Question

Given complexes are low spin complexes,

$$\begin{gathered} \left[\mathrm{V}{\left(\mathrm{CN}\right)}_6{]}^{4-} \\ \right[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6{]}^{2+} \\ \left[\mathrm{Ru}{\left({\mathrm{NH}}_3\right)}_6{]}^{3+} \\ \right[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-} \end{gathered}$$

Then the order of magnetic moments for ${\mathrm{V}}^{2+},{\mathrm{Fe}}^{2+},{\mathrm{Cr}}^{3+},{\mathrm{Ru}}^{3+}$is:

• A. ${\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Fe}}^{2+}>{\mathrm{Ru}}^{3+}$
• B. ${\mathrm{Fe}}^{2+}>{\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Ru}}^{3+}$
• C. ${\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Ru}}^{3+}>{\mathrm{Fe}}^{2+}$
• D. ${\mathrm{Fe}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{V}}^{2+}>{\mathrm{Ru}}^{3+}$

Answer 1

The correct option is C

${\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Ru}}^{3+}>{\mathrm{Fe}}^{2+}$

The explanation for the correct option:

C. ${\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Ru}}^{3+}>{\mathrm{Fe}}^{2+}$

The magnetic moment of the complexes are:

$$\begin{gathered} [\mathrm{V}{\left(\mathrm{CN}\right)}_6{]}^{4-} \\ \mathrm{Magnetic}\mathrm{moment}=\sqrt{3(3+2)}=\sqrt{15}\mathrm{BM}=3.9\mathrm{BM} \end{gathered}$$

$$\begin{gathered} [\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6{]}^{2+} \\ \mathrm{Magnetic}\mathrm{moment}=\sqrt{2(2+2)}=\sqrt{8}\mathrm{BM}=2.9\mathrm{BM} \end{gathered}$$

$$\begin{gathered} [\mathrm{Ru}{\left({\mathrm{NH}}_3\right)}_6{]}^{3+} \\ \mathrm{Magnetic}\mathrm{moment}=\sqrt{1(1+2)}=\sqrt{3}\mathrm{BM}=1.73\mathrm{BM} \end{gathered}$$

$$\begin{gathered} [\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{4-} \\ \mathrm{Magnetic}\mathrm{moment}=\sqrt{\mathrm{no}\mathrm{unpaired}\mathrm{electron})}=0 \end{gathered}$$

The explanation for the incorrect options:

A.${\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Fe}}^{2+}>{\mathrm{Ru}}^{3+}$

This statement is wrong as ${\mathrm{Ru}}^{3+}$has a high magnetic moment than ${\mathrm{Fe}}^{2+}$. The magnetic moment of ${\mathrm{Fe}}^{2+}$is zero because there is no unpaired electron.

B. ${\mathrm{Fe}}^{2+}>{\mathrm{V}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{Ru}}^{3+}$

Among all the elements ${\mathrm{Fe}}^{2+}$ has the lowest magnetic moment.

D.${\mathrm{Fe}}^{2+}>{\mathrm{Cr}}^{3+}>{\mathrm{V}}^{2+}>{\mathrm{Ru}}^{3+}$

The magnetic moment of ${\mathrm{Fe}}^{2+}$is zero and the magnetic moment of ${\mathrm{Cr}}^{3+}$is smaller than ${\mathrm{V}}^{2+}$.

Hence, option C is correct. As it is in the correct series of magnetic moments.