Question: 61: \mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \xrightarrow{\text { heat }} \mathrm{X}+\mathrm{NaBO}_{2}
in the above reaction the product ” X ” is :
(1) \mathrm{H}_{3} \mathrm{BO}_{3}
(2) \mathrm{B}_{2} \mathrm{O}_{3}
(3) \mathrm{Na}_{2} \mathrm{~B}_{2} \mathrm{O}_{5}
(4) \mathrm{NaB}_{3} \mathrm{O}_{5}
Answer: Option (2)
Explanation:
Sodium tetraborate \mathrm{Na}_{2}\mathrm{B}_{4}\mathrm{O}_{7}
on heating undergoes thermal decomposition.
On heating, borax loses a part of its borate structure and forms sodium metaborate
\mathrm{NaBO}_{2} along with boron trioxide.
The reaction can be written as:
\mathrm{Na}_{2}\mathrm{B}_{4}\mathrm{O}_{7} \xrightarrow{\text{heat}} 2\,\mathrm{NaBO}_{2} + \mathrm{B}_{2}\mathrm{O}_{3}From the balanced reaction, the compound formed along with \mathrm{NaBO}_{2} is
\mathrm{B}_{2}\mathrm{O}_{3}.
Therefore, the correct product X is \mathrm{B}_{2}\mathrm{O}_{3}.