Question
Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration.
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Answer
The transition elements show a large number of oxidation states. The various oxidation states are related to the electronic configuration of their atoms. The various oxidation states of a transition metal is due to the involvement of ( $\mathrm{n}-1) \mathrm{d}$ and outer ns electrons in bonding. The lower oxidation state is generally shown when ns electrons participate and higher oxidation states are exhibited when $n s$ and $(n-1) d$ electrons take part in bonding. For example, manganese, electronic configuration $(n-1) d^5 n s^2$, can show $+2,+3,+4,+6$ and +7 oxidation states. In the first five elements of the first transition series, i.e., up to manganese, the maximum oxidation state is equal to the sum of 4 s and 3 d electrons. For remaining five elements, the maximum state is not related to their electronic configurations. The non-transition elements mainly the p-block elements can show a number of oxidation states from +n to $(\mathrm{n}-8$ ) where n is the number of electrons present in the outermost shell. For example, phosphorus can show $-3,+3$ and +5 oxidation states while sulphur can show $-2,+2,+4$ and +6 oxidation states. lodine can show $-1,+1,+3,+5$ and +7 oxidation states. Lower oxidation states are ionic as the atom accepts the electron or electrons to achieve stable configuration while higher oxidation states are achieved by unpairing the paired orbitals and shifting the electrons to vacant d-orbitals.
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