Structure of Atom — Telangana (SSC) Class 10 Science Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Science solutions for "Structure of Atom" — important questions with detailed answers, download PDF for board exam preparation.

Structure of Atom — Telangana (SSC) Class 10 Science Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Science solutions for "Structure of Atom" — important questions with detailed answers, download PDF for board exam preparation.

✓ 100% Free ✓ No Login Needed ✓ NCERT / CBSE Aligned ✓ Download as PDF

TL;DR: Free step-by-step Telangana (SSC) Class 10 Science solutions for "Structure of Atom" — important questions with detailed answers, download PDF for boa…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated

🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.

Structure of Atom — Textbook Questions Answered (7)

Q1: State Dalton's Atomic Theory. Which of its points were later found to need modification, and why?

Dalton's Atomic Theory (1808) proposed: 1. All matter is made of tiny, indivisible particles called atoms. 2. Atoms of the same element are identical in mass and properties; atoms of different elements differ. 3. Compounds form when atoms of different elements combine in fixed, simple whole-number ratios. 4. Chemical reactions involve only the rearrangement of atoms — atoms are neither created nor destroyed. What needed modification: Point 1 (atoms are indivisible) was disproved by the discovery of the electron (J.J. Thomson, 1897), proton, and neutron — atoms are themselves made of smaller subatomic particles. Point 2 (atoms of the same element are identical) was refined by the discovery of isotopes — atoms of the same element CAN have different masses (different numbers of neutrons) while remaining chemically identical. Despite these corrections, points 3 and 4 remain fundamentally correct and are still the basis of the laws of chemical combination taught today.

Q2: Describe Rutherford's atomic model, and explain the key experimental observation that led to it.

Rutherford's alpha-particle scattering experiment (1911) fired a beam of positively charged alpha particles at a very thin sheet of gold foil. Most particles passed straight through, but a small fraction were deflected at large angles, and a very few bounced almost straight back. What this revealed: - Most particles passing straight through meant most of the atom is empty space. - The occasional large deflection meant the particles were repelled by something small, dense, and positively charged — the nucleus. - The very rare direct bounce-back meant the nucleus, though tiny, concentrates almost all of the atom's mass. Rutherford's model, built from this: - A dense, positively charged nucleus sits at the atom's centre, containing protons (and later, neutrons). - Electrons orbit the nucleus in circular paths at relatively large distances, held in orbit by electrostatic attraction to the nucleus. - Most of the atom's volume is empty space between the nucleus and the orbiting electrons. The model's drawback: according to classical electromagnetic theory, an orbiting (accelerating) charged particle should continuously radiate energy and spiral into the nucleus within a fraction of a second — yet atoms are stable. Rutherford's model had no explanation for this, which is the problem Bohr's model was built to solve.

Q3: How did Bohr's model resolve the stability problem that Rutherford's model could not explain?

Bohr's model (1913) kept Rutherford's nuclear structure but added a crucial restriction: electrons can only occupy certain fixed, specific orbits (called energy levels or shells), each with a definite, quantized energy — not just any orbit at any distance. Bohr's key postulates: 1. Electrons revolve only in specific allowed orbits (numbered n = 1, 2, 3, ... called K, L, M, N shells) without radiating energy while in these orbits — resolving the stability problem directly, since an electron in an allowed orbit simply does not lose energy. 2. An electron can jump between allowed orbits only by absorbing or emitting energy exactly equal to the energy difference between them, which is why atoms absorb and emit light only at specific, discrete wavelengths (spectral lines) rather than a continuous range. This explained the hydrogen atom's line spectrum with high precision, which Rutherford's model could not do at all. Bohr's model does have its own limits — it does not accurately predict the spectra of atoms with more than one electron — but it correctly introduced the idea that electron energy is quantized, which later, more complete quantum-mechanical models retained and built upon.

Q4: What is valency, and how is it determined from an atom's electron configuration?

Valency is the combining capacity of an element — the number of electrons an atom must lose, gain, or share to achieve a stable outer (valence) shell configuration, usually 8 electrons (the octet rule), or 2 electrons for elements close to helium. How to find it from electron configuration: - If the valence shell has fewer than 4 electrons, the atom tends to LOSE those electrons; valency = number of valence electrons. Example: Sodium (Na), electron configuration 2, 8, 1 — has 1 valence electron, loses it to form Na⁺, so valency = 1. - If the valence shell has more than 4 electrons, the atom tends to GAIN electrons to complete the octet; valency = 8 − (number of valence electrons). Example: Chlorine (Cl), electron configuration 2, 8, 7 — has 7 valence electrons, needs 1 more, so valency = 1 (by gaining). - Elements with exactly 4 valence electrons (like Carbon) typically share electrons (covalent bonding) rather than clearly losing or gaining, and can show a valency of 4 in either direction.

Q5: What is an isotope? Give an example, and explain one practical use of isotopes.

Isotopes are atoms of the same element that have the same number of protons (same atomic number, so the same chemical identity) but a different number of neutrons — and therefore a different mass number. Example: Carbon has two common isotopes — Carbon-12 (₆C¹²), with 6 protons and 6 neutrons, and Carbon-14 (₆C¹⁴), with 6 protons and 8 neutrons. Both behave identically in chemical reactions, since chemical behaviour depends on the number of electrons (which equals the number of protons), not neutrons. Practical use — radiocarbon dating: Carbon-14 is radioactive and decays at a known, constant rate (its half-life is about 5,730 years). Living organisms continuously take in carbon-14 along with ordinary carbon-12, keeping a roughly constant ratio between them. Once an organism dies, it stops taking in new carbon, and its carbon-14 steadily decays while carbon-12 stays constant. By measuring how much carbon-14 remains relative to carbon-12 in a fossil or ancient artefact, scientists can calculate how long ago the organism died — this is the basis of carbon dating, used to date archaeological remains up to about 50,000 years old.

Q6: How many electrons can the third shell (M shell) hold at most? Show the calculation and give the correct electron sub-shell notation.

The maximum number of electrons any shell can hold is given by the formula 2n², where n is the shell number. For the third shell (M shell), n = 3: Maximum electrons = 2n² = 2 × 3² = 2 × 9 = 18 electrons The M shell's own sub-shells (3s, 3p, 3d — NOT the 2s/2p sub-shells, which belong to the second, L shell) hold this total as: 3s² 3p⁶ 3d¹⁰, which adds up to 2 + 6 + 10 = 18 ✓. An important exception at Class 10 board level: while the M shell can hold up to 18 electrons in principle, when it is the OUTERMOST (valence) shell of a lighter element, it is only filled up to 8 electrons before the next shell (N) begins filling — this is the Bohr-Bury rule that the outermost shell of an atom cannot hold more than 8 electrons. The full 18-electron capacity of the M shell is only used once it is no longer the outermost shell, in heavier elements.

Q7: Write the electron configuration of Chlorine (atomic number 17) and Calcium (atomic number 20) using the shell-filling (2, 8, 8...) rule.

Using the Bohr-Bury shell-filling rule (K shell max 2, L shell max 8, M shell max 8 as an outermost shell, then a new shell starts): Chlorine (atomic number 17, so 17 electrons): K shell: 2 electrons L shell: 8 electrons Remaining: 17 − 2 − 8 = 7 electrons go into the M shell Configuration: 2, 8, 7 Calcium (atomic number 20, so 20 electrons): K shell: 2 electrons L shell: 8 electrons M shell: since it is not yet the outermost shell once the N shell starts, it fills to 8 (not the full 18, because the outermost-shell-of-8 rule applies once a new shell begins) Remaining: 20 − 2 − 8 − 8 = 2 electrons go into the N shell Configuration: 2, 8, 8, 2 This matches the general pattern: Chlorine, with 7 valence electrons, has valency 1 (gains 1 electron); Calcium, with 2 valence electrons, has valency 2 (loses 2 electrons) — consistent with Calcium's well-known +2 ion, Ca²⁺.

Related