respectively.
The d- orbital configuration of ${{\text{V}}^{+2}}$ can be represented as: $\begin{matrix} \uparrow & \uparrow & \uparrow & {} & {} \\\ \end{matrix}$. Thus, the number of unpaired electrons are 3. So, the spin magnetic will be $\sqrt{\text{n(n}+\text{2)}}$, $\sqrt{3\times 5}$which is equal to $\sqrt{15}$ (3.87 B.M).
Thus, the increasing order of spin magnetic moment will be $\text{F}{{\text{e}}^{+2}}>{{\text{V}}^{+2}}>\text{T}{{\text{i}}^{+2}}>\text{C}{{\text{u}}^{+2}}$. So, the correct option is option‘d’ which is III, II, IV, I .
So, the correct answer is “Option D”.

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