Question

The increasing order of magnetic moment values of the given compounds is:
$\left(I\right)\left[Fe\right(CN{)}_6{]}^{4-}$
$\left(II\right)\left[Fe\right(CN{)}_6{]}^{3-}$
$\left(III\right)\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
$\left(IV\right)\left[Ni\right(H_{2}O{)}_6{]}^{2+}$

• A. $I<II<III<IV$
• B. $IV<III<II<I$
• C. $II<III<I<IV$
• D. $I<II<IV<III$

Answer 1

The correct option is D $I<II<IV<III$

Magnetic moment can be calculated using the formula $\sqrt{n(n+2)}$.
Where, $n$ is the number of unpaired electrons.

In $\left[Fe\right(CN{)}_6{]}^{4-}$, the oxidation state of $Fe$ is $+2$ and $C{N}^{-}$ is a strong field ligand. Thus, the electronic configuration of $Fe$ would be $t_{2g}^6{e}_g^0$. Hence, the number of unpaired electrons is $0$.

In $\left[Fe\right(CN{)}_6{]}^{3-}$, the oxidation state of $Fe$ is $+3$ and $C{N}^{-}$ is a strong field ligand. Thus, the electronic configuration of $Fe$ would be $t_{2g}^5{e}_g^0$. Hence, the number of unpaired electrons is $1$.

In $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$, the oxidation state of $Ni$ is $+2.$ Thus, the electronic configuration of $Ni$ would be $t_{2g}^6{e}_g^2$. Hence, the number of unpaired electrons is $2$.

In $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$, the oxidation state of $Cr$ is $+3.$ Thus, the electronic configuration of $Cr$ would be $t_{2g}^3{e}_g^0.$ Hence, the number of unpaired electrons is $3$.

As the number of unpaired electrons increases, magnetic moment increases. Thus, the correct order is $I<II<IV<III$