Question

The paramagnetic species among the following is

• A. $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• B. $\left[Ni\right(CN{)}_4{]}^{2-}$
• C. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• D. $\left[Ni\right(CO{)}_4]$

Answer 1

The correct option is C $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

In $\left[Co\right(N{H}_3{)}_6{]}^{3+}$, $N{H}_3$ acts as a strong field ligand. So, the $d^6$ electrons of $C{o}^{3+}$ get paired up. Thus $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is diamagnetic.


In $\left[Ni\right(CN{)}_4{]}^{2-}$, $C{N}^{-}$ acts as a strong field ligand. So, the $d^8$ electrons of $N{i}^{2+}$ get paired up. Thus $\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic.


In $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$, $H_{2}O$ acts as a weak field ligand. So, the $d^6$ electrons of $F{e}^{2+}$ don't get paired. Thus $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$ is paramagnetic.


In $\left[Ni\right(CO{)}_4]$, $CO$ acts as a strong field ligand and the oxidation state of $Ni$ is $0$. So, the $d^{10}$ electrons of $N{i}^0$ are paired. Thus $\left[Ni\right(CO{)}_4]$ is diamagnetic.