Language of Chemistry

A chemical equation is a shorthand form for representing the result of a chemical change.

It is represented with the help of formulas and symbols of the reactants and products.

Information provided by a chemical equation :

1. The formulas & symbols of the reactants & products.
2. The physical state of the substance whether it is solid, liquid or gas.
3. The special conditions required such as heat, catalyst, direction of reaction.
4. Tells the ratio in which the substances react in a balanced equation.

Chemical equation do not tell us about:

1. The physical states of the reactants & products - hence [(s) for solids, (g) for gas may be added]
2. The conditions such as temperature, pressure or catalyst which affect the reaction.
3. The concentration of the reactants & products hence [(dil.) dilute & (conc) concentrated may be added.]
4. The nature of the chemical reaction.
5. The speed of the reaction.
6. The heat changes accompanying the reaction.
7. The completion of the reaction.

A balanced equation is one in which the number of atoms of each element of the reactant is equal to the number of atoms of each element of the products.

Zn + H₂SO₄ ⟶ ZnSO₄ + H₂

An equation must be balanced to comply with the 'law of conservation of matter' because according to this law : matter is neither created nor destroyed during the chemical change.

(a) 2Ca + O₂ ⟶ 2CaO

(b) Ca + 2H₂O ⟶ Ca(OH)₂ + H₂

(c) Zn + H₂SO₄ ⟶ ZnSO₄ + H₂

(d) PbSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Pb(OH)₂

(e) Cu(OH)₂ + 2HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O

(f) PbNO₃ + NaCl ⟶ NaNO₃ + PbCl

(a) 4P + 5O₂ ⟶ 2P₂O₅

(b) Na₂O + H₂O ⟶ 2NaOH

(c) 2K + 2H₂O ⟶ 2KOH + H₂

(d) 3Fe + 4H₂O ⇌ Fe₃O₄ + 4H₂

(e) CaO + 2HCl ⟶ CaCl₂ + H₂O

(f) 2Fe + 3Cl₂ + 2FeCl₃

(g) 2Al + 3H₂O ⟶ Al₂O₃ + 3H₂

(h) 2Al + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 3H₂

(i) Fe₂O₃ + 3H₂ ⟶ 2Fe + 3H₂O

(j) C + 2H₂SO₄ ⟶ + CO₂ + 2H₂O + 2SO₂

(k) 2Pb₃O₄ ⟶ 6PbO + O₂

(l) 4Al + 3O₂ ⟶ 2Al₂O₃

(m) 2NO + O₂ ⟶ 2NO₂

(n) 2ZnS + 3O₂ ⟶ 2ZnO + 2SO₂

(o) Pb₃O₄ + 8HCl ⟶ 3PbCl₂ + 4H₂O + Cl₂

(p) ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O

(q) H₂S + Cl₂ ⟶ S + 2HCl

(r) FeCl₃ + 3NaOH ⟶ 3NaCl + Fe(OH)₃

(s) Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

(t) 2KHCO₃ ⟶ K₂CO₃ + H₂O + CO₂

(u) 3CuO + 2NH₃ ⟶ 3Cu + 3H₂O + N₂

(a) Potassium chloride

Step 1 — Write each symbol with its valency

$\text{K}^{1+} \phantom{\nearrow} \text{Cl}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{Cl}}$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Potassium chloride : $\bold{KCl}$

(b) Sodium bromide

Step 1 — Write each symbol with its valency

$\text{Na}^{1+} \phantom{\nearrow} \text{Br}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Br}} \Rightarrow \underset{\phantom{1}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{Br}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Sodium bromide : $\bold{NaBr}$

(c) Potassium nitrate

Step 1 — Write each symbol with its valency

$\text{K}^{1+} \phantom{\nearrow} \text{NO}_3^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{NO}_3} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Potassium nitrate : $\bold{KNO}_\bold{3}$

(d) Calcium hydroxide

Step 1 — Write each symbol with its valency

$\text{Ca}^{2+} \phantom{\nearrow} \text{OH}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Calcium hydroxide : $\bold{Ca(OH)}_\bold{2}$

(e) Calcium bicarbonate

Step 1 — Write each symbol with its valency

$\text{Ca}^{2+} \phantom{\nearrow} \text{HCO}_3^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{HCO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{HCO}_3} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Calcium bicarbonate : $\bold{Ca(HCO_\bold{3})_\bold{2}}$

(f) Sodium bisulphate

Step 1 — Write each symbol with its valency

$\text{Na}^{1+} \phantom{\nearrow} \text{HSO}_4^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{HSO}_4} \Rightarrow \underset{\phantom{1}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{HSO}_4} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Sodium bisulphate : $\bold{NaHSO}_4$

(g) Potassium sulphate

Step 1 — Write each symbol with its valency

$\text{K}^{1+} \phantom{\nearrow} \text{SO}_4^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{SO}_4} \Rightarrow \underset{\phantom{1}{2}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{SO}_4} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Potassium sulphate : $\bold{K_2SO_4}$

(h) Zinc hydroxide

Step 1 — Write each symbol with its valency

$\text{Zn}^{2+} \phantom{\nearrow} \text{OH}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Zn}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Zn}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Zinc hydroxide : $\bold{Zn(OH)_2}$

(i) Potassium permanganate

Step 1 — Write each symbol with its valency

$\text{K}^{1+} \phantom{\nearrow} \text{MnO}_4^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{MnO}_4} \Rightarrow \underset{\phantom{1}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{MnO}_4} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Potassium permanganate : $\bold{KMnO_4}$

(j) Potassium dichromate

Step 1 — Write each symbol with its valency

$\text{K}^{1+} \phantom{\nearrow} \text{Cr}_2\text{O}_7^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{Cr}_2\text{O}_7} \Rightarrow \underset{\phantom{1}{2}}{\text{K}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{Cr}_2\text{O}_7} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Potassium dichromate : $\bold{K_2Cr_2O_7}$

(k) Aluminium hydroxide

Step 1 — Write each symbol with its valency

$\text{Al}^{3+} \phantom{\nearrow} \text{OH}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{OH}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Aluminium hydroxide : $\bold{Al(OH)_3}$

(l) Magnesium nitride

Step 1 — Write each symbol with its valency

$\text{Mg}^{2+} \phantom{\nearrow} \text{N}^{3-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{3}{\text{N}} \Rightarrow \underset{\phantom{1}{3}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{N}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Magnesium nitride : $\bold{Mg_3N_2}$

(m) Sodium zincate

Step 1 — Write each symbol with its valency

$\text{Na}^{1+} \phantom{\nearrow} \text{ZnO}_2^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{ZnO}_2} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{ZnO}_2} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Sodium zincate : $\bold{Na_2ZnO_2}$

(n) Copper [II] oxide

Step 1 — Write each symbol with its valency

$\text{Cu}^{2+} \phantom{\nearrow} \text{O}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{2}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Copper [II] oxide : $\bold{CuO}$

(o) Copper [I] sulphide

Step 1 — Write each symbol with its valency

$\text{Cu}^{1+} \phantom{\nearrow} \text{S}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{1}{1}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}} \Rightarrow \underset{\phantom{1}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{S}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Copper [I] sulphide : $\bold{Cu_2S}$

(p) Iron [III] chloride

Step 1 — Write each symbol with its valency

$\text{Fe}^{3+} \phantom{\nearrow} \text{Cl}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{3}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Iron [III] chloride : $\bold{FeCl_3}$

(q) Iron [II] hydroxide

Step 1 — Write each symbol with its valency

$\text{Fe}^{2+} \phantom{\nearrow} \text{OH}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Iron [II] hydroxide : $\bold{Fe(OH)_2}$

(r) Iron [III] sulphide

Step 1 — Write each symbol with its valency

$\text{Fe}^{3+} \phantom{\nearrow} \text{S}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{3}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}} \Rightarrow \underset{\phantom{1}{2}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{S}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Iron [III] sulphide : $\bold{Fe_2S_3}$

(s) Iron [III] oxide

Step 1 — Write each symbol with its valency

$\text{Fe}^{3+} \phantom{\nearrow} \text{O}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{3}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Fe}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{O}} \\[0.5em]$

Step 3 — Write the interchanged number & hence the formula

Therefore, we get

Formula of Iron [III] oxide : $\bold{Fe_2O_3}$

Symbol — An atom of any element is denoted by a symbol. It is the short form or abbreviated name of the element. It represents

• a specific element or one atom of an element. e.g. 'H' represents one atom of the element hydrogen.
• the weight of the element equal to its atomic weight.
• distinguishes one element from another and is characteristic of that element only.

The symbol 'Ca' for calcium is taken from the first two letters of its name "Calcium" whereas the symbol 'Cu' for Copper is taken from the first two letters of its latin name "Cuprum".

1. The formula of silver [I] chloride is AgCl.

2. The basic unit of an element is a/an atom.

3. Atom contains nucleus, with positively charged protons.

4. Element lead has the symbol derived from its Latin name 'plumbum'.

5. From the elements - He, Br, Pt & O ; the element which forms a polyatomic molecule is O & which is liquid at room temperature is Br

6. The valency of iron in FeO is 2⁺ of chlorine [chloride] in CaCl₂ is 1^- and of dichromate in K₂Cr₂O₇ is 2^-.

1. 2Na + 2H₂O ⟶ 2NaOH + H₂

2. 4P + 5O₂ ⟶ 2P₂O₅

3. Fe₂O₃ + 3H₂ ⟶ 2Fe + 3H₂O

4. 2Al + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 3H₂

5. N₂ + 3H₂ ⇌ 2NH₃

6. ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O

7. FeCl₃ + 3NH₄OH ⟶ 3NH₄Cl + Fe(OH)₃

8. FeS + 2HCl ⟶ FeCl₂ + H₂S

9. 2NH₃ + H₂SO₄ ⟶ (NH₄)₂SO₄

10. PbO₂ + 4HCl ⟶ PbCl₂ + 2H₂O + Cl₂

Valency is the number of hydrogen atoms which can combine with [or displace] one atom of the element [or radical] forming a compound.

In case of water, two atoms of hydrogen combine with one atom of oxygen. Hence, valency of oxygen is 2.

In case of ammonia, three atoms of hydrogen combine with one atom of nitrogen. Hence, valency of nitrogen is 3.

Valency is also defined as the number of electrons an atom can donate or accept so as to achieve stable electronic configuration of the nearest noble gas. In case of magnesium [2, 8, 2], it donates two electrons to achieve a stable electronic configuration of the nearest noble gas [Neon (2,8)], hence its valency is 2⁺

Variable valency refers to the capacity of certain elements to have multiple valencies or combining abilities. These elements can lose electrons beyond their valence shell, including the penultimate shell. This characteristic allows them to display variable valency.

The outermost shell of copper has 1 electron & the penultimate shell [last but one] has 18 electrons. The penultimate shell has not attained stability & one or more electrons sometimes jumps to the outermost shell. The valency electron thus increases & Cu has new configuration: Cu = 29 [2, 8, 17, 2]. Therefore copper exhibits: Cu¹⁺ & Cu²⁺ valency.

The compound CuCl is Copper [I] Chloride or Cuprous Chloride & CuCl₂ Copper [II] Chloride or Cupric Chloride.