Writing Chemical Formulae – Long Answer Questions
Medium Level (Application & Explanation)
Q1. What is valency? Explain its role in writing chemical formulae with suitable examples.
Answer:
- Valency is the combining power of an element.
- It tells how many bonds an atom can make.
- Think of valency as arms that hold other atoms.
- We use valency to balance charges in a compound.
- Example: Mg²⁺ and Cl⁻ combine as MgCl₂ by the crossover method.
- Example: Al³⁺ and O²⁻ combine as Al₂O₃ to neutralize total charge.
Q2. List the steps to write a chemical formula from a compound’s name. Use calcium hydroxide and ammonium sulphate as examples.
Answer:
- Step 1: Write the symbols/ions with their valencies/charges.
- Step 2: Place the metal or positive ion first.
- Step 3: Cross over the valencies to balance charges.
- Step 4: Use brackets for a polyatomic ion if it repeats.
- Example: Calcium hydroxide → Ca²⁺ and OH⁻ → Ca(OH)₂.
- Example: Ammonium sulphate → NH₄⁺ and SO₄²⁻ → (NH₄)₂SO₄.
Q3. Differentiate between monatomic and polyatomic ions. How does this affect formula writing?
Answer:
- Monatomic ions have one atom, like Na⁺, Cl⁻, O²⁻.
- Polyatomic ions have many atoms, like OH⁻, NO₃⁻, CO₃²⁻, SO₄²⁻, PO₄³⁻.
- In formulas, monatomic ions do not need brackets.
- Polyatomic ions need brackets when more than one is present.
- Example: Mg with hydroxide is Mg(OH)₂, not MgOH₂.
- Brackets protect the whole ion group during crossover.
Q4. Why do we write Ca(NO₃)₂ and Mg(OH)₂ with brackets? When are brackets not needed?
Answer:
- Brackets are used when a polyatomic ion repeats.
- In Ca(NO₃)₂, nitrate is NO₃⁻, and it appears twice.
- In Mg(OH)₂, hydroxide is OH⁻, and it also appears twice.
- Brackets show that the entire ion is multiplied.
- Brackets are not needed if the ion appears once, like NaNO₃.
- Brackets are also not used with single-atom ions, like CaCl₂.
Q5. How do we name compounds from their formulas? Explain with Na₂S, Al₂(SO₄)₃, and NH₄Cl.
Answer:
- First, say the cation (positive ion) name.
- Then, say the anion (negative ion) name.
- For binary compounds, use the -ide ending, like sulphide, chloride.
- For compounds with oxygen-containing ions, use -ate, like sulphate, nitrate.
- Na₂S is sodium sulphide; Al₂(SO₄)₃ is aluminium sulphate.
- NH₄Cl is ammonium chloride, showing a polyatomic cation.
High Complexity (Analysis & Scenario-Based)
Q6. Write and justify the formulas for aluminium sulphide, sodium phosphate, and calcium nitride using the crossover method.
Answer:
- Aluminium sulphide: Al³⁺ and S²⁻ cross to give Al₂S₃.
- The total positive charge is +6; the total negative charge is −6.
- So the compound is electrically neutral.
- Sodium phosphate: Na⁺ and PO₄³⁻ cross to give Na₃PO₄.
- Calcium nitride: Ca²⁺ and N³⁻ cross to give Ca₃N₂.
- Each formula shows balanced charges and follows the metal-first rule.
Q7. A student wrote these incorrect formulas: MgOH₂, NaCO₃₂, and CaNO₃. Correct them and explain the mistakes.
Answer:
- MgOH₂ should be Mg(OH)₂. Hydroxide is a polyatomic ion.
- Without brackets, only H seems doubled, which is wrong.
- NaCO₃₂ should be Na₂CO₃. The 2 belongs to Na, not to CO₃.
- We cross valencies: Na⁺ and CO₃²⁻ → two sodium ions.
- CaNO₃ should be Ca(NO₃)₂. Nitrate is NO₃⁻ and appears twice.
- We balance Ca²⁺ with two nitrates to get a neutral compound.
Q8. A compound is made of Ca²⁺ and PO₄³⁻ ions. Find its formula. Count total atoms and justify charge balance.
Answer:
- Ions are Ca²⁺ and PO₄³⁻ (phosphate).
- Cross over the valencies: 3 for Ca and 2 for PO₄.
- The formula is Ca₃(PO₄)₂.
- Total atoms: Ca = 3; P = 2; O = 8; so 13 atoms in one unit.
- Total positive charge: 3 × (+2) = +6.
- Total negative charge: 2 × (−3) = −6. Net charge is zero.
Q9. Compare writing formulas for binary compounds and compounds with polyatomic ions. Use K₂O, CCl₄, and Al(NO₃)₃ as cases.
Answer:
- Binary compounds have two elements only.
- Example: K⁺ and O²⁻ give K₂O by crossover.
- Example: C and Cl form CCl₄ using valency (C = 4, Cl = 1).
- Compounds with polyatomic ions include ion groups like NO₃⁻.
- Example: Al³⁺ and NO₃⁻ give Al(NO₃)₃ with brackets.
- In both cases, we balance valencies to get neutral formulas.
Q10. Element X forms X₂Y₃ with Y. If Y has valency 2, find X’s valency. Then suggest two real pairs that fit this pattern.
Answer:
- In X₂Y₃, total positive charge must equal total negative charge.
- Y has valency 2, and there are 3 Y atoms → total −6.
- So the two X atoms together must give +6.
- Therefore, X has valency 3 (since 2 × 3 = +6).
- Real pairs: Aluminium oxide fits as Al₂O₃ (Al³⁺, O²⁻).
- Another pair: Aluminium sulphide as Al₂S₃ (Al³⁺, S²⁻).