Question

explain the observation:- Fluorine does not exhibit any positive oxidation state.

Answer 1

This is an answer which is subject to change due to constant research performed by scientists everyday. Fluorine does exibit a positive oxidation state despite its high electronegativity(~3.9). It happens to be shown in one of its isotopes F{19}. Just to give a brief background, F{19} is the only stable isotope and its abundance is ~99.9℅. It is NMR-active, I.e. nuclear magnetic resonance active and is used in NMR-spectroscopy.

Nucleosynthesis calculations have been made for F{19} stellar yields for a large set of models with different masses and metallicity. During the asymptotic giant branch (AGB) phase,which is a period of stellar evolution undertaken by all low- to intermediate-mass stars (0.6–10 solar masses) late in their lives,F{19} is produced as a consequence of nucleosynthesis occurring during the convective thermal pulses and also during the interpulse periods if protons from the envelope are partially mixed in the top layers of the He intershell, which happen to act as partial intermixing zones.

Answer 2

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Fluorine is relatively smaller in size and is the element with high electro-negative property in periodic table.

It also has very high positive ionization enthalpy.

For fluorine to exhibit any positive oxidation state it has to be bonded to a negative element even higher than it which is not possible, thus fluorine does not exhibit any positive oxidation state.

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