Explain the following cases giving appropriate reasons : (i) Nickel does not form low spin octahedral complexes (ii) Co^(2+) is easily oxidized to Co^(3+) in the presence of a strong ligand. (iii) CO is a stronger ligand than NH_(3) for many metals.

Explain the following cases giving appropriate reasons : (i) Nickel d

1.	Explain the following cases giving appropriate reasons : (i) Nickel does not form low spin octahedral complexes (ii) Co^(2+) is easily oxidized to Co^(3+) in the presence of a strong ligand. (iii) CO is a stronger ligand than NH_(3) for many metals.
Answer» Solution : For low spin, ELECTRONS should pair up. This will produce only one empty d-orbital. HENCE, `d^(2)sp^(3)` hybridisation is not possible to form OCTAHEDRAL COMPLEXES. (ii) Co(II) has the configuration `3d^(7)`, i.e., it has three unpaired electron. Water being a weak ligand, the unpaired electrons do not pair up. In the presence of strong ligand and air, two unpaired electrons in 3d pair up and the third unpaired electron shifts tohigher energy subshell from where it can be easily lost and hence shows an oxidation state of III. (iii) Ligands such as CO, `CN^(-)` and `NO^(+)` have empty `pi`-orbitals which overlap with the filled d-orbital (`t_(2g)`-orbitals) of transition metals forming `pi`-BONDS (back bonding). These `pi`-interactions increase the value of `Delta_(0)`. This accounts for the position of these ligands as strong field ligands. `NH_(3)` cannot form `pi` bonds by back bonding.