In Borax bead test which compound is formed:
(A) Ortho borax
(B) meta borax
(C) double oxide
(D) tetra borax
Answer
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Hint: For this problem, we have to study the Borax bead test and its equation. As it is a hydrous compound so firstly it will undergo heating and produce anhydrous form. After this, it will melt to give two compounds which will be the correct answer.
Complete step by step solution:
- In the given question, we have to choose the product which will be formed in the borax bead test in which mostly coloured salts are used.
- As we know that borax is also known as sodium pyroborate which has the molecular formula of $\text{N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}.\text{10}{{\text{H}}_{2}}\text{O}$.
- When borax is heated it yields sodium tetraborate and 10 molecules of water as shown below:
$\text{N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}.\text{10}{{\text{H}}_{2}}\text{O }\to \text{ N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}\text{ + 10}{{\text{H}}_{2}}\text{O}$
- This process of separation of the water molecule is also termed as a loss of water of crystallization because the water molecules from the lattice structure are lost.
- After this, the compound will swell into a white coloured and porous mass which is melted to give a colourless liquid.
- Later, this liquid will form transparent bead in the loop of the platinum wire that consists of two compounds that are boric anhydride and sodium metaborate as shown below:
$\text{N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}\text{ }\to \text{ }{{\text{B}}_{2}}{{\text{O}}_{3}}\ \text{+ 2NaB}{{\text{O}}_{2}}$
Here, ${{\text{B}}_{2}}{{\text{O}}_{3}}$ is boric anhydride which is non-volatile whereas $\text{2NaB}{{\text{O}}_{2}}$ is known as sodium metaborate.
Therefore, option (B) is the correct answer.
Note: The borax bead test is usually used in the detection of the colour of the cation. When the newly formed bead comes in contact with coloured salt, it gives a characteristic colour for each cation. For example, copper salt gives a blue colour.
Complete step by step solution:
- In the given question, we have to choose the product which will be formed in the borax bead test in which mostly coloured salts are used.
- As we know that borax is also known as sodium pyroborate which has the molecular formula of $\text{N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}.\text{10}{{\text{H}}_{2}}\text{O}$.
- When borax is heated it yields sodium tetraborate and 10 molecules of water as shown below:
$\text{N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}.\text{10}{{\text{H}}_{2}}\text{O }\to \text{ N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}\text{ + 10}{{\text{H}}_{2}}\text{O}$
- This process of separation of the water molecule is also termed as a loss of water of crystallization because the water molecules from the lattice structure are lost.
- After this, the compound will swell into a white coloured and porous mass which is melted to give a colourless liquid.
- Later, this liquid will form transparent bead in the loop of the platinum wire that consists of two compounds that are boric anhydride and sodium metaborate as shown below:
$\text{N}{{\text{a}}_{2}}{{\text{B}}_{4}}{{\text{O}}_{7}}\text{ }\to \text{ }{{\text{B}}_{2}}{{\text{O}}_{3}}\ \text{+ 2NaB}{{\text{O}}_{2}}$
Here, ${{\text{B}}_{2}}{{\text{O}}_{3}}$ is boric anhydride which is non-volatile whereas $\text{2NaB}{{\text{O}}_{2}}$ is known as sodium metaborate.
Therefore, option (B) is the correct answer.
Note: The borax bead test is usually used in the detection of the colour of the cation. When the newly formed bead comes in contact with coloured salt, it gives a characteristic colour for each cation. For example, copper salt gives a blue colour.
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