$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O.S.{\text{ of }}S \cr & {\text{Dithionic acid}}\,{\text{ - }}\,{H_2}{S_2}{O_6}\,\,\,\,\,\, = + 5 \cr & {\text{Sulphurous acid}}\,{\text{ - }}\,{H_2}S{O_3}\,\,\, = + 4 \cr & {\text{Sulphuric acid}}\,{\text{ - }}\,{H_2}S{O_4}\,\,\,\,\,\,\, = + 6 \cr & {\text{Disulphuric acid}}\,{\text{ - }}\,{H_2}{S_2}{O_7} = + 6 \cr} $$

\[\begin{align} & \left( \text{A} \right)Ba{{\left( {{N}_{3}} \right)}_{2}}\xrightarrow{\Delta }Ba+3{{N}_{2}} \\ & \left( \text{B} \right){{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }C{{r}_{2}}{{O}_{3}}+{{N}_{2}}+4{{H}_{2}}O \\ & \left( \text{C} \right)N{{H}_{4}}N{{O}_{2}}\xrightarrow{\Delta }{{N}_{2}}+2{{H}_{2}}O \\ & \left( \text{D} \right){{\left( N{{H}_{4}} \right)}_{2}}S{{O}_{4}}\xrightarrow{\Delta }2N{{H}_{3}}+{{H}_{2}}S{{O}_{4}} \\ & N{{H}_{3}}\,\text{ is evolved in reaction (C)}\text{. } \\ \end{align}\]

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$$Cs{I_3}$$  dissociates as
$$Cs{I_3} \to C{s^ + } + I_3^ - $$

\[N{{H}_{4}}Cl+KN{{O}_{2}}\xrightarrow{\Delta }\left[ N{{H}_{4}}N{{O}_{2}} \right]\]       \[\xrightarrow{\Delta }{{N}_{2}}+2{{H}_{2}}O\]

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As we move from $$B$$ to $$Al,$$  the sum of $${\Delta _i}{H_1} + {\Delta _i}{H_2} + {\Delta _i}{H_3}$$     decreases substantially $$\left( {6887\,kJ\,mo{l^{ - 1}}\,{\text{to}}\,5137\,kJ\,mo{l^{ - 1}}} \right)$$       due to increase in the atomic size and hence $$Al$$  has a high tendency to lose electrons. Since the electrode potentials $$E_{\frac{{{M^{3 + }}}}{M}}^ \circ $$  increase from $$Al$$  to $$Tl,$$  therefore, their electropositive character decreases, i.e., $$Al\left( { - 1.66\,V} \right)$$   to $$Ga\left( { - 0.56\,V} \right)$$   to $$In\left( { - 0.34\,V} \right)$$   to $$Tl\left( { + 1.26\,V} \right)$$   accordingly.

\[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to \underset{\begin{smallmatrix} \text{Sodium} \\ \text{tetrathionate} \end{smallmatrix}}{\mathop{N{{a}_{2}}{{S}_{4}}{{O}_{6}}}}\,+2NaI\]

Anhydrous $$AlC{l_3}$$  is hydrolysed in moist air and fumes of $$HCl$$  are given out.

Planar $$B{O_3}$$  units are joined by hydrogen bonds to give a polymeric layer structure in boric acid.