Question

$N{a}_2{B}_4{O}_7.10H_{2}O\stackrel{\Delta }{\to }N{a}_2{B}_4{O}_7++10H_{2}O\stackrel{{740}^{o}C}{\to }2NaB{O}_2+A$, $A+C{r}_2{O}_3\stackrel{\Delta }{\to }B$
$A$ and $B$ respectively are:

• A. $N{a}_{3}B{O}_3,M{n}_3(B{O}_3{)}_2$
• B. $N{a}_2(B{O}_2{)}_2,Mn(B{O}_2{)}_2$
• C. $B_2{O}_3,Cr(B{O}_2{)}_3$
• D. None is correct

Answer 1

The correct option is C $B_2{O}_3,Cr(B{O}_2{)}_3$

On heating, it first swells due to​ loss of water in the form of steam​.
$N{a}_2{B}_4{O}_7.10H_{2}O\to N{a}_2{B}_4{O}_7+10H_{2}O\uparrow$​
At ${740}^{\circ }C$, it is converted into​ colourless transparent borax bead.
$N{a}_2{B}_4{O}_7\stackrel{{740}^{\circ }C}{\to }2NaB{O}_2+B_2{O}_3\text{(borax bead)}$​
$B_2{O}_3+C{r}_2{O}_3\stackrel{\Delta }{\to }Cr(B{O}_2{)}_3$
Hence, the correct answer is option (c).