Question

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

(ii) On the basis of crystal field teory, write the electronic configuration for $d^4$ 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

$\left(ii\right)t_{2g}^4{e}_g^0$

(iii) In $[NiC{l}_4{]}^{2-},Ni$ is in +2 O.S. with the configuration $3d^{8}4s^0$. $C{l}^{-}$ is a weak ligand. It cannot pair up the electrons in 3d orbitals. Hence, it is paramagnetic. In $\left[Ni\right(CO{)}_4]$, Ni is in Zero O.S. with the configuration $3d^{8}4s^2$. In presence of CO ligand, the 4s electrons shift to 3d to pair up 3d electrons. Thus, there is no unpaired electron present.
Hence, it is diamagnetic