The ratio of sp, sp2 and sp3 orbitals in the product ‘B’ formed from the reaction belowCH3−CH=CH−C≡CH→−NH3NaNH2A→−NaBrCH3−BrB

[Step 1]: terminal alkyne on reaction with sodamide  forms sodium acetylide, [Step 2]: Sodium acetylide on reaction with methyl halide  forms compound B.  The steps are respectivelyCompound ‘B’ is having 2 sp carbons, 2 sp2 carbons and 2 sp3 carbons Total sp orbitals are 2*2=4, total sp2 orbitals are 2*3=6 and total sp3 orbitals are 2*4=8  Therefore the ratio is 2:3:4