Q69 / 200 · NEET 2022 (Phase 2) · Chemistry

$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\text { heat }}{\longrightarrow} \mathrm{X}+\mathrm{NaBO}_2$ in the above reaction the product " $\mathrm{X}$ " is :

  • A
    $\mathrm{H}_3 \mathrm{BO}_3$
  • Correct answer
    B
    $\mathrm{B}_2 \mathrm{O}_3$
  • C
    $\mathrm{Na}_2 \mathrm{~B}_2 \mathrm{O}_5$
  • D
    $\mathrm{NaB}_3 \mathrm{O}_5$

Solution

$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} \underset{-10 \mathrm{H}_2 \mathrm{O}}{\stackrel{\Delta}{\longrightarrow}} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$ $\begin{array}{r} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3 \\ (\mathrm{X}) \end{array}$ $\mathrm{X}=$ Boric anhydride

Hence, the correct answer is option: 'B'.

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