A for precipitate formation reaction.

B for precipitate dissolution reaction.

C for precipitate exchange reaction.

D for no reaction.

${P_4} + NaOH \longrightarrow P{H_3} \uparrow  + Na{H_2}P{O_2}$

$Ba{\left( {{N_3}} \right)_2}\xrightarrow{\Delta }Ba + {N_2} \uparrow $

${H_2}{O_2}\xrightarrow{{R.T}}{H_2}O + \frac{1}{2}{O_2}$

${N_2}{O_3}\xrightarrow{{R.T.}}NO + N{O_2}$

${P_4} + 10C{l_2}\xrightarrow{\Delta }PC{l_5}$

${P_4} + 6Cl_2\xrightarrow{\Delta }PC{l_3}$

$S{e_2}C{l_2}\xrightarrow{\Delta }SeC{l_4} + Se$

$2KMn{O_4}\,\,\, + \,\,\,\,2KOH\,\,\,\,\,\,\, \to \,\,\,\,\,2{K_2}Mn{O_4}\,\,\, + \,\,\,{H_2}O\,\,\, + \,\,\,O$ તો $KMnO_4$ નો તુલ્યભાર કેટલો થાય ?

(પ.ભાર : $K = 39, Mn = 55, O = 16$)

${H_2}O\,\, + \,\,\,B{r_2}\,\,\,\, \to \,\,\,HOBr\,\,\, + \,\,\,HBr$

$Fe\, + \,{H_2}O\, \to \,F{e_3}{O_4}\, + \,{H_2}$

$5{H_2}{O_2} + \,xCl{O_2}\, + 2O{H^ - }\, \to \,xC{l^ - }\, + \,y{O_2} + 6{H_2}O$

$xC{l_2}\, + 6O{H^ - }\, \to \,Cl{O_3}^ - \, + \,yC{l^ - } + 3{H_2}O$

$MnO_4^ - \,\, + \,\,{C_2}O_4^{2 - }\,\, + \,\,{H^ + }\, \rightarrow \,\,M{n^{2 + }}\, + \,C{O_2} + \,{H_2}O$

$S{O_2}\, + \,\,2{H_2}S\,\,\, \to \,\,3S\,\, + \,\,2{H_2}O$

$14{H^ + } + C{r_2}O_7^{2 - } + 3Ni \to 2C{r^{3 + }} + 7{H_2}O + 3N{i^{2 + }}$

${H_2}\mathop S\limits^{ } {O_4}\,\,\,\,\, \to \,\,\,\,\,\mathop S\limits^{ } {O_2}$ 

કારણ : બે $S$ પરમાણુ સીધે સીધા $O$-પરમાણુ સાથે જોડાયેલા નથી.