The geometry around boron in the product ' B ' formed from the following reaction is BF3+NaH450KA+NaF A+NMe3→B
(A)trigonal planar
(B)tetrahedral✓ Correct
(C)pyramidal
(D)square planar
Answer: (B)
Solution
BF3 reacts with NaH at 450 K to form diborane (B2H6) as product A. Diborane undergoes symmetrical cleavage with NMe3 to form the adduct H3B⋅NMe3. In this adduct, boron is sp³ hybridized with tetrahedral geometry.
JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option