how to find conventional oxidation no and maximum oxidation no
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Answer:
Variable oxidation states are mainly shown by many nonmetals and most transition metals.
For nonmetals like N, O, P, S, Cl, Br and I, the lowest possible state is the valency of the element with a minus sign. Thus, the minimum oxidation number of nitrogen and phosphorus is -3 (as in NH3 and PH3), for oxygen and sulphur it is -2 (as in H2O and H2S), and for chlorine, bromine, iodine it is -1 (as in HCl, NaCl, HBr, KI).
The highest possible state for the nonmetals is equal to the total number of electrons present in the valence (outermost) shell of the atom with a plus sign. Thus, the maximum oxidation states of nitrogen, sulphur and chlorine are +5 (as in HNO3), +6 (as in H2SO4) and +7 ( as in HClO4) respectively. Two exceptions here are fluorine and oxygen, the two most electronegative elements. Fluorine, by convention, always has the -1 state in its compounds, and for oxygen the maximum state is only +2 (in F2O).
For transition metals, the minimum state is zero (free, elemental form), as in the case of non-transition metals. And the maximum oxidation number possible is same as the total number of electrons in both the outer ns and (n-1)d orbitals with a plus sign, but it very rarely exceeds +8. Thus, vanadium, with an outer configuration of 4s2 3d3, shows a maximum oxidation number of +5, as in V2O5, and chromium, with an outer configuration of 4s1 3d5, shows a maximum state of +6, as in CrO3 and K2Cr2O7. Only ruthenium and osmium seem to have the maximum state of +8 in certain compounds. But, in all the three transition series, only the metals in the middle of the series have the tendency to form stable compounds in very high oxidation states.
The highly electropositive metals in groups 1 and 2 of the periodic table generally exhibit only one positive oxidation number (+1 and +2 respectively) in compound formation which is numerically equal to the normal valency.