See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(i) 2 2 3 Borax NaBO B O    (ii) 4 2 3 2 Borax NH Cl BN B O NaCl H O       (iii) 3 3 Borax HCl H BO   (iv)   2 3 3 4 Borax H O Na B OH H BO        (v) 2 6 3 3 3 6 2 3B H 6NH 2B N H 12H     (vi) THF 3 3 3 3 4 3 3 6 3 B N H Cl 3LiBH B N H 3LiCl 3BH     (vii) 2 6 4 3 LiH B H 2LiBH sp    (viii)   3 6 3 4 3 3 K AlF BF 3KBF AlF sp     Except (vii) and (viii) all options can give sp2 hybridised Boron centre as one of the product.