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Answer:
The smaller size, higher electronegativity, and the absence of vacant d-orbital make fluorine atom exist in only one oxoacid, HOF·
Explanation:
Fluorine is an element that belongs to the seventeenth group, the halogen family· As we know going down the group, the atomic size decreases· So fluorine is the smallest element in the halogen family·
Fluorine atom will have an electronic configuration of [tex]1s^{2} \ 2s^{2}\ 2p^{5}[/tex]· From this configuration, it is clear that fluorine atom does not have any vacant d-orbitals· So it is not possible to exist in higher oxidation states such as [tex]+3[/tex],[tex]+5[/tex], etc· The oxidation state of fluorine in HOF is [tex]+1[/tex] ·But other halogens have vacant d orbitals in its configuration which helps them to have [tex]+3,+5[/tex] and [tex]+7[/tex] oxidation states, respectively· So it is possible to them to have various oxoacids·
Also, fluorine has a higher electronegativity· It has an electronegative value of [tex]3.98.[/tex] The valence [tex]2[/tex]p orbital will have [tex]7[/tex] electrons which are one electron less than the ideal electronic configuration· Due to this higher electronegativity, fluorine is reluctant to form various positive oxidation states and not able to form any other oxoacids·
So the smaller atomic size, higher electronegativity, and the absence of vacant d-orbital make fluorine form only one oxoacid, HOF·
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Author:
pantherglass
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