Question
Using valence bond theory, predict the shape and magnetic character of [Ni(CO)4] [Ni = 28].
Solution
Ni(28) has outer electronic configuration 4s23d8. The oxidation state of Ni is zero . Thus electronic configuration is 4s°3d10.

Since there is no any unpaired electron therefore it is diamagnetic in nature and tetrahedral in shape due to sp3 hybridization.

Since there is no any unpaired electron therefore it is diamagnetic in nature and tetrahedral in shape due to sp3 hybridization.