Final answer:
In the reaction Sr(s) + F2 -> SrF2(s), the oxidation number of fluorine increased.
Step-by-step explanation:
In the reaction Sr(s) + F2 -> SrF2(s), the statement that is true is (C) The oxidation number of fluorine increased. In this reaction, fluorine is being reduced as its oxidation number increases from 0 to -1 in F2 to 0 in SrF2. Strontium, on the other hand, is being oxidized as its oxidation number decreases from 0 in Sr to +2 in SrF2.