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Diborane reacts with HCl in the presence of $AlCl_3$ and liberates:

\[(a)\;H_2 \quad (b)\;Cl_2 \quad (c)\;BCl_3 \quad (d)\;Cl_2\;  \&\; BCl_3 \]

1 Answer

$(a)\;H_2$
answered Nov 7, 2013 by pady_1
 
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Tag:MathPhyChemBioOther
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