Using valence bond theory, predict the hybridization and magnetic character of the following :
(a) [Co(NH3)6]3+
(b) [Ni(CO)4]
[At. no. : Co = 27, Ni = 28]
(a) Oxidation of Co in[Co(NH3)6]3+ is +3 as NH3 is a neutral ligand.
Co3+ in [Co(NH3)6]3+is in 3d6 oxidation state (electronic configuration of Co is [Ar] 3d7 4s2] . As –NH3 is a strong field ligand, the electrons pair up inside the orbital. Due to which, the first 3 orbitals of the ‘d’ gets occupied by pairing and the remaining 2 gets filled by the incoming electrons of –NH3 .1 s orbital is also filled by –NH3 and also 3 p orbitals are filled by –NH3. So, we get the hybridisation of d2sp3. The geometry is octahedral and as no unpaired electrons are present, it is diamagnetic.
(b) Oxidation state of Ni in [Ni(CO)4] is 0 as CO is a neutral ligand.
Ni has oxidation state of 0 in [Ni(CO)4]and its configuration is 3d8 4s2.As –CO is a strong field ligand, the electrons pair up inside the orbital. Due to which, all the5 orbitals of the ‘d’ gets occupied and the remaining ‘s’ and ‘p’ orbital gets filled by the incoming electrons of the ligand –CN (2 in ‘s’ and 6 in ‘p’) . 1 ‘s’ orbital and 3 ‘p’ orbitals are filled by –CO as there are 4 CO atoms in the complex. So, we get the hybridisation of sp3. The geometry is tetrahedral and as no unpaired electrons are present, it is diamagnetic.