It is said in d block that compound with less oxidation state forms ionic bond but higher oxidation state means that compound is made by covalent bonding for example Cr2O4^2-. Pls explain how. Why will it be covalent when higher oxidation state and ionic when lower oxidation state? In 1st electron will be removed from s and in 2nd it will be shared from both s and d to form covalent I think.
Dear Student,
This is mainly due to Fajan's rule. Transition metals are electropositive in nature, and we know that according to Fajan's rule smaller the size of cation and higher the charge on cation, greater will be its polarising power and so the bond formed will be covalent. Higher oxidation state means higher charge and smaller size, therefore, the cations have higher tendency to form covalent bond. Lower oxidation state means lower charge and larger size, and so ionic bond is formed.
This is mainly due to Fajan's rule. Transition metals are electropositive in nature, and we know that according to Fajan's rule smaller the size of cation and higher the charge on cation, greater will be its polarising power and so the bond formed will be covalent. Higher oxidation state means higher charge and smaller size, therefore, the cations have higher tendency to form covalent bond. Lower oxidation state means lower charge and larger size, and so ionic bond is formed.