Boron of highest purity (∼99.9 %) is obtained by

Question

Boron of highest purity (∼99.9 %) is obtained by (A) reduction of B2O3 by Mg (B) electrolysis of fused tetrafluoroborates (KBF4) in molten KF (C) reduction of BCl3 with H2 at 1270 K (D) thermal decomposition of boranes at 1173 K

Tardigrade Admin · Accepted Answer

Boron of maximum purity (∼99.9 %) is obtained by the reduction of BCl3 with H2 at 1270 K

Q. Boron of highest purity $(\sim 99.9%)$ is obtained by

reduction of $B_{2} O_{3}$ by Mg

electrolysis of fused tetrafluoroborates $(K B F_{4})$ in molten $K F$

reduction of $BC l_{3}$ with $H_{2}$ at $1270 K$

thermal decomposition of boranes at $1173 K$

Boron of maximum purity $(\sim 99.9%)$ is obtained by the reduction of $BC l_{3}$ with $H_{2}$ at $1270 K$

Q. Boron of highest purity (∼99.9%) is obtained by

reduction of B2​O3​ by Mg

electrolysis of fused tetrafluoroborates (KBF4​) in molten KF

reduction of BCl3​ with H2​ at 1270K

thermal decomposition of boranes at 1173K