Write the hybridisation and number of unpaired electrons in the complex [ CoF 6 ] 3 - . (Atomic No. of Co = 27 ).

The oxidation number of the given complex ion, CoF 6 3 - is x + 6 × - 1 = - 3 ⇒ x = - 3 + 6 ⇒ x = + 3 . The electronic configuration of the central metal ion is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 6 . The fluoride ligand is the weak ligand so, the coordination compound is the outer-orbital complex. Therefore, the hybridization of the compound is sp 3 d 2 . The number of unpaired electron is 4 .

What type of isomerism is exhibited by the complex [ Co ( NH 3 ) 5 Cl ] SO 4 ?

Two or more compounds that have the same chemical formula but, a different arrangement of atoms are known as isomers. In these complexes, there is ionisation isomerism, because both the isomers are identical except for a ligand has exchanged places with an anion that was originally outside the coordination complex, the central metal and other ligands are identical. The isomers are [ Co ( NH 3 ) 5 Cl ] SO 4 and [ Co ( NH 3 ) 5 SO 4 ] Cl .

Explain [ Co ( NH 3 ) 6 ] 3 + is an inner orbital complex whereas [ Ni ( NH 3 ) 6 ] 2 + is an outer orbital complex.

In the presence of NH 3 , the 3 d electrons pair up leaving two d orbitals empty to be involved in d 2 sp 3 hybridisation forming inner orbital complex in case of [ Co ( NH 3 ) 6 ] 3 + . The ligand acts as the strong ligand in the cobalt complex. In Ni ( NH 3 ) 6 2 + , Ni is in + 2 oxidation state and has d 8 configuration, the hybridisation involved is sp 3 d 2 forming outer orbital complex. Here, the ligand acts as a weak ligand.