In the following reaction,identify the nature of the product formed by the underlined atom upon complete hydrolysis at $R.T$.:
$H_4\underline{P}_2O_8 + H_2O \longrightarrow H_3PO_4 + H_2O_2$
$H_4\underline{P}_2O_8 + H_2O \longrightarrow H_3PO_4 + H_2O_2$
- AIf the product is an oxyacid with an $-ic$ suffix.
- BIf the product is an oxyacid with an $-ous$ suffix.
- CIf the products are two oxyacids,one with an $-ic$ suffix and the other with an $-ous$ suffix.
- DIf the product is not an oxyacid,neither with an $-ic$ suffix nor with an $-ous$ suffix.
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The correct code for stability of oxidation states for given cations is:
$(i)$ $Pb^{2+} > Pb^{4+}, Tl^{+} < Tl^{3+}$
$(ii)$ $Bi^{3+} < Sb^{3+}, Sn^{2+} < Sn^{4+}$
$(iii)$ $Pb^{2+} > Pb^{4+}, Bi^{3+} > Bi^{5+}$
$(iv)$ $Tl^{3+} < In^{3+}, Sn^{2+} > Sn^{4+}$
$(v)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} > Pb^{4+}$
$(vi)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} < Pb^{4+}$
$(i)$ $Pb^{2+} > Pb^{4+}, Tl^{+} < Tl^{3+}$
$(ii)$ $Bi^{3+} < Sb^{3+}, Sn^{2+} < Sn^{4+}$
$(iii)$ $Pb^{2+} > Pb^{4+}, Bi^{3+} > Bi^{5+}$
$(iv)$ $Tl^{3+} < In^{3+}, Sn^{2+} > Sn^{4+}$
$(v)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} > Pb^{4+}$
$(vi)$ $Sn^{2+} < Pb^{2+}, Sn^{4+} < Pb^{4+}$
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