Question

When a microcosmic salt is heated, a transparent bead formed is:
A) $NaP{O_3}$
B) $Na(N{H_4})HP{O_4}$
C) $NaB{O_2}$
D) ${B_2}{O_3}$

Answer 1

Hint: In the microscopic salt bead test we use the tetra hydrated ammonium sodium phosphate which produces the transparent glass-like bead. One can think about the reaction and guess the possible products and select the correct choice.

Complete step by step answer:
1) First of all we will learn about the microscopic salt bead test where the reaction is conducted by making a small loop on a platinum metal wire.

2) The platinum metal wire is then immersed in the solution ${\text{Conc}} \cdot {\text{HCl}}$ so as to remove impurities and deposits which previously remained on the wire. The concentrated hydrochloric acid removes the impurities by converting them into the volatile chlorides which really become gases into the flame of a Bunsen burner.

3) The above process is repeated more than once until we do not get the colorful emission in flame. Now a bit of chemical $N{H_4}NaHP{O_4} \cdot 4{H_2}O$ is added to the loop which was previously practiced on the top of our wire.

4) The reaction scheme of the process is as follows,
$N{H_4}NaHP{O_4} \cdot 4{H_2}O \rightleftarrows N{H_3} + NaP{O_3} + 5{H_2}O$
Where the ammonia and water molecules are released in the gas form. The molecule $NaP{O_3}$ formed is a transparent bead.

5) Therefore, when a microcosmic salt is heated, a transparent bead formed is $NaP{O_3}$

Hence, option A is the correct answer.

Note:
The monohydrogen phosphate which is $N{H_4}NaHP{O_4} \cdot 4{H_2}O$ loses ammonia and crystallization water in gas form and melts to give a colorless and clear glass-like bead of sodium metaphosphate i.e. $NaP{O_3}$. The formed bead is then reacted with copper and then it is used for the recognition of halide present based on the color of the flame around the bead.