Let $x$ be oxidation state of oxygen. The oxidation state of $K$ is $+1.$ Hence
$+1+2(x)=0$
$2x=-1$
$x=-\frac {1}{2}$
કારણ : $S{O_2}$ એ રિડક્શનકર્તા છે.
${P_4} + 6Cl_2\xrightarrow{\Delta }PC{l_3}$
${P_4} + 10C{l_2}\xrightarrow{\Delta }PC{l_5}$
A for precipitate formation reaction.
B for precipitate dissolution reaction.
C for precipitate exchange reaction.
D for no reaction.
${P_4} + NaOH \longrightarrow P{H_3} \uparrow + Na{H_2}P{O_2}$