Assertion (A) : Al forms ${\left[{\mathrm{AlF}}_6\right]}^{-3}$but ‘B’ does not form ${\left[{\mathrm{BF}}_6\right]}^{-3}$

Reason (R) : B does not react with $F_2$

Al forms ${\left[{\mathrm{AlF}}_6\right]}^{-3}$ but not B, because it lacks d orbitals. Hence, it cannot have more than 8 electrons in its valence shell. 
B reacts with fluorine to form [BF₃​].