Structure of Atom | ISC Class 11 Chemistry Notes

This note covers subatomic particles, atomic numbers and isotopes, atomic models, electromagnetic radiation, quantum theory, atomic spectra, matter waves, uncertainty, quantum numbers, orbital shapes, electron filling rules and electronic configurations.

What do experiments reveal about subatomic particles?

How are cathode and anode rays produced?

An atom contains smaller constituents called subatomic particles. The electron is negatively charged, the proton positively charged and the neutron electrically neutral. Their discovery disproved atomic indivisibility.

A discharge tube contains gas at very low pressure and two metal electrodes. The cathode is the negative electrode; the anode is the positive electrode. At sufficiently high voltage, cathode rays travel from cathode towards anode.

• The rays themselves are invisible, but produce a glow when they strike suitable fluorescent or phosphorescent materials.
• In the absence of electric or magnetic fields, they travel in straight lines.
• Their deflection in electric and magnetic fields shows that they consist of negatively charged particles.
• Their characteristics do not depend on the gas or electrode material, showing that electrons are constituents of all atoms.

Canal rays, also called anode rays, contain positive gaseous ions. An ion is an atom or group carrying a net electric charge. In a modified discharge tube, positive ions pass through holes in the cathode and form a beam behind it.

Unlike cathode rays, their masses and charge-to-mass ratios depend on the gas. Some carry multiples of the fundamental charge. Their deflection is opposite to that of electrons. The lightest positive ion, obtained from hydrogen, is the proton.

What are the measured properties of these particles?

Let e denote the magnitude of the electron's charge and mₑ its mass. Thomson measured e/mₑ = 1.758820 × 10¹¹ C kg⁻¹. Millikan's oil-drop experiment determined the electron's charge. Here C means coulomb, the unit of charge, and kg means kilogram, the unit of mass.

Chadwick bombarded beryllium with alpha particles, written α-particles, which are helium nuclei carrying two positive charges. A nucleus is the small central region of an atom. The emitted neutral particles had masses slightly greater than proton masses and were named neutrons. The neutral radiation ejected protons from paraffin wax. Analysis supported neutral particles of approximately proton mass, rather than an explanation using gamma radiation alone.

How do atomic number, mass number and isotopes differ?

The nucleus contains protons and neutrons. Atomic number, represented by Z, is the number of protons in the nucleus and identifies the element.

Mass number, represented by A, is the total number of protons and neutrons, collectively called nucleons. If N represents neutron number, then A = Z + N and N = A − Z. Mass number is a particle count, not a measured mass in kilograms.

A neutral atom has Z electrons. A cation is positively charged because electrons have been lost; an anion is negatively charged because electrons have been gained. Changing electron number changes charge, while changing proton number changes the element.

In the nuclear symbol ⁸⁰₃₅Br, Br is bromine, the upper left number is A and the lower left number is Z. Thus this neutral atom has 35 protons, 35 electrons and 80 − 35 = 45 neutrons.

Which atoms are isotopes or isobars?

Hydrogen's isotopes are protium, with one proton and no neutron; deuterium, with one proton and one neutron; and tritium, with one proton and two neutrons. Their neutral atoms each contain one electron.

Chemical properties are controlled by electron number and arrangement. Neutron number has very little effect on an element's chemical properties, so isotopes show the same chemical behaviour.

Note: Before counting electrons, check whether the species is neutral or charged. Neutron number remains A − Z for both atoms and ions; electron number equals Z only for a neutral atom.

How did Thomson's and Rutherford's models explain the atom?

What changed from a positive sphere to a nucleus?

Thomson's model pictured a sphere of uniformly distributed positive charge with electrons embedded in it. It explained overall electrical neutrality. Its distributed mass and charge, however, were inconsistent with the later scattering observations.

In Rutherford's scattering experiment, a stream of energetic α-particles struck a thin gold foil. A surrounding zinc sulphide screen produced flashes when particles hit it. Scattering means a change in a particle's direction of motion.

The nuclear model placed positive charge and most of the mass in a very small nucleus. Electrons moved around it in circular paths called orbits, held by electrostatic attraction. Atomic radius is about 10⁻¹⁰ m; nuclear radius is about 10⁻¹⁵ m, where m denotes metre.

The upper drawing labels an alpha-particle source, lead plate, gold foil and a surrounding photographic plate. The lower drawing shows a parallel incoming beam, many straight paths and a few deflected paths near the centres in the foil.

See Fig. 2.5 in your NCERT textbook

Why could Rutherford's model not explain stability?

Circular electron motion involves acceleration because direction changes. Classical electromagnetic theory predicts that an accelerating charged particle emits radiation. It would lose energy, follow a shrinking orbit and fall into the nucleus, contrary to the stability of atoms.

Simply making the electrons stationary does not solve this problem: electrostatic attraction would pull them towards the nucleus. Rutherford's model also gave no account of the distribution of electrons or their energies.

How are electromagnetic waves and photon energies described?

Electromagnetic radiation consists of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel. It needs no material medium. In vacuum, its speed, c, is approximately 3.0 × 10⁸ m s⁻¹, where s denotes second.

Wavelength λ is the distance between successive corresponding points of a wave. Frequency ν is the number of waves passing a point per second. Wavenumber ν̄ is the number of wavelengths per unit length.

c = νλ; therefore ν = c/λ and ν̄ = 1/λ. At constant speed, shorter wavelength means higher frequency. The electromagnetic spectrum includes radio waves, microwaves, infrared, visible, ultraviolet, X-rays and gamma rays.

Which units and conversions are needed?

• The SI unit of wavelength is metre, m. SI means International System of Units.
• The SI unit of frequency is hertz, Hz; 1 Hz = 1 s⁻¹.
• The SI unit of wavenumber is reciprocal metre, m⁻¹.
• The SI unit of energy is joule, J; 1 J = 1 kg m² s⁻².
• The SI unit of power is watt, W; power is energy transferred per second, and 1 W = 1 J s⁻¹.

A nanometre, nm, is 10⁻⁹ m; a picometre, pm, is 10⁻¹² m; an ångström, Å, is 10⁻¹⁰ m. A kilohertz, kHz, is 10³ Hz.

Planck proposed discrete energy packets called quanta. A quantum of light is a photon. Its energy E is E = hν, where h is Planck's constant, 6.626 × 10⁻³⁴ J s. Quantisation means restriction to discrete permitted values.

Derivation: How does photon energy depend on wavelength?

1. Start with the wave relation c = νλ.
2. Rearrange it to obtain ν = c/λ.
3. Substitute this frequency into Planck's relation E = hν.

E = hc/λ. Thus a photon of shorter wavelength has greater energy.

Worked example 1. Find the wavelength of radiation at 1,368 kHz, using c = 3.00 × 10⁸ m s⁻¹ and 1 kHz = 10³ Hz.

Formula: λ = c/ν. Substitute: ν = 1.368 × 10⁶ s⁻¹; λ = (3.00 × 10⁸)/(1.368 × 10⁶). Answer: approximately 219.3 m, in the radio-wave region.

Worked example 2. Find the frequency of violet light of wavelength 400 nm. Use c = 3.00 × 10⁸ m s⁻¹ and 1 nm = 10⁻⁹ m.

Formula: ν = c/λ. Substitute: ν = (3.00 × 10⁸)/(400 × 10⁻⁹). Answer: frequency = (7.50 Hz) × 10¹⁴.

Why were black-body radiation and the photoelectric effect important?

How does quantum theory explain thermal radiation?

A black body is an ideal perfect absorber and radiator. In practice no such body exists; carbon black approximates it fairly closely. Radiation entering a cavity through a tiny opening undergoes repeated internal reflections, making the cavity a useful approximation.

The intensity and spectral distribution of black-body radiation depend on temperature. At a fixed temperature, intensity rises with wavelength to a maximum and then falls. Increasing temperature shifts that maximum towards shorter wavelengths. Classical wave theory could not satisfactorily explain this distribution.

Intensity is plotted vertically and wavelength horizontally. The curve labelled T₂ has a higher peak at a shorter wavelength than the curve labelled T₁; T₁ and T₂ denote temperatures, with T₂ greater than T₁.

See Fig. 2.8 in your NCERT textbook

Planck's quantum theory allowed atoms and molecules to emit or absorb energy in discrete quantities. For a frequency ν, the permitted exchanged amounts are whole-number multiples of hν.

What determines photoelectron emission?

The photoelectric effect is the ejection of electrons from a metal surface by suitable incident light. A photoelectron is an electron ejected in this way. Emission begins without time lag when light of sufficient frequency reaches the surface.

The threshold frequency ν₀ is the minimum frequency required for emission from a given metal. The work function W₀ is the minimum energy needed to remove an electron: W₀ = hν₀. Below this frequency, increasing brightness does not produce emission.

For light above threshold, increasing intensity increases the number of emitted electrons. Increasing frequency increases their maximum kinetic energy, the energy of motion. If v is electron speed and Kₘₐₓ is maximum kinetic energy, Kₘₐₓ = ½mₑv² = hν − W₀.

Light thus has dual behaviour. Its wave properties explain interference, the combination of waves, and diffraction, bending around obstacles. Its photon properties explain energy exchange in the photoelectric effect.

Worked example 3. Find the energy per mole of photons of frequency 5 × 10¹⁴ Hz. A mole contains 6.022 × 10²³ specified particles. Use h = 6.626 × 10⁻³⁴ J s.

Formula: E = hν; Q = ENₐ, where Q is energy per mole and Nₐ is Avogadro's constant, 6.022 × 10²³ mol⁻¹. Substitute: E = 3.313 × 10⁻¹⁹ J. Answer: Q = 199,510 J mol⁻¹, or 199.51 kJ mol⁻¹; one kilojoule is 10³ J.

Worked example 4. A metal has threshold frequency 7.0 × 10¹⁴ s⁻¹. Find the maximum photoelectron kinetic energy for incident light of frequency 1.0 × 10¹⁵ s⁻¹, using h = 6.626 × 10⁻³⁴ J s.

Formula: Kₘₐₓ = h(ν − ν₀). Substitute: Kₘₐₓ = (6.626 × 10⁻³⁴)(3.0 × 10¹⁴). Answer: Kₘₐₓ ≈ (1.988 J) × 10⁻¹⁹.

What do emission, absorption and hydrogen spectra show?

A spectrum displays radiation separated according to wavelength or frequency. Spectroscopy is the study of emission or absorption spectra. A continuous spectrum contains an uninterrupted range of wavelengths, with neighbouring colours merging into one another.

An emission spectrum records radiation given out by an energised substance. An absorption spectrum records missing wavelengths when a continuous beam passes through an absorbing sample. These appear as dark spaces against the bright continuous background.

A line spectrum contains discrete spectral lines. Gas-phase atoms emit at particular wavelengths rather than across a continuous range. Each element has a characteristic line spectrum, allowing its identification by comparison with known spectra.

A band spectrum contains groups of closely spaced lines that appear as bands. Such spectra are characteristic of molecules, whereas separated atomic lines are characteristic of isolated atoms.

How does Rydberg's formula organise hydrogen lines?

Excited hydrogen atoms, raised above their lowest-energy state, release radiation as electrons move to lower energy states. A transition is a change between allowed energy states. A spectral series groups transitions ending at a common lower level.

Let n₁ be the lower level number, n₂ the upper level number, and R the Rydberg wavenumber constant for hydrogen, 109,677 cm⁻¹ or 1.09677 × 10⁷ m⁻¹. Here cm is centimetre, equal to 10⁻² m.

1/λ = R(1/n₁² − 1/n₂²), with n₂ greater than n₁. Use a positive difference when calculating emitted photon wavelength.

Worked example 5. Calculate the wavelength for hydrogen emission from n₂ = 5 to n₁ = 2. Use R = 1.09677 × 10⁷ m⁻¹.

Formula: ν̄ = R(1/n₁² − 1/n₂²); λ = 1/ν̄. Substitute: ν̄ = (1.09677 × 10⁷)(1/4 − 1/25) = 2.303217 × 10⁶ m⁻¹. Answer: λ ≈ (4.342 m) × 10⁻⁷. This transition belongs to the Balmer series.

How does Bohr's model explain hydrogen and its limitations?

Bohr's model introduced permitted circular orbits with fixed energies. A stationary state is an allowed state whose energy does not change with time. The ground state is the lowest-energy state; a higher-energy state is an excited state.

1. An electron moves in a permitted orbit of fixed radius and energy.
2. It does not continuously radiate energy while remaining in that stationary state.
3. Absorption raises it to a higher state; emission accompanies a transition to a lower state.
4. The photon energy equals the energy difference between the two states.
5. Angular momentum, the rotational quantity mₑvr for circular motion of radius r, is quantised: mₑvr = nh/(2π). Here n is the positive integer orbit number and π is the circle constant, approximately 3.1416.

Which formulas apply to one-electron species?

A hydrogen-like species contains one electron, as in hydrogen, the helium ion He⁺ or the lithium ion Li²⁺. For nuclear charge number Z and level n, the Bohr radius rₙ, energy Eₙ and speed vₙ are:

• rₙ = (52.9 pm)n²/Z.
• Eₙ = −(2.18 × 10⁻¹⁸ J)Z²/n².
• vₙ ≈ (2.19 × 10⁶ m s⁻¹)Z/n.

The energy zero is a free electron at rest infinitely far from the nucleus. Negative energy therefore signifies a bound electron. As n increases, radius increases, speed decreases and energy becomes less negative. Increasing Z at fixed n contracts the orbit and binds the electron more strongly.

Let Eᵢ and E𝒻 denote initial and final atomic energies. The atom's change is ΔE = E𝒻 − Eᵢ, where Δ means change. Emission has negative ΔE, but photon energy is the positive magnitude |ΔE|. Consequently ν = |ΔE|/h.

Worked example 6. Find the first-orbit energy, radius and speed of He⁺, with Z = 2 and n = 1. Use the coefficients 2.18 × 10⁻¹⁸ J, 52.9 pm and 2.19 × 10⁶ m s⁻¹ in the Bohr formulas.

Formula: Eₙ = −(2.18 × 10⁻¹⁸)Z²/n² J; rₙ = 52.9n²/Z pm; vₙ = 2.19 × 10⁶Z/n m s⁻¹. Substitute: Z²/n² = 4 and n²/Z = ½. Answer: energy = −8.72 × 10⁻¹⁸ J, radius = 26.45 pm and speed ≈ 4.38 × 10⁶ m s⁻¹.

Where does the model fail?

Bohr's model explains hydrogen-like energies and line spectra, but not multi-electron spectra or the finer details of hydrogen lines. It does not explain Zeeman splitting, spectral splitting in a magnetic field, or Stark splitting, splitting in an electric field.

It also fails to explain chemical bonding. Its definite electron paths ignore matter's wave character and conflict with the uncertainty principle.

How do matter waves and uncertainty change the electron picture?

De Broglie's relation assigns a wavelength to a moving material particle. If m is its mass, v its speed and p its linear momentum, then p = mv and λ = h/p = h/(mv). The latter expression applies to non-relativistic motion, where speeds are well below light speed.

Electron diffraction confirms wave behaviour. Ordinary moving objects also have associated wavelengths, but their large masses make those wavelengths too small for their wave properties to be detected.

Derivation: How is wavelength obtained from kinetic energy?

Let K denote the particle's kinetic energy. Use the non-relativistic relation between kinetic energy, mass and speed.

1. Begin with K = ½mv².
2. Multiply by 2m to obtain 2mK = m²v² = p².
3. Take the positive momentum magnitude p = √(2mK) and insert it into λ = h/p. The symbol √ means square root.

λ = h/√(2mK).

Worked example 7. An electron has mass 9.1 × 10⁻³¹ kg and kinetic energy 3.0 × 10⁻²⁵ J. Find its speed and wavelength using h = 6.626 × 10⁻³⁴ J s.

Formula: v = √(2K/m); p = mv; λ = h/p. Substitute: v = √[(2 × 3.0 × 10⁻²⁵)/(9.1 × 10⁻³¹)] ≈ 812 m s⁻¹. Answer: λ ≈ (8.967 m) × 10⁻⁷, or 896.7 nm.

What does Heisenberg's uncertainty principle state?

Exact position and exact momentum cannot be determined simultaneously. Let Δx be uncertainty in position along the x-direction, and Δpₓ and Δvₓ the uncertainties in momentum and velocity along that same direction. Then Δx Δpₓ ≥ h/(4π), or Δx Δvₓ ≥ h/(4πm).

Here ≥ means greater than or equal to. Reducing one uncertainty increases the minimum possible value of the other. The restriction is significant microscopically and negligible macroscopically. It rules out the precise electron trajectories assumed by Bohr.

Worked example 8. An electron of mass 9.11 × 10⁻³¹ kg is located within Δx = 0.1 Å. Find the minimum velocity uncertainty. Use 1 Å = 10⁻¹⁰ m, h = 6.626 × 10⁻³⁴ J s and π = 3.1416.

Formula: Δvₓ ≥ h/(4πmΔx). Substitute: Δx = 1.0 × 10⁻¹¹ m. Answer: Δvₓ ≥ (5.79 m s⁻¹) × 10⁶. Equality gives the minimum bound, not a universally exact uncertainty.

What does the quantum mechanical model say about orbitals?

Quantum mechanics describes microscopic systems while accounting for wave-particle duality. It replaces a definite electron path with a mathematical description that predicts possible energies and probabilities of finding the electron in different regions.

For a system whose energy does not change with time, Schrödinger's equation is written ĤΨ = EΨ. Here Ψ, pronounced psi, is the wave function; E is the allowed energy; and Ĥ is the Hamiltonian operator, a mathematical operation representing the system's total energy.

An atomic orbital is a one-electron wave function in an atom. Ψ itself has no direct physical meaning. Its squared magnitude, |Ψ|², is the probability density, meaning probability per unit volume. Multiplying it by a sufficiently small volume gives the probability of finding the electron there.

How do probability density and radial probability differ?

Radial probability concerns finding an electron anywhere in a thin spherical layer at a distance r from the nucleus. It includes the volume of that layer. Thus it is different from probability density at a single location.

For a spherically symmetric orbital, P(r) = 4πr²|Ψ(r)|², where P(r) is radial probability per unit radial distance. Then P(r)dr is probability in a thin layer of thickness dr. For hydrogen 1s, the lowest-energy spherical orbital, density is greatest at the nucleus, but radial probability is zero there and peaks at the Bohr radius.

The upper plots show Ψ against distance r for 1s and 2s; the 2s curve crosses zero. The lower plots show squared wave functions, with a zero and a second maximum for 2s. These are density plots, not radial-probability plots.

See Fig. 2.12 in your NCERT textbook

Solving Schrödinger's equation for hydrogen gives allowed energies and orbitals naturally. Exact solutions are not available for multi-electron atoms, so approximate methods are used. Their orbital energies depend on both shell, a group of orbitals with the same principal quantum number, and subshell, a subdivision distinguished by orbital shape, because electron-electron repulsions accompany attraction to the nucleus.

What information do the four quantum numbers provide?

Quantum numbers label permitted electron states. Three specify an orbital: principal quantum number n, azimuthal quantum number l and magnetic orbital quantum number mₗ. Spin is intrinsic electron angular momentum; the spin quantum number mₛ distinguishes its two orientations.

A shell contains orbitals sharing n. Shells n = 1, 2, 3 and 4 are labelled K, L, M and N respectively. A subshell contains orbitals sharing both n and l; l = 0, 1, 2 and 3 correspond to s, p, d and f.

How are orbital and electron capacities counted?

A shell has n subshells and n² orbitals. A subshell has 2l + 1 orbitals. Since each orbital accommodates at most two electrons, maximum shell capacity = 2n². The s, p, d and f subshell capacities are respectively 2, 6, 10 and 14 electrons.

For n = 3, permitted l values are 0, 1 and 2. The 3s, 3p and 3d subshells contain one, three and five orbitals respectively: nine in total.

There is no 2d subshell because n = 2 permits only l = 0 and 1. In hydrogen-like species, energy depends only on n; in multi-electron atoms, both n and l influence energy.

How do s, p and d orbitals differ in shape and nodes?

A boundary surface encloses a region where the electron is very likely to be found, say 90%. It is not an electron path or a rigid wall. Orbital drawings represent this probability distribution rather than the shape of an electron.

s orbitals are spherically symmetric. At a given distance from the nucleus, probability density is the same in every direction. Their size increases in the sequence 1s, 2s, 3s, 4s.

Each p orbital has two lobes on opposite sides of the nucleus. A lobe is a region enclosed by the drawn orbital surface. The three p orbitals, pₓ, pᵧ and p𝓏, point along the mutually perpendicular coordinate axes x, y and z.

The s-orbital drawings show spherical boundaries, with 2s larger than 1s. The three 2p drawings show paired lobes directed along labelled x, y and z axes.

See Figs. 2.13 and 2.14 in your NCERT textbook

The five d orbitals are dₓᵧ, dᵧ𝓏, dₓ𝓏, dₓ²₋ᵧ² and d𝓏². The first three have lobes between the relevant axes; dₓ²₋ᵧ² has lobes along x and y. The d𝓏² shape has two axial lobes and a ring around the centre.

Four drawings show four-lobed shapes on coordinate axes. The fifth, d𝓏², shows two lobes along z and a ring in the central plane. The labelled axes distinguish their orientations.

See Fig. 2.15 in your NCERT textbook

Where is the probability density zero?

A node is a region where orbital probability density is zero. A radial node occurs at a particular distance from the nucleus. An angular node is associated with direction; a nodal plane is a plane on which density vanishes.

Radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1. Thus 2s has one radial node, while 2p has no radial node and one angular node. For p𝓏, the xy-plane is the nodal plane.

The three p orbitals in a subshell have identical size, shape and energy but different orientations. There is no simple one-to-one correspondence between the three mₗ values and the x, y and z directions.

How are electrons arranged and why are some configurations especially stable?

Electronic configuration describes the distribution of electrons among orbitals. In 1s², the initial number identifies the shell, the letter the subshell and the superscript the number of electrons. Orbitals of equal energy are called degenerate.

What are the electron filling rules?

• Aufbau principle: ground-state electrons occupy available orbitals in order of increasing energy.
• The (n + l) rule: lower n + l usually identifies the lower-energy orbital. For equal sums, the orbital with lower n fills first.
• Pauli's exclusion principle: no two electrons in an atom have identical sets of all four quantum numbers. Two electrons sharing an orbital must have opposite spins.
• Hund's rule: degenerate orbitals become singly occupied with parallel spins before pairing begins.

A useful filling sequence is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s. It is a rough guide: there is no single ordering universally correct for all atoms, and exceptions may occur.

Shielding is the reduction of nuclear attraction experienced by outer electrons because of inner electrons. The net positive charge experienced after shielding is called effective nuclear charge. Different orbital penetration towards the nucleus helps explain why subshell energies differ in multi-electron atoms.

How are configurations written for atoms and ions?

Nitrogen has configuration 1s² 2s² 2p³. Its three 2p electrons occupy separate orbitals with parallel spins, shown as [↑] [↑] [↑]. Each bracket represents one orbital and each arrow an electron's spin orientation.

Square-bracket notation such as [Ne] abbreviates the neon configuration 1s² 2s² 2p⁶. Likewise, [Ar] abbreviates the argon configuration 1s² 2s² 2p⁶ 3s² 3p⁶. Sodium is [Ne] 3s¹; its ion Na⁺ is [Ne]. Chlorine is [Ne] 3s² 3p⁵; chloride, Cl⁻, is [Ar].

For transition-metal cations, remove outer ns electrons before (n − 1)d electrons. Here ns means an s subshell in shell n. Iron, Fe, is [Ar] 3d⁶ 4s²; Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵.

Why do chromium and copper depart from the simple sequence?

Chromium, Cr, is [Ar] 3d⁵ 4s¹, and copper, Cu, is [Ar] 3d¹⁰ 4s¹. Their 4s and 3d subshell energies differ slightly. Half-filled and completely filled subshells have extra stability, associated with symmetrical electron distribution and exchange energy.

Exchange energy is the stabilising energy associated with exchanges between same-spin electrons in degenerate orbitals. Relatively small shielding, smaller electrostatic repulsion energy and larger exchange energy support these stable arrangements. This explanation is not a universal instruction to promote electrons: exceptions do exist.

Glossary

• Nucleons — Protons and neutrons considered together as the constituents of an atomic nucleus.
• Isotopes — Atoms of the same element having identical atomic numbers but different mass numbers.
• Isobars — Atoms with the same mass number but different numbers of protons in their nuclei.
• Photon — A quantum of electromagnetic radiation whose energy equals Planck's constant multiplied by frequency.
• Threshold frequency — The minimum incident light frequency required to eject electrons from a particular metal.
• Work function — The minimum energy required to remove an electron from a particular metal surface.
• Stationary state — An allowed atomic state in which the electron's energy does not change with time.
• Atomic orbital — A one-electron wave function that describes an electron in an atom.
• Probability density — Probability per unit volume, given by the squared magnitude of the wave function.
• Radial probability — Probability associated with finding an electron in a thin spherical layer around the nucleus.
• Node — A region where the orbital wave function and its probability density are zero.
• Degenerate orbitals — Orbitals with equal energies, such as the three p orbitals within a subshell.

Common errors and misconceptions

• Misconception: Mass number gives the number of electrons. Correct: It counts protons plus neutrons; electron number also depends on ionic charge.
• Misconception: Every positive canal-ray particle is a proton. Correct: Canal-ray particles are positive gaseous ions whose properties depend on the gas.
• Misconception: Brighter light must eject electrons below threshold frequency. Correct: Each photon must supply sufficient energy; brightness cannot replace the frequency requirement.
• Misconception: Negative atomic energy means negative photon energy. Correct: A bound state's energy uses a chosen zero; an emitted photon carries the positive energy difference.
• Misconception: An orbital is a circular electron track. Correct: It is a wave function used to describe probability distributions, not a definite trajectory.
• Misconception: Maximum 1s probability density at the nucleus means maximum radial probability there. Correct: Radial probability includes the spherical layer's volume and vanishes at the nucleus.
• Misconception: Electrons pair immediately in degenerate orbitals. Correct: Hund's rule gives single occupation with parallel spins before pairing.
• Misconception: The listed filling order is universal. Correct: It is a rough guide; chromium and copper illustrate the importance of total electronic energy.

Exam-style questions with model answers

Q1. Distinguish isotopes from isobars in terms of atomic number Z and mass number A. [2 marks]

1. Isotopes have the same atomic number Z but different mass numbers A, so they differ in neutron number.
2. Isobars have the same mass number A but different atomic numbers Z, so they are atoms of different elements.

Q2. A neutral bromine atom has mass number A = 80 and atomic number Z = 35. Calculate its proton, neutron and electron numbers, explaining each result. [3 marks]

1. The atom contains 35 protons because atomic number is defined as the number of protons in the nucleus. Therefore proton number equals Z.
2. It contains 45 neutrons. Mass number counts protons and neutrons together, so neutron number equals A − Z = 80 − 35.
3. It contains 35 electrons because it is explicitly neutral. Its negative electron charge must balance the positive charge of its 35 protons.

Q3. State three observations from Rutherford's gold-foil experiment and give the structural conclusion supported by each. [3 marks]

1. Most alpha particles passed through the foil without deflection. This showed that most of the space within an atom is empty.
2. A small fraction were deflected through small angles. This showed that the positive charge was concentrated rather than spread uniformly throughout the atom.
3. A very few particles turned back. Such strong deflection supported a very small central region containing concentrated positive charge and most atomic mass.

Q4. For light of wavelength 400 nm, calculate its frequency and photon energy. Use c = 3.00 × 10⁸ m s⁻¹, h = 6.626 × 10⁻³⁴ J s and 1 nm = 10⁻⁹ m. [4 marks]

1. Convert wavelength to metres before substitution: λ = 400 × 10⁻⁹ m = 4.00 × 10⁻⁷ m.
2. Use the wave relation ν = c/λ. Thus ν = (3.00 × 10⁸)/(4.00 × 10⁻⁷) = 7.50 × 10¹⁴ Hz.
3. Apply Planck's relation E = hν to one photon, using the calculated frequency and the supplied Planck constant.
4. Substitution gives E = (6.626 × 10⁻³⁴)(7.50 × 10¹⁴) = 4.9695 × 10⁻¹⁹ J, approximately 4.97 × 10⁻¹⁹ J per photon.

Q5. State five postulates of Bohr's hydrogen model, including its rules for photon energy and angular momentum. Define the symbols used in your answer. [5 marks]

1. The electron can occupy permitted circular orbits around the nucleus. Each orbit has a fixed radius and a definite energy, rather than an arbitrary energy.
2. While an electron remains in a stationary orbit, its energy does not change with time. It does not continuously lose energy by radiation.
3. Absorption transfers an electron from a lower-energy state to a higher-energy state. Emission accompanies a transition from a higher-energy state to a lower-energy state.
4. The photon frequency is ν = |E𝒻 − Eᵢ|/h, where Eᵢ and E𝒻 are initial and final atomic energies and h is Planck's constant.
5. Angular momentum satisfies mₑvr = nh/(2π): mₑ is electron mass, v speed, r radius, n a positive integer and π the circle constant.

Q6. For the principal shell n = 3, determine its subshells, their orbital counts and the maximum electron capacity. Use l = 0, 1, …, n − 1; l = 0, 1, 2 denote s, p, d; each subshell contains 2l + 1 orbitals; each orbital holds at most two electrons. [4 marks]

1. The allowed azimuthal quantum numbers are l = 0, 1 and 2. They identify the three subshells 3s, 3p and 3d.
2. For 3s, l = 0, so there is one orbital. For 3p, l = 1, so there are three orbitals.
3. For 3d, l = 2, so there are five orbitals. The total is therefore 1 + 3 + 5 = 9 orbitals.
4. Each orbital accommodates at most two electrons, giving a maximum of 18 electrons in the shell; paired electrons must have opposite spins.

Q7. Chromium has Z = 24 and copper Z = 29. The argon core [Ar] contains 18 electrons. Write their observed ground-state configurations and explain their stability using half-filled or filled subshells and exchange energy. [5 marks]

1. Chromium has six electrons beyond the argon core. Its observed configuration is [Ar] 3d⁵ 4s¹, giving a half-filled set of five d orbitals.
2. Copper has eleven electrons beyond that core. Its observed configuration is [Ar] 3d¹⁰ 4s¹, giving a completely filled set of five d orbitals.
3. The 4s and 3d subshells differ slightly in energy. The ground state corresponds to the arrangement with the lowest total electronic energy.
4. Half-filled and completely filled subshells have symmetrical electron distributions. Their relatively small mutual shielding contributes to the extra stability of these arrangements.
5. Exchange energy stabilises arrangements with same-spin electrons in degenerate orbitals. Greater exchange stabilisation contributes to the stability of half-filled and completely filled subshells.

Q8. An electron is located within position uncertainty Δx = 0.1 Å. Calculate the minimum velocity uncertainty using Δx Δvₓ ≥ h/(4πm), m = 9.11 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ J s, π = 3.1416 and 1 Å = 10⁻¹⁰ m. [3 marks]

1. Convert the position uncertainty: Δx = 0.1 × 10⁻¹⁰ m = 1.0 × 10⁻¹¹ m. The velocity uncertainty refers to the same direction as this position uncertainty.
2. Rearrange the supplied inequality to Δvₓ ≥ h/(4πmΔx). Use equality to calculate the smallest value allowed by the uncertainty relation.
3. Substituting the given quantities gives Δvₓ ≥ 5.79 × 10⁶ m s⁻¹. Thus the minimum velocity uncertainty is approximately 5.79 × 10⁶ m s⁻¹.

Key takeaways

• Atomic number counts protons, mass number counts nucleons, and ionic charge determines how electron number differs from proton number.
• Rutherford established a concentrated nucleus, but classical circular electron motion could not account for the observed stability of atoms.
• Radiation links wavelength and frequency through c = νλ, while each photon carries the discrete energy E = hν.
• Bohr's model explains hydrogen-like energies and spectra, but its definite paths conflict with wave behaviour and the uncertainty principle.
• Orbitals are wave functions; probability density and radial probability answer different questions about where an electron may be found.
• Quantum numbers specify shell, subshell, orientation and spin, while Pauli's principle limits each orbital to two opposite-spin electrons.
• Aufbau ordering, Hund's rule and the extra stability of half-filled or filled subshells together explain electronic arrangements and important exceptions.

Test yourself

Why do canal rays differ when the gas is changed?

They consist of positive gaseous ions, so their masses and charge-to-mass ratios depend on the gas present.

What happens to photon energy when wavelength decreases?

It increases because photon energy equals hc/λ, with Planck's constant and light speed fixed.

Why does stronger light below threshold frequency fail to eject electrons?

Each photon still has insufficient energy to overcome the metal's work function, regardless of the number of photons.

Which hydrogen spectral series ends at level n = 2?

The Balmer series contains transitions from higher levels to the level with principal quantum number two.

Why is a bound electron's Bohr energy negative?

The zero is assigned to a free electron at rest infinitely far away; the bound electron has lower energy.

What is the difference between an orbit and an orbital?

An orbit is a proposed definite path; an orbital is a one-electron wave function describing an atomic state.

For 3p, n = 3 and l = 1. How many radial and angular nodes occur?

There is one radial node from n − l − 1 and one angular node from l.

How are three electrons distributed among three degenerate p orbitals?

Each orbital receives one electron, and the three electrons have parallel spins before pairing begins.