Language of Chemistry — Question 1

(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 : $\bold{Fe_2O_3}$