Question

(i) Draw the geometrical isomers of complex $\left[Co\right(en{)}_{2}C{l}_2{]}^{+}$ .

(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if ${\Delta }_0>P$.

(iii) $[NiC{l}_4{]}^{2-}$ is paramagnetic, while $\left[Ni\right(CO{)}_4]$ is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)

Answer 1

(i) Geometrical isomers of $\left[Co\right(en{)}_{2}C{l}_2{]}^{+}$ :

(ii)If ${\Delta }_0>P$, it becomes more energetically favourable for the fourth electron to occupy a $t_{2g}$ orbital with configuration te.

(iii) In $[NiC{l}_4{]}^{2-}$ , Ni is in the +2 state. $C{l}^{-}$ is a ligand which is a weak field ligand which does not cause pairing of unpaired 3d electrons. Hence, it is paramagnetic.

In $\left[Ni\right(CO{)}_4]$, Ni has 0 oxidation state. CO is a strong field ligand, which causes pairing of unpaired 3d electrons. No unpaired electrons are present in this case. So, it is diamagnetic.