[Baksu888]

I was asked to correct the following hypothesis:

The number of F atoms that will bond to a nonmetal is always equal to 8 minus the number of valence electrons in the nonmetal atom.

This seems to apply to several binary compounds, such as CF₄, NF₃, OF₂, etc.

However, this does not apply to compounds that are formed between fluorine and halogens.

For example: ClF, ClF₃; BrF, BrF₃, BrF₅; IF, IF₃, IF₅, IF₇, etc.

I am supposed to propose a modification to the hypothesis based on the concept's of atomic structure and periodicity, but I don't understand why the compounds form the way they do.

The only pattern I was able to observe was that the period of the halogen minus 1 would equal to the number of compounds that are possible to form with fluorine. Also, the atoms of fluorine bonded with the halogen seems to increase by two with each possible compound starting with one.

For example: Br is in period 4, so there are three possible compounds that may form between Br and F: BrF, BrF₃, and BrF₅. The fluorine atoms increase by two with each compound.

[mjc123]

Not just halogens, e.g. PF₃ and PF₅; SF₂, SF₄ and SF₆. And noble gas fluorides.
The rule implies that there are a maximum of 8 electrons in the valence shell. This applies to first-row elements, that only have 2s and 2p orbitals available, but elements lower down the table have low-energy d orbitals available for bonding, so they can "expand the octet" and form higher-valency compounds.
One might restate the rule as predicting the lowest-oxidation-state fluoride for a particular element (though you would then have to consider, say, Kr as "KrF₀"), but allowing that there may be higher fluorides. But then what about boron and BF₃?

[AWK]

There are also chemical compounds that are bound without the participation of valence electrons