Home › Chemistry Class 11 › Classification of Elements and Periodicity
Classification of Elements and Periodicity
Chemistry · Class 11NCERT pp. 74–952 sets × 25 questionsFree
Key facts and formulas for Classification of Elements and Periodicity, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Dobereiner triad | Middle weight = (first + third) / 2 |
| Group valence (Mendeleev) | Higher oxide valence = group number; hydride valence = 8 − group number |
| Orbitals per subshell | s = 1, p = 3, d = 5, f = 7, g = 9; elements in a period = 2 × orbitals filled |
| IUPAC digit roots | 0 nil, 1 un, 2 bi, 3 tri, 4 quad, 5 pent, 6 hex, 7 sept, 8 oct, 9 enn; add ium |
| Slater's rules | Zeff = Z − shielding; same group 0.35, shell n−1 0.85, deeper shells 1.00 |
| Isoelectronic ions | Electrons = Z − charge for cations, Z + charge for anions; more protons → smaller radius |
| Ionization enthalpy dips | B < Be and O < N; successive IE jumps after valence electrons are removed |
| Mulliken electronegativity | EN = (IE + Ae) / 2 |
| Polarising power | Charge / radius: higher charge and smaller size → larger value |
| Bond length | Single covalent bond length = r₁ + r₂ |
Three sample questions
Q1. Lithium and potassium have atomic weights 6.9 and 39.1. According to Dobereiner's triads, the atomic weight of sodium should be close to
- 46.0
- 23.0
- 16.1
- 32.2
Hint and solution
Hint: In a triad the middle element has an atomic weight equal to the average of the other two.
- Triad: middle = (first + third) / 2 — Dobereiner said that in a triad the middle element has a weight equal to the mean of the other two.
- sum = 6.9 + 39.1 = 46.0 — Add the atomic weights of lithium and potassium.
- mean = 46.0 / 2 = 23.0 — Divide the sum by 2 to get the middle element's weight.
- Na ≈ 23.0 — The expected weight of sodium is 23.0, which is also its real value. Option B.
Answer: B. The middle element is the mean of the other two: (6.9 + 39.1) / 2 = 23.0. Actual sodium is 23.
Q2. The correct chronological order of the early classification attempts is
- Newlands, Dobereiner, Mendeleev, Chancourtois
- Chancourtois, Dobereiner, Mendeleev, Newlands
- Dobereiner, Newlands, Chancourtois, Mendeleev
- Dobereiner, Chancourtois, Newlands, Mendeleev
Hint and solution
Hint: Recall the years: triads 1829, telluric screw 1862, law of octaves 1865, periodic table 1869.
- Dobereiner 1829 (triads) — The earliest attempt grouped elements in threes.
- Chancourtois 1862 (telluric screw) — He arranged elements on a cylinder in order of atomic weight.
- Newlands 1865 (octaves) — Every eighth element repeated properties, like musical notes.
- Mendeleev 1869 comes last — So the order is Dobereiner, Chancourtois, Newlands, Mendeleev. Option D.
Answer: D. Triads (1829), cylinder (1862), octaves (1865), then Meyer and Mendeleev (1869).
Q3. In Mendeleev's table an element of Group VI forms a higher oxide and a gaseous hydride of formulae
- RO₃ and RH₂
- R₂O₅ and RH₃
- RO₂ and RH₄
- R₂O₇ and RH
Hint and solution
Hint: Group number gives the oxide's oxygen count as RO with the group number over 2. Hydride valence is 8 minus the group number.
- Group VI → highest valence 6 towards oxygen — The higher oxide uses the group number as the valence, so R shows valence 6 with oxygen.
- Oxide: valence 6 / O valence 2 → RO₃ — Six units of positive valence need three oxygen atoms, giving RO₃.
- Hydride valence = 8 − 6 = 2 — Towards hydrogen the valence is 8 minus the group number, so 2.
- Hydride = RH₂ — So the pair is RO₃ and RH₂. Option A.
Answer: A. Group VI oxide is RO₃. Valence to hydrogen is 8 − 6 = 2, so the hydride is RH₂.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.