The dissolution of $Al(OH)_3$ by a solution of $NaOH$ results in the formation of

The correct statements from the following are:
$(A)$ The decreasing order of atomic radii of group $13$ elements is $Tl > In > Ga > Al > B$.
$(B)$ Down the group $13$ electronegativity decreases from top to bottom.
$(C)$ $Al$ dissolves in dil. $HCl$ and liberates $H_2$,but conc. $HNO_3$ renders $Al$ passive by forming a protective oxide layer on the surface.
$(D)$ All elements of group $13$ exhibit a highly stable $+1$ oxidation state.
$(E)$ Hybridisation of $Al$ in $[Al(H_2O)_6]^{3+}$ ion is $sp^3d^2$.
Choose the correct answer from the options given below:

In which of the following reactions is hydrogen one of the products?
$i$. $2NaBH_4 + I_2 \longrightarrow$
$ii$. $2BF_3 + 6NaH \xrightarrow{450 \ K}$
$iii$. $4BF_3 + 3LiAlH_4 \longrightarrow$
$iv$. $3B_2H_6 + 6NH_3 \xrightarrow{\text{heat}} 2B_3N_3H_6 + 12H_2$

Draw the structures of $BCl_3.NH_3$ and $AlCl_3$ (dimer).

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

$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\text{Heat}} X + NaBO_2 + H_2O$,$X + Cr_2O_3 \xrightarrow{\text{Heat}} \underset{(\text{Green coloured})}{Y}$. $X$ and $Y$ are