KVPY SX SB Exam 20141 Question Paper

Ni exist in zero oxidation state so its configuration is -
its configuration is -
${}_{28}^{}\mathit{N}\mathit{i}=\left[\mathit{A}\mathit{r}\right]3{\mathit{d}}^{8}4{\mathit{s}}^2$ – CO is strong ligand so pairing of electron possible and configuration will be – Number of unpaired electron are zero hence it is diamagnetic in nature.