Question

Which of the following is correct about the following?

$Ni{\left(CO\right)}_4$, $Ni{\left(CN\right)}_4^{2-}$ and ${\left[Ni{\left(Cl\right)}_4\right]}^{2-}$:

• A. $Ni{\left(CO\right)}_4$ and $Ni{Cl}_4^{2-}$ are diamagnetic and $Ni{\left(CN\right)}_4^{2-}$ is paramagnetic
• B. $Ni{Cl}_4^{2-}$ and $Ni{\left(CN\right)}_4^{2-}$ are diamagnetic and $Ni{\left(CO\right)}_4$ is paramagnetic
• C. $Ni{\left(CO\right)}_4$ and $Ni{\left(CN\right)}_4^{2-}$ are diamagnetic and $Ni{Cl}_4^{2-}$ is paramagnetic
• D. $Ni\left({CO}_4\right)$ is diamagneitc and $Ni{Cl}_4^{2-}$ and $Ni{\left(CN\right)}_4^{2-}$ are paramagnetic

Answer 1

The correct option is C $Ni{\left(CO\right)}_4$ and $Ni{\left(CN\right)}_4^{2-}$ are diamagnetic and $Ni{Cl}_4^{2-}$ is paramagnetic

In the complex $Ni(CO{)}_4$ the oxidation state of Ni is 0. hence the electronic configuration of the complex is $3d^{8}4s^2$ but the CO which is a strong field ligand pair up the electrons in 4s orbital to 3d hence the configuration becomes $3d^{10}4s^0$ which implies the complex is diamagnetic. In complex $\left[Ni\right(CN{)}_4{]}^{2-}$ the oxidation state of Ni is (+2) and hence electronic configuration of the complex is $3d^{8}4s^0$ and since the CN ligand is a strong field ligand pair up the electrons in #d orbitals such that no electron is unpaired Hence it is also a diamagnetic complex. Now, In complex $\left[Ni\right(Cl{)}_4{]}^{2-}$ the oxidation state of Ni is (+2) and hence electronic configuration of the complex is $3d^{8}4s^0$ and the Cl ligand is a weak field ligand which does not have the power to pair up the electrons hence the 2 electrons remained unpaired hence, option C is correct.