H_3BO_3 xrightarrow{T_1} X xrightarrow{T_2} Y | vedclass

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(A) The thermal decomposition of orthoboric acid $(H_3BO_3)$ occurs in steps:$1$. At $100^{\circ}C$ $(T_1)$,$H_3BO_3$ loses water to form metaboric ac

Vedclass · Accepted Answer

$X = 	ext{Metaboric acid and } Y = 	ext{Tetraboric acid}$

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(A) The thermal decomposition of orthoboric acid $(H_3BO_3)$ occurs in steps:$1$. At $100^{\circ}C$ $(T_1)$,$H_3BO_3$ loses water to form metaboric acid $(HBO_2)$:$H_3BO_3 \xrightarrow{100^{\circ}C} HBO_2 + H_2O$$2$. At $160^{\circ}C$ $(T_2)$,metaboric acid $(HBO_2)$ converts into tetraboric acid $(H_2B_4O_7)$:$4HBO_2 \xrightarrow{160^{\circ}C} H_2B_4O_7 + H_2O$$3$. On further strong heating (red hot),tetraboric acid decomposes into boron trioxide $(B_2O_3)$:$H_2B_4O_7 \xrightarrow{\Delta} 2B_2O_3 + H_2O$Thus,$X$ is metaboric acid and $Y$ is tetraboric acid.

$H_3BO_3$ $\xrightarrow{T_1} X$ $\xrightarrow{T_2} Y$ $\xrightarrow{\text{red hot}} B_2O_3$
If $T_1 < T_2$,then $X$ and $Y$ respectively are:

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$H_3BO_3$ $\xrightarrow{T_1} X$ $\xrightarrow{T_2} Y$ $\xrightarrow{\text{red hot}} B_2O_3$If $T_1 < T_2$,then $X$ and $Y$ respectively are: