Amorphous boron of 95% to 98% purity is ...

Amorphous boron of `95%` to `98%` purity is obtained by:

heating `B_(2)O_(3)` with `H_(2)`

heating `B_(2)O_(3)` with `Mg` or `Na` at high temperature.

heating `KBF_(4)` with `Na` or `K`.

heating `BBr_(3)` with `H_(2)` in presence of a catalyst.

Verified by Experts

The correct Answer is:

`B_(2)O_(3)+3Mg` or `Naunderset(Delta)overset("High temperature")to2B+3MgO //Na_(2)O`

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

reduction of `B_(2)O_(2)` by `Mg`.

electrolysis of fused tetrafluoroborates `(KBF_(4))` in molten `KF`.

reduction of `BCl_(3)` with `H_(2)` at `1270 K`.

thermal decomposition of boranes at `1173 K`.

`B(OH)_(3)`

Mixture of `B_(2)O_(3)` and `BN`

Only `B_(2)O_(3)`

Only `BN`

Electrolysis of pure water

Electrolysis of warm aqueous Barium hydroxide

Action of Zn on NaOH

Electrolysis of acidulated water