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| 1. |
Fluorine does not exhibit any positive oxidation states. Explain. |
| Answer» Solution :The electron configuration of F is `1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(2)2p_(z)^(1)`. It has ONE unpaired electron. Since F is the most electronegative element KNOWN, THEREFORE, the possibility of sharing its unpaired electron with an element more electronegative than itself does not arise and hence it cannot show an OXIDATION state of +I. Further, F does not have d-orbitals in its valence shell. Therefore, it cannot expand its valence shell and hence does not show higher positive oxidation states of +3, +5 and +7. Thus, in nutshell, F cannot show any positive oxidation states. | |