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Explain the difference between covalency and oxidation state by taking the example of N_(2)O_(5). |
Answer» Solution :Covalency and oxidation states 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 form a N=O bond, one electron with the second oxygen atom to form a N-O bond the LONE pair of electrons with the THIRD oxygen atom to form a coordinate bond (N `rarr` O). For a coordinate bond in which donor atom is LESS electronegative (e.g., N in `N_(2)O_(5)`) and the acceptor atom is more electronegative (e.g., O in `N_(2)O_(5)`), the donor atom (i.e., N atom) is assigned an oxidation state of +2. THUS, the total oxidation state of N in `N_(2)O_(5)` is +5 as calculated below : `{:((+2),+,(+1),+,(+2),=,+5),((N = O),,(N - O),,(N rarr O),,):}` But for a coordinate bond irrespective of the nature of donor atom whether more or less electronegative, covalency is always 1 (for each shared pair of elctrons, covalency is counted 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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