Question

(a) write the symbols/formulae of two simple ions and two compound ions (or polyatomic ions).
(b) an element Y has a valency of 4. Write the formula for its
(i) chloride (ii) oxide (iii) sulphate (iv) carbonate (v) nitrate

Answer 1

Valency of Carbon is +4

Explanation:

Answer (a)

Symbol of two simple ions− $Na^+, Al^3^+$

Symbol of two compound ions- $NH^4^+, NO_3^2^-$

Answer (b)

Valency equal to 4 (Y, Let it be Carbon) is being there with Carbon (C)

So, On reacting carbon with -

(i) Chloride (Valency is -1)($Cl^-$)

We get Carbon Tetrachloride $CCl_4$

(ii) Oxide (Valency is -2)($O^-^2$)

We get $CO_2$ Carbon di-Oxide

(iii) Sulphate ( -2)  ($SO_3^-^2$)

We get Carbon di-Sulphate $CS_2$

(iv) Carbonate (valency -2)($CO_3^-^2$)

If we react calcium (valency +2)($Ca^+^2$)

We get, Calcium Carbonate ($CaCO_3$)

(v) Nitrate (valency -3)($NO^3^-$)  

We get Trinitrooxymethyl nitrate $(CN_4O_1_2)$

*Reacting in presence of Oxygen

Answer 2

hope it will helpful for you