Question

Which of the following ions show higher spin only magnetic moment value?

• A. $T{i}^{3+}$
• B. $C{o}^{3+}$
• C. $Mn$
• D. $F{e}^{2+}$

Answer 1

Answer: (C), (D)

$F{e}^{2+}$

Solution
Analyzing the given options
The formula to find spin -only magnetic
moment is:

$\mu =\sqrt{\left(n\right(n+2)}$
Where, n is the number of unpaired electrons
present in the ions.

Option{A}
The electronic configuration of $T{i}^{3+}$ is
[Ar]$3d{}^1\left(t_2{g}^1{e}_g^0\right)$.

Thus, number of unpaired electrons in $T{i}^3$+ is 1
Therefore, magnetic moment of $T{i}^3$+ will be:

$\mu =\sqrt{n(n+2)}=\sqrt{1(1+2)}=\sqrt{3}$=$1.73B.M.$

Option{B}
The electronic configuration of $M{n}^{2+}$ is
[Ar] $3d{}^5\left(t_2{g}^3{e}_g^2\right)$

Thus, number of unpaired electrons $M{n}^{2+}$is 5.
Therefore, magnetic moment $M{n}^{2+}$ will be:
$\mu =\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{5\left(7\right)}=5.91B.M.$

Option(C)
The electronic configuration of $F{e}^{2+}$ is
$\left[Ar\right]3d{}^6\left(t_2{g}^4{e}_g^2\right)$

Thus, the number of unpaired electrons in $M{n}^{2+}$ is 5.
Therefore, magnetic moment of $M{n}^{2+}$will be:

$\mu =\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{5\left(7\right)}=5.91B.M$

Option (D)
The electronic configuration of $C{o}^{3+}$ is

$\left[Ar\right]3d{}^6\left(t_2{g}^6{e}_g^0\right)$



Thus, number of unpaired electrons $C{o}^{3+}iso$

Therefore, magnetic moment $C{o}^{3+}$ will be;

$\mu =\sqrt{n(n+2)}=\sqrt{0(0+2)}=0$.

Hence, only $F{e}^{2+}andM{n}^{2+}$ show higher spin magnetic moments values.
So correct answers are option (B) and Option (C).