Question

Transition elements show magnetic moments due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?

• A. $C{r}^{2+}$
• B. $M{n}^{2+}$
• C. $C{r}^{3+}$
• D. $C{o}^{2+}$

Answer 1

Answer: (C), (D)

$C{o}^{2+}$

Analyzing the given options :

The magnetic moment can be calculated by using the formula:

$\mu =\sqrt{n(n+2)}$ where, $n$ is the number of unpaired electrons.

The, electronic confriguration of $Co$ in its $+2$ state $\left(C{o}^{2+}\right)$ is $\left[Ar\right]3d^7.$

Thus, the number of unpaired electrons in $C{o}^{2+}$ is $3.$

Therefore, magnetic moment of $C{o}^{2+}$ will be:

$\mu =\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{3\left(5\right)}$

$=3.87B.M$

Option$\left(B\right)$

The, electronic confriguration of $Cr$ in its $+2$ state $\left(C{r}^{2+}\right)$ is $\left[Ar\right]3d^4.$

Thus, the number of unpaired electrons in $C{r}^{2+}$ is $4.$

Therefore, magnetic moment of $C{r}^{2+}$ will be:

$\mu =\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{4\left(6\right)}$

$=4.89B.M$

Option$\left(C\right)$

The, electronic confriguration of $Mn$ in its $+2$ state $\left(M{n}^{2+}\right)$ is $\left[Ar\right]3d^5.$

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$\left(D\right)$

The, electronic confriguartion of $Cr$ in its $+3$ state $\left(C{r}^{3+}\right)$ is $\left[Ar\right]3d^3.$

Thus, the number of unpaired electrons in $C{r}^{3+}$ is $3.$

Therefore, magnetic moment of $C{r}^{3+}$ will be:

$\mu =\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{3\left(5\right)}$

$=3.87B.M$

Thus $C{o}^{2+}$ and $C{r}^{3+}$ ions have the same value of spin only magnetic moments.

Hence, the correct options are $\left(A\right)$ and $\left(D\right).$