NIOS Pure HTML Study Hub

Chemistry — Class 12 — L2: Atomic Structure

NIOS Code 313 · Module 2 · Atomic Structure and Chemical Bonding

Notes extracted from NIOS Chemistry Course (313), Lesson 2 — Atomic Structure (313_Chemistry_Eng_Lesson2.pdf). Content covers sections 2.1–2.11.
Study timer: 00:00:00

Overview — Module 2: Atomic Structure

Chemistry is defined as the study of matter in terms of its structure, composition and properties. Matter is made up of atoms, and therefore an understanding of the internal structure of the atom is fundamental to chemistry. Ancient Indian and Greek philosophers (600–400 BC) proposed the earliest concept of the atom as the smallest indivisible part of matter — without experimental evidence, based on thought experiments about continuous subdivision. John Dalton revived the atomic concept in the nineteenth century through his atomic theory, which explained the laws of chemical combination. Later experiments showed that the atom is not indivisible but possesses an internal structure.

In this lesson you will learn about the internal structure of an atom, which helps you understand correlations between structure and properties studied in later lessons. Topics span fundamental particles, atomic and mass numbers, isotopes, historical atomic models (Thomson, Rutherford, Bohr), electromagnetic radiation, the hydrogen line spectrum, wave–particle duality, Heisenberg's uncertainty principle, the quantum mechanical model, quantum numbers, orbital shapes, and rules for electronic configuration including the stability of half-filled and completely filled subshells.

Section 1: Fundamental Particles and Atomic Notation (2.1–2.2)

2.1 Discovery of Fundamental Particles

In 1897, J.J. Thomson discovered the electron as a constituent of the atom. He determined that an electron carries a negative charge and has very little mass compared to the whole atom. Since atoms are electrically neutral, a source of positive charge must exist within the atom. This led to the experimental discovery of the proton — a positively charged subatomic particle approximately 1840 times heavier than an electron.

Further experiments revealed that atomic masses exceeded what protons and electrons alone could account for. For example, helium's mass was expected to be double hydrogen's but was found to be almost four times greater. This suggested neutral particles with mass comparable to protons. In 1932, Sir James Chadwick discovered the neutron. Atoms are therefore composed of three fundamental particles:

  • Electron (e): mass 9.109 × 10⁻³¹ kg, charge −1.602 × 10⁻¹⁹ C (relative charge −1)
  • Proton (p): mass 1.673 × 10⁻²⁷ kg, charge +1.602 × 10⁻¹⁹ C (relative charge +1)
  • Neutron (n): mass 1.675 × 10⁻²⁷ kg, charge 0

Because atoms contain still smaller particles, they must have an internal structure — the subject of the rest of this chapter. Comparing masses: an electron is about 1/1840 the mass of a proton, yet both are dwarfed by the atom as a whole because most atomic mass resides in the nucleus (protons + neutrons). This mass discrepancy for helium (four times hydrogen, not twice) was the crucial clue that forced scientists beyond the proton–electron picture.

2.2 Atomic Number, Mass Number, Isotopes and Isobars

All atoms are identified by the number of protons and neutrons they contain. The atomic number (Z) is the number of protons in the nucleus of each atom of an element. In a neutral atom, protons equal electrons, so Z also gives the electron count. Chemical identity depends solely on Z — every atom with 7 protons is nitrogen.

The mass number (A) is the total number of protons and neutrons in the nucleus: A = Z + N, where N is the number of neutrons. Except for ordinary hydrogen (one proton, no neutrons), all nuclei contain both protons and neutrons. Neutrons = A − Z. For fluorine (A = 19, Z = 9), neutrons = 10.

A = Z + N  |  Neutrons N = A − Z
A = mass number · Z = atomic number (protons) · N = neutrons · Neutral atom: electrons = Z

Atoms of the same element can differ in mass. Most elements have isotopes — same Z, different A. Hydrogen has three isotopes: ¹H (protium, 0 neutrons), ²H (deuterium, 1 neutron), ³H (tritium, 2 neutrons). Uranium-235 and uranium-238 share Z = 92 but differ in properties — U-235 is used in reactors; U-238 lacks those properties. Isotopes are named by mass number (except hydrogen's special names). Chemical properties depend on protons and electrons; neutrons do not participate in ordinary chemical change, so isotopes of an element have similar chemistry.

Isobars are atoms of different elements with the same mass number A but different Z — for instance ⁴⁰Ar₁₈ and ⁴⁰Ca₂₀ both have A = 40 but are different elements. Do not confuse isotopes (same Z) with isobars (same A).

Isotope notation places mass number A as superscript and atomic number Z as subscript on the element symbol X. Hydrogen isotopes are written ¹H₁, ²H₁, ³H₁ (protium, deuterium, tritium). For uranium: ²³⁵U₉₂ versus ²³⁸U₉₂ — same chemistry (both uranium) but very different nuclear behaviour. Example 2.1: In ¹⁷O₈: 8 protons, 9 neutrons, 8 electrons. In ¹⁹⁹Hg₈₀: 80 protons, 119 neutrons, 80 electrons. ²⁰⁰Hg₈₀ is a chemically similar isotope with 120 neutrons. All three values Z, N and A must be positive integers.

Section 2: Earlier Models of the Atom (2.3)

2.3.1 Thomson's Plum Pudding Model

After establishing that atoms are divisible, scientists proposed models for internal structure. J.J. Thomson, from discharge tube experiments, suggested atoms as a large positively charged body with small negative electrons scattered throughout — the plum pudding model (electrons = plums in positive pudding) or watermelon model (pulp = positive charge, seeds = electrons).

2.3.2 Rutherford's Gold Foil Experiment

Ernest Rutherford tested Thomson's model with the α-ray scattering experiment (1908 Nobel Prize in Chemistry). A beam of fast-moving α-particles (He²⁺ ions) passed through very thin gold foil. Most particles passed straight through, but some were deflected — a few through large angles, and about 1 in 10,000 rebounded.

Rutherford α-Scattering Experiment α source Au foil Most pass through Small deflection Large angle Rebound ~1/10⁴ + Dense positive nucleus at centre; electrons in empty space
Fig 2.2–2.3 — α-particles mostly pass through gold foil; rare large deflections imply a tiny, dense, positively charged nucleus.

Rutherford's conclusions: (1) Atom contains a dense, positively charged nucleus at the centre. (2) Nearly all positive charge and mass reside in the nucleus. (3) Remaining volume is mostly empty space containing small negative electrons.

Rutherford's model explained scattering data elegantly: undeflected α-particles pass through empty space; near-miss deflections occur when α-particles approach the nucleus; direct collision causes rebound. Yet the model could not explain why electrons do not radiate while orbiting.

Failure of Rutherford's model: Maxwell's electromagnetic theory states that an accelerating charged particle radiates energy continuously. An orbiting electron is accelerating (centripetal force) and should continuously lose energy, spiralling inward (Fig. 2.5) until the atom collapses in about 10⁻⁸ s — but atoms are stable and emit line spectra, not continuous radiation. This contradiction motivated Niels Bohr, a student of Rutherford, to introduce quantisation of electron energy in 1913.

Section 3: Electromagnetic Radiation (2.4)

Before explaining atomic spectra, we need electromagnetic radiation (EMR). EMR is energy transmitted through space as oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation — no medium required. Examples: visible light, heat, radio waves, X-rays, gamma rays. EMR travels at c = 3.0 × 10⁸ m s⁻¹ in vacuum.

Electromagnetic Wave Electric field (E) Magnetic field (B) Propagation λ c = νλ = 3.00 × 10⁸ m s⁻¹
Fig 2.6 — EM wave: E and B fields oscillate perpendicular to each other and to propagation direction.

2.4.1 Characteristic Parameters

  • Amplitude: maximum height of crest or depth of trough
  • Wavelength (λ): distance between consecutive crests; units m, cm, nm, Å (1 Å = 10⁻¹⁰ m)
  • Frequency (ν): wave crests passing a point per second; unit Hz (s⁻¹)
  • Wave number (ν̄): waves per unit length = 1/λ; SI unit m⁻¹, often cm⁻¹
  • Velocity (c): distance travelled per second; c = νλ
c = νλ  |  ν̄ = 1/λ
c = 3.00×10⁸ m s⁻¹ · ν = frequency (Hz) · λ = wavelength · Link: ν = c/λ

EM radiation also exhibits particle nature. Energy is carried in bundles called quanta; a quantum of visible light is a photon. Photon energy is proportional to frequency:

E = hν = hc/λ
E = photon energy (J) · h = 6.626×10⁻³⁴ J s (Planck's constant) · Also E = hcν̄ when ν̄ in m⁻¹

Example 2.2: 12 GHz microwave: E = 6.626×10⁻³⁴ × 1.2×10¹⁰ = 7.95×10⁻²⁴ J. Example 2.3: Green light λ = 535 nm: E = hc/λ = 3.71×10⁻¹⁹ J.

The electromagnetic spectrum spans radio waves through gamma rays; visible light is a tiny portion (Fig. 2.7). Radiation with shorter wavelength (higher frequency) carries more energy per photon — gamma rays at one extreme, radio waves at the other. Wave number ν̄ is especially convenient in spectroscopy because it is directly proportional to energy: E = hcν̄.

Quantisation of energy was revolutionary: unlike classical waves that can have any energy, photons come in discrete packets proportional to ν. This particle picture complements the wave picture — together they foreshadow the wave–particle duality developed later in this chapter.

Section 4: Line Spectrum and Bohr's Model (2.5–2.6)

2.5 Line Spectrum

Sunlight through a prism gives a continuous spectrum (VIBGYOR) — wavelengths vary without break. Flame tests (Na = yellow, Cu = green, Sr = crimson) produce light that, when passed through a prism, splits into discrete lines — a line spectrum (Fig. 2.8).

2.5.1 Hydrogen Line Spectrum

Electric discharge through low-pressure H₂ emits light that forms discrete lines in UV, visible and IR. The general Rydberg formula:

1/λ = RH(1/n₁² − 1/n₂²)   [ν̄ in cm⁻¹]
RH = 109677 cm⁻¹ · n₁, n₂ positive integers, n₁ < n₂ · Lyman n₁=1; Balmer n₁=2; Paschen n₁=3

Named series (Table 2.2): Lyman (UV, n₁=1, n₂=2,3,4…), Balmer (visible, n₁=2, n₂=3,4,5…), Paschen (IR, n₁=3), Brackett (IR, n₁=4), Pfund (IR, n₁=5). Johann Balmer first found a simple formula for visible lines. Example 2.4: For Balmer n₂=3: ν̄ = 109677(1/4 − 1/9) = 109677 × 5/36 cm⁻¹; λ = 1/ν̄ = 656 nm (red line). The discrete lines prove that electron energies in hydrogen are quantised, not continuous.

Bohr Energy Levels — Hydrogen Atom nucleus n=1 n=2 n=3 hν emitted absorption E₃ E₂ E₁ En = −RH/n² — photon on transition: ΔE = hν
Fig 2.11–2.12 — Electron transitions between Bohr stationary states emit or absorb photons matching ΔE = hν.

2.6 Bohr's Model (1913)

Niels Bohr proposed electrons moving in definite circular orbits around the nucleus. Postulates:

  • Electrons in a given orbit have fixed energy — stationary or non-radiating orbits.
  • Electron changes orbit by absorbing or emitting a photon: E = hν = Ef − Ei.
  • Angular momentum is quantised: mvr = nh/2π where n = 1, 2, 3… (principal quantum number).
En = −RH/n²  |  mvr = nh/2π
Negative En = bound electron · Higher n → less negative energy · RH correlates to atomic constants (eq. 2.9)

Bohr explained the hydrogen line spectrum: transitions between stationary states give hν = Ei − Ef, yielding the Rydberg formula when combined with En = −RH/n². Energy levels are inversely proportional to n² — E₁ is most negative (most stable); as n increases, energy approaches zero (ionisation limit). Bohr correlated RH to fundamental constants: RH = mez²e⁴/(8ε₀²h³) for nuclear charge z (Z for hydrogen).

Bohr won the 1922 Nobel Prize in Physics. His model was a landmark but limited: it works well for one-electron species (H, He⁺, Li²⁺) but fails for multi-electron atoms (no account of electron–electron repulsion), cannot explain fine structure (multiple closely spaced lines), Zeeman effect (magnetic field splitting) or the varying intensities of spectral lines. These failures pointed toward a more complete quantum mechanical treatment.

Section 5: Wave–Particle Duality and Uncertainty (2.7–2.8)

2.7 de Broglie Hypothesis

Light shows both wave properties (diffraction, interference) and particle properties (photoelectric effect). In 1923, Louis de Broglie proposed that matter particles also have wave nature. A particle of mass m and velocity v has wavelength:

λ = h/mv = h/p
λ = de Broglie wavelength · p = mv = momentum · Macroscopic objects have immeasurably small λ

Example 2.5: A 380 g cricket ball at 140 km/h (38.89 m s⁻¹) has λ = h/(mv) ≈ 4.48×10⁻³⁵ m — far too small to measure. Electrons, with tiny mass, have measurable wavelengths. In 1927, G.P. Thomson and C.J. Davisson demonstrated electron diffraction by nickel crystals (Fig. 2.13) — the diffraction pattern proved electrons behave as waves under appropriate conditions. de Broglie received the 1929 Nobel Prize in Physics for his 1924 PhD thesis proposing this duality.

Wave–particle duality is not contradiction but complementarity: diffraction and interference demand wave models; photoelectric effect and α-scattering demand particle models. The same electron can display either behaviour depending on the experiment.

2.8 Heisenberg's Uncertainty Principle (1927)

Wave–particle duality implies we cannot simultaneously measure position and momentum of an electron with perfect precision:

Δx · Δp ≥ h/4π
More precise position → less precise momentum, and vice versa · Relevant only for microscopic particles (h is tiny)

If Δx = 0 (electron located exactly), Δp → ∞ (momentum completely unknown). Conversely, precise momentum means position is entirely uncertain. In practice both have finite uncertainty. Because h = 6.626×10⁻³⁴ J s is extraordinarily small, the principle is negligible for cars, cricket balls and aeroplanes — but dominant for electrons.

Heisenberg (Nobel Prize 1932) showed Bohr's simultaneous exact radius and velocity for each orbit is impossible. You cannot trace a definite path for an electron; only speak of probability of finding it in a region. This directly motivated Schrödinger's wave mechanical model replacing orbits with orbitals.

Section 6: Quantum Mechanical Model and Orbitals (2.9)

Erwin Schrödinger (1926) proposed the wave mechanical model. Electron motion is described by a wave function ψ obtained from the Schrödinger wave equation (SWE). |ψ|² gives the probability of finding the electron in 3D space around the nucleus. The region of maximum probability is an atomic orbital — not a fixed path (orbit) but a probability cloud.

2.9.1 Quantum Numbers

Each electron has a unique set of four quantum numbers:

  • Principal quantum number (n): energy level / shell; n = 1, 2, 3… Shell capacity = 2n² electrons (n=1→2, n=2→8, n=3→18).
  • Azimuthal quantum number (l): subshell shape; l = 0 to (n−1). l=0 → s (spherical); l=1 → p (dumb-bell, 3 orbitals); l=2 → d (cloverleaf, 5); l=3 → f (7 orbitals).
  • Magnetic quantum number (ml): orbital orientation; ml = −l to +l. For l=1: ml = −1, 0, +1 (px, py, pz).
  • Spin quantum number (ms): +½ or −½ (clockwise / anticlockwise); introduced via Pauli's principle.
Shapes of Atomic Orbitals s spherical p (nodal plane) d cloverleaf n−l−1 spherical nodes
Fig 2.14–2.18 — s: sphere; p: two lobes with nodal plane; d: cloverleaf shapes; spherical nodes = n−l−1.

2.9.2 Shapes of Orbitals

1s orbital: radial probability peaks at 52.9 pm for H; boundary surface encloses 95% probability — drawn as a sphere. 2s orbital: larger sphere with one spherical node (region of zero probability). Number of spherical nodes = n − l − 1. A nodal plane is a flat region of zero probability (p orbitals have one nodal plane). 3s has 2 spherical nodes.

Orbit vs orbital: Bohr's circular path = orbit (definite trajectory). Quantum orbital = 3D probability region — no fixed path. The wave function ψ itself has no direct physical meaning; only |ψ|² (probability density) is observable.

For n = 3 shell, Table 2.3 lists all allowed combinations: l = 0 (one 3s), l = 1 (three 3p with ml = −1,0,+1), l = 2 (five 3d). Each orbital holds max 2 electrons (Pauli), giving 2 + 6 + 10 = 18 total — matching 2n² = 2(9) = 18. Every electron in the atom has a unique set of four quantum numbers — no two electrons are identical in all four.

px, py, pz differ in orientation (ml) but have equal energy in the absence of external fields (degenerate). Five d orbitals (dxy, dyz, dxz, dx²−y², d) are also degenerate. Nodal planes in p and d orbitals arise where wave function changes sign — zero probability of finding the electron in that plane.

Section 7: Electronic Configuration (2.10–2.11)

2.10.1 Aufbau Principle

Electrons fill orbitals in order of increasing energy so the atom has minimum energy. For multi-electron atoms, use the (n + l) rules:

  • Rule 1: Lower (n + l) fills first — 4s (n+l=4) before 3d (n+l=5).
  • Rule 2: Same (n+l) → lower n first — 3d before 4p.

Order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s …

2.10.2 Pauli's Exclusion Principle

No two electrons in an atom can have all four quantum numbers identical. Since three quantum numbers define an orbital, only two electrons per orbital with opposite spins (+½ and −½).

2.10.3 Hund's Rule of Maximum Multiplicity

Electrons occupy degenerate orbitals singly with parallel spins before pairing. Carbon: 1s² 2s² 2px¹ 2py¹ 2pz⁰ (not paired in one p orbital). Electrons repel — they spread across orbitals.

Notation methods: superscript (1s² 2s² 2p¹x 2p¹y 2p¹z for N) and orbital box diagram with ↑↓ arrows. Shorthand: [He] 2s¹ for Li; core electrons in noble gas brackets, valence electrons outside.

2.11 Stability of Half-Filled and Completely Filled Subshells

Aufbau sometimes fails when subshell energies are close (4s/3d, 5s/4d). Chromium: predicted [Ar] 4s² 3d⁴, actual [Ar] 4s¹ 3d⁵. Copper: predicted 3d⁹ 4s², actual 3d¹⁰ 4s¹. Reasons:

  • Symmetry: half-filled or fully-filled d subshell is more symmetrical and stable.
  • Exchange energy: more electrons with parallel spin → more exchange possibilities → greater stability. 3d⁵ 4s¹ has 10 exchange ways vs 6 for 3d⁴ 4s².

These exceptions are exam favourites — memorise Cr and Cu configurations. Similar reasoning applies to other half-filled d⁵ and fully-filled d¹⁰ cases (e.g. Mo, Ag) though NIOS highlights Cr and Cu specifically.

Electronic configuration determines valence electrons and hence chemical behaviour. Core electrons ([noble gas] configuration) are buried in inner shells; valence electrons participate in bonding. Nitrogen 1s² 2s² 2px¹ 2py¹ 2pz¹ has 5 valence electrons (2s + 2p); lithium [He] 2s¹ has 1. Understanding filling order explains periodic trends studied in later modules.

Intext checkpoints from the textbook reinforce key skills: comparing e⁻ and p⁺ masses (Intext 2.1), listing Rutherford conclusions (2.2), computing photon energies (2.3–2.4), distinguishing line vs continuous spectra (2.4), calculating de Broglie λ for electrons (2.5), defining wave function and quantum numbers (2.6), counting spherical nodes in 3s (answer: 2, from n−l−1 = 3−0−1) (2.7), and applying (n+l) rules to decide 4s before 3d (2.8).

Chapter Summary

Atoms contain electrons, protons and neutrons. Z defines the element; isotopes share Z but differ in N. Thomson's plum pudding gave way to Rutherford's nuclear model, which failed on stability. Bohr quantised orbits and explained the H spectrum via E = hν and En = −RH/n². de Broglie and Heisenberg established wave–particle duality and uncertainty, leading to Schrödinger's quantum model with orbitals defined by quantum numbers n, l, ml, ms. Electron filling follows Aufbau, Pauli and Hund's rules, with Cr and Cu as notable exceptions due to half-filled/full d-subshell stability.

MCQ Quiz — L2 Atomic Structure

0 / 10 correct

Flashcards — L2

1 / 18

Golden Rules — L2 Atomic Structure

Most exam-important points from this chapter:

Fundamental particles

Atom = electrons (−1), protons (+1), neutrons (0). Proton mass ≈ 1840 × electron. Chadwick discovered neutron (1932).

Z, A and isotopes

Z = protons = electrons (neutral). A = Z + N. Isotopes: same Z, different N. Chemistry depends on Z, not N.

Rutherford vs Bohr

Rutherford: nucleus proved by α-scattering but electron should spiral in (Maxwell). Bohr: quantised orbits + photon transitions fix H spectrum.

EM radiation trinity

c = νλ links wave properties; E = hν links particle properties. Know Rydberg for H lines: Balmer = visible (n₁=2).

Filling electrons

Aufbau (n+l), Pauli (max 2 e⁻/orbital, opposite spin), Hund (single occupancy first). Cr = 3d⁵ 4s¹, Cu = 3d¹⁰ 4s¹ — half/full d stability.

A = Z + N
c = νλ
E = hν
1/λ = RH(1/n₁² − 1/n₂²)
En = −RH/n²
mvr = nh/2π
λ = h/mv
Δx·Δp ≥ h/4π
2n² electrons per shell
1s < 2s < 2p < 3s < 3p < 4s < 3d

Section 1: Formulas with Derivations

All key equations from NIOS Chemistry 313, Module 2 — Atomic Structure (sections 2.1–2.11). Unlock for full derivations, examples, and exam prep.

Atomic Notation — A = Z + N

Formula: Mass number A = atomic number Z + number of neutrons N

Where

• Z = protons in nucleus (defines element)
• N = neutrons = A − Z
• Neutral atom: electrons = Z

When to use

Find protons, neutrons, electrons from isotope notation e.g. ¹⁷O₈, ¹⁹⁹Hg₈₀ (Example 2.1).

Derived from

Nuclear composition: nearly all mass is protons + neutrons; electrons contribute negligible mass.

Memory aid: "A is the sum — Add Z and neutrons."

Derivation

By definition, mass number counts nucleons (protons + neutrons). Z counts protons only, so N = A − Z. Example: ¹⁹F₉ → Z=9, N=10, e⁻=9.

❌ Confusing mass number with atomic mass (decimal u).
✓ A is always a whole number (protons + neutrons).

❌ Calling ⁴⁰Ar and ⁴⁰Ca isotopes — they are isobars (same A, different Z).
✓ Isotopes = same Z, different A.

Worked Examples

Basic

Q: Neutrons in ¹⁹F₉?

N = 19 − 9

Answer: 10 neutrons

Intermediate

Q: Electrons in ²³⁵U₉₂ ion U⁴⁺?

Neutral U has 92 e⁻; U⁴⁺ lost 4

Answer: 88 electrons

Advanced

Q: Compare ¹H, ²H, ³H — same element?

All Z=1 (hydrogen isotopes); N = 0, 1, 2

Answer: Isotopes — same chemistry, different mass

Exam

Q: Protons, neutrons, electrons in ¹⁹⁹Hg₈₀?

Z=80, N=199−80=119, e⁻=80

Answer: 80, 119, 80 (Example 2.1)

Wave Equation — c = νλ

Formula: c = νλ  |  ν = c/λ  |  ν̄ = 1/λ

Where

• c = speed of light = 3.00 × 10⁸ m s⁻¹ (vacuum)
• ν = frequency (Hz or s⁻¹)
• λ = wavelength (m, nm, Å)
• ν̄ = wave number (m⁻¹ or cm⁻¹)

When to use

Convert between wavelength and frequency of EM radiation; link to photon energy via E = hν.

Memory aid: "Crazy Velocity Lambda" — c = νλ.

Basic

Q: Wave number if λ = 500 nm?

ν̄ = 1/(500×10⁻⁹) = 2×10⁶ m⁻¹ = 2×10⁴ cm⁻¹

Answer: 2.0 × 10⁴ cm⁻¹

Intermediate

Q: Frequency of light with λ = 600 nm?

ν = c/λ = 3×10⁸ / (600×10⁻⁹)

Answer: 5.0 × 10¹⁴ Hz

Exam

Q: Which has higher energy — red (700 nm) or blue (450 nm)?

Shorter λ → higher ν → higher E = hν

Answer: Blue light

Photon Energy — E = hν = hc/λ

Formula: E = hν = hc/λ   (h = 6.626 × 10⁻³⁴ J s)

When to use

Calculate energy of a quantum/photon from frequency or wavelength (Examples 2.2, 2.3). Planck's quantum theory: energy is quantised in packets.

Real-world application

Photoelectric effect, spectroscopy, laser energy, UV damage to DNA.

❌ Using nm without converting to metres in hc/λ.
✓ Always convert λ to metres: 535 nm = 535×10⁻⁹ m.

Basic

Q: Energy of 12 GHz microwave photon?

E = 6.626×10⁻³⁴ × 1.2×10¹⁰

Answer: 7.95 × 10⁻²⁴ J (Example 2.2)

Intermediate

Q: Energy of green light λ = 535 nm?

E = hc/λ

Answer: 3.71 × 10⁻¹⁹ J (Example 2.3)

Advanced

Q: Photons per second from 1 W green LED?

N/s = Power/Ephoton = 1 / 3.71×10⁻¹⁹

Answer: ~2.7 × 10¹⁸ photons s⁻¹

Exam

Q: Minimum energy to eject electron if threshold ν = 5×10¹⁴ Hz?

E₀ = hν

Answer: 3.31 × 10⁻¹⁹ J

Rydberg Formula — Hydrogen Spectrum

Formula: 1/λ = RH(1/n₁² − 1/n₂²)   (ν̄ in cm⁻¹; RH = 109677 cm⁻¹)

Where

• n₁, n₂ = positive integers, n₁ < n₂
• Lyman: n₁=1 (UV); Balmer: n₁=2 (visible); Paschen: n₁=3 (IR)

When to use

Calculate wavelength of spectral lines; Balmer visible lines (Example 2.4: n₂=3 → 656 nm red).

Intermediate

Q: Balmer line for n₂ = 4?

ν̄ = 109677(1/4 − 1/16); λ = 1/ν̄ (convert cm to nm)

Answer: 486 nm (blue-green)

Exam

Q: Balmer line for n₂ = 3?

ν̄ = 109677(1/4 − 1/9); λ = 1/ν̄

Answer: 656 nm (red) — Example 2.4

Bohr Model — En = −RH/n²  |  mvr = nh/2π

Also: ΔE = hν = Ef − Ei

Where

• n = principal quantum number (1, 2, 3…)
• Negative E means electron bound to nucleus
• Higher n → less negative (higher) energy

When to use

Energy levels of H atom; explain emission/absorption lines; ionisation energy from n=1.

History: Bohr (1913) quantised angular momentum to fix Rutherford's collapsing atom problem.

❌ Thinking electron spirals into nucleus in Bohr model (that's Rutherford's failure).
✓ Bohr: fixed orbits, photon emitted only on transition.

❌ Applying Bohr model to multi-electron atoms.
✓ Bohr works only for one-electron species (H, He⁺, Li²⁺).

Intermediate

Q: Energy of n=3 level in H?

E₃ = −13.6/9 eV

Answer: −1.51 eV

Exam

Q: Photon energy for n=3 → n=2 transition?

ΔE = E₃ − E₂ = (−1.51) − (−3.4)

Answer: 1.89 eV (Balmer series)

de Broglie Wavelength — λ = h/mv = h/p

Formula: λ = h/(mv) where p = mv = momentum

When to use

Wave nature of matter; electron diffraction (Davisson–Germer 1927). Macroscopic objects have negligible λ.

Deep insight: If electrons have wave nature, precise Bohr orbits are impossible → quantum mechanical model.

Advanced

Q: λ for electron at 100 km/s (me = 9.1×10⁻³¹ kg)?

λ = h/(mev)

Answer: ~7.3 × 10⁻⁹ m

Exam

Q: Why don't cricket balls show diffraction?

λ = h/mv → λ ∝ 1/m; large m makes λ immeasurably small

Answer: Wave nature significant only for microscopic particles

Heisenberg Uncertainty — Δx·Δp ≥ h/4π

Cannot simultaneously measure position and momentum of an electron with perfect accuracy. Invalidates precise Bohr orbits → leads to quantum mechanical model (orbitals, not orbits).

Consequence

Electrons described by probability distributions |ψ|² — orbital shapes s, p, d, f.

Section 2: Detailed Definitions

DEFINITION: Atomic Number (Z)

Meaning: Number of protons in nucleus; defines element identity.
Symbol: Z
Unit: dimensionless
Real-life example: All carbon atoms have Z = 6.
Connects to: A = Z + N, periodic table position
Common confusion: Z changes only in nuclear reactions, not chemical reactions.

DEFINITION: Isotope

Meaning: Atoms of same element (same Z) with different mass number A.
Symbol:
Real-life example: ²³⁵U and ²³⁸U — same chemistry, different nuclear properties.
Connects to: N = A − Z
Common confusion: Isotopes vs isobars (same A, different Z).

DEFINITION: Quantum Number n

Meaning: Principal quantum number — shell, size, energy level.
Values: 1, 2, 3…
Max electrons: 2n² per shell
Connects to: Bohr orbits, Aufbau filling order
Common confusion: Shell number ≠ always same as period for d-block.

DEFINITION: Orbital

Meaning: Region of space with high probability (|ψ|²) of finding electron.
Shapes: s spherical, p dumbbell, d cloverleaf
Connects to: l, ml quantum numbers
Common confusion: Orbit (Bohr path) vs orbital (probability cloud).

DEFINITION: Photoelectric Effect

Meaning: Emission of electrons when light above threshold frequency hits metal surface.
Key point: E ∝ ν, not intensity — proves particle nature of light.
Connects to: E = hν
Common confusion: Below threshold ν, no electrons regardless of intensity.

Section 3: Diagrams & Visuals

Atomic Structure Formula Map — L2 A = Z + N Isotopes / Isobars c = νλ E = hν Rydberg / Bohr transitions λ = h/mv (de Broglie) Δx·Δp ≥ h/4π Aufbau: 1s 2s 2p 3s 3p 4s 3d · Hund · Pauli · Cr/Cu exceptions

Classical notation → EM radiation → quantisation → wave-particle duality → quantum model

ELECTRON FILLING ORDER (NIOS L2) ═══════════════════════════════════════ Aufbau: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p … (n+l) rule: lower (n+l) first; tie → lower n Pauli: max 2 e⁻ per orbital (opposite spins) Hund: maximise parallel spins in degenerate orbitals Exceptions: Cr [Ar]3d⁵4s¹ · Cu [Ar]3d¹⁰4s¹ Max per shell: 2n² (n=1→2, n=2→8, n=3→18) ═══════════════════════════════════════

Section 5: Comprehensive Q&A (12 Questions)

Q1: Why did Rutherford's model fail?

Accelerating electrons should radiate energy and spiral into nucleus (Maxwell). Atoms are stable and emit line spectra — contradiction resolved by Bohr's quantisation and later quantum mechanics.

Q2: What is the difference between isotopes and isobars?

Isotopes: same Z, different A (¹H, ²H). Isobars: same A, different Z (⁴⁰Ar, ⁴⁰Ca). Chemical properties follow Z; nuclear properties follow A.

Q3: How does E = hν explain the photoelectric effect?

One photon ejects one electron if hν exceeds work function. Below threshold frequency, no emission — regardless of light intensity. Energy depends on ν, not brightness.

Q4: Which series of hydrogen spectrum is in visible region?

Balmer series (n₁ = 2). Lyman is UV (n₁=1); Paschen is IR (n₁=3). Red line at 656 nm is n=3→2 transition.

Q5: State the four quantum numbers and their significance.

n — shell/size/energy; l — subshell shape (0=s…3=f); ml — orbital orientation; ms — spin ±½. Pauli: no two electrons share all four.

Q6: Why is 4s filled before 3d?

(n+l) rule: 4s has n+l=4, 3d has n+l=5. Lower sum fills first. After filling, 3d becomes lower energy for transition metals — hence 4s empties in Cr, Cu.

Q7: What does negative Bohr energy mean?

Electron is bound to nucleus. E = 0 at infinite separation (ionised). More negative = more stable (closer to nucleus). Ionisation requires +13.6 eV from n=1.

Q8: de Broglie wavelength of a moving car — measurable?

No. λ = h/mv; macroscopic mass makes λ ~10⁻³⁵ m — far below any detector. Wave nature matters for electrons, neutrons, atoms.

Q9: How many orbitals in a d subshell?

l = 2 → ml = −2, −1, 0, +1, +2 → 5 orbitals → max 10 electrons. p has 3 orbitals (6 e⁻); f has 7 (14 e⁻).

Q10: Thomson vs Rutherford model — key difference?

Thomson: positive charge spread throughout atom. Rutherford: dense positive nucleus at centre, mostly empty space — proved by α-scattering.

Q11: Write electronic configuration of Fe and explain unpaired electrons.

Fe (Z=26): [Ar] 3d⁶ 4s². Hund's rule → 3d has 4 unpaired e⁻ (one pair + four singles). Magnetic properties depend on unpaired spins.

Q12: Exam — Calculate ν̄ and λ for Lyman line n₂ = 2.

ν̄ = 109677(1/1 − 1/4) = 82257.75 cm⁻¹. λ = 1/ν̄ = 1.22×10⁻⁵ cm = 122 nm (UV).

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • c = νλ: "Crazy Velocity Lambda"
  • E = hν: "Energy has frequency"
  • Aufbau: "Orbitals fill low (n+l) first"
  • Pauli: "Two max per orbital, opposite spins"
  • Hund: "One electron per box before pairing"

Exam Tips

  • Convert nm → m before hc/λ calculations
  • Rydberg: n₁ < n₂ always; identify series by n₁
  • Bohr model: H-like species only (H, He⁺, Li²⁺)
  • Cr/Cu exceptions: half-filled/full d subshell stability
  • Include units: Hz, J, nm, cm⁻¹ as appropriate

Common Mistakes:
❌ Confusing orbit with orbital
✓ Orbit = fixed path (Bohr); orbital = probability region

❌ Writing 2,8,8,18 without Aufbau exceptions
✓ Use (n+l) rule; 4s before 3d

❌ Applying Rydberg to multi-electron atoms
✓ Rydberg exact for H; other atoms need corrections

Section 7: Connections & Relationships

This chapter builds on: L1 mole concept, basic SI units, periodic table from L3.
This chapter leads to: L4 Chemical Bonding (valence electrons, quantum numbers), spectroscopy, periodic trends.
Related formulas: Z determines valence e⁻ → Lewis structures; orbital shapes → VSEPR and hybridisation.

MODEL EVOLUTION MAP Dalton (indivisible atom) │ Thomson (plum pudding) ──► Rutherford (nucleus) │ │ │ Bohr (quantised orbits) │ │ └──────────► Quantum mechanics (ψ, orbitals) │ Aufbau + Pauli + Hund │ L4: valence electrons bond

Section 8: Complete Quick Reference

Formulas at a glance:

• A = Z + N — nuclear composition

• c = νλ — wave relation (c = 3×10⁸ m s⁻¹)

• E = hν = hc/λ — photon energy (h = 6.626×10⁻³⁴ J s)

• 1/λ = RH(1/n₁² − 1/n₂²) — H spectrum (RH = 109677 cm⁻¹)

• En = −13.6/n² eV — Bohr H atom

• λ = h/mv — de Broglie wavelength

• Δx·Δp ≥ h/4π — uncertainty principle

Quantum numbers: n · l · ml · ms

Filling rules: Aufbau · Pauli · Hund · Cr/Cu exceptions

Decision tree: Nuclear question? → A=Z+N. Light/energy? → c=νλ then E=hν. H spectrum? → Rydberg. Electron config? → Aufbau + exceptions.

Remember: ✓ Isotopes same Z ✓ Isobars same A ✓ Bohr for H only ✓ 4s before 3d ✓ Balmer = visible

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L2 — Atomic Structure only. Use Model Answer for marking points; Explanation for concept clarity.

L2 — Atomic Structure

10 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. 1 a.m.u. is equal to — (A) 12 th of mass of one C–12 atom   (B) 14 th of mass of one C–12 atom   (C) 16 th of mass of one O–16 atom   (D) mass of one H atom 1 a.m.u

1 mark · Q2 · 313/MAY/205A

Model Answer

Model approach (select the best option):

  • (A) 12 th of mass of one C–12 atom
  • (B) 14 th of mass of one C–12 atom
  • (C) 16 th of mass of one O–16 atom
  • (D) mass of one H atom 1 a.m.u

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L2. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205A · Q2.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ2. Which statement is not correct about quantum? — (A) It is a bundle of energy   (B) A quantum of visible light is called a photon   (C) The energy of the quantum is proportional to the frequency of the radiation   (D) The energy of the quantum is proportional to the wavelength of the radiation. ¹

1 mark · Q4 · 313/MAY/205A

Model Answer

Model approach (select the best option):

  • (A) It is a bundle of energy
  • (B) A quantum of visible light is called a photon
  • (C) The energy of the quantum is proportional to the frequency of the radiation
  • (D) The energy of the quantum is proportional to the wavelength of the radiation. ¹

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L2. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205A · Q4.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ3. Read the passage given below and answer the following questions : Electromagnetic radiations travel with the velocity of light. These do not require any medium to propagate. These travel as waves in the planes perpendicular to each other and also to the direction of propagation. Depict the amplitude and wavelength () of an electromagnetic wave in the form of a diagram. Define a photon. Give its mathematical expression.

2 marks · Q18 · 313/MAY/205A

Model Answer

State the precise definition from the L2 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q18 · 2 mark(s) · L2.

PYQ4. 1 a.m.u. is equal to — (A) 12 th of mass of one C–12 atom   (B) 14 th of mass of one C–12 atom   (C) 16 th of mass of one O–16 atom   (D) mass of one H atom 1 a.m.u

1 mark · Q9 · 313/MAY/205B

Model Answer

Model approach (select the best option):

  • (A) 12 th of mass of one C–12 atom
  • (B) 14 th of mass of one C–12 atom
  • (C) 16 th of mass of one O–16 atom
  • (D) mass of one H atom 1 a.m.u

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L2. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205B · Q9.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ5. Which statement is not correct about quantum? — (A) It is a bundle of energy   (B) A quantum of visible light is called a photon   (C) The energy of the quantum is proportional to the frequency of the radiation   (D) The energy of the quantum is proportional to the wavelength of the radiation. ¹

1 mark · Q10 · 313/MAY/205B

Model Answer

Model approach (select the best option):

  • (A) It is a bundle of energy
  • (B) A quantum of visible light is called a photon
  • (C) The energy of the quantum is proportional to the frequency of the radiation
  • (D) The energy of the quantum is proportional to the wavelength of the radiation. ¹

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L2. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205B · Q10.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ6. Read the passage given below and answer the following questions : Electromagnetic radiations travel with the velocity of light. These do not require any medium to propagate. These travel as waves in the planes perpendicular to each other and also to the direction of propagation. Depict the amplitude and wavelength () of an electromagnetic wave in the form of a diagram. Define a photon. Give its mathematical expression.

2 marks · Q20 · 313/MAY/205B

Model Answer

State the precise definition from the L2 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q20 · 2 mark(s) · L2.

PYQ7. 1 a.m.u. is equal to — (A) 12 th of mass of one C–12 atom   (B) 14 th of mass of one C–12 atom   (C) 16 th of mass of one O–16 atom   (D) mass of one H atom 1 a.m.u

1 mark · Q7 · 313/MAY/205C

Model Answer

Model approach (select the best option):

  • (A) 12 th of mass of one C–12 atom
  • (B) 14 th of mass of one C–12 atom
  • (C) 16 th of mass of one O–16 atom
  • (D) mass of one H atom 1 a.m.u

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L2. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205C · Q7.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ8. Which statement is not correct about quantum? — (A) It is a bundle of energy   (B) A quantum of visible light is called a photon   (C) The energy of the quantum is proportional to the frequency of the radiation   (D) The energy of the quantum is proportional to the wavelength of the radiation. ¹

1 mark · Q8 · 313/MAY/205C

Model Answer

Model approach (select the best option):

  • (A) It is a bundle of energy
  • (B) A quantum of visible light is called a photon
  • (C) The energy of the quantum is proportional to the frequency of the radiation
  • (D) The energy of the quantum is proportional to the wavelength of the radiation. ¹

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L2. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205C · Q8.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ9. Read the passage given below and answer the following questions : Electromagnetic radiations travel with the velocity of light. These do not require any medium to propagate. These travel as waves in the planes perpendicular to each other and also to the direction of propagation. Depict the amplitude and wavelength () of an electromagnetic wave in the form of a diagram. Define a photon. Give its mathematical expression.

2 marks · Q23 · 313/MAY/205C

Model Answer

State the precise definition from the L2 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q23 · 2 mark(s) · L2.

Section B — Short / Long answer (from papers)

PYQ10. Draw the shapes of d-orbitals. d-H$jH$m|

2 marks · Q30 · 313/TUS/105A

Model Answer

Answer using key concepts from L2 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L2 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q30 · 2 mark(s) · L2.

Problem Solving — L2 Atomic Structure

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6Bohr

State Bohr’s angular momentum quantisation. Write ground-state energy of H atom. Draw a sketch of n=1,2,3 orbits.

mvr = nh/2π
E_n = −13.6/n² eV (H)

Pencil sketch (labelled)

Atomic structure (schematic) nucleus e⁻ protons + neutrons in nucleus
Pencil sketch: nucleus and electron shells

Solution — step by step with formulas

  1. mvr = nh/2π.
  2. For n=1, E = −13.6 eV.

Final answer: L = nh/2π; E₁ = −13.6 eV

Formulas used in this problem

mvr = nh/2π
E_n = −13.6/n² eV (H)

Textbook formal language

Bohr assumed stationary orbits with quantised angular momentum; hydrogen energies follow E_n = −13.6/n² eV.

Working formulas: mvr = nh/2π; E_n = −13.6/n² eV (H). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Only certain circular orbits allowed; lowest hydrogen energy is −13.6 eV.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Bohr model

Explains line spectrum via ΔE = hf between levels.

Linked to chapter notes (L2). Remember: mvr = nh/2π; E_n = −13.6/n² eV (H). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write mvr = nh/2π; E_n = −13.6/n² eV (H) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Quantum numbers

List the four quantum numbers and state what each describes.

n, l, m_l, m_s

Solution — step by step with formulas

  1. n: shell/energy; l: subshell/shape; m_l: orbital orientation; m_s: electron spin.

Final answer: n, l, m_l, m_s as above

Formulas used in this problem

n, l, m_l, m_s

Textbook formal language

Four quantum numbers uniquely label an electron in an atom (Pauli framework).

Working formulas: n, l, m_l, m_s. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

n = which floor; l = room shape; m_l = which door; m_s = spin up/down.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Quantum numbers

l = 0…n−1; m_l = −l…+l; m_s = ±½.

Linked to chapter notes (L2). Remember: n, l, m_l, m_s. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write n, l, m_l, m_s before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Orbitals

How many orbitals are in a p subshell? Maximum electrons in 3p?

s: 1 orbital; p: 3; d: 5

Solution — step by step with formulas

  1. 3 orbitals in p.
  2. Each holds 2 e⁻ ⇒ 6 electrons in 3p.

Final answer: 3 orbitals; 6 electrons

Formulas used in this problem

s: 1 orbital; p: 3; d: 5

Textbook formal language

Subshell capacity is 2(2l+1) electrons.

Working formulas: s: 1 orbital; p: 3; d: 5. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Three p boxes, two electrons each → six.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — s,p,d orbitals

3p means n=3, l=1.

Linked to chapter notes (L2). Remember: s: 1 orbital; p: 3; d: 5. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write s: 1 orbital; p: 3; d: 5 before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Config

Write the electronic configuration of ₇N and explain Hund’s rule for 2p.

Aufbau, Pauli, Hund

Solution — step by step with formulas

  1. N: 1s² 2s² 2p³.
  2. Three 2p electrons occupy three different orbitals with parallel spins (Hund).

Final answer: 1s² 2s² 2p³; parallel spins in 2p

Formulas used in this problem

Aufbau, Pauli, Hund

Textbook formal language

Hund’s rule: degenerate orbitals fill singly with parallel spin before pairing.

Working formulas: Aufbau, Pauli, Hund. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Don’t pair p electrons until each p orbital has one—like seats on a bus.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Electronic configuration

Pauli: no two electrons share all four quantum numbers.

Linked to chapter notes (L2). Remember: Aufbau, Pauli, Hund. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write Aufbau, Pauli, Hund before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Spectrum

Name the series for transitions ending at n=2.

1/λ = R(1/n₁² − 1/n₂²)

Solution — step by step with formulas

  1. Balmer series (visible region largely).

Final answer: Balmer series

Formulas used in this problem

1/λ = R(1/n₁² − 1/n₂²)

Textbook formal language

Discrete lines arise from electronic transitions between Bohr levels.

Working formulas: 1/λ = R(1/n₁² − 1/n₂²). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Jumps down to level 2 make the coloured Balmer lines.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Hydrogen spectrum

Lyman → n=1 (UV); Paschen → n=3 (IR).

Linked to chapter notes (L2). Remember: 1/λ = R(1/n₁² − 1/n₂²). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write 1/λ = R(1/n₁² − 1/n₂²) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Duality

State de Broglie’s relation for a material particle.

λ = h/p

Solution — step by step with formulas

  1. λ = h/p = h/(mv).

Final answer: λ = h/p

Formulas used in this problem

λ = h/p

Textbook formal language

Matter waves have wavelength h/p; supported Bohr quantisation via standing waves.

Working formulas: λ = h/p. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Faster/heavier particles have shorter wavelengths.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — de Broglie / dual nature intro

Electron diffraction confirms wave nature of electrons.

Linked to chapter notes (L2). Remember: λ = h/p. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write λ = h/p before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.