\(\mathop {KCl{O_3}}\limits^{ + 1 + 5 - 2} {\mkern 1mu} + \mathop {{H_2}{C_2}{O_4}}\limits^{ + 1 + 3 - 2} {\mkern 1mu} + \mathop {{H_2}S{O_4}}\limits^{ + 1 + 6 - 2} {\mkern 1mu} \)\( \rightarrow \mathop {{K_2}S{O_4}}\limits^{ + 1 + 6 - 2} {\mkern 1mu} + \mathop {KCl}\limits^{ + 1 - 1} {\mkern 1mu} + \mathop {C{O_2}}\limits^{ + 4 - 2} {\mkern 1mu} + \mathop {{H_2}O}\limits^{ + 4 - 2} {\mkern 1mu} \)
Thus, \(Cl\) is the element which undergoes maximum change in the oxidation state.
$PC{l_5}\xrightarrow{\Delta }PC{l_3} + C{l_2}$
$H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$
વિધાન $II$: $\mathrm{ClO}_4^{-}$એ એસીડીક પરિસ્થિતિમાં વિષમીકરણ પામે છે. તો યોગ્ય વિકલ્પ પસંદ કરો.