A compound $(A)$ of boron reacts with $NMe_3$ to give an adduct $(B)$ which on hydrolysis gives a compound $(C) $ and hydrogen gas. Compound $(C)$ is an acid. Identify the compounds $A, B$ and $C.$ Give the reactions involved.
Compound $(A)$ of boron reacts with NMe $_{3}$ and gives an adduct $(B)$ thus compound $(A)$ is Lewis
acid. Since $(B)$ on hydrolysis gives an acid $(C)$ and $\mathrm{H}_{2}$ gas, therefore (A) is $\mathrm{B}_{2} \mathrm{H}_{6}$, [B] isan
adduct $2 \mathrm{BH}_{3} \mathrm{NM} e_{3}$ and $(C)$ is boric acid.
Reactions are as follows:
$\quad \mathrm{B}_{2} \mathrm{H}_{6}+2 \mathrm{NM} e_{3} \rightarrow 2 \mathrm{BH}_{3} \mathrm{NMe}_{3}$
Diborane $(A)$ Adduct $(B)$ $\quad$ Adduct $(B)$
$\mathrm{BH}_{3} \mathrm{NM} e_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{BO}_{3}+\mathrm{NM} e_{3}+6 \mathrm{H}_{2}$
Write balanced equations for:
$(i)$ $BF _{3}+ LiH \rightarrow$
$(ii)$ $B _{2} H _{6}+ H _{2} O \rightarrow$
$(\text { iii }) NaH + B _{2} H _{6} \rightarrow$
$(i v) H_{3} B O_{3} \stackrel{\Delta}{\longrightarrow}$
$( v ) Al + NaOH \rightarrow$
$( v i ) B _{2} H _{6}+ NH _{3} \rightarrow$
$A{l_2}{O_3}$ formation involves evolution of a large quantity of heat, which makes its use in
The incorrect statement regarding $'X'$ in given reaction is $B{F_3} + LiAl{H_4}\xrightarrow{{Ether}}\left( X \right) + LiF + Al{F_3}$
Explain ionization enthalpy and electro-negativity for elements of Boron family.
Lithium aluminium hydride reacts with silicon tetrachloride to form