Explain what happens when boric acid is heated.
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 $\mathrm{B}_{2} \mathrm{O}_{3}$.
$\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{370 \mathrm{~K}}{\longrightarrow} \mathrm{HBO}_{2} \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}$
$BCl_3$ does not exist as dimer but $BH_3$ exist as dimer $(B_2H_6)$ becuase
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:
Give increasing order of group $13$ elements for atomic radius.
Taking stability as the factor, which one of the following represents correct relationship?
In Goldschmidt aluminothermic process, thermite contains