Explain what happens when boric acid is heated. 

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On heating orthoboric acid $\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)$ at $370 \mathrm{~K}$ or above, it changes to metaboric acid $\left(\mathrm{HBO}_{2}\right)$. On further heating, this yields boric oxide $\mathrm{B}_{2} \mathrm{O}_{3}$.

$\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{370 \mathrm{~K}}{\longrightarrow} \mathrm{HBO}_{2} \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}$

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The correct statements from the following are :

$(A)$ The decreasing order of atomic radii of group $13$ elements is $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$.

$(B)$ Down the group $13$ electronegativity decreases from top to bottom.

$(C)$ $\mathrm{Al}$ dissolves in dil. $\mathrm{HCl}$ and liberate $\mathrm{H}_2$ but conc. $\mathrm{HNO}_3$ renders Al passive by forming a protective oxide layer on the surface.

$(D)$ All elements of group 13 exhibits highly stable +1 oxidation state.

$(E)$ Hybridisation of $\mathrm{Al}$ in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion is $\mathrm{sp}^3 \mathrm{~d}^2$.

Choose the correct answer from the options given below:

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Give increasing order of group $13$ elements for atomic radius. 

Taking stability as the factor, which one of the following represents correct relationship?

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In Goldschmidt aluminothermic process, thermite contains