Match List-I with List-II : \(\begin{array}{l|l} \text{List-I} & \text{List-II} \\ \text{(Compounds)} &…

Match List-I with List-II : \(\begin{array}{l|l}
\text{List-I} & \text{List-II} \\
\text{(Compounds)} & \text{(Molecular formula)} \\
\hline \text { (a) Borax } & \text { (i) } \mathrm{NaBO}_2 \\
\text { (b) Kernite } & \text { (ii) } \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 4 \mathrm{H}_2 \mathrm{O} \\
\text { (c) Orthoboric acid } & \text { (iii) } \mathrm{H}_3 \mathrm{BO}_3 \\
\text { (d) Borax bead } & \text { (iv) } \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}
\end{array}\) Choose the correct answer from the options given below :
  1. (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
  2. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
  3. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
  4. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

Solution

(a) Borax $-\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}$ or $\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{OH})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}$ (b) Kernite $-\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 4 \mathrm{H}_2 \mathrm{O}$ (c) Orthoboric acid $-\mathrm{H}_3 \mathrm{BO}_3$ (d) Borax bead $-\mathrm{NaBO}_2$ (obtained by heating of borax) $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\stackrel{-10 \mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3$ Sodium metaborate

Asked in: NEET 2022 (Phase 2)

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