Question

$\left[NI\right(Cl{)}_2\left(P\right(C{H}_3{)}_3{)}_2]$ is a paramagnetic complex of Ni(II), Analgous Pd(II) complex is diamagnetic. How many geometrical isomers will be possible for Ni(II) and Pd(II) complexes? Also explain their magnetic behaviour.

Answer 1

NI(II) complex is paramagentic, so it has unpaired electrons while complex of Pd(II) is diamagnetic without any unpaired electron. In both NI(II) and Pd(II), there is $d^8$ configuration. In NI(II), value of crystal field splitting energy $\left(\Delta \right)$ is less than in Pd(II). So in NI(II) pairing is less favoured while in Pd(II), all electrons are paired because unpairing is unfavoured owing to high value of CFSE. Thus the geometry and hybridization in two complexes is explained as follows: