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Explain the difference between covelency and oxidation state by taking the example of N_(2)O_(5. |
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Answer» Solution :Covalency and oxidation stats are TWO different concepts and should not be used interchangebly. Although N cannot have a covalency of 5, it can have an oxidation state of +5 in its compounds with oxygen, i.e., `N_(2)O_(5)` In `N_(2)O_(5)`, each N atom shares two of its valence electrons with an oxygen atom to forom a N=O BOND, one electron with the second oxygen atom to form a N-O bond and the lone pair of electrons with the THIRD oxygen atom to from a coordinate bond `(NtoO)`. For a coordinate bond in which donor atom is less ELECTRONEGATIVE (e.g., N in `N_(2)O_(5)`) and the acceptor atom is moer eletronegative (e.g., O in `N_(2)O_(5)`), the donor atom (i.e., N atom) is assigend an oxidation state of `+2`. Thus, the total oxidation state of N in `N_(2)O_(5)` is +5 as caluclated below: `{:((+2),+,(+1),+,(+2),=+5),((N=O),,(N-O),,(NtoO),):}` But for a coordinate bond irrespctive of the nature of donor atom whether more or less electronegative, covalency is always 1 (for each shared pair of electrons, covalency is conuted as one). Thus, N can have an oxidation state of +5 but cannot have a covalency of 5. At the maximum, it can have a covalency of 4. 
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