Search any question & find its solution
Question:
Answered & Verified by Expert
Borax is converted into crystalline boron by the following steps :
$\text { Borax } \underset{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{H}_3 \mathrm{BO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{B}_2 \mathrm{O}_3 \underset{\Delta}{\stackrel{Y}{\longrightarrow}} B$
Identify $X$ and $Y$ respectively.
Options:
$\text { Borax } \underset{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{H}_3 \mathrm{BO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{B}_2 \mathrm{O}_3 \underset{\Delta}{\stackrel{Y}{\longrightarrow}} B$
Identify $X$ and $Y$ respectively.
Solution:
2243 Upvotes
Verified Answer
The correct answer is:
$\mathrm{HCl}$ and $\mathrm{Mg}$
In the given reaction :
$(X)=\mathrm{HCl}$
$(Y)=\mathrm{Mg}$
The reaction occurs as follows :
Borax $\underset{\mathrm{H}_2 \mathrm{O}}{\stackrel{\mathrm{HCl}(X)}{\longrightarrow}} \mathrm{H}_3 \mathrm{BO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{BrO}_3$$\underset{\Delta}{\stackrel{\operatorname{Mg}(Y)}{\longrightarrow}}$(B) Product
$\left\{\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{HO})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}\right\}$
(i) $\begin{aligned} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}+2 \mathrm{HCl} \longrightarrow & 4 \mathrm{~B}(\mathrm{OH})_3 \\ & +2 \mathrm{NaCl}+5 \mathrm{H}_2 \mathrm{O}\end{aligned}$
(ii) $\mathrm{H}_3 \mathrm{BO}_3+\mathrm{Mg} \stackrel{\Delta}{\longrightarrow} \mathrm{Mg}\left(\mathrm{BO}_2\right)_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2$
$(X)=\mathrm{HCl}$
$(Y)=\mathrm{Mg}$
The reaction occurs as follows :
Borax $\underset{\mathrm{H}_2 \mathrm{O}}{\stackrel{\mathrm{HCl}(X)}{\longrightarrow}} \mathrm{H}_3 \mathrm{BO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{BrO}_3$$\underset{\Delta}{\stackrel{\operatorname{Mg}(Y)}{\longrightarrow}}$(B) Product
$\left\{\mathrm{Na}_2\left[\mathrm{~B}_4 \mathrm{O}_5(\mathrm{HO})_4\right] \cdot 8 \mathrm{H}_2 \mathrm{O}\right\}$
(i) $\begin{aligned} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}+2 \mathrm{HCl} \longrightarrow & 4 \mathrm{~B}(\mathrm{OH})_3 \\ & +2 \mathrm{NaCl}+5 \mathrm{H}_2 \mathrm{O}\end{aligned}$
(ii) $\mathrm{H}_3 \mathrm{BO}_3+\mathrm{Mg} \stackrel{\Delta}{\longrightarrow} \mathrm{Mg}\left(\mathrm{BO}_2\right)_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.