The valency of an element A is 4. Write the formula of its:

(a) oxide

(b) nitrate

(c) phosphate

(a) Chemical formula of Oxide of A:

$\text{A}^{4+} \space \text{O}^{2-} \\[1em] \text{Dividing by H.C.F. it becomes} \\[1em] \text{A}^{2+} \space \text{O}^{1-}\\[1em] \overset{{2}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{O}} \Rightarrow \underset{\phantom{2}{1}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{2}}{\text{O}} \\[1em] \text{AO}_2$

∴ Chemical formula of Oxide of A is AO₂

(b) Chemical formula of Nitrate of A:

$\text{A}^{4+} \space \text{NO}$3^{1-} \\[1em] \overset{{4}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{1}{\text{NO}3} \Rightarrow \underset{\phantom{2}{1}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{4}}{\text{NO}3} \\[1em] \text{A(NO}3)_4

∴ Chemical formula of Nitrate of A is A(NO₃)₄

(c) Chemical formula of Phosphate of A:

$\text{A}^{4+} \space \text{PO}$4^{3-} \\[1em] \overset{{4}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \space \overset{3}{\text{PO}4} \Rightarrow \underset{\phantom{2}{3}}{\text{A}} \space {\swarrow}\mathllap{\searrow} \underset{\phantom{2}{4}}{\text{PO}4} \\[1em] \text{A}3(\text{PO}4)4

∴ Chemical formula of Phosphate of A is A₃(PO₄)₄