BF3→EtherLiAlH4A→473 KNH3B. ‘A’ and ‘B’ are Boron compounds. Incorrect statement about ‘B’ is

4BF3+3LiAlH4→2B2H6+3LiF+3AlF3; 3B2H6+6NH3→473 K2B3N3H6+12H2;Borazine is inorganic benzene. It is aromatic with 12σ and 3π bonds (all the three pi bonds are dative bonds). Both boron and nitrogen atoms are sp2 hybridized. It is a planar molecule.