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WBB • Class XI • Chemistry • Ch 2
Estimated Time: 90 minutes
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Structure of an Atom

The exploration of atomic structure represents one of the greatest triumphs in the history of physical science, transitioning chemistry from classical mechanical billiard-ball concepts to modern quantum mechanics. Chapter 2 deconstructs the architecture of the atom starting from the experimental discoveries of the fundamental subatomic particles: the electron via cathode rays and Thomson e/m charge-to-mass ratio measurements, the proton via canal rays, and the neutron through Chadwick alpha-bombardment. Students analyze the milestones and failures of early classical models, notably Thomson plum-pudding model and Rutherford nuclear model, which established the dense positive nucleus but succumbed to Maxwell classical electromagnetic instability. The narrative then shifts to the quantum revolution, driven by black body radiation, Planck quantum theory, Einstein explanation of the photoelectric effect, and the line emission spectrum of hydrogen. Niels Bohr groundbreaking model introduces stationary orbits and quantized angular momentum, providing exact quantitative expressions for orbital radii, electron velocities, and energy levels for single-electron species. Finally, the chapter addresses the fundamental breakdowns of Bohr theory, leading directly to wave-particle duality (de Broglie hypothesis), Heisenberg Uncertainty Principle, the Schrödinger wave equation, the physical interpretation of the wavefunction and electron probability density, the four quantum numbers, orbital boundary surfaces and nodes, and the fundamental governing rules of electronic configuration: the Aufbau principle, Pauli exclusion principle, and Hund rule of maximum multiplicity.

Why This Chapter Matters

Understanding atomic structure is the bedrock upon which all modern chemistry, solid-state physics, materials science, and semiconductor technology are built. The quantum mechanical distribution of electrons in orbitals dictates the entirety of the periodic table, chemical bonding, molecular geometry, reactivity, color, and magnetism. In pharmaceutical engineering, molecular recognition and drug-receptor binding are governed by orbital interactions and electronic charge distributions. In contemporary materials science and nanotechnology, engineering quantum dots, solar cells, lasers, and graphene transistors relies directly on the wave mechanics of electrons and discrete energy bandgaps. For students preparing for WBCHSE Class 11 examinations, WBJEE, JEE Main, and NEET, this chapter represents a massive scoring pillar that connects basic physical principles with subsequent chapters, including periodic classification, chemical bonding, and transition metal coordination chemistry.

Before You Begin (Prerequisites)

  • Class 10 Physical Science: Basic atomic structure, protons, neutrons, electrons, and simple valency concepts
  • Basic mathematics: Exponential notations, algebra of powers, and basic trigonometry

Chapter Roadmap & Progression

1 Module 1: Sub-atomic Particles, Ear...
2 Module 2: Dual Nature of Radiation,...
3 Module 3: Bohr Model of the Hydroge...
4 Module 4: de Broglie Wavelength & H...
5 Module 5: Quantum Mechanical Model,...
6 Module 6: Rules for Filling Orbital...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Sub-atomic Particles, Early Atomic Models & Nuclear Structure

The classical view of the atom as an indivisible sphere, postulated by John Dalton in 1808, was challenged toward the close of the nineteenth century by a succession of electrical discharge experiments proving the existence of subatomic particles.

1. Discovery of Subatomic Particles
  • The Electron (Cathode Rays): In 1897, J.J. Thomson investigated electrical discharge through rarefied gases in Crookes tubes at low pressure (~0.01 mm Hg) and high voltage (~10,000 V). Cathode rays traveled in straight lines, cast sharp shadows of opaque objects, exerted mechanical force on paddle wheels, and deflected toward the positive plate in electric and magnetic fields. Thomson determined their universal charge-to-mass ratio: \[\frac{e}{m_e} = 1.758820 \times 10^{11}\text{ C/kg}\] Robert A. Millikan (1909) determined the fundamental charge of the electron via the Oil Drop Experiment: \(e = -1.602176 \times 10^{-19}\text{ C}\). Consequently, the rest mass of the electron was calculated as: \[m_e = \frac{e}{e/m_e} = \frac{1.602176 \times 10^{-19}\text{ C}}{1.758820 \times 10^{11}\text{ C/kg}} = 9.109383 \times 10^{-31}\text{ kg} \approx \frac{1}{1837}\text{ u}\]
  • The Proton (Anode / Canal Rays): E. Goldstein (1886) utilized a perforated cathode in discharge tubes and observed canal rays traveling toward the cathode. Unlike cathode rays, their \(e/m\) ratio depended upon the residual gas, reaching a maximum value when hydrogen was used: \(e/m = 9.58 \times 10^7\text{ C/kg}\). The particle was identified as the proton: charge \(+1.6022 \times 10^{-19}\text{ C}\), mass \(m_p = 1.67262 \times 10^{-27}\text{ kg} = 1.00727\text{ u}\).
  • The Neutron: James Chadwick (1932) bombarded thin sheets of beryllium with high-velocity \(\alpha\)-particles and detected neutral penetrating radiation consisting of neutral particles with mass slightly greater than a proton: \[^9_4\text{Be} + ^4_2\alpha \rightarrow ^{12}_6\text{C} + ^1_0\text{n}\] Mass of neutron \(m_n = 1.67493 \times 10^{-27}\text{ kg} = 1.00866\text{ u}\).
2. Thomson Plum-Pudding Model & Rutherford Scattering Experiment

Thomson (1898) proposed that an atom is a positively charged spherical cloud (~\(10^{-10}\text{ m}\) radius) with electrons embedded like raisins in a pudding. This model could not explain large-angle scattering of charged particles.

In 1911, Ernest Rutherford, Hans Geiger, and Ernest Marsden directed a collimated beam of \(5.5\text{ MeV}\) \(\alpha\)-particles (\(^4_2\text{He}^{2+}\)) onto an ultra-thin gold foil (~\(100\text{ nm}\) thick) surrounded by a fluorescent zinc sulfide (ZnS) screen:

  • Observations: Most \(\alpha\)-particles (~99.9%) passed through undeflected; a small fraction (~1 in 8,000) was deflected by angles \(\theta > 90^\circ\); and extremely rarely (~1 in 20,000), an \(\alpha\)-particle bounced almost directly backward (\(\theta \approx 180^\circ\)).
  • Conclusions & Nuclear Model: Most of the atom is empty space. All positive charge and virtually all mass are concentrated in an extraordinarily minute, dense central core termed the nucleus (radius \(r_N \sim 10^{-15}\text{ m}\) compared to atomic radius \(r_A \sim 10^{-10}\text{ m}\); volume fraction \(\sim 10^{-15}\)). Electrons orbit the nucleus in circular paths, held by electrostatic Coulombic attraction.
  • Limitations of Rutherford Model:
    1. Instability under Classical Electrodynamics: According to Maxwell electromagnetic theory, an accelerating charged particle (in circular orbit) must continuously radiate energy. The electron would lose kinetic energy, spiral inward in \(\sim 10^{-8}\text{ s}\), and collapse into the nucleus.
    2. Failure to Explain Atomic Spectra: If the spiral is continuous, the emitted radiation must form a continuous spectrum; experimentally, atoms emit sharp, discrete line spectra.
3. Atomic Nomenclature & Nuclear Species
TermDefinitionKey Relationships / Examples
Atomic Number (\(Z\))Number of protons in the nucleus of an atom\(Z = p = e\) (in a neutral atom)
Mass Number (\(A\))Total number of nucleons (protons + neutrons)\(A = Z + n \implies n = A - Z\)
IsotopesNuclides of the same element with identical \(Z\) but different \(A\)\(^1_1\text{H}, ^2_1\text{H}, ^3_1\text{H}\) and \(^{35}_{17}\text{Cl}, ^{37}_{17}\text{Cl}\)
IsobarsNuclides of different elements with identical \(A\) but different \(Z\)\(^{40}_{18}\text{Ar}, ^{40}_{19}\text{K}, ^{40}_{20}\text{Ca}\)
IsotonesNuclides containing identical number of neutrons (\(A - Z\))\(^{14}_6\text{C}, ^{15}_7\text{N}, ^{16}_8\text{O}\) (all have 8 neutrons)
Isoelectronic SpeciesAtoms or ions possessing an identical number of electrons\(\text{N}^{3-}, \text{O}^{2-}, \text{F}^-, \text{Ne}, \text{Na}^+, \text{Mg}^{2+}\) (all 10 electrons)

Module 2: Dual Nature of Radiation, Planck Quantum Theory & Photoelectric Effect

To overcome the limitations of classical mechanics, physicists investigated the interaction of electromagnetic radiation with matter, uncovering the dual wave-particle nature of light.

1. Wave Characteristics of Electromagnetic Radiation

James Clerk Maxwell (1870) demonstrated that accelerating electric charges generate oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation. All electromagnetic waves travel through vacuum at the speed of light \(c = 2.997925 \times 10^8\text{ m/s} \approx 3.0 \times 10^8\text{ m/s}\).

  • Wavelength (\(\lambda\)): Distance between two adjacent crests or troughs. Units: m, cm, nm (\(10^{-9}\text{ m}\)), or Ångström (\(1\text{ \AA} = 10^{-10}\text{ m} = 0.1\text{ nm}\)).
  • Frequency (\(\nu\)): Number of wave cycles passing a fixed point per second. Units: Hertz (\(\text{Hz} = \text{s}^{-1}\)). \[c = \nu \lambda \implies \nu = \frac{c}{\lambda}\]
  • Wavenumber (\(\bar{\nu}\)): Number of wavelengths per unit length. Units: \(\text{m}^{-1}\) or \(\text{cm}^{-1}\). \[\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}\]
2. Black Body Radiation & Planck Quantum Hypothesis

Classical wave theory failed to explain the spectral distribution of thermal radiation emitted by an ideal black body—an ideal absorber and radiator of all electromagnetic frequencies. Classical physics (Rayleigh-Jeans law) predicted infinite energy emission at high frequencies (the ultraviolet catastrophe).

In 1900, Max Planck resolved this discrepancy by proposing that radiant energy is not emitted or absorbed continuously, but in discrete packets called quanta (photons for light). The energy \(E\) of a quantum is directly proportional to its frequency:

\[E = h \nu = \frac{h c}{\lambda} = h c \bar{\nu}\]

where \(h = 6.62607015 \times 10^{-34}\text{ J}\cdot\text{s}\) is Planck constant. Total radiant energy emitted or absorbed is quantized in integer multiples: \(E_{\text{total}} = n h \nu\) (where \(n = 1, 2, 3, \dots\)).

3. The Photoelectric Effect & Einstein Quantum Formulation

Heinrich Hertz (1887) observed that when ultraviolet light illuminates clean metal surfaces (e.g., Cs, K, Na), electrons are instantly ejected. Classical wave theory predicted that increasing light intensity would increase kinetic energy and that a time lag should occur for low-intensity light. Observations contradicted this:

  1. Ejection of electrons is instantaneous (time lag \(< 10^{-9}\text{ s}\)).
  2. For each metal, there exists a characteristic threshold frequency (\(\nu_0\)) below which no photoelectrons are ejected, regardless of light intensity.
  3. The number of ejected electrons (photocurrent) is proportional to light intensity (photon flux).
  4. The maximum kinetic energy (\(E_{k,\max}\)) of ejected electrons depends linearly on the frequency of incident radiation and is independent of intensity.

Albert Einstein (1905) explained this phenomenon by treating light as a stream of photons. When a photon of energy \(h\nu\) collides with a bound electron, a fixed portion of energy, termed the work function (\(W_0\) or \(\Phi\)), is consumed to liberate the electron from the surface lattice. The surplus energy is converted into electron kinetic energy:

\[h \nu = W_0 + E_{k,\max} = h \nu_0 + \frac{1}{2} m_e v_{\max}^2\] \[E_{k,\max} = h (\nu - \nu_0) = h c \left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) = e V_0\]

where \(\nu_0\) is the threshold frequency, \(\lambda_0 = c/\nu_0\) is the threshold wavelength, and \(V_0\) is the stopping potential.

Module 3: Bohr Model of the Hydrogen Atom & Line Spectra

Niels Bohr (1913) merged Rutherford nuclear model with Planck quantum concept to construct a quantitative model for single-electron species (\(\text{H}, \text{He}^+, \text{Li}^{2+}, \text{Be}^{3+}\)).

1. Postulates of Bohr Atomic Theory
  • Postulate I (Stationary Orbits): An electron revolves around the nucleus only in certain stable, non-radiating circular orbits called stationary states. While in these orbits, the electron does not emit electromagnetic radiation.
  • Postulate II (Quantization of Angular Momentum): The orbital angular momentum \(L = m_e v r\) of the revolving electron is an integral multiple of \(h / (2\pi)\): \[m_e v r = n \frac{h}{2\pi} = n \hbar \quad (n = 1, 2, 3, \dots)\] where \(n\) is the principal quantum number.
  • Postulate III (Quantum Energy Transitions): Radiation is emitted or absorbed only when an electron jumps from one stationary state to another. The frequency \(\nu\) of the photon is given by Bohr frequency condition: \[\Delta E = E_2 - E_1 = h \nu = \frac{h c}{\lambda}\]
2. Derivation of Radius, Velocity, and Energy Levels

Balancing the Coulombic electrostatic attraction with the centrifugal requirement: \(\frac{Z e^2}{4\pi \varepsilon_0 r^2} = \frac{m_e v^2}{r}\). Combining this with \(v = \frac{n h}{2\pi m_e r}\) yields:

  • Radius of \(n\)-th Orbit: \[r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m_e Z e^2} = \frac{n^2}{Z} a_0 = 0.529 \times \frac{n^2}{Z} \text{ \AA} = 52.9 \times \frac{n^2}{Z} \text{ pm}\] where \(a_0 = 0.529\text{ \AA}\) is the Bohr radius for hydrogen (\(n=1, Z=1\)).
  • Velocity in \(n\)-th Orbit: \[v_n = \frac{Z e^2}{2 \varepsilon_0 n h} = 2.188 \times 10^6 \times \frac{Z}{n} \text{ m/s}\]
  • Total Energy of \(n\)-th Orbit: Total energy is the sum of kinetic energy (\(E_k = \frac{1}{2}m_e v^2 = -\frac{1}{2}E_p\)) and electrostatic potential energy (\(E_p = -\frac{Z e^2}{4\pi \varepsilon_0 r}\)): \[E_n = -\frac{m_e Z^2 e^4}{8 \varepsilon_0^2 n^2 h^2} = -R_H \left(\frac{Z^2}{n^2}\right)\] \[E_n = -2.18 \times 10^{-18} \times \left(\frac{Z^2}{n^2}\right) \text{ J/atom} = -13.6 \times \left(\frac{Z^2}{n^2}\right) \text{ eV/atom} = -\frac{1312}{n^2} \times Z^2 \text{ kJ/mol}\]

    Significance of the Negative Sign: The negative sign indicates that the electron is electrostatically bound to the nucleus. When the electron is infinitely distant (\(n = \infty\)), attraction is zero and \(E_\infty = 0\). As it approaches the nucleus, potential energy drops, releasing energy.

3. The Hydrogen Spectral Series & Rydberg Formula

When atomic hydrogen gas is excited in an electrical discharge tube, emitted photons produce discrete line series described by Johannes Rydberg empirical formula:

\[\bar{\nu} = \frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]

where \(R_H = 109,677\text{ cm}^{-1} = 1.09677 \times 10^7\text{ m}^{-1}\) is the Rydberg constant.

Spectral SeriesLower Level (\(n_1\))Upper Levels (\(n_2\))Electromagnetic Region
Lyman Series\(n_1 = 1\)\(n_2 = 2, 3, 4, \dots\)Ultraviolet (UV)
Balmer Series\(n_1 = 2\)\(n_2 = 3, 4, 5, \dots\)Visible (\(H_\alpha, H_\beta, H_\gamma, H_\delta\))
Paschen Series\(n_1 = 3\)\(n_2 = 4, 5, 6, \dots\)Near Infrared (NIR)
Brackett Series\(n_1 = 4\)\(n_2 = 5, 6, 7, \dots\)Mid Infrared (MIR)
Pfund Series\(n_1 = 5\)\(n_2 = 6, 7, 8, \dots\)Far Infrared (FIR)
Humphreys Series\(n_1 = 6\)\(n_2 = 7, 8, 9, \dots\)Far Infrared
Limitations of Bohr Model:
  • It applies exclusively to single-electron species (\(\text{H}, \text{He}^+, \text{Li}^{2+}\)) and fails completely for multi-electron atoms due to inter-electronic repulsions.
  • It cannot account for the fine structure (closely spaced spectral doublets/triplets) resolved by high-resolution spectrometers.
  • It fails to explain the splitting of spectral lines under external magnetic fields (Zeeman Effect) and external electric fields (Stark Effect).
  • It violates wave-particle duality and the Heisenberg Uncertainty Principle by assuming precise simultaneous circular orbits and velocities.

Module 4: de Broglie Wavelength & Heisenberg Uncertainty Principle

The classical concept of a particle having a definite trajectory was replaced by quantum mechanics through two revolutionary physical principles.

1. Dual Nature of Matter — de Broglie Hypothesis (1924)

Louis de Broglie proposed that if electromagnetic radiation exhibits both wave and particle characteristics, material particles (such as electrons, protons, and atoms) must also possess wave-like properties. Equating Planck photon energy \(E = h \nu = h c / \lambda\) with Einstein mass-energy equivalence \(E = m c^2\) for a photon of momentum \(p = m c\):

\[\lambda = \frac{h}{p} = \frac{h}{m v}\]

For a particle of mass \(m\) and kinetic energy \(E_k = \frac{1}{2}m v^2 = \frac{p^2}{2m} \implies p = \sqrt{2m E_k}\):

\[\lambda = \frac{h}{\sqrt{2 m E_k}}\]

When a charged particle of charge \(q\) is accelerated through a potential difference \(V\) volts, \(E_k = q V\):

\[\lambda = \frac{h}{\sqrt{2 m q V}}\]

For an electron (\(m = 9.11 \times 10^{-31}\text{ kg}, q = 1.602 \times 10^{-19}\text{ C}\)):

\[\lambda_e = \frac{1.227}{\sqrt{V}} \text{ nm} = \sqrt{\frac{150}{V}} \text{ \AA}\]
2. Experimental Confirmation & Macroscopic Significance
  • Davisson and Germer Experiment (1927): Directed a beam of electrons accelerated at 54 V onto a nickel crystal. The scattered electrons formed distinct diffraction rings matching Bragg X-ray diffraction, verifying \(\lambda_{\text{exp}} = 0.165\text{ nm}\) in agreement with de Broglie prediction (\(0.167\text{ nm}\)).
  • Macroscopic vs Microscopic: A macroscopic object (e.g., a 0.1 kg cricket ball at 20 m/s) has \(\lambda = \frac{6.63 \times 10^{-34}}{0.1 \times 20} \approx 3.3 \times 10^{-34}\text{ m}\), which is smaller than nuclear dimensions and physically undetectable. For an electron (\(m \sim 10^{-30}\text{ kg}\)), \(\lambda \sim 10^{-10}\text{ m}\), comparable to atomic spacing and measurable via electron diffraction.
3. Heisenberg Uncertainty Principle (1927)

Werner Heisenberg formulated that it is fundamentally impossible to measure simultaneously and accurately both the position (\(x\)) and linear momentum (\(p\)) of a subatomic particle:

\[\Delta x \cdot \Delta p_x \ge \frac{h}{4\pi} = \frac{\hbar}{2}\] \[\Delta x \cdot (m \Delta v_x) \ge \frac{h}{4\pi} \implies \Delta x \cdot \Delta v_x \ge \frac{h}{4\pi m}\]

where \(\Delta x\) is the uncertainty in position and \(\Delta p_x\) (or \(\Delta v_x\)) is the uncertainty in momentum (or velocity). Alternative formulations include energy-time uncertainty: \(\Delta E \cdot \Delta t \ge \frac{h}{4\pi}\).

Why Electrons Cannot Exist in the Nucleus:

The nuclear radius is approximately \(10^{-14}\text{ m}\). If an electron were confined inside the nucleus, the maximum uncertainty in its position would be \(\Delta x \approx 10^{-14}\text{ m}\). By Heisenberg principle:

\[\Delta v \ge \frac{h}{4\pi m_e \Delta x} = \frac{6.626 \times 10^{-34}}{4\pi \times (9.11 \times 10^{-31}) \times 10^{-14}} \approx 5.79 \times 10^{10}\text{ m/s}\]

This calculated velocity exceeds the speed of light (\(3 \times 10^8\text{ m/s}\)), which is physically impossible. Hence, electrons cannot reside within the nucleus.

Module 5: Quantum Mechanical Model, Wavefunction, Orbitals & Quantum Numbers

In 1926, Erwin Schrödinger formulated the quantum mechanical wave equation for matter waves, replacing deterministic orbits with probabilistic 3D electron clouds called orbitals.

1. The Schrödinger Wave Equation & Probability Density

For an electron moving in three dimensions under a potential energy field \(V(x,y,z)\):

\[\frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} + \frac{8\pi^2 m}{h^2}(E - V)\psi = 0 \quad \text{or} \quad \hat{H}\psi = E\psi\]

where \(\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V\) is the Hamiltonian operator, \(E\) is the total energy eigenvalue, and \(\psi\) is the wavefunction.

  • Physical Significance of \(\psi\): The amplitude \(\psi\) has no direct physical meaning and can be positive, negative, or complex.
  • Physical Significance of \(|\psi|^2\) (Max Born Interpretation): The square of the absolute value, \(|\psi|^2\) (or \(\psi^*\psi\)), represents the probability density of locating the electron per unit volume around a specific point \((x,y,z)\).
  • Atomic Orbital: A three-dimensional boundary surface in space around the nucleus within which the probability of finding the electron is approximately 90% to 95%.
2. The Four Quantum Numbers
Quantum NumberSymbolPermissible ValuesPhysical Meaning & Significance
Principal\(n\)\(1, 2, 3, 4, \dots\) (K, L, M, N...)Determines shell, main energy level, orbital size (\(r_n \propto n^2\)), and maximum electrons per shell (\(2n^2\)).
Azimuthal (Orbital Angular Momentum)\(l\)\(0, 1, 2, \dots, (n-1)\) (\(n\) values)Determines subshell, 3D geometric shape of orbital, and orbital angular momentum: \(L = \sqrt{l(l+1)}\frac{h}{2\pi}\). Notation: \(l=0\,(s)\), \(l=1\,(p)\), \(l=2\,(d)\), \(l=3\,(f)\).
Magnetic\(m_l\)\(-l, \dots, 0, \dots, +l\) (\(2l + 1\) values)Determines spatial orientation of orbital along coordinate axes. Gives number of orbitals in subshell: \(s=1, p=3, d=5, f=7\).
Spin\(m_s\)\(+\frac{1}{2}\,(\uparrow), -\frac{1}{2}\,(\downarrow)\)Specifies intrinsic spin angular momentum: \(S = \sqrt{s(s+1)}\hbar = \frac{\sqrt{3}}{2}\hbar\). Two degenerate orientations.
3. Shapes of Orbitals & Nodal Surfaces
  • \(s\)-Orbitals (\(l = 0\)): Spherically symmetrical; probability density depends only on distance \(r\) from nucleus, non-directional.
  • \(p\)-Orbitals (\(l = 1\)): Dumbbell-shaped consisting of two lobes separated by a nodal plane passing through the nucleus. Three mutually perpendicular degenerate orbitals: \(p_x, p_y, p_z\).
  • \(d\)-Orbitals (\(l = 2\)): Cloverleaf / double dumbbell shape. Five degenerate orbitals: four with four lobes in planes (\(d_{xy}, d_{yz}, d_{zx}\) between axes; \(d_{x^2-y^2}\) along axes) and one donut-shaped orbital (\(d_{z^2}\) with a central collar ring in the xy-plane).
4. Node Calculations

A node is a region where the probability density \(|\psi|^2\) drops to zero.

  • Radial Nodes (Spherical Nodes): Spherical surfaces around the nucleus where \(|\psi|^2 = 0\): \[\text{Radial Nodes} = n - l - 1\]
  • Angular Nodes (Nodal Planes / Cones): Planar or conical surfaces passing through the nucleus where \(|\psi|^2 = 0\): \[\text{Angular Nodes} = l\]
  • Total Nodes: \[\text{Total Nodes} = \text{Radial} + \text{Angular} = (n - l - 1) + l = n - 1\]

Module 6: Rules for Filling Orbitals & Electronic Configurations

The ground-state distribution of electrons among available orbitals is governed by three fundamental quantum principles.

1. The Aufbau Principle & The \((n + l)\) Rule

The word Aufbau is German for 'building up'. In the ground state of an atom, electrons occupy orbitals in order of increasing orbital energies:

\[1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s\]

Bohr-Bury \((n + l)\) Rule:

  • Rule 1: An orbital with a lower value of \((n + l)\) has lower energy and is filled first. For example, for \(4s\), \(n+l = 4+0 = 4\); for \(3d\), \(n+l = 3+2 = 5\). Hence, \(4s\) fills before \(3d\).
  • Rule 2: If two orbitals possess identical \((n + l)\) values, the orbital with the smaller value of \(n\) has lower energy and fills first. For example, for \(3d\) (\(3+2 = 5\)) and \(4p\) (\(4+1 = 5\)), \(3d\) fills before \(4p\).
2. Pauli Exclusion Principle (1925)

Statement: No two electrons in an atom can have the same set of all four quantum numbers (\(n, l, m_l, m_s\)).

If two electrons have identical \(n, l,\) and \(m_l\) (meaning they reside in the same orbital), they must have different spin quantum numbers \(m_s\) (one \(+1/2\) and the other \(-1/2\)). Consequently, an orbital can accommodate a maximum of two electrons, and they must have antiparallel (opposite) spins (\(\uparrow\downarrow\)).

3. Hund's Rule of Maximum Multiplicity

Statement: Pairing of electrons in degenerate orbitals (orbitals belonging to the same subshell, such as \(p, d, f\)) does not occur until each available orbital is singly occupied with electrons having parallel spins.

Physical Rationale: Electrons repel each other electrostatically. By occupying separate spatial orbitals with parallel spins, electrons maximize their average distance, minimizing inter-electronic repulsion. Parallel spins also maximize exchange energy.

4. Anomalous Configurations of Chromium (\(Z = 24\)) & Copper (\(Z = 29\))
  • Chromium (\(Z = 24\)):
    • Expected: \([\text{Ar}]\, 3d^4 4s^2\)
    • Actual Ground State: \([\text{Ar}]\, 3d^5 4s^1\)
  • Copper (\(Z = 29\)):
    • Expected: \([\text{Ar}]\, 3d^9 4s^2\)
    • Actual Ground State: \([\text{Ar}]\, 3d^{10} 4s^1\)

Underlying Causes for Extra Stability:

  1. Symmetrical Distribution of Charge: Spherically symmetrical and completely degenerate half-filled (\(d^5, p^3, f^7\)) and fully-filled (\(d^{10}, p^6, f^{14}\)) subshells minimize inter-electronic shielding and repulsion.
  2. Maximum Exchange Energy: Electrons with parallel spins in degenerate orbitals can exchange positions. Each exchange releases stabilizing energy. The number of exchange pairs for \(n\) electrons with parallel spins is \(K = \frac{n(n-1)}{2}\). For \(d^4\), \(K = \frac{4(3)}{2} = 6\); for \(d^5\), \(K = \frac{5(4)}{2} = 10\). The 4 extra exchange interactions substantially lower the ground-state energy.

Key Formulas, Reactions & Definitions

Planck Quantum Relation & Einstein Photoelectric Equation
$$E = h\nu = \frac{hc}{\lambda}, \quad h\nu = W_0 + \frac{1}{2}m_e v_{\max}^2 = h\nu_0 + eV_0$$
Bohr Quantized Radius, Velocity & Energy Formulas
$$r_n = 0.529\frac{n^2}{Z}\text{ \AA}, \quad v_n = 2.188 \times 10^6\frac{Z}{n}\text{ m/s}, \quad E_n = -13.6\frac{Z^2}{n^2}\text{ eV} = -2.18 \times 10^{-18}\frac{Z^2}{n^2}\text{ J}$$
Rydberg Wavenumber Formula for Hydrogen Spectral Series
$$\bar{\nu} = \frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad R_H = 109,677\text{ cm}^{-1} = 1.09677 \times 10^7\text{ m}^{-1}$$
de Broglie Matter Wave Equations
$$\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2m E_k}} = \frac{h}{\sqrt{2mqV}} = \sqrt{\frac{150}{V}}\text{ \AA}$$
Heisenberg Uncertainty Principle
$$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}, \quad \Delta x \cdot \Delta v \ge \frac{h}{4\pi m}, \quad \Delta E \cdot \Delta t \ge \frac{h}{4\pi}$$
Orbital Nodes & Angular Momentum Equations
$$\text{Radial Nodes} = n - l - 1, \quad \text{Angular Nodes} = l, \quad L = \sqrt{l(l+1)}\frac{h}{2\pi}$$

Conceptual Solved Examples & Case Studies

Example 1
(a) Calculate the energy in Joules of one mole of photons of radiation whose frequency is 5.0 x 10^14 Hz. (b) Calculate the de Broglie wavelength of an electron traveling at 2.05 x 10^6 m/s. (h = 6.626 x 10^-34 J s, me = 9.11 x 10^-31 kg, NA = 6.022 x 10^23 mol^-1).
Step-by-Step Solution:

Step 1: Part (a) — Energy of One Mole of Photons

Energy of a single photon: \(E_1 = h \nu = (6.626 \times 10^{-34}\text{ J}\cdot\text{s}) \times (5.0 \times 10^{14}\text{ s}^{-1}) = 3.313 \times 10^{-19}\text{ J}\)

Energy of 1 mole of photons (Einstein of energy):

\[E = E_1 \times N_A = (3.313 \times 10^{-19}\text{ J}) \times (6.022 \times 10^{23}\text{ mol}^{-1}) = 1.995 \times 10^5\text{ J/mol} = 199.5\text{ kJ/mol}\]

Step 2: Part (b) — de Broglie Wavelength of the Electron

\[\lambda = \frac{h}{m_e v} = \frac{6.626 \times 10^{-34}\text{ kg}\cdot\text{m}^2\text{/s}}{(9.109 \times 10^{-31}\text{ kg}) \times (2.05 \times 10^6\text{ m/s})}\] \[\lambda = \frac{6.626 \times 10^{-34}}{1.867 \times 10^{-24}} = 3.55 \times 10^{-10}\text{ m} = 0.355\text{ nm} = 3.55\text{ \AA}\]

Final Answer: (a) \(1.995 \times 10^5\text{ J/mol}\); (b) \(\lambda = 3.55 \times 10^{-10}\text{ m}\) (or \(0.355\text{ nm}\)).

Example 2
The threshold frequency nu0 for potassium metal is 5.0 x 10^14 s^-1. Calculate the kinetic energy and maximum velocity of the ejected photoelectron when radiation of wavelength lambda = 300 nm strikes the potassium surface. (h = 6.626 x 10^-34 J s, c = 3.0 x 10^8 m/s, me = 9.11 x 10^-31 kg).
Step-by-Step Solution:

Step 1: Calculate Incident Photon Frequency and Energy

\[\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{300 \times 10^{-9}\text{ m}} = 1.0 \times 10^{15}\text{ s}^{-1}\] \[E_{\text{photon}} = h \nu = (6.626 \times 10^{-34}) \times (1.0 \times 10^{15}) = 6.626 \times 10^{-19}\text{ J}\]

Step 2: Calculate Work Function (W0)

\[W_0 = h \nu_0 = (6.626 \times 10^{-34}) \times (5.0 \times 10^{14}) = 3.313 \times 10^{-19}\text{ J}\]

Step 3: Calculate Kinetic Energy (Ek,max)

\[E_{k,\max} = E_{\text{photon}} - W_0 = 6.626 \times 10^{-19} - 3.313 \times 10^{-19} = 3.313 \times 10^{-19}\text{ J}\]

In electron-volts: \(E_{k,\max} = \frac{3.313 \times 10^{-19}\text{ J}}{1.602 \times 10^{-19}\text{ J/eV}} = 2.068\text{ eV}\)

Step 4: Calculate Maximum Velocity (v_max)

\[\frac{1}{2} m_e v_{\max}^2 = E_{k,\max} \implies v_{\max} = \sqrt{\frac{2 E_{k,\max}}{m_e}}\] \[v_{\max} = \sqrt{\frac{2 \times 3.313 \times 10^{-19}\text{ J}}{9.11 \times 10^{-31}\text{ kg}}} = \sqrt{7.273 \times 10^{11}} = 8.528 \times 10^5\text{ m/s}\]

Final Answer: Kinetic Energy = \(3.313 \times 10^{-19}\text{ J}\) (\(2.07\text{ eV}\)); Maximum Velocity = \(8.53 \times 10^5\text{ m/s}\).

Example 3
Calculate the wavelength and wavenumber of the spectral line emitted when an electron in a hydrogen atom undergoes a transition from the n = 4 level to the n = 2 level. State the name of the spectral series and the region of the electromagnetic spectrum in which this line appears. (RH = 1.097 x 10^7 m^-1).
Step-by-Step Solution:

Step 1: Identify Energy Levels and Spectral Series

Lower level \(n_1 = 2\), upper level \(n_2 = 4\). Because \(n_1 = 2\), this transition belongs to the Balmer Series and falls in the Visible Region (specifically, the green-blue \(H_\beta\) line).

Step 2: Apply the Rydberg Equation

\[\bar{\nu} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = (1.097 \times 10^7\text{ m}^{-1}) \left(\frac{1}{2^2} - \frac{1}{4^2}\right)\] \[\bar{\nu} = 1.097 \times 10^7 \times \left(\frac{1}{4} - \frac{1}{16}\right) = 1.097 \times 10^7 \times \frac{3}{16} = 2.0569 \times 10^6\text{ m}^{-1}\]

Step 3: Calculate Wavelength (lambda)

\[\lambda = \frac{1}{\bar{\nu}} = \frac{1}{2.0569 \times 10^6\text{ m}^{-1}} = 4.8617 \times 10^{-7}\text{ m} = 486.2\text{ nm} = 4862\text{ \AA}\]

Final Answer: Wavenumber \(\bar{\nu} = 2.057 \times 10^6\text{ m}^{-1}\); Wavelength \(\lambda = 486.2\text{ nm}\); Series = Balmer Series; Region = Visible Spectrum.

Example 4
(a) A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Angstrom (0.1 x 10^-10 m). What is the uncertainty involved in the measurement of its velocity? (b) If a bullet of mass 0.02 kg has its position measured with the same accuracy of 0.1 Angstrom, calculate the uncertainty in its velocity and comment on its significance. (h = 6.626 x 10^-34 J s, me = 9.11 x 10^-31 kg).
Step-by-Step Solution:

Step 1: Part (a) — Velocity Uncertainty of the Electron

Given \(\Delta x = 0.1\text{ \AA} = 1.0 \times 10^{-11}\text{ m}\), \(m_e = 9.11 \times 10^{-31}\text{ kg}\):

\[\Delta v_e \ge \frac{h}{4\pi m_e \Delta x} = \frac{6.626 \times 10^{-34}\text{ kg}\cdot\text{m}^2\text{/s}}{4 \times 3.1416 \times (9.11 \times 10^{-31}\text{ kg}) \times (1.0 \times 10^{-11}\text{ m})}\] \[\Delta v_e \ge \frac{6.626 \times 10^{-34}}{1.1448 \times 10^{-40}} = 5.788 \times 10^6\text{ m/s}\]

Significance: This uncertainty (\(\sim 5.79 \times 10^6\text{ m/s}\)) is on the order of the actual electron velocity in Bohr first orbit (\(2.188 \times 10^6\text{ m/s}\)). Thus, defining a deterministic trajectory for the electron is physically meaningless.

Step 2: Part (b) — Velocity Uncertainty of the Bullet

Given \(m = 0.020\text{ kg}\), \(\Delta x = 1.0 \times 10^{-11}\text{ m}\):

\[\Delta v_{\text{bullet}} \ge \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.020 \times 1.0 \times 10^{-11}} = \frac{6.626 \times 10^{-34}}{2.513 \times 10^{-12}} = 2.637 \times 10^{-22}\text{ m/s}\]

Significance: An uncertainty of \(2.64 \times 10^{-22}\text{ m/s}\) is completely negligible and undetectable by any real-world instrument. Macroscopic objects obey classical Newtonian trajectories without perceptible quantum fuzziness.

Final Answer: (a) Electron \(\Delta v \ge 5.79 \times 10^6\text{ m/s}\); (b) Bullet \(\Delta v \ge 2.64 \times 10^{-22}\text{ m/s}\) (insignificant).

Example 5
(i) Which of the following sets of quantum numbers are not permissible and why? (a) n = 2, l = 2, ml = 0, ms = +1/2; (b) n = 3, l = 1, ml = -2, ms = -1/2; (c) n = 4, l = 0, ml = 0, ms = -1/2. (ii) For the 4d orbital, calculate: (a) the number of radial nodes, (b) the number of angular nodes, and (c) the total number of nodes.
Step-by-Step Solution:

Step 1: Part (i) — Evaluation of Quantum Number Sets

  • Set (a) \(n = 2, l = 2, m_l = 0, m_s = +1/2\): NOT PERMISSIBLE. The azimuthal quantum number \(l\) can only take integer values from \(0\) to \(n - 1\). For \(n = 2\), maximum \(l = 1\). \(l = 2\) is impossible.
  • Set (b) \(n = 3, l = 1, m_l = -2, m_s = -1/2\): NOT PERMISSIBLE. The magnetic quantum number \(m_l\) can only take values from \(-l\) to \(+l\). For \(l = 1\), \(m_l\) can only be \(-1, 0, +1\). \(m_l = -2\) is impossible.
  • Set (c) \(n = 4, l = 0, m_l = 0, m_s = -1/2\): PERMISSIBLE. Corresponds to a valid \(4s\) electron.

Step 2: Part (ii) — Nodal Surfaces for 4d Orbital

For a \(4d\) orbital: Principal quantum number \(n = 4\), Azimuthal quantum number \(l = 2\).

  • (a) Radial Nodes: \(n - l - 1 = 4 - 2 - 1 = 1\text{ radial node}\).
  • (b) Angular Nodes: \(l = 2\text{ angular nodes (nodal planes)}\).
  • (c) Total Nodes: \(n - 1 = 4 - 1 = 3\text{ total nodes}\) (or \(1 + 2 = 3\)).

Final Answer: (i) (a) and (b) are not permissible; (ii) For 4d: Radial nodes = 1, Angular nodes = 2, Total nodes = 3.

Example 6
(a) Write the ground-state electronic configurations of Chromium (Z = 24) and Copper (Z = 29), and explain why the 3d^5 4s^1 and 3d^10 4s^1 configurations are preferred over 3d^4 4s^2 and 3d^9 4s^2. (b) Calculate the number of exchange interactions possible for the d-electrons in a 3d^4 configuration versus a 3d^5 configuration. (c) Calculate the spin-only magnetic moment mu in Bohr Magnetons (BM) for the Cr3+ ion. [2 + 2 + 1 = 5]
Step-by-Step Solution:

Step 1: Part (a) — Electronic Configurations and Stability Rationale

  • \(\text{Cr } (Z = 24): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1 \equiv [\text{Ar}]\, 3d^5 4s^1\)
  • \(\text{Cu } (Z = 29): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1 \equiv [\text{Ar}]\, 3d^{10} 4s^1\)

Physical Rationale: Half-filled (\(3d^5\)) and completely-filled (\(3d^{10}\)) subshells possess extra stability due to:

  1. Symmetrical Distribution of Charge: Spherically symmetrical arrangements result in less shielding and minimal inter-electronic repulsion.
  2. Maximum Exchange Energy: Electrons with parallel spins can exchange their positions, releasing significant stabilizing exchange energy.

Step 2: Part (b) — Exchange Interactions Calculation

The number of exchange pairs for \(m\) electrons with parallel spin is given by \(K = \frac{m(m - 1)}{2}\):

  • For \(3d^4\) (4 parallel spin electrons): \[K(3d^4) = \frac{4(4 - 1)}{2} = \frac{12}{2} = 6\text{ exchange interactions}\]
  • For \(3d^5\) (5 parallel spin electrons): \[K(3d^5) = \frac{5(5 - 1)}{2} = \frac{20}{2} = 10\text{ exchange interactions}\]

The extra 4 exchange interactions provide an energy stabilization that easily offsets the small energy penalty of shifting one electron from the \(4s\) orbital to the \(3d\) orbital.

Step 3: Part (c) — Spin-Only Magnetic Moment of Cr3+

Formation of \(\text{Cr}^{3+}\): Lose 1 electron from \(4s\) and 2 electrons from \(3d\):

\[\text{Cr}^{3+} = [\text{Ar}]\, 3d^3 4s^0\]

Number of unpaired electrons \(n = 3\). Spin-only magnetic moment formula:

\[\mu = \sqrt{n(n + 2)}\text{ BM} = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87\text{ BM (Bohr Magnetons)}\]

Final Answer: (a) Cr: \([\text{Ar}]\,3d^5 4s^1\), Cu: \([\text{Ar}]\,3d^{10} 4s^1\); (b) \(3d^4 = 6\) exchanges, \(3d^5 = 10\) exchanges (\(\Delta K = +4\)); (c) \(\mu(\text{Cr}^{3+}) = 3.87\text{ BM}\).

Common Misconceptions & Examiner Traps

Common Misconception

Confusing Bohr Orbits with Quantum Mechanical Orbitals: assuming an orbital is a circular or elliptical planetary path along which the electron travels.

Scientific Reality & Correction

A Bohr orbit is a planar, 2D circular path with fixed radius and momentum (which directly violates the Heisenberg Uncertainty Principle). A quantum mechanical orbital is a three-dimensional region of space around the nucleus where the probability density |psi|^2 of finding the electron is maximal (~90-95%). An electron has no defined trajectory.

Common Misconception

Inverting n1 and n2 in the Rydberg equation, leading to negative wavelengths or misidentifying which series belongs to the visible region.

Scientific Reality & Correction

In emission spectra, the electron falls from a higher level (n2) to a lower level (n1). Hence, n2 > n1 always. Remember that the ONLY series falling in the visible range is the Balmer series (n1 = 2). The Lyman series (n1 = 1) is strictly in the ultraviolet region, while Paschen (n1 = 3), Brackett (n1 = 4), and Pfund (n1 = 5) are in the infrared region.

Common Misconception

Writing [Ar] 3d^4 4s^2 for Chromium (Z = 24) or [Ar] 3d^9 4s^2 for Copper (Z = 29) based on strict Aufbau sequence.

Scientific Reality & Correction

A half-filled d-subshell (d^5) and a completely-filled d-subshell (d^10) have exceptional stability due to symmetrical charge distribution and maximized exchange energy. One electron shifts from 4s to 3d, giving Cr: [Ar] 3d^5 4s^1 and Cu: [Ar] 3d^10 4s^1.

Common Misconception

Conflating radial nodes with angular nodes or confusing total nodes (n - 1) with azimuthal quantum number l.

Scientific Reality & Correction

Remember the distinct formulas: Radial nodes = n - l - 1; Angular nodes = l; Total nodes = (n - l - 1) + l = n - 1. For a 3d orbital (n = 3, l = 2): Radial nodes = 3 - 2 - 1 = 0; Angular nodes = 2; Total nodes = 2.

Common Misconception

Writing Delta x · Delta p >= h / (2pi) by confusing Heisenberg inequality with Bohr angular momentum quantization.

Scientific Reality & Correction

Bohr angular momentum is quantized in units of h / (2pi) (or h-bar), but Heisenberg uncertainty principle strictly has h / (4pi) (or h-bar / 2) on the right-hand side: Delta x · Delta p >= h / (4pi). Using h / (2pi) halves the resulting uncertainty and leads to incorrect numerical solutions.

Atomic Structure, Quantum Mechanics, Energy Levels & Orbital Architecture

ATOMIC STRUCTURE & QUANTUM ARCHITECTURE WBCHSE Class 11 Chemistry • Unit II: Structure of an Atom 1. BOHR ATOM & SPECTRAL SERIES Quantized Orbits: rn = 0.529 n²/Z Å • En = -13.6 Z²/n² eV +Ze n=1 n=2 n=3 Lyman (UV, n1=1) Balmer (Vis, n1=2) Paschen (IR, n1=3) Rydberg: ν̄ = RH [1/n1² - 1/n2²] 2. WAVE-PARTICLE DUALITY & UNCERTAINTY de Broglie Matter Wave: λ = h / p = h / (m v) p = √(2m·Ek) Heisenberg Uncertainty: Δx · Δp ≥ h / (4π) Δx · Δv ≥ h/(4πm) Wavefunction Interpretation: |ψ|² = Probability Density Nodes = n - 1 3. THE FOUR QUANTUM NUMBERS n n (Principal): Shell, Energy & Size (n = 1, 2, 3...) l l (Azimuthal): Subshell & 3D Shape (0 to n-1: s, p, d, f) m ml (Magnetic): Spatial Orientation (-l to +l) s ms (Spin): Electron Spin (+1/2, -1/2) 4. AUFBAU PRINCIPLE & (n + l) ENERGY LADDER 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s Pauli: Max 2 e⁻ per orbital with opposite spins (↑↓) Hund: Max multiplicity; singly occupy degenerate orbitals first Stability: Half-filled (d⁵) & Fully-filled (d¹⁰) Cr [Ar]3d⁵4s¹, Cu [Ar]3d¹⁰4s¹ Symmetrical Charge & Maximum Exchange Energy Stabilization

Chapter Summary & 10 Key Takeaways

Takeaway 1
Cathode rays consist of negatively charged electrons with a universal charge-to-mass ratio e/m = 1.758820 x 10^11 C/kg, independent of the gas or electrode material.
Takeaway 2
Rutherford alpha-scattering experiment established the nuclear model of the atom: nearly all mass and positive charge reside in a dense nucleus of radius ~10^-15 m, surrounded by electrons at ~10^-10 m.
Takeaway 3
Planck Quantum Theory posits that radiant energy is emitted or absorbed in discrete packets called quanta (photons), with energy E = h nu, where h = 6.626 x 10^-34 J s.
Takeaway 4
Einstein photoelectric equation h nu = h nu0 + 1/2 m v_max^2 proves the particle nature of light; kinetic energy depends on radiation frequency, while photocurrent depends on light intensity.
Takeaway 5
Bohr model quantizes electron angular momentum (m v r = n h / 2pi); the radius of the n-th orbit is rn = 0.529 (n^2 / Z) Angstroms, and energy is En = -13.6 (Z^2 / n^2) eV/atom.
Takeaway 6
The negative sign of Bohr atomic energy signifies that the electron is electrostatically bound to the nucleus, with zero potential energy assigned at infinite separation.
Takeaway 7
The de Broglie relation lambda = h / p = h / (m v) assigns wave properties to all moving matter; wave behavior is dominant for microscopic electrons but imperceptible for macroscopic objects.
Takeaway 8
Heisenberg Uncertainty Principle states that Delta x · Delta p >= h / (4 pi); it is impossible to determine simultaneously and precisely both the position and momentum of a microscopic particle.
Takeaway 9
The square of the wavefunction |psi|^2 represents the probability density of finding an electron in a given volume element; orbitals are 3D regions with 90-95% electron probability.
Takeaway 10
The four quantum numbers (n, l, ml, ms) uniquely specify an electron; stability of half-filled (d^5) and fully-filled (d^10) subshells stems from symmetrical charge distribution and maximized exchange energy.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Why is the energy of an electron in a hydrogen atom assigned a negative value?
Reveal Answer & Explanation
Answer: At infinite distance from the nucleus (n = infinity), the electrostatic attraction is zero, which is taken as the reference state of zero potential energy (E = 0). As the electron approaches the positive nucleus, electrostatic attractive forces do work, decreasing the potential energy and releasing energy. Hence, the bound state energy is negative, indicating stability relative to a free electron.
Consider the electrostatic reference point at infinite distance.
2
How many electrons in an atom can have the quantum numbers n = 4, ms = -1/2?
Reveal Answer & Explanation
Answer: For the n = 4 shell, the total number of orbitals is n^2 = 4^2 = 16 orbitals (1 in 4s, 3 in 4p, 5 in 4d, 7 in 4f). Each orbital can hold exactly one electron with spin ms = -1/2. Therefore, exactly 16 electrons can share the quantum numbers n = 4, ms = -1/2.
Calculate the total number of orbitals in the n = 4 shell.
3
Why is the 4s orbital filled before the 3d orbital in the ground state of an atom?
Reveal Answer & Explanation
Answer: According to the Bohr-Bury (n + l) rule, orbitals with lower (n + l) values possess lower energy and are occupied first. For the 4s orbital, n + l = 4 + 0 = 4. For the 3d orbital, n + l = 3 + 2 = 5. Because 4 < 5, the 4s orbital has lower energy than 3d and fills first.
Apply the (n + l) rule for orbital energies.
4
Calculate the total number of radial and angular nodes present in a 4p orbital.
Reveal Answer & Explanation
Answer: For a 4p orbital, n = 4 and l = 1. Radial nodes = n - l - 1 = 4 - 1 - 1 = 2 radial nodes. Angular nodes = l = 1 angular node (one nodal plane). Total nodes = 2 + 1 = 3 nodes.
Use Radial = n - l - 1 and Angular = l.
5
State the physical significance of psi and |psi|^2 in quantum mechanics.
Reveal Answer & Explanation
Answer: The wavefunction psi represents the amplitude of the electron matter wave and has no direct physical meaning; it can be positive, negative, or complex. In contrast, |psi|^2 (or psi* psi) represents the probability density, which is the probability of finding the electron in an infinitesimal volume element dV around a given point in space.
Recall Max Born statistical interpretation.
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