Oxidation states of P in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{H}_{4} \mathrm{P}_{2

Question

Oxidation states of P in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}, \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}, \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}$ are respectively\

Professor PC Thomas Classes · Answer

(d) : The oxidation state can be calculated as:\ $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}:+4+2 x+5(-2)=0 \Rightarrow 2 x-6=0 \Rightarrow x=+3$\ $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}:+4+2 x+6(-2)=0 \Rightarrow 2 x-8=0 \Rightarrow x=+4$\ $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}:+4+2 x+7(-2)=0 \Rightarrow 2 x-10=0 \Rightarrow x=+5$\