Question

The number of diamagnetic species among the following is ______$Ni(CO{)}_4,[NiC{l}_4{]}^{2-},[Co{F}_6{]}^{3-},[Cu(N{H}_3{)}_4{]}^{2+},[Ni(CN{)}_4{]}^{2-}$

Answer 1

Answer: 2

$[Ni(CO{)}_4]:$ In this complex nickel is in $0$ oxidation state and valence electronic configuration is $4s^{2}3d^8$ . In presence of strong field ligand $CO$ , two electrons present in $4s$ orbital get shifted to $3d$ orbital. Due to this, d orbital is completely filled in complex and all electrons are paired due to which it is diamagnetic.
$[NiC{l}_4{]}^{2-}$ : In this complex nickel is in $+2$ oxidation state and outermost electronic configuration is $3d^8$ and since $Cl$ is weak field ligand it can't pair electrons in orbital of nickel due to which it is paramagnetic with two unpaired electrons.
$[Co{F}_6{]}^{3-}$ : In this complex cobalt is in $+3$ oxidation state and outermost electronic configuration is $3d^6$ and since fluoride is weak field ligand due to which it can't pair the electrons present in orbital of cobalt and complex is paramagnetic with four unpaired electrons.
$[Cu({\left(NH\right)}_3{)}_4{\left(]\right)}^{2+}$ : In this complex copper is in $+2$ oxidation state and configuration is $3d^9$ and these electrons can't be paired anyhow therefore complex is paramagnetic with one unpaired electron.
$[Ni(CN{]}^{2-}:$ : In this complex nickel is in $+2$ oxidation state and outermost electronic configuration is $3d^8$ and since $CN$ is strong field ligand it will pair electrons in orbital of nickel due to which it is diamagnetic with two unpaired electrons.
Thus, $[Ni(CO{)}_4]\text{and}[Ni(CN{)}_4{\left(]\right)}^{2-}$ are diamagnetic.