CN02 · Quiz — Atomic Structure

Question 1 of 20

The de Broglie wavelength of a particle moving with momentum p is given by λ = h/p. This implies that:

AOnly photons (light particles) have wave-like properties BAny moving particle with mass has an associated wavelength CWavelength increases with increasing momentum DOnly electrons in atoms have wave properties

de Broglie (1924): λ = h/mv (h = Planck's constant, m = mass, v = velocity). Matter waves: every moving particle has an associated wavelength. For macroscopic objects (cricket ball), λ is immeasurably small. For electrons (small mass, high speed), λ is measurable and comparable to atomic dimensions. Confirmed by electron diffraction (Davisson-Germer experiment, 1927).

Question 2 of 20

The Heisenberg Uncertainty Principle states that:

AThe exact position of an electron can always be determined if we know its energy BIt is impossible to simultaneously determine both the exact position and exact momentum of a particle: Δx·Δp ≥ h/4π CElectrons move in perfectly circular orbits with precisely defined positions DThe energy and time can both be measured with arbitrary precision simultaneously

Heisenberg Uncertainty Principle (1927): Δx·Δp ≥ ℏ/2 (where ℏ = h/2π). The more precisely position is known (small Δx), the more uncertain the momentum (large Δp) — and vice versa. This is a fundamental quantum mechanical property — not a limitation of instruments. This is why Bohr's model (precise circular orbits with defined positions) is incorrect.

Question 3 of 20

The four quantum numbers (n, l, ml, ms) that describe an electron in an atom — which combination violates the Pauli Exclusion Principle?

An=2, l=1, ml=0, ms=+1/2 and n=2, l=1, ml=0, ms=-1/2 (different ms — allowed) Bn=3, l=2, ml=-1, ms=+1/2 and n=3, l=2, ml=-1, ms=+1/2 (identical set — FORBIDDEN) Cn=1, l=0, ml=0, ms=+1/2 (single electron — allowed) Dn=4, l=3, ml=+2, ms=-1/2 (valid quantum numbers — allowed)

Pauli Exclusion Principle: no two electrons in the same atom can have the same four quantum numbers. An orbital is defined by (n, l, ml) — it can hold maximum 2 electrons with OPPOSITE spins (ms = +1/2 and -1/2). Two electrons with IDENTICAL quantum numbers (n=3, l=2, ml=-1, ms=+1/2 for both) is forbidden.

Question 4 of 20

The shape of a 2p orbital is best described as:

ASpherical (like all orbitals) BDumbbell-shaped — two lobes on opposite CCloverleaf (four-lobed) shape DToroidal (donut-shaped) along one axis

2p orbitals (l=1, ml = -1, 0, +1): three dumbbell-shaped orbitals oriented along x, y, and z axes (2px, 2py, 2pz). Each has two lobes with a node at the nucleus. 3d orbitals (l=2): mainly cloverleaf (four lobes) except dz², which has two lobes + a ring. s orbitals (l=0): spherical. f orbitals (l=3): complex multi-lobed shapes.

Question 5 of 20

In the emission spectrum of hydrogen, the Balmer series corresponds to transitions:

ATransitions from any level (n≥2) down to n=1 — Lyman series, UV region BTransitions from any level (n≥3) down to n=2 — visible light region CTransitions from any level (n≥4) down to n=3 — Paschen series, IR region DOnly the single transition from n=2 down to n=1

Hydrogen emission series: Lyman (n≥2 → n=1, UV); Balmer (n≥3 → n=2, visible: red Hα at 656nm, blue-green Hβ at 486nm, etc.); Paschen (n≥4 → n=3, near-IR); Brackett (→n=4, IR); Pfund (→n=5, far-IR). Balmer series is visible — historically first discovered (1885). Energy of photon: E = hν = 13.6(1/n1² - 1/n2²) eV.

Question 6 of 20

The energy of an electron in the nth orbit of the hydrogen atom according to Bohr's model is:

AEn = +13.6/n² eV BEn = -13.6/n² eV (negative CEn = 13.6 × n² eV DEn = -13.6 × n eV

Bohr's energy for hydrogen: En = -13.6/n² eV (or -2.18 × 10⁻¹⁸/n² J). Negative sign = bound state (energy is lower than free electron). Ground state (n=1): -13.6 eV. First excited state (n=2): -3.4 eV. Ionisation energy = energy needed to remove electron to n=∞ (E=0) from ground state = +13.6 eV = 1312 kJ/mol.

Question 7 of 20

Which quantum number determines the orientation of an orbital in space?

APrincipal quantum number (n) — determines energy level BAzimuthal quantum number (l) — determines shape CMagnetic quantum number (ml) — determines orientation in space DSpin quantum number (ms) — determines electron spin direction

Magnetic quantum number (ml): takes integer values from -l to +l (including 0). For l=1 (p subshell): ml = -1, 0, +1 → 3 orientations (px, py, pz). For l=2 (d): ml = -2, -1, 0, +1, +2 → 5 orientations. ml determines how each orbital is oriented relative to an external magnetic field — explains splitting of spectral lines in a magnetic field (Zeeman effect).

Question 8 of 20

The effective nuclear charge (Zeff) experienced by an outer electron is less than the actual nuclear charge (Z) because of:

AThe negative charge of the nucleus repelling outer electrons BShielding (screening) by inner electrons CThe uncertainty principle making Zeff impossible to define precisely DRelativistic effects at high electron speeds

Effective nuclear charge: Zeff = Z - σ (σ = shielding constant). Inner electrons shield outer electrons from the full nuclear charge. More inner electrons = more shielding = lower Zeff = outer electrons less tightly held = higher atomic radius, lower ionisation energy. Slater's rules calculate σ. Explains periodic trends in IE, EA, atomic radius, and electronegativity.

Question 9 of 20

According to the modern quantum mechanical model, an orbital is best defined as:

AA fixed circular path in which electrons travel around the nucleus BA region of space where the probability of finding an electron is highest CThe energy level at which an electron is found DThe speed at which an electron moves around the nucleus

Modern quantum mechanical model (Schrödinger equation, 1926): orbitals are mathematical functions (wave functions ψ) whose square (|ψ|²) gives the probability density of finding an electron at any point in space. No definite orbit — only probability. The 90% probability boundary gives the 'shape' of the orbital. Replaces Bohr's deterministic circular orbits.

Question 10 of 20

In multi-electron atoms, the energy of orbitals in the same shell depends on:

AOnly the principal quantum number n — same as in hydrogen (one-electron system) BBoth n and l — the (n+l) rule determines filling order in multi-electron atoms CBoth n and ml — the magnetic quantum number primarily determines orbital energy DOnly the spin quantum number ms — the sole remaining degree of freedom

In multi-electron atoms, electron-electron repulsion and shielding cause energy to depend on BOTH n and l. The (n+l) rule (Madelung/Klechkowsky): lower (n+l) = lower energy. 4s (n+l=4) fills before 3d (n+l=5). 3d (n+l=5) fills before 4p (n+l=5, but lower n=3 < 4 → 3d fills first). This explains the anomalous electron configurations of Cr and Cu.

Question 11 of 20

The anomalous electronic configuration of chromium (Z=24) is [Ar]3d⁵4s¹ instead of [Ar]3d⁴4s² because:

AThe 4s orbital is always filled last BHalf-filled d subshells are particularly stable due to C3d has lower energy than 4s for Z=24 DThe Pauli exclusion principle requires this configuration

Cr: [Ar]3d⁵4s¹ (not expected 3d⁴4s²). Cu: [Ar]3d¹⁰4s¹ (not 3d⁹4s²). Reason: extra stability of completely half-filled (d⁵) or completely filled (d¹⁰) d subshells due to: (1) maximum exchange energy (all parallel spins maximise exchange interactions), (2) symmetrical electron distribution reduces repulsion. One electron 'migrates' from 4s to 3d to achieve this stable configuration.

Question 12 of 20

The photoelectric effect proved the particle nature of light because:

ALight waves are reflected off the metal surface without any electron emission BElectron kinetic energy depends on light frequency, not on light intensity CElectrons are emitted from any metal regardless of the frequency of light used DThe number of emitted electrons depends on the wavelength of the incident light

Photoelectric effect (Einstein, 1905, Nobel 1921): KE_max = hν - φ (work function). Key observations explained by photon model: (1) electrons are only emitted above a threshold frequency (hν ≥ φ) — intensity doesn't matter if ν < threshold, (2) KE depends on ν (not intensity), (3) instantaneous emission. Wave theory couldn't explain the threshold frequency — particle (photon) model required.

Question 13 of 20

The Schrödinger wave equation ĤΨ = EΨ is significant because:

AIt describes the exact position and momentum of electrons simultaneously BIts solutions (wave functions Ψ) describe the quantum states of electrons CIt only applies to the hydrogen atom and cannot be generalised DIt replaces quantum mechanics with classical wave theory

Schrödinger equation (1926): ĤΨ = EΨ (Ĥ = Hamiltonian operator, Ψ = wave function, E = energy). Solutions: (1) wave functions Ψnlml (orbitals) automatically emerge, (2) |Ψ|² = probability density, (3) energy eigenvalues E are quantised — arise naturally without Bohr's arbitrary quantum condition. Works for multi-electron atoms (approximately), molecules, and forms the basis of all modern chemistry and materials science.

Question 14 of 20

Spin quantum number (ms) can only take values +½ or -½. This explains why:

AElectrons can only exist in two energy states in any atom BEach orbital can hold a maximum of two electrons with opposite spins CElectrons spin at exactly half the speed of light DThere are only two types of orbitals: s and p

ms = +½ or -½: represents the two possible spin states of an electron (spin-up ↑ and spin-down ↓). Each orbital (defined by n, l, ml) can hold at most 2 electrons — they must have OPPOSITE spins (ms = +½ and -½) to satisfy Pauli exclusion (no two identical quantum numbers). This is the fundamental reason for the 2n² rule for maximum electrons per shell.

Question 15 of 20

The energy emitted when an electron in hydrogen transitions from n=3 to n=1 is:

A12.1 eV = 13.6 × 8/9 ≈ 12.1 eV) B1.51 eV C3.4 eV D10.2 eV

E = 13.6(1/n1² - 1/n2²) eV = 13.6(1/1² - 1/3²) = 13.6(1 - 1/9) = 13.6 × 8/9 = 12.09 eV ≈ 12.1 eV. This photon falls in the UV region (Lyman series, n=3→n=1). Compare: n=2→n=1 = 13.6(1 - 1/4) = 10.2 eV (Lyman alpha); n=3→n=2 = 13.6(1/4 - 1/9) = 1.89 eV (Balmer series, visible red).

Question 16 of 20

The radius of the nth Bohr orbit of hydrogen is given by rn = n² × 0.529 Å (Bohr radius a0). The radius of the 3rd orbit is:

A0.529 Å B1.587 Å C4.761 Å D2.116 Å

r3 = 3² × 0.529 = 9 × 0.529 = 4.761 Å. General formula: rn = n²a0 where a0 = 0.529 Å (Bohr radius). r1 = 0.529 Å (ground state); r2 = 4 × 0.529 = 2.116 Å; r3 = 9 × 0.529 = 4.761 Å. Orbital radius increases as n² — outer shells are much larger. For hydrogen-like ions: r = n²a0/Z.

Question 17 of 20

Which of the following transitions in hydrogen emits the highest energy photon?

An=2 → n=1 Bn=3 → n=2 Cn=4 → n=3 Dn=5 → n=4

Energy emitted: ΔE = 13.6(1/n1² - 1/n2²) eV. n=2→1: 13.6(1-0.25) = 10.2 eV. n=3→2: 13.6(0.25-0.111) = 1.89 eV. n=4→3: 13.6(0.111-0.0625) = 0.66 eV. n=5→4: 13.6(0.0625-0.04) = 0.306 eV. The n=2→n=1 transition (Lyman alpha) emits the highest energy photon (UV). Transitions to lower n emit more energy — greater energy difference.

Question 18 of 20

The wave function Ψ for the 1s orbital of hydrogen has its maximum value at:

AAt large distances from the nucleus where electron density approaches zero BAt r = 0 (the nucleus) where the wave function ψ itself is at maximum CAt r = a₀ (Bohr radius) where the radial distribution function 4πr²|ψ|² peaks DAt the radial node where ψ = 0 and electron probability is zero

For 1s orbital: Ψ is maximum at r=0 (nucleus). BUT the radial probability distribution function (RDF) = 4πr²|Ψ|² — the probability of finding the electron in a spherical shell of radius r — has its maximum at r = a0 (Bohr radius = 0.529 Å). The r² term dominates at small r even though |Ψ|² is largest at nucleus. The RDF is what tells us where the electron is most likely to be found physically.

Question 19 of 20

The number of radial nodes in a wave function is equal to:

Al (the azimuthal quantum number) Bn - l - 1 (where n is principal and Cn - 1 (total nodes = n - 1) Dml (the magnetic quantum number)

Radial nodes = n - l - 1. Angular nodes = l. Total nodes = n - 1. Examples: 1s (n=1, l=0): 0 radial, 0 angular, 0 total. 2s (n=2, l=0): 1 radial, 0 angular. 2p (n=2, l=1): 0 radial, 1 angular (nodal plane). 3s: 2 radial, 0 angular. 3p: 1 radial, 1 angular. 3d: 0 radial, 2 angular. Nodes are surfaces where Ψ=0 (probability of finding electron = 0).

Question 20 of 20

Zeff increases across a period for atoms in the same period because:

AShielding by inner electrons increases dramatically across a period BNuclear charge increases while same-shell electrons shield each other poorly CThe number of inner electron shells increases across a period DElectron-electron repulsion between valence electrons decreases across a period

Across a period: Z increases by +1 for each element (proton added). Electrons added to same valence shell — valence electrons shield each other POORLY (σ ≈ 0.35 per same-shell electron in Slater's rules). Net: Zeff = Z - σ increases across period. Increased Zeff → smaller atomic radius, higher ionisation energy, higher electronegativity. This is the mechanistic explanation for all periodic trends across a period.

CN02 — Atomic Structure