Class 9 Science Chapter 4: Important Questions with Answers

J.J. Thomson.

E. Goldstein (canal rays).

The Christmas pudding / watermelon model.

Charge +2, mass 4 u.

E. Rutherford (by α-particle scattering).

The discrete orbits (shells) in which electrons revolve.

n.

2 × 3² = 18.

2, 5.

The combining capacity of an atom — the number of electrons gained, lost or shared to complete the octet.

8 − 6 = 2.

The total number of protons and neutrons in the nucleus.

Protium, deuterium, tritium.

¹²₆C and ¹⁴₆C.

⁴⁰₂₀Ca and ⁴⁰₁₈Ar.

Electron: e⁻, charge −1, negligible mass (about 1/2000 of a hydrogen atom), outside the nucleus; Thomson. Proton: p⁺, charge +1, mass 1 u, in the nucleus; from canal rays (Goldstein).

(i) The atom is a positive sphere with electrons embedded; (ii) positive and negative charges are equal, so the atom is neutral. Limitation: it could not explain the results of Rutherford's scattering (rebounding); no idea of a nucleus.

Most went straight → most of the atom is empty; some deflected by small angles → positive charge occupies little space; 1 in 12000 rebounded → all positive charge and mass are in a very small volume (nucleus).

Every stone hits a wall, but most pass through the gaps in a fence — likewise most α-particles pass through the empty space of the atom and only rarely hit the solid part (nucleus).

An electron in a circular orbit is accelerating and would radiate energy; losing energy it would fall into the nucleus and the atom would be unstable — but atoms are stable.

(i) Only discrete orbits are allowed; (ii) electrons in them do not radiate energy. Sketch: nucleus at the centre with three concentric circles K (n=1), L (n=2), M (n=3).

(i) At most 2n² electrons in a shell; (ii) at most 8 in the outermost shell; (iii) electrons do not enter a shell until the inner ones are full — step-wise filling.

Na (11) = 2, 8, 1. Losing one electron gives Na⁺ with 10 electrons = 2, 8 — K (2) and L (8) both full.

Z = 3 → lithium, configuration 2, 1; one outer electron → valency 1.

X: 12, Y: 14. Same protons (carbon), different mass numbers — isotopes.

An isotope of uranium as nuclear-reactor fuel; an isotope of cobalt in cancer treatment (iodine for goitre).

No single atom has a fractional mass; a sample has ³⁵Cl and ³⁷Cl in a 3 : 1 mix, so the average is 35 × 3/4 + 37 × 1/4 = 35.5 u.

Thomson: electrons embedded in a positive sphere; explained neutrality, not scattering. Rutherford: tiny, heavy, positive nucleus, electrons in circular paths, mostly empty; explained scattering, not stability. Bohr: electrons only in discrete orbits (K, L, M…), no radiation there — explained stability.

Fast α-particles (He²⁺, 4 u) on gold foil about 1000 atoms thick. Observations: most straight; some slightly deflected; 1 in 12000 rebounded. Conclusions: mostly empty; positive charge in little space; mass and positive charge in a tiny centre, nucleus ~10⁵ times smaller. Model: positive nucleus with nearly all the mass, electrons in circular paths, nucleus very small. Drawback: stability.

Rules: 2n²; at most 8 in the outer shell; step-wise filling. H 1; C 2, 4; O 2, 6; Na 2, 8, 1; Cl 2, 8, 7; Ar 2, 8, 8.

Valency = electrons gained, lost or shared to get an octet. Si (14) = 2, 8, 4 — shares 4 → valency 4. O (8) = 2, 6 — gains 2 → 8 − 6 = 2.

Z = protons (C: 6); A = protons + neutrons (C: 12). Isotopes: same Z, different A — ¹H, ²H, ³H; ³⁵Cl, ³⁷Cl. Isobars: same A, different Z — ⁴⁰Ca, ⁴⁰Ar. Uses: uranium — reactor fuel; cobalt — cancer; iodine — goitre.

Br: 79 × 0.497 + 81 × 0.503 = 39.263 + 40.743 = 80.006 u ≈ 80 u. X: let ¹⁶X = x%, then 16x + 18(100 − x) = 1620 → 1800 − 2x = 1620 → x = 90. ¹⁶X 90%, ¹⁸X 10%.

Positively charged radiations in a gas discharge, discovered by Goldstein (1886); they led to the discovery of the proton.

No — the +1 and −1 charges balance; the atom is neutral.

The atom is a positive sphere with electrons embedded; the negative and positive charges are equal in magnitude, so the atom is neutral.

The proton — the nucleus is positively charged.

Nucleus at the centre with three concentric circles — K (n = 1), L (n = 2), M (n = 3); see the Lesson 5 figure.

If the foil is thin enough, the observations would be much the same — most straight, some deflected, very few rebounding — because every atom has the same structure (tiny nucleus, mostly empty). A metal with lighter nuclei may give fewer large-angle deflections.

Electron, proton and neutron.

4 − 2 = 2 neutrons.

C (6): K 2, L 4 → 2, 4. Na (11): K 2, L 8, M 1 → 2, 8, 1.

2 + 8 = 10 (neon).

Cl 2, 8, 7 → 8 − 7 = 1; S 2, 8, 6 → 8 − 6 = 2; Mg 2, 8, 2 → loses 2 → 2.

(a) 8 (b) zero — the atom is neutral.

O: 8 + 8 = 16; S: 16 + 16 = 32.

Protons / neutrons / electrons — H: 1/0/1; D: 1/1/1; T: 1/2/1.

Isotopes ¹²₆C, ¹⁴₆C: both 2, 4. Isobars ⁴⁰₂₀Ca: 2, 8, 8, 2 and ⁴⁰₁₈Ar: 2, 8, 8.

Electron: charge −1, negligible mass (~1/2000 u), outside the nucleus. Proton: +1, 1 u, in the nucleus. Neutron: 0, ~1 u, in the nucleus (except ordinary hydrogen).

It could not explain the scattering results — most particles going straight and a few rebounding; it had no nucleus and assumed positive charge spread through the sphere.

An electron in a circular orbit would accelerate, radiate energy and fall into the nucleus — the atom would be unstable; the model could not explain the stability of the atom.

A positive nucleus at the centre; electrons revolve only in certain discrete orbits in which they do not radiate energy; these orbits are energy levels K, L, M, N (n = 1, 2, 3, 4).

Thomson: electrons in a positive sphere; explained neutrality. Rutherford: tiny heavy positive nucleus, electrons in circular paths, mostly empty; did not explain stability. Bohr: discrete orbits with no radiation — explained stability.

(i) At most 2n² (K 2, L 8, M 18); (ii) at most 8 in the outermost shell; (iii) not into a shell until the inner ones are full — step-wise.

Valency = combining capacity, electrons gained/lost/shared to get an octet. Si 2, 8, 4 → 4; O 2, 6 → 8 − 6 = 2.

Z = protons (C 6); A = protons + neutrons (C 12); isotopes ¹²C/¹⁴C; isobars ⁴⁰Ca/⁴⁰Ar. Uses: uranium — reactor fuel; cobalt — cancer treatment; iodine — goitre.

Na = 2, 8, 1; losing one electron gives Na⁺ = 2, 8 — 2 in K and 8 in L, both full.

79 × 49.7/100 + 81 × 50.3/100 = 39.263 + 40.743 = 80.006 u ≈ 80 u.

Let ¹⁶X = x%: 16x + 18(100 − x) = 1620 → 2x = 180 → x = 90. ¹⁶X 90%, ¹⁸X 10%.

Configuration 2, 1 → valency 1; the element is lithium.

X = 12, Y = 14; same atomic number (carbon) — isotopes.

(a) F — Thomson's model had no nucleus (b) F (c) T (d) F — an iodine isotope is used to treat goitre.

(a) atomic nucleus ✓.

(c) different number of neutrons ✓.

(b) 8 ✓ — Cl⁻ = 2, 8, 8.

(d) 2, 8, 1 ✓.

(1) 9, 19, 10, 9, 9 — fluorine (2) 16, 32, 16, 16, 16 — sulphur (3) 12, 24, 12, 12, 12 — magnesium (4) 1, 2, 1, 1, 1 — deuterium (5) 1, 1, 0, 1, 0 — hydrogen ion (H⁺, a proton).