A. i.sodium hydride ii.sodium iii.methanol
B. i.sodium Ii.methanol iii.sodium hydride
C. i.sodium ii.sodium hydride iii.methanol
D. i.sodium hydride ii.methanol iii.sodium
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This key question is from Inorganic Chemical in portion Inorganic Chemical Industries of Chemical Technology
To elaborate: The reaction is 7NaH + B(OCH3)3 → NaBH4 + 6Na + 3CH3OH, hence compound i is sodium hydride, compound ii is sodium and compound iii is methanol.