Periodic Classification of Elements – Case-based Questions with Answers
Class 10
Chemistry — Chapter 14
Board: CBSE
Chapter Standard: NCERT Class 10
Overview
How to use: Each case describes a real-world or examination-style scenario followed by focused questions (a, b, ...) and model answers that are concise and NCERT-aligned. Use them for classroom practice, tests or board exam revision.
Case 1 — Predicting Reactivity
Riya notices that a piece of potassium reacts more vigorously with water than a piece of sodium under identical conditions.
Questions:
- Why is potassium more reactive than sodium?
- Predict the product formed when potassium reacts with water.
Answers:
- Reactivity in Group 1 increases down the group because valence electron is farther from nucleus and more easily lost. Potassium has a larger atomic radius and lower ionization energy than sodium, so it reacts more vigorously.
- Products: K + H₂O → KOH + 1/2 H₂ (overall: 2K + 2H₂O → 2KOH + H₂), forming potassium hydroxide and hydrogen gas.
Case 2 — Noble Gas Properties
An experiment shows that neon lamps do not undergo chemical change under normal operating conditions.
Questions:
- Why are noble gases like neon chemically inert?
- How does electronic configuration explain this inertness?
Answers:
- Noble gases are inert because they have complete valence shells, making them stable and unlikely to gain, lose or share electrons under normal conditions.
- Neon has electronic configuration 1s²2s²2p⁶ — a complete n=2 shell (octet). Full valence shells mean no tendency to form bonds, explaining chemical inertness.
Case 3 — Displacement Reaction
During a lab demo, chlorine gas was bubbled into a potassium bromide solution and a change was observed.
Questions:
- What change would you expect and why?
- Write the chemical equation for the reaction.
Answers:
- Chlorine is more reactive than bromine (higher ability to gain an electron). Chlorine displaces bromide ions from solution, producing bromine which may appear as a brown colour.
- Cl₂ + 2KBr → 2KCl + Br₂.
Case 4 — Position from Electronic Configuration
Given an element X with electronic configuration 1s²2s²2p⁶3s²3p⁴, a student is asked to determine its group and period.
Questions:
- Identify the element and its group and period.
- Predict its typical valency.
Answers:
- Configuration corresponds to sulphur (S, Z=16). Highest shell n=3 → Period 3. Valence electrons in 3s²3p⁴ → total 6 → Group 16 (Chalcogens).
- Typical valency is 2 (to gain 2 electrons to attain octet) though it can show other valencies in covalent bonding (e.g., 2, 4, 6 in compounds).
Case 5 — Anomalies in Order
Mendeleev placed iodine (I) after tellurium (Te) even though iodine has a lower atomic mass. This puzzled some students.
Questions:
- Why did Mendeleev arrange I after Te despite mass anomaly?
- How does modern periodic law resolve such anomalies?
Answers:
- Mendeleev prioritized chemical properties over strict atomic mass ordering to keep elements with similar properties in the same group, hence he placed iodine after tellurium.
- Modern periodic law arranges elements by atomic number (number of protons). Since iodine has a higher atomic number than tellurium, it correctly follows Te, resolving the anomaly.
Case 6 — Predicting Ion Formation
A student is asked why aluminium commonly forms a +3 ion while chlorine forms a −1 ion.
Questions:
- Explain using group positions and electronic configurations.
- Write the electron transfer in forming Al³⁺ and Cl⁻.
Answers:
- Aluminium (Group 13) has configuration …3s²3p¹ and tends to lose three electrons to attain noble gas configuration (Al³⁺). Chlorine (Group 17) has …3s²3p⁵ and tends to gain one electron to complete octet (Cl⁻).
- Al → Al³⁺ + 3e⁻; Cl + e⁻ → Cl⁻.
Case 7 — Trends in Atomic Radius
Comparing elements Na, Mg and Al in Period 3, a student measures their atomic radii and notes a decrease from Na to Al.
Questions:
- Explain the observed trend.
- How does effective nuclear charge play a role?
Answers:
- Across a period from Na → Mg → Al, atomic radius decreases because increasing nuclear charge pulls the electron cloud closer without additional shielding.
- Effective nuclear charge (Z_eff) increases across the period as protons are added but shielding doesn't increase proportionally, resulting in stronger attraction on valence electrons and smaller radius.
Case 8 — Ionization Energy Irregularity
When comparing first ionization energies, oxygen has a slightly lower value than nitrogen, which surprises some learners.
Questions:
- Why is O's first ionization energy slightly lower than N's?
- What electronic configuration explanation accounts for this anomaly?
Answers:
- Oxygen has a lower first ionization energy than nitrogen because of electron-electron repulsion in paired electrons.
- Nitrogen has a half-filled 2p³ configuration (stable), whereas oxygen is 2p⁴ with one paired electron. The additional repulsion from pairing makes it slightly easier to remove an electron from O.
Case 9 — Reactivity of Halogens
In a school lab, fluorine reacts explosively with many substances while iodine reacts much less vigorously.
Questions:
- Explain the trend in reactivity down Group 17.
- Predict which halogen would displace another in a solution and give an example.
Answers:
- Reactivity decreases down the group because atomic size increases and the ability to attract an additional electron reduces; fluorine (smallest, highest electronegativity) is most reactive.
- A more reactive halogen displaces a less reactive halide: e.g., Cl₂ will displace Br⁻ from KBr: Cl₂ + 2KBr → 2KCl + Br₂.
Case 10 — Use of Periodic Table in Predicting Compounds
A chemist wants to predict the formula of the compound formed between magnesium and chlorine.
Questions:
- Use periodic trends to predict the formula of the compound formed.
- Explain why this formula is expected.
Answers:
- Magnesium (Group 2) forms Mg²⁺, chlorine (Group 17) forms Cl⁻. To balance charges, formula is MgCl₂.
- Mg loses two electrons to attain noble gas configuration; each Cl gains one electron. Two Cl⁻ are needed to balance the +2 charge of Mg²⁺, producing MgCl₂.
Case 11 — Positioning Hydrogen
Students debate whether hydrogen belongs to Group 1 or Group 17.
Questions:
- Give arguments for placing hydrogen in Group 1.
- Give arguments for showing hydrogen separately or near halogens.
Answers:
- Like Group 1, hydrogen has one valence electron (1s¹) and can lose one electron to form H⁺, resembling alkali metals in some reactions.
- Hydrogen also needs one electron to achieve noble gas configuration and can form H⁻ in certain compounds, resembling halogens; due to its unique properties it is often placed separately at the top of the table.
Case 12 — Lanthanoids & Actinoids Placement
A textbook shows lanthanoids and actinoids as separate rows below the main table.
Questions:
- Why are they placed separately?
- Mention one characteristic property of these series.
Answers:
- They fill f-orbitals (4f and 5f) and including them within the main body would make the table unwieldy. Placing them separately keeps the table compact while acknowledging their related properties.
- Characteristic: exhibit similar chemical properties within the series and show gradual changes (lanthanoid contraction), often used in high-tech and nuclear applications (actinoids are often radioactive).
Case 13 — Colour of Transition Metal Compounds
A student observes that many transition metal salts are coloured while salts of Group 1 metals are usually white.
Questions:
- Explain why transition metal compounds are often coloured.
- Why are alkali metal salts typically colourless?
Answers:
- Transition metals have partially filled d-orbitals; d–d electronic transitions absorb visible light, producing colours in their compounds.
- Alkali metals have no partially filled d-orbitals (s-block) and form ionic compounds with colourless ions like Na⁺, K⁺; thus their salts are often white or colourless.
Case 14 — Amphoteric Oxides
Aluminium oxide reacts both with acids and bases.
Questions:
- Explain why Al₂O₃ is called amphoteric.
- Write balanced reactions showing its amphoteric nature.
Answers:
- Amphoteric substances react with both acids and bases. Al₂O₃ behaves as a metal oxide (basic) with acids and as a non-metal oxide (acidic) with bases.
- With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O. With base: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄] (or Al(OH)₄⁻ formation).
Case 15 — Predicting Product of Reaction
When calcium metal is burnt in air, a white ash remains.
Questions:
- Identify the product and justify its formation.
- Is the oxide basic or acidic? Explain.
Answers:
- Product: calcium oxide (CaO), formed by 2Ca + O₂ → 2CaO. The white ash is due to the ionic lattice of CaO.
- CaO is a basic oxide; it reacts with water to form Ca(OH)₂ (a strong base): CaO + H₂O → Ca(OH)₂.
Case 16 — Electronegativity & Bonding
Hydrogen chloride (HCl) is a gas that dissolves in water to form hydrochloric acid.
Questions:
- Explain the bond type in HCl using electronegativity values.
- Why does HCl dissolve in water to give acidic solution?
Answers:
- Chlorine is more electronegative than hydrogen, so the H–Cl bond is polar covalent with chlorine bearing a partial negative charge and hydrogen partial positive.
- In water, HCl ionises: HCl → H⁺ + Cl⁻. The release of H⁺ (protons) makes the solution acidic (hydrochloric acid).
Case 17 — Metallic Character & Applications
Sodium is used in street lamps (sodium vapour lamps) and copper for electrical wiring.
Questions:
- Relate the choice of these metals to their periodic properties.
- Why isn't sodium used for wiring despite being metallic?
Answers:
- Cu is chosen for wiring because of high electrical conductivity, ductility and corrosion resistance — properties of transition metals. Sodium vapour is used in lamps because its emission spectrum (bright yellow) arises from electronic transitions and vapour pressure properties at lamp temperatures.
- Sodium is highly reactive (especially with water and air) and soft, making it unsafe and impractical for wiring despite being a metal.
Case 18 — Reactivity Series & Displacement
Iron filings are added to a solution of copper sulfate and a reddish-brown deposit appears on iron.
Questions:
- Explain the observation using periodic/activity series.
- Write the chemical equation for the reaction.
Answers:
- Iron is more reactive than copper and displaces copper from its salt solution, producing copper metal which appears reddish-brown.
- Fe + CuSO₄ → FeSO₄ + Cu.
Case 19 — Predicting Compound Properties
Compare sodium chloride (NaCl) and silicon dioxide (SiO₂) in terms of bonding and typical physical properties.
Questions:
- Describe the type of bonding in each compound.
- Explain differences in melting point and solubility in water.
Answers:
- NaCl is ionic (Na⁺ and Cl⁻ lattice). SiO₂ is covalent network (strong covalent bonds in a giant structure).
- SiO₂ has very high melting point due to strong covalent network; NaCl has high but lower melting point due to ionic lattice. NaCl is soluble in water (ions separate and hydrate), whereas SiO₂ is insoluble due to strong covalent bonding in the network.
Case 20 — Using Periodic Table for Chemical Reasoning
You are given two unknown elements A and B. A is soft, reacts violently with water to form a strongly basic solution; B is brittle, forms acidic oxides and is a poor conductor.
Questions:
- Which groups are A and B likely to belong to?
- Suggest one probable compound for each and its use.
Answers:
- A resembles an alkali metal (Group 1). B resembles a non-metal (p-block, likely near right side such as sulphur or phosphorus).
- Possible compound for A: NaOH (produced when Na reacts with water), used in soap and drain cleaners. For B: SO₂ or SO₃ (acidic oxides of sulphur) — SO₂ used in making sulphuric acid and as a preservative in food industry.