What happens when
$(a)$ Borax is heated strongly,
$(b)$ Boric acid is added to water,
$(c)$ Aluminium is treated with dilute $NaOH$,
$(d)$ $BF_3$ is reacted with ammonia?
What happens when
$(a)$ Borax is heated strongly,
$(b)$ Boric acid is added to water,
$(c)$ Aluminium is treated with dilute $NaOH$,
$(d)$ $BF_3$ is reacted with ammonia?
$(a)$ When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.
$\mathop {N{a_2}{B_4}{O_7}.10{H_2}O}\limits_{Borax} \xrightarrow{\Delta }\mathop {N{a_2}{B_4}{O_7}}\limits_{Sodium\,metaborate} \xrightarrow{\Delta }2NaB{O_2} + \mathop {{B_2}{O_3}}\limits_{Boric\,anhydride} $
$(b)$ When boric acid is added to water, it accepts electrons from $-OH$ ion.
$B ( OH )_{3}+2 HOH \longrightarrow\left[ B ( OH )_{4}\right]^{-}+ H _{3} O ^{+}$
$(c)$ $Al$ reacts with dilute $NaOH$ to form sodium tetrahydroxoaluminate $(III)$. Hydrogen gas is liberated in the process.
$2 Al _{(s)}+2 NaOH _{(aq)}+6 H _{2} O _{(l)} \longrightarrow 2 Na ^{+}\left[ Al ( OH )_{4}\right]_{(aq)}+3 H _{2( g )}$
$(d)$ $BF_3$ (a Lewis acid) reacts with $NH_ 3$ (a Lewis base) to form an adduct. This results in a complete octet around $B$ in $ BF_3$.
$F _{3} B +: NH _{3} \longrightarrow F _{3} B \leftarrow : NH _{3}$
Similar Questions
In the electrolysis of alumina, cryolite is added to
In the electrolysis of alumina, cryolite is added to
- [IIT 1986]
Match List $I$ with List $II$. Choose the correct answer from the options given below:
List $-I$
List $-II$
$A.$ Melting point $[\mathrm{K}]$
$I.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$
$B.$ Ionic Radius $\left[\mathrm{M}^{+3} / \mathrm{pm}\right]$
$II.$ $\mathrm{B}>\mathrm{Tl}>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In}$
$C.$ $\Delta_{\mathrm{i}} \mathrm{H}_1 $ $ [\mathrm{~kJ} \mathrm{~mol}^{-1}]$
$III.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B}$
$D.$ Atomic Radius $[pm]$
$IV.$ $\mathrm{B}>\mathrm{Al}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}$
Match List $I$ with List $II$. Choose the correct answer from the options given below:
| List $-I$ | List $-II$ |
|
$A.$ Melting point $[\mathrm{K}]$ |
$I.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$ |
|
$B.$ Ionic Radius $\left[\mathrm{M}^{+3} / \mathrm{pm}\right]$ |
$II.$ $\mathrm{B}>\mathrm{Tl}>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In}$ |
| $C.$ $\Delta_{\mathrm{i}} \mathrm{H}_1 $ $ [\mathrm{~kJ} \mathrm{~mol}^{-1}]$ | $III.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B}$ |
| $D.$ Atomic Radius $[pm]$ | $IV.$ $\mathrm{B}>\mathrm{Al}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}$ |
- [JEE MAIN 2024]
The type of hybridisation of boron in diborane is
The type of hybridisation of boron in diborane is
$H_3BO_3$ $\xrightarrow{{{T_1}}}X\xrightarrow{{{T_2}}}Y\xrightarrow{{red\,hot}}$ $B_2O_3$
if $T_1 < T_2$ then $X$ and $Y$ respectively are
$H_3BO_3$ $\xrightarrow{{{T_1}}}X\xrightarrow{{{T_2}}}Y\xrightarrow{{red\,hot}}$ $B_2O_3$
if $T_1 < T_2$ then $X$ and $Y$ respectively are
Select the incorrect statement
$(a)$ $PCl_5$ form weak monobasic acid during hydrolysis
$(b)$ In $B_3N_3H_6$ nucleophile attack on $B-$ atom
$(c)$ $Al_2Cl_6$ is a polar and planar molecule
$(d)$ In $AlF_3$, hybridisation of $Al$ is $sp^2$
Select the incorrect statement
$(a)$ $PCl_5$ form weak monobasic acid during hydrolysis
$(b)$ In $B_3N_3H_6$ nucleophile attack on $B-$ atom
$(c)$ $Al_2Cl_6$ is a polar and planar molecule
$(d)$ In $AlF_3$, hybridisation of $Al$ is $sp^2$