In $$Si{O_2}$$ (quartz), each of $$O$$ - atom is shared between two $$SiO_4^{4 - }$$ tetrahedra.
382.
A metal $$X$$ reacts with aqueous $$NaOH$$ solution to form $$Y$$ and a highly inflammable gas. Solution $$Y$$ is heated and $$C{O_2}$$ is poured through it. $$Z$$ precipitates out and $$N{a_2}C{O_3}$$ is formed. $$Z$$ on heating gives $$A{l_2}{O_3}.$$ Identify $$X, Y$$ and $$Z.$$
$$\mathop {2Al}\limits_{\left( X \right)} + 2NaOH + 2{H_2}O \to $$ $$\mathop {2NaAl{O_2}}\limits_{\left( Y \right)} + 3{H_2}$$
$$\mathop {2NaAl{O_2}}\limits_{\left( Y \right)} + C{O_2} + 3{H_2}O \to $$ $$\mathop {2Al{{\left( {OH} \right)}_3}}\limits_{\left( Z \right)} + N{a_2}C{O_3}$$
\[\underset{\left( Z \right)}{\mathop{2Al{{\left( OH \right)}_{3}}}}\,\xrightarrow{\Delta }A{{l}_{2}}{{O}_{3}}+3{{H}_{2}}O\]
383.
Phosphorous acid on heating gives the following products : \[4{{H}_{3}}P{{O}_{3}}\xrightarrow{\Delta }3{{H}_{3}}P{{O}_{4}}+P{{H}_{3}}\]
The above reaction is an example of
\[4{{H}_{3}}P{{O}_{3}}\xrightarrow{\Delta }3{{H}_{3}}P{{O}_{4}}+P{{H}_{3}}\]
In the reaction, the acid in + 3 oxidation state of $$P$$ tends to disproportionate to higher (+ 5) and lower (- 3) oxidation state in $${H_3}P{O_4}$$ and $$P{H_3}$$ respectively.
384.
Aluminium chloride exists as dimer, $$A{l_2}C{l_6}$$ in solid state as
well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives
A
$${\left[ {Al{{\left( {OH} \right)}_6}} \right]^{3 - }} + 3HCl$$
385.
Which of the following statements are incorrect?
(i) Dry slaked lime reacts with $$C{l_2}$$ to give calcium chlorate.
(ii) With excess chlorine ammonia forms nitrogen trichloride.
(iii) Sulphur reacts with $$C{l_2}$$ to give $$SC{l_4}.$$
(iv) With hot and cone. $$NaOH,C{l_2}$$ gives sodium chlorate.
If chlorine is passed through dry slaked lime, bleaching powder is produced.
$$\mathop {Ca{{\left( {OH} \right)}_2}}\limits_{{\text{Slaked lime}}} + C{l_2} \to $$ $$\mathop {Ca\left( {OCl} \right)Cl}\limits_{{\text{Bleaching powder}}} + {H_2}O$$
$${S_8} + 4C{l_2} \to \mathop {4{S_2}C{l_2}}\limits_{{\text{Sulphur monochloride}}} $$
386.
$$NaB{H_4} + {I_2} \to X + Y + Z$$
\[B{{F}_{3}}+NaH\xrightarrow{450\,K}X+P\]
$$B{F_3} + LiAl{H_4} \to X + Q + R$$
$$X, Y, Z, P, Q$$ and $$R$$ in the reactions are
$$2NaB{H_4} + {I_2} \to \mathop {{B_2}{H_6}}\limits_{\left( X \right)} + \mathop {2NaI}\limits_{\left( Y \right)} + \mathop {{H_2}}\limits_{\left( Z \right)} $$
\[2B{{F}_{3}}+6NaH\xrightarrow{450\,K}\underset{\left( X \right)}{\mathop{{{B}_{2}}{{H}_{6}}}}\,+\underset{\left( P \right)}{\mathop{6NaF}}\,\]
$$4B{F_3} + 3LiAl{H_4} \to $$ $$\mathop {2{B_2}{H_6}}\limits_{\left( X \right)} + \mathop {3LiF}\limits_{\left( Q \right)} + \mathop {3Al{F_3}}\limits_{\left( R \right)} $$
387.
In the following reaction sequence in aqueous solution, the species $$X,Y\,{\text{and}}\,Z,$$ respectively, are
\[{{S}_{2}}O_{3}^{2-}\xrightarrow{A{{g}^{+}}}\underset{\begin{smallmatrix}
\text{clear} \\
\text{solution}
\end{smallmatrix}}{\mathop{X}}\,\xrightarrow{A{{g}^{+}}}\underset{\begin{smallmatrix}
\text{white} \\
\text{precipitate }
\end{smallmatrix}}{\mathop{Y}}\,\] \[\xrightarrow{\text{with time }}\underset{\text{black precipitate }}{\mathop{Z}}\,\]
A
$${\left[ {Ag{{\left( {{S_2}{O_3}} \right)}_2}} \right]^{3 - }},A{g_2}{S_2}{O_3},A{g_2}S$$
B
$${\left[ {Ag{{\left( {{S_2}{O_3}} \right)}_3}} \right]^{5 - }},A{g_2}S{O_3},A{g_2}S$$
C
$${\left[ {Ag{{\left( {S{O_3}} \right)}_2}} \right]^{3 - }},A{g_2}{S_2}{O_3},Ag$$
D
$${\left[ {Ag{{\left( {S{O_3}} \right)}_3}} \right]^{3 - }},A{g_2}S{O_4},Ag$$
Structure of graphite consist of a two dimensional sheet like network joined together in hexagonal rings. These layers are held together by weak van der Waals’ forces. In graphite each carbon atom is bonded to three others, forming $$s{p^2}$$ hybrid bonds. The fourth electron forms a $$\pi $$-bond.
Graphite is a conductor of electricity which is due to
the fact that all the carbon bonds being not
satisfied. Thus, some of the electrons are free to
move through the crystal.
389.
An aqueous solution of sodium carbonate absorbs $$NO$$ and $$N{O_2}$$ to give
The key step in the manufacture of $${H_2}S{O_4}$$ is catalytic oxidation of $$S{O_2}$$ with $${O_2}$$ to give $$S{O_3}$$ in presence of $${V_2}{O_5}.$$