
Amorphous boron of 95% to 98% purity is obtained by?
A. Heating ${{B}_{2}}{{O}_{3}}$ with ${{H}_{2}}$
B. Heating ${{B}_{2}}{{O}_{3}}$ with Mg or Na at high temperature
C. Heating $KB{{F}_{4}}$ with Na or K
D. Heating $BB{{r}_{3}}$ with ${{H}_{2}}$ in presence of a catalyst
Answer
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Hint: Amorphous boron is not a pure form of boron and is present as a brown powder. It can be obtained through the reduction of boric acid with metals such as aluminium and magnesium and metal boride is present as a contamination in products that are formed.
Complete step by step solution:
Boron, a non-metallic element, exists in several allotropic forms. It normally occurs in the form of orthoboric acid or borate$({{H}_{3}}B{{O}_{3}})$ . Amorphous boron exists as a dark brown powder and is the more common allotrope of boron. Amorphous remain unreactive towards alkalis, oxygen, water and acids.
In amorphous boron variable amounts of impurities are present which depends on the method of purification and preparation.
The main process through which we can prepare amorphous boron is known as magnesiothermic reduction of boric acid (${{B}_{2}}{{O}_{3}}$ ) . Thermic word is used for heat or energy.
From this result it has been that 95% to 98% purity is gotten with the help of stoichiometric ratio of $\dfrac{{{B}_{2}}{{O}_{3}}}{Mg}$that is 3.0, at the temperature of 850${}^\circ C$ ${}^\circ C$ and at the constant temperature of 30 min.
The ${{B}_{2}}{{O}_{3}}$ particle size of 148-180 µm and the Mg particle size of 74-148 µm takes place in this process.
During this process the excess boric acid combines with a combustion product that is MgO into compounds such as $xMgO.{{B}_{2}}{{O}_{3}}$.
Amorphous boron is obtained by the reduction of boric acid with Mg or Na at high temperature and thus the boron that is obtained is 95%-98% pure and has contamination of metal borides and is black in color.
\[{{H}_{3}}B{{O}_{3}}\xrightarrow{\Delta }{{B}_{2}}{{O}_{3}}\xrightarrow{Mg\,or\,Na}2B+MgO\]
Thus the correct option will be (B).
Note: Amorphous boron of low purity is defined as Moissan boron. Moissan boron is also formed through the reduction of boric anhydride. The impurities of this compound are present in the form of Na particles. The color of this compound is black with 95-98% purity. If sodium is reacting with boric acid itr will produce sodium oxide.
Complete step by step solution:
Boron, a non-metallic element, exists in several allotropic forms. It normally occurs in the form of orthoboric acid or borate$({{H}_{3}}B{{O}_{3}})$ . Amorphous boron exists as a dark brown powder and is the more common allotrope of boron. Amorphous remain unreactive towards alkalis, oxygen, water and acids.
In amorphous boron variable amounts of impurities are present which depends on the method of purification and preparation.
The main process through which we can prepare amorphous boron is known as magnesiothermic reduction of boric acid (${{B}_{2}}{{O}_{3}}$ ) . Thermic word is used for heat or energy.
From this result it has been that 95% to 98% purity is gotten with the help of stoichiometric ratio of $\dfrac{{{B}_{2}}{{O}_{3}}}{Mg}$that is 3.0, at the temperature of 850${}^\circ C$ ${}^\circ C$ and at the constant temperature of 30 min.
The ${{B}_{2}}{{O}_{3}}$ particle size of 148-180 µm and the Mg particle size of 74-148 µm takes place in this process.
During this process the excess boric acid combines with a combustion product that is MgO into compounds such as $xMgO.{{B}_{2}}{{O}_{3}}$.
Amorphous boron is obtained by the reduction of boric acid with Mg or Na at high temperature and thus the boron that is obtained is 95%-98% pure and has contamination of metal borides and is black in color.
\[{{H}_{3}}B{{O}_{3}}\xrightarrow{\Delta }{{B}_{2}}{{O}_{3}}\xrightarrow{Mg\,or\,Na}2B+MgO\]
Thus the correct option will be (B).
Note: Amorphous boron of low purity is defined as Moissan boron. Moissan boron is also formed through the reduction of boric anhydride. The impurities of this compound are present in the form of Na particles. The color of this compound is black with 95-98% purity. If sodium is reacting with boric acid itr will produce sodium oxide.
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