Statement I : The oxidation number of boron in BF₃ is +3
Statement II : Boron cannot form a cation

${\mathrm{BF}}_3$

$\mathrm{x}+3\left(-1\right)=0$

$\mathrm{x}=+3$

Therefore, the oxidation state of B in BF₃ is +3.

→Boron can't form cation, due to its small size and high sum of first three ionization energies. Therefore, it does not lose its three valency electrons and does not form B⁺³ cation.