Summary and Quick Revision

How to Use This Section

This section is for the night before the paper, and again in the queue outside the hall. Nothing new is taught. Every card compresses something Sections 1 to 15 worked through, in the same notation and with the same numbers. If a line surprises you, go back and reread that section instead of memorising the line.

Eight cards, one mistake checklist, one 60-second list. Screenshot the three figures.

Card	Topic	Sections it compresses	Who needs it most
1	Discovery of sub-atomic particles, the fundamental-particle table, the atomic-model timeline	1, 2	Board, NEET
2	Atomic number, mass number, isotopes, isobars, isoelectronic species	2	Board, NEET
3	Electromagnetic radiation, Planck's quantum, photoelectric effect	3, 4	Everyone
4	Atomic spectra, Rydberg formula, the five hydrogen series	5	Everyone
5	Bohr model formula sheet and its limitations	6, 12	Everyone
6	de Broglie wavelength and Heisenberg's uncertainty principle	7	JEE, NEET
7	Quantum numbers, orbital shapes, nodes, capacities	8, 9	Everyone
8	Orbital energies, aufbau order, filling rules, configurations, magnetic moment	9, 10, 14	Everyone

Key Point: The chapter is one story in three acts. Act 1: the atom has parts (electron, proton, neutron) arranged around a tiny nucleus. Act 2: light comes in packets ($E = h\nu$) and so do electron energies ($E_n = -13.6\,Z^2/n^2$ eV) — hence line spectra. Act 3: electrons are waves too, so orbits become orbitals, labelled by four quantum numbers and filled by three rules. Every formula on the cards belongs to one of those acts.

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Card 1 — Discovery of Sub-atomic Particles and the Atomic-Model Timeline

Who found what, and how

Particle	Experiment	Key observation	Scientist and year
Electron	Cathode-ray discharge tube (low pressure, high voltage)	Rays travel from cathode to anode, cast shadows, deflected by electric and magnetic fields like negative charge; $e/m_e$ independent of the gas and the electrode metal	J. J. Thomson, 1897 (name "electron" from Stoney)
Electron charge	Oil-drop experiment	Charge on every drop is an integral multiple of $1.6 \times 10^{-19}$ C	R. A. Millikan
Proton	Canal rays (anode rays) in a perforated-cathode tube	Positive rays; $e/m$ depends on the gas; smallest and lightest positive ion is from hydrogen	Goldstein (canal rays), named proton by Rutherford
Neutron	Bombarding a thin beryllium sheet with $\alpha$-particles	Electrically neutral particles of mass slightly greater than the proton	James Chadwick, 1932

Key Point: Cathode rays are the same whatever the gas (they are electrons); canal rays differ for each gas (they are the leftover positive ions).

The fundamental-particle table

Particle	Symbol	Absolute charge / C	Relative charge	Mass / kg	Mass / u	Approx. mass / u
Electron	$e$	$-1.602176 \times 10^{-19}$	$-1$	$9.109382 \times 10^{-31}$	0.00054	0
Proton	$p$	$+1.602176 \times 10^{-19}$	$+1$	$1.6726216 \times 10^{-27}$	1.00727	1
Neutron	$n$	0	0	$1.674927 \times 10^{-27}$	1.00867	1

Three numbers to carry: $e/m_e = 1.758820 \times 10^{11}\ \mathrm{C\ kg^{-1}}$ (Thomson), $e = 1.602 \times 10^{-19}$ C (Millikan), $m_e = 9.1094 \times 10^{-31}$ kg (the ratio of the two). A proton is about 1836 times heavier than an electron.

The four atomic models in one table

Model (year)	Picture	What it explained	Where it failed
Thomson (1898) — plum pudding / watermelon	Positive charge spread uniformly through a sphere, electrons embedded in it	Overall electrical neutrality of the atom	Could not explain the $\alpha$-scattering results — most of the atom turned out to be empty
Rutherford (1911) — nuclear model	Tiny, dense, positive nucleus (radius $\sim 10^{-15}$ m) at the centre; electrons circle it; atom radius $\sim 10^{-10}$ m	Most $\alpha$-particles pass straight through, a few deflect, about 1 in 20,000 bounces back	An orbiting electron should radiate continuously and spiral into the nucleus in about $10^{-8}$ s; also gave no account of line spectra
Bohr (1913) — quantised orbits	Electron in fixed stationary orbits with $mvr = nh/2\pi$; energy is emitted or absorbed only on jumping between orbits	Stability of the atom; the hydrogen line spectrum; $\mathrm{He^+}$, $\mathrm{Li^{2+}}$, $\mathrm{Be^{3+}}$; Rydberg's constant from first principles	Fine structure (doublets), multi-electron atoms, Zeeman and Stark effects, chemical bonding; contradicts de Broglie and Heisenberg
Quantum mechanical (Schrodinger, 1926)	Electron is a wave; $\hat{H}\psi = E\psi$ gives allowed energies and orbitals ($\psi^2$ = probability density)	Everything Bohr explained plus multi-electron atoms, shapes, bonding	Cannot be solved exactly for more than one electron — approximations are used

[NEET] The $\alpha$-scattering conclusions in three lines: (1) most of the atom is empty space; (2) all the positive charge and almost all the mass sit in a very small nucleus; (3) the nucleus radius is about $10^{-5}$ of the atomic radius.

Card 2 — Atomic Number, Mass Number, Isotopes, Isobars and Isoelectronic Species

The two numbers on the symbol

$$^{A}_{Z}\mathrm{X} \qquad Z = \text{atomic number} = \text{number of protons}, \qquad A = \text{mass number} = \text{protons} + \text{neutrons}$$

Key Point (Definition): Atomic number $Z$ fixes the identity of the element (Moseley); mass number $A$ is the total count of nucleons. Neither is a mass in grams — $A$ is a whole-number count.

The counting recipe

Want	For a neutral atom	For an ion of charge $q$ (positive for cations, negative for anions)
Protons	$Z$	$Z$ (charge never changes the proton count)
Neutrons	$A - Z$	$A - Z$
Electrons	$Z$	$Z - q$: cation loses electrons, anion gains them

Species	$Z$	$A$	$p$	$n$	$e$
$^{35}_{17}\mathrm{Cl}$	17	35	17	18	17
$^{56}_{26}\mathrm{Fe^{3+}}$	26	56	26	30	23
$^{16}_{8}\mathrm{O^{2-}}$	8	16	8	8	10
$^{80}_{35}\mathrm{Br}$	35	80	35	45	35

The "iso" family

Term	Same	Different	Examples
Isotopes	$Z$ (protons, hence chemistry)	$A$ (neutrons, hence mass)	$^{1}\mathrm{H}$ protium, $^{2}\mathrm{H}$ deuterium, $^{3}\mathrm{H}$ tritium; $^{12}\mathrm{C}$, $^{13}\mathrm{C}$, $^{14}\mathrm{C}$; $^{35}\mathrm{Cl}$, $^{37}\mathrm{Cl}$
Isobars	$A$	$Z$ (different elements)	$^{14}_{6}\mathrm{C}$ and $^{14}_{7}\mathrm{N}$; $^{40}_{18}\mathrm{Ar}$, $^{40}_{19}\mathrm{K}$, $^{40}_{20}\mathrm{Ca}$
Isotones (extra)	number of neutrons	$Z$ and $A$	$^{14}_{6}\mathrm{C}$ and $^{16}_{8}\mathrm{O}$ (8 neutrons each)
Isoelectronic	number of electrons	nuclear charge	$\mathrm{N^{3-}}$, $\mathrm{O^{2-}}$, $\mathrm{F^-}$, $\mathrm{Ne}$, $\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$, $\mathrm{Al^{3+}}$ (10 electrons each)

Key Point: Isotopes of an element have identical chemical properties, because chemistry is decided by electrons and all isotopes share the same $Z$. They differ in mass-dependent physical properties (rate of diffusion, boiling point of $\mathrm{D_2O}$).

[JEE/NEET] In an isoelectronic series size falls as nuclear charge rises: $\mathrm{N^{3-}} > \mathrm{O^{2-}} > \mathrm{F^-} > \mathrm{Ne} > \mathrm{Na^+} > \mathrm{Mg^{2+}} > \mathrm{Al^{3+}}$. The same ten electrons are pulled by 7, 8, 9, 10, 11, 12, 13 protons.

Card 3 — Electromagnetic Radiation, Planck's Quantum and the Photoelectric Effect

The wave formula sheet

Quantity	Symbol	Formula	SI unit	Notes
Wavelength	$\lambda$	crest-to-crest distance	m (also nm, $\text{\AA}$, pm)	$1\ \mathrm{nm} = 10^{-9}$ m; $1\ \text{\AA} = 10^{-10}$ m $= 0.1$ nm; $1\ \mathrm{pm} = 10^{-12}$ m
Frequency	$\nu$	waves per second	Hz $= \mathrm{s^{-1}}$	fixed by the source; does not change when light enters a medium
Speed	$c$	$c = \nu\lambda$	$\mathrm{m\ s^{-1}}$	$3.0 \times 10^8\ \mathrm{m\ s^{-1}}$ in vacuum, same for all frequencies
Wavenumber	$\bar{\nu}$	$\bar{\nu} = \dfrac{1}{\lambda} = \dfrac{\nu}{c}$	$\mathrm{m^{-1}}$ (usually $\mathrm{cm^{-1}}$)	number of waves per unit length; $1\ \mathrm{cm^{-1}} = 100\ \mathrm{m^{-1}}$

Maxwell (1870): light is oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, needing no medium.

The spectrum in order of increasing wavelength (decreasing frequency and energy)

$$\gamma\text{-rays} < \text{X-rays} < \text{UV} < \text{visible} < \text{IR} < \text{microwaves} < \text{radio waves}$$

Visible light spans about 400 nm (violet) to 750 nm (red): VIBGYOR from short to long wavelength, so violet has the highest frequency and energy, red the lowest. Sample numbers: FM radio $\sim 10^8$ Hz; microwaves $\sim 10^{10}$ Hz; visible $\sim 10^{14}$ to $10^{15}$ Hz; X-rays $\sim 10^{18}$ Hz.

Where waves failed and quanta won

Phenomenon	What the wave theory could not explain
Black-body radiation	The intensity-versus-wavelength curve has a maximum that shifts to shorter wavelength as $T$ rises; classical physics predicted ever-rising intensity at short wavelengths
Photoelectric effect	Instant emission, a threshold frequency, and kinetic energy depending on frequency not intensity
Line spectra	Atoms emit only certain discrete wavelengths

Planck (1900) and Einstein (1905)

Key Point (Definition): Energy is exchanged in discrete packets called quanta (photons for light). The energy of one quantum is proportional to its frequency: $$E = h\nu = \frac{hc}{\lambda} = hc\bar{\nu}, \qquad h = 6.626 \times 10^{-34}\ \mathrm{J\ s}$$ A beam of frequency $\nu$ can carry only $nh\nu$, with $n$ a whole number.

[JEE/NEET] The shortcut: $E\ (\text{in eV}) = \dfrac{1240}{\lambda\ (\text{in nm})}$. A 620-nm photon carries 2.0 eV, a 400-nm photon 3.1 eV, a 124-nm photon 10 eV. In joules, $E = 6.626 \times 10^{-34} \times 3.0 \times 10^8 / \lambda$: a 580-nm photon carries $3.43 \times 10^{-19}$ J, and one mole of them $206$ kJ.

The photoelectric effect — three observations, one equation

Observation	Meaning
Electrons are ejected without time lag as soon as light strikes the metal	Energy arrives in packets, not as a slowly accumulating wave
Number of electrons ejected is proportional to intensity	More photons per second, more electrons per second
Emission needs a minimum (threshold) frequency $\nu_0$; below it, no electrons whatever the intensity; above it, kinetic energy rises linearly with $\nu$ and is independent of intensity	One photon gives all its energy to one electron

$$h\nu = W_0 + \frac{1}{2}m_e v^2, \qquad W_0 = h\nu_0 = \frac{hc}{\lambda_0}, \qquad KE_{max} = h(\nu - \nu_0)$$

Metal	Li	Na	K	Mg	Cu	Ag
$W_0$ / eV	2.42	2.3	2.25	3.7	4.8	4.3
Threshold $\lambda_0 = 1240/W_0$ / nm	512	539	551	335	258	288

[Board] In words: red light (1.8 eV or less) cannot eject electrons from potassium because a single red photon carries less than $W_0 = 2.25$ eV; yellow light (2.1 eV) is just below; violet (3.1 eV) ejects electrons with about 0.85 eV of kinetic energy. Doubling the intensity doubles the number of electrons, never their energy.

Key Point: Light is a wave when it travels (interference, diffraction) and a particle when it interacts with matter (photoelectric effect, black-body radiation). Both are true — dual behaviour.

Card 4 — Atomic Spectra, the Rydberg Formula and the Five Hydrogen Series

Continuous versus line spectra

Spectrum	How it arises	Looks like
Continuous	White light through a prism; hot solid	all colours merging, violet to red, no gaps
Emission line spectrum	Excited atoms (heated or in an electric discharge) drop to lower energy and radiate	bright lines on a dark background
Absorption line spectrum	White light passed through a cool gas; the gas absorbs the same wavelengths it would emit	dark lines on a bright continuous background — the photographic negative of the emission spectrum

Each element has a unique line spectrum, a fingerprint used in spectroscopy. Rubidium, caesium, thallium, indium, gallium and scandium were discovered spectroscopically; helium was found in the Sun's spectrum before it was found on Earth.

The Rydberg formula

Key Point: Every line of the hydrogen spectrum fits one formula: $$\bar{\nu} = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad R_H = 109{,}677\ \mathrm{cm^{-1}} = 1.09677 \times 10^7\ \mathrm{m^{-1}}, \qquad n_2 > n_1$$ Balmer's original (1885) formula is the $n_1 = 2$ case, $\bar{\nu} = 109{,}677\,(1/4 - 1/n^2)$, the only series in the visible region.

The five series in one table

Series	$n_1$	$n_2$	Region	First line ($n_2 = n_1 + 1$)	Series limit ($n_2 \to \infty$)
Lyman	1	2, 3, 4, …	Ultraviolet	121.6 nm ($\Delta E = 10.2$ eV)	91.2 nm (13.6 eV)
Balmer	2	3, 4, 5, …	Visible	656.3 nm, red $\mathrm{H_\alpha}$ (1.89 eV)	364.6 nm (3.4 eV)
Paschen	3	4, 5, 6, …	Infrared	1875 nm	820.4 nm
Brackett	4	5, 6, 7, …	Infrared	4051 nm	1458 nm
Pfund	5	6, 7, 8, …	Infrared	7458 nm	2279 nm

Where the numbers come from: series limit $\lambda_\infty = n_1^2 / R_H$ (Lyman, $1/109{,}677$ cm $= 91.2$ nm); first-line wavenumber $= R_H\,(1/n_1^2 - 1/(n_1+1)^2)$ (Balmer, $109{,}677 \times 5/36 = 15{,}233\ \mathrm{cm^{-1}}$, so $\lambda \approx 656$ nm).

Words that mean specific things

Phrase in the question	Translate to
Longest wavelength / least energetic / first line of a series	$n_2 = n_1 + 1$
Shortest wavelength / series limit / most energetic line	$n_2 = \infty$
Second line of Balmer	$4 \to 2$ ($\mathrm{H_\beta}$, 486.1 nm, blue-green)
Lines in the Balmer series that are visible	$\mathrm{H_\alpha}$ 656.3, $\mathrm{H_\beta}$ 486.1, $\mathrm{H_\gamma}$ 434.0, $\mathrm{H_\delta}$ 410.2 nm
Number of lines when electrons fall from level $n$ to the ground state	$\dfrac{n(n-1)}{2}$; from $n$ to $n_1$: $\dfrac{(n-n_1)(n-n_1+1)}{2}$

[NEET] Lyman UV, Balmer visible, everything from Paschen onward infrared. The first line of any series is its longest-wavelength line; the limit is its shortest.

Card 5 — Bohr's Model: The Formula Sheet

The postulates (1913)

#	Postulate
1	The electron moves in circular orbits of fixed radius and energy (stationary states) around the nucleus
2	In a stationary state the electron does not radiate; energy changes only when it jumps between orbits
3	Frequency of the radiation absorbed or emitted is $\nu = \dfrac{\Delta E}{h} = \dfrac{E_2 - E_1}{h}$ (Bohr's frequency rule)
4	Angular momentum is quantised: $mvr = n\dfrac{h}{2\pi}$, $n = 1, 2, 3, \ldots$ — only orbits with a whole-number multiple of $h/2\pi$ are allowed

The formulas for hydrogen and hydrogen-like ions ($\mathrm{He^+}$, $\mathrm{Li^{2+}}$, $\mathrm{Be^{3+}}$; nuclear charge $Z$)

Quantity	Formula	Numbers to carry	Scaling
Radius	$r_n = a_0\dfrac{n^2}{Z}$	$a_0 = 52.9\ \mathrm{pm} = 0.529\ \text{\AA}$; $r_2(\mathrm{H}) = 211.6$ pm	$r \propto n^2/Z$
Energy (J)	$E_n = -R_H\dfrac{Z^2}{n^2}$	$R_H = 2.18 \times 10^{-18}$ J per atom $= 1312\ \mathrm{kJ\ mol^{-1}}$	$E \propto -Z^2/n^2$
Energy (eV)	$E_n = -13.6\dfrac{Z^2}{n^2}$ eV	H: $-13.6, -3.40, -1.51, -0.85, -0.54$ eV for $n = 1$ to 5; $E_2 = -5.45 \times 10^{-19}$ J, $E_3 = -2.42 \times 10^{-19}$ J
Velocity	$v_n = 2.18 \times 10^6\dfrac{Z}{n}\ \mathrm{m\ s^{-1}}$	$v_1(\mathrm{H}) \approx c/137$	$v \propto Z/n$
Transition energy	$\Delta E = 2.18 \times 10^{-18}\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ J $= 13.6\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ eV	$2 \to 1$ in H: $1.635 \times 10^{-18}$ J $= 10.2$ eV
Frequency, wavenumber	$\nu = \dfrac{\Delta E}{h}$; $\bar{\nu} = \dfrac{\Delta E}{hc} = 109{,}677\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\ \mathrm{cm^{-1}}$	$R_H/hc = 1.09677 \times 10^7\ \mathrm{m^{-1}}$
Ionisation energy	$IE = 0 - E_1 = +13.6\,Z^2$ eV	H 13.6 eV; $\mathrm{He^+}$ 54.4 eV; $\mathrm{Li^{2+}}$ 122.4 eV	$\propto Z^2$
Number of lines	$\dfrac{n(n-1)}{2}$ for a fall from $n$ to $n = 1$	$n = 4$: 6 lines; $n = 5$: 10 lines

Key Point: The negative sign means the electron is bound: $E_\infty = 0$ is a free electron at rest, and every bound state lies below zero. As $n$ increases $E_n$ becomes less negative, orbits get farther apart in radius ($\propto n^2$) but closer in energy ($\propto 1/n^2$). Energy is released when the electron falls inward and absorbed when it climbs.

[JEE Main] For hydrogen-like ions keep every hydrogen number and multiply: energies by $Z^2$, radii by $1/Z$, velocities by $Z$. $\mathrm{He^+}$ has $E_1 = -54.4$ eV, $r_1 = 26.45$ pm, and its $4 \to 2$ line has the same wavenumber as the $2 \to 1$ line of hydrogen ($Z^2 = 4$ compensates the $1/4$ from the doubled quantum numbers).

Why Bohr's model had to go

Limitation	What it means
Fine structure	Each hydrogen line is actually a doublet of closely spaced lines; Bohr predicts one
Multi-electron atoms	Fails even for helium — electron-electron repulsion is not in the model
Zeeman and Stark effects	Splitting of lines in a magnetic (Zeeman) or electric (Stark) field is unexplained
Chemical bonding	Cannot say why atoms combine into molecules
Wave nature of the electron	A definite orbit needs exact position and momentum at the same time, forbidden by Heisenberg; Bohr ignores de Broglie waves entirely

Card 6 — de Broglie Waves and Heisenberg's Uncertainty Principle

de Broglie (1924): matter has a wavelength

Key Point (Definition): Every moving particle has a wave associated with it, of wavelength $$\lambda = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$$ Wave character shows only when $\lambda$ is comparable to the size of the obstacle, which is why it matters for electrons ($10^{-10}$ m) and is invisible for cricket balls ($10^{-34}$ m).

Form	Formula	When to use
From velocity	$\lambda = \dfrac{h}{mv}$	mass and speed given
From kinetic energy	$\lambda = \dfrac{h}{\sqrt{2m\,KE}}$	energy given in J or eV
Electron accelerated through $V$ volts	$\lambda = \dfrac{h}{\sqrt{2m_e eV}} = \dfrac{12.27}{\sqrt{V}}\ \text{\AA}$	$V = 100$ V gives 1.227 $\text{\AA}$; 10 kV gives 0.123 $\text{\AA}$
Bohr orbit as a standing wave	$2\pi r_n = n\lambda$	the $n$th orbit fits exactly $n$ de Broglie wavelengths — Bohr's $mvr = nh/2\pi$ derived

Worked numbers: an electron at $2.05 \times 10^7\ \mathrm{m\ s^{-1}}$ has $\lambda = \dfrac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.05 \times 10^7} = 3.55 \times 10^{-11}$ m; a 0.1-kg ball at 10 $\mathrm{m\ s^{-1}}$ has $\lambda = 6.626 \times 10^{-34}$ m, unobservable.

[JEE Main] Same kinetic energy, lighter particle, longer wavelength ($\lambda \propto 1/\sqrt{m}$); same velocity, $\lambda \propto 1/m$; same wavelength, $KE \propto 1/m$. Electron beams diffract (Davisson and Germer) — the experimental proof, and the basis of the electron microscope.

Heisenberg (1927): you cannot pin both down

Key Point (Definition): It is impossible to determine simultaneously the exact position and the exact momentum (or velocity) of a microscopic particle: $$\Delta x \cdot \Delta p_x \geq \frac{h}{4\pi}, \qquad \text{i.e.}\qquad \Delta x \cdot \Delta v_x \geq \frac{h}{4\pi m}$$ $h/4\pi = 5.27 \times 10^{-35}$ J s.

Object	Why the principle matters or not
Electron ($m = 9.1 \times 10^{-31}$ kg)	$\Delta x \cdot \Delta v \geq 5.79 \times 10^{-5}\ \mathrm{m^2\ s^{-1}}$: fix $\Delta x$ at $10^{-10}$ m and $\Delta v$ is already $5.8 \times 10^5\ \mathrm{m\ s^{-1}}$ — an orbit is meaningless
Milligram dust particle ($10^{-6}$ kg)	$\Delta x \cdot \Delta v \geq 5.27 \times 10^{-29}$; pin $\Delta x$ at $10^{-10}$ m and $\Delta v$ is $5 \times 10^{-19}\ \mathrm{m\ s^{-1}}$ — nothing
Physical reason	To see an electron you must hit it with a photon of $\lambda$ smaller than the electron's size; such a photon carries enough momentum to change the electron's velocity unpredictably

Why these two ideas killed the orbit

A Bohr orbit needs a definite radius and a definite velocity at the same instant, which the uncertainty principle forbids for something as light as an electron. A particle that is also a wave cannot follow a single line in space. The replacement is the orbital, a region where the probability of finding the electron ($\lvert\psi\rvert^2$) is high — usually the surface enclosing 90% probability.

Card 7 — Quantum Numbers, Orbital Shapes, Nodes and Capacities

The quantum mechanical model in three sentences

Schrodinger's equation $\hat{H}\psi = E\psi$ is solved for the atom; each acceptable solution is a wave function $\psi$ (an orbital) with a definite energy. $\psi$ itself has no physical meaning; $\lvert\psi\rvert^2$ is the probability density of finding the electron at that point. Energy in an atom is quantised, and the electron has no fixed path.

The four quantum numbers

Name	Symbol	Allowed values	What it tells you	Decided by
Principal	$n$	1, 2, 3, … (K, L, M, N shells)	Size and energy of the orbital; number of orbitals in a shell $= n^2$, electrons $= 2n^2$	Schrodinger equation
Azimuthal (angular momentum, subsidiary)	$l$	0 to $n - 1$ ($s, p, d, f$ for $l = 0, 1, 2, 3$)	Shape of the orbital and the subshell; orbital angular momentum $= \sqrt{l(l+1)}\,\dfrac{h}{2\pi}$; number of subshells in a shell $= n$	Schrodinger equation
Magnetic	$m_l$	$-l$ to $+l$ including 0, i.e. $2l + 1$ values	Orientation of the orbital in space; number of orbitals in a subshell $= 2l + 1$	Schrodinger equation
Spin	$m_s$	$+\tfrac{1}{2}$ or $-\tfrac{1}{2}$	Intrinsic spin of the electron; two electrons per orbital, opposite spins	Added (Uhlenbeck and Goudsmit); not from Schrodinger

Shell	$n$	Subshells ($l$)	Orbitals per subshell	Orbitals in shell ($n^2$)	Max electrons ($2n^2$)
K	1	$1s$	1	1	2
L	2	$2s$, $2p$	1, 3	4	8
M	3	$3s$, $3p$, $3d$	1, 3, 5	9	18
N	4	$4s$, $4p$, $4d$, $4f$	1, 3, 5, 7	16	32

Key Point: A subshell holds $2(2l + 1)$ electrons: $s$ 2, $p$ 6, $d$ 10, $f$ 14. Combinations that do not exist: $1p$, $2d$, $3f$ (because $l \leq n - 1$); $m_l = +2$ in a $p$ subshell; $n = 0$.

Shapes

Orbital	Shape	Lobes / orientation	Sign of $\psi$
$s$	Spherical, non-directional	one; $\lvert \psi \rvert^2$ falls off with distance; $2s$ larger than $1s$	positive everywhere in $1s$; $2s$ changes sign at its radial node
$p$	Dumbbell, two lobes either side of the nucleus	$p_x$, $p_y$, $p_z$ along the three axes; nucleus at the nodal plane	opposite sign in the two lobes
$d$	Double dumbbell (cloverleaf) for four of them	$d_{xy}$, $d_{yz}$, $d_{xz}$ between axes; $d_{x^2 - y^2}$ along $x$ and $y$; $d_{z^2}$ along $z$ with a ring (doughnut) in the $xy$ plane	alternate lobes opposite

All three $p$ orbitals of a subshell have the same energy and size (degenerate), and so do the five $d$ orbitals; they differ only in orientation.

Nodes — the counting rule

Key Point: A node is where $\psi = 0$ (probability zero). Radial (spherical) nodes $= n - l - 1$; angular (planar) nodes $= l$; total $= n - 1$.

Orbital	Radial nodes	Angular nodes	Total
$1s$	0	0	0
$2s$	1	0	1
$2p$	0	1	1
$3s$	2	0	2
$3p$	1	1	2
$3d$	0	2	2
$4d$	1	2	3
$4f$	0	3	3

[NEET] Orbitals with no radial node are those with $l = n - 1$: $1s$, $2p$, $3d$, $4f$. Two radial nodes and one angular node means $n - l - 1 = 2$ with $l = 1$, so $n = 4$: the $4p$ orbital.

Card 8 — Orbital Energies, the Filling Rules, Configurations and Magnetic Moment

Orbital energies

Atom	What fixes the energy of an orbital	Order
Hydrogen (one electron)	only $n$	$1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f$
Multi-electron atoms	$n + l$; electron-electron repulsion and shielding split the subshells of one shell	$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p$

Key Point: Lower $(n + l)$, lower energy; equal $(n + l)$, lower $n$ is lower. $4s\ (4 + 0 = 4)$ fills before $3d\ (3 + 2 = 5)$; $3d\ (5)$ fills before $4p\ (4 + 1 = 5)$ because $n$ is smaller. Within a shell, $s$ electrons penetrate closest to the nucleus and are shielded least, so $E_s < E_p < E_d < E_f$ and the effective nuclear charge $Z_{eff}$ felt by an $s$ electron is larger. Orbital energies fall with increasing $Z$ for the same orbital: $E_{2s}(\mathrm{H}) > E_{2s}(\mathrm{Li}) > E_{2s}(\mathrm{Na}) > E_{2s}(\mathrm{K})$.

The three filling rules

Rule	Statement	One-line consequence
Aufbau principle	Orbitals are filled in order of increasing energy ($n + l$ order above)	$\mathrm{K}$ is $[\mathrm{Ar}]\,4s^1$, not $[\mathrm{Ar}]\,3d^1$
Pauli exclusion principle	No two electrons in an atom can have the same set of four quantum numbers; an orbital holds at most two electrons with opposite spin	Shell capacity $2n^2$
Hund's rule of maximum multiplicity	Pairing in degenerate orbitals ($p$, $d$, $f$) begins only after each is singly occupied, and the single electrons have parallel spins	N is $\uparrow\ \uparrow\ \uparrow$ in $2p$, giving 3 unpaired electrons; O has 2, F 1, Ne 0

Configurations to know cold

Element ($Z$)	Configuration	Unpaired electrons
H (1), He (2)	$1s^1$; $1s^2$	1; 0
Li (3), Be (4), B (5)	$[\mathrm{He}]\,2s^1$; $[\mathrm{He}]\,2s^2$; $[\mathrm{He}]\,2s^2 2p^1$	1; 0; 1
C (6), N (7), O (8)	$[\mathrm{He}]\,2s^2 2p^2$; $2s^2 2p^3$; $2s^2 2p^4$	2; 3; 2
F (9), Ne (10)	$[\mathrm{He}]\,2s^2 2p^5$; $2s^2 2p^6$	1; 0
Na (11), Mg (12), Al (13)	$[\mathrm{Ne}]\,3s^1$; $3s^2$; $3s^2 3p^1$	1; 0; 1
Si (14), P (15), S (16), Cl (17), Ar (18)	$[\mathrm{Ne}]\,3s^2 3p^2$; $3p^3$; $3p^4$; $3p^5$; $3p^6$	2; 3; 2; 1; 0
K (19), Ca (20)	$[\mathrm{Ar}]\,4s^1$; $[\mathrm{Ar}]\,4s^2$	1; 0
Sc (21), Ti (22), V (23)	$[\mathrm{Ar}]\,3d^1 4s^2$; $3d^2 4s^2$; $3d^3 4s^2$	1; 2; 3
Cr (24)	$[\mathrm{Ar}]\,3d^5 4s^1$ (not $3d^4 4s^2$)	6
Mn (25), Fe (26), Co (27), Ni (28)	$[\mathrm{Ar}]\,3d^5 4s^2$; $3d^6 4s^2$; $3d^7 4s^2$; $3d^8 4s^2$	5; 4; 3; 2
Cu (29)	$[\mathrm{Ar}]\,3d^{10} 4s^1$ (not $3d^9 4s^2$)	1
Zn (30), Ga (31) to Kr (36)	$[\mathrm{Ar}]\,3d^{10} 4s^2$; $[\mathrm{Ar}]\,3d^{10} 4s^2 4p^{1}$ to $4p^6$	0; 1, 2, 3, 2, 1, 0

For ions, remove from the highest $n$ first. Cations of transition metals lose the $4s$ electrons before $3d$:

Ion	Configuration	Unpaired	$\mu = \sqrt{n(n+2)}$ BM
$\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$, $\mathrm{F^-}$, $\mathrm{O^{2-}}$	$1s^2 2s^2 2p^6$	0	0
$\mathrm{Fe^{2+}}$	$[\mathrm{Ar}]\,3d^6$	4	4.90
$\mathrm{Fe^{3+}}$	$[\mathrm{Ar}]\,3d^5$	5	5.92
$\mathrm{Cu^+}$; $\mathrm{Cu^{2+}}$	$[\mathrm{Ar}]\,3d^{10}$; $[\mathrm{Ar}]\,3d^9$	0; 1	0; 1.73
$\mathrm{Cr^{3+}}$	$[\mathrm{Ar}]\,3d^3$	3	3.87
$\mathrm{Zn^{2+}}$	$[\mathrm{Ar}]\,3d^{10}$	0	0

Why Cr and Cu break the pattern — stability of half-filled and fully filled subshells

Cause	Meaning
Symmetry	$p^3$, $d^5$, $f^7$ (half-filled) and $p^6$, $d^{10}$, $f^{14}$ (full) have a symmetrical charge distribution, which lowers the energy
Exchange energy	Electrons of the same spin in degenerate orbitals can exchange positions; more such pairs, more stabilisation. $d^5$ has $\binom{5}{2} = 10$ exchanges, $d^4$ only 6
Small $3d$-$4s$ gap	Promoting one $4s$ electron costs little and the gain in exchange plus symmetry pays for it

The same idea explains why $\mathrm{Fe^{3+}}$ ($d^5$) is more stable than $\mathrm{Fe^{2+}}$ ($d^6$), and why $\mathrm{Cu^+}$ ($d^{10}$) is diamagnetic while $\mathrm{Cu^{2+}}$ ($d^9$) is paramagnetic.

Magnetic behaviour

$$\mu = \sqrt{n(n+2)}\ \text{BM (Bohr magneton)}, \qquad n = \text{number of unpaired electrons}$$

$n$	1	2	3	4	5
$\mu$ / BM	1.73	2.83	3.87	4.90	5.92

Paramagnetic — has unpaired electrons, attracted into a magnetic field (O, N, $\mathrm{Fe^{3+}}$, $\mathrm{Cu^{2+}}$). Diamagnetic — all paired, weakly repelled (Ne, Zn, $\mathrm{Cu^+}$, $\mathrm{Zn^{2+}}$).

[Board] Two checks on any configuration you write: the superscripts must add up to the electron count, and the highest-$n$ subshell must be the one you empty first for a cation. $\mathrm{Fe^{2+}}$ is $3d^6$, never $3d^4 4s^2$.

The Mistakes That Cost the Most Marks

Each of these was flagged somewhere in Sections 1 to 15, ordered roughly by how often they show up in answer scripts.

1. Writing $\mathrm{Cr}$ as $3d^4 4s^2$ and $\mathrm{Cu}$ as $3d^9 4s^2$. Chromium is $[\mathrm{Ar}]\,3d^5 4s^1$ and copper $[\mathrm{Ar}]\,3d^{10} 4s^1$ — half-filled and fully filled $d$ subshells are extra stable (symmetry and exchange energy).

2. Removing $3d$ electrons before $4s$ when making a cation. Electrons leave from the highest $n$ first. $\mathrm{Fe^{2+}}$ is $[\mathrm{Ar}]\,3d^6$ (4 unpaired, 4.90 BM), $\mathrm{Fe^{3+}}$ is $[\mathrm{Ar}]\,3d^5$ (5 unpaired, 5.92 BM). Writing $3d^4 4s^2$ for $\mathrm{Fe^{2+}}$ also wrecks the unpaired count and the magnetic moment.

3. Forgetting the negative sign on Bohr energies, or ranking them upside down. $E_n = -13.6/n^2$ eV. $E_1 = -13.6$ eV is the lowest energy, $E_2 = -3.4$ eV is higher (less negative). Energy is released on falling inward, absorbed on climbing out. Ionisation energy is $+13.6$ eV, not $-13.6$ eV.

4. Leaving out $Z^2$ for $\mathrm{He^+}$ and $\mathrm{Li^{2+}}$. Energies scale as $Z^2$, radii as $1/Z$, velocities as $Z$. $\mathrm{He^+}$ ground state is $-54.4$ eV and its first orbit has radius 26.45 pm. The Rydberg constant for $\mathrm{He^+}$ is $4 \times 109{,}677\ \mathrm{cm^{-1}}$.

5. Mixing up "first line" with "series limit". The first (longest-wavelength, least-energy) line is $n_2 = n_1 + 1$; the limit (shortest wavelength, most energy) is $n_2 = \infty$. Balmer: first line 656.3 nm, limit 364.6 nm. Lyman is UV, Balmer visible, the rest infrared.

6. Treating intensity as if it changed the kinetic energy of photoelectrons. Intensity changes the number of electrons per second; frequency alone sets $KE_{max} = h\nu - W_0$. Below $\nu_0$ no electrons come out however bright the light. The equation needs $h\nu$ and $W_0$ in the same unit — do not subtract eV from joules.

7. Unit slips in $E = hc/\lambda$. $\lambda$ must be in metres: 580 nm $= 5.80 \times 10^{-7}$ m, giving $3.43 \times 10^{-19}$ J. With $\lambda$ in nm use $1240/\lambda$ eV instead. Wavenumber in $\mathrm{cm^{-1}}$ needs $\lambda$ in cm: $109{,}677\ \mathrm{cm^{-1}} = 1.09677 \times 10^7\ \mathrm{m^{-1}}$.

8. Illegal quantum-number sets. $l$ runs from 0 to $n - 1$, $m_l$ from $-l$ to $+l$. So $n = 2, l = 2$ is impossible ($2d$ does not exist); $n = 3, l = 1, m_l = +2$ is impossible ($p$ orbitals have only $m_l = -1, 0, +1$); $n = 1, l = 0, m_l = 0, m_s = 0$ is impossible ($m_s$ is $\pm\frac{1}{2}$).

9. Counting nodes as $n$ instead of $n - 1$. Total nodes $= n - 1$, angular $= l$, radial $= n - l - 1$. $3p$: 1 radial, 1 angular. $2s$: 1 radial, 0 angular. $3d$: 0 radial, 2 angular.

10. Filling $3d$ before $4s$ (or ranking $3d$ below $4s$ in hydrogen). In multi-electron atoms $(n + l)$ decides: $4s\ (4) < 3d\ (5)$, so potassium is $[\mathrm{Ar}]\,4s^1$. In hydrogen all subshells of a shell are degenerate — $3s = 3p = 3d$ — because there is no shielding.

11. Using $\Delta x \cdot \Delta p \geq h/2\pi$ (or $h$) instead of $h/4\pi$. The relation is $\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}$, so $\Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}$; $h/4\pi = 5.27 \times 10^{-35}$ J s. The de Broglie formula is $\lambda = h/mv$ — mass times velocity, never $h/m$ or $h/v$ alone.

12. Confusing isobars with isotopes, and cations with anions in electron counts. Isotopes share $Z$ (same element), isobars share $A$ (different elements). For an ion, electrons $= Z - \text{charge}$: $\mathrm{Fe^{3+}}$ has $26 - 3 = 23$ electrons; $\mathrm{O^{2-}}$ has $8 + 2 = 10$. Neutrons are always $A - Z$ and never change with charge.

Key Point: Two more single-mark slips: writing "orbit" when the question is about the quantum mechanical model (an orbital is a probability region, an orbit is a fixed Bohr path), and quoting the photoelectric threshold as a wavelength above which emission happens — it is a minimum frequency (maximum wavelength).

The 60-Second Revision

The irreducible minimum, for the queue outside the hall.

Particles. Electron: cathode rays, Thomson 1897, $e/m_e = 1.76 \times 10^{11}\ \mathrm{C\ kg^{-1}}$; charge $1.602 \times 10^{-19}$ C from Millikan's oil drops; mass $9.1 \times 10^{-31}$ kg. Proton: canal rays, $e/m$ depends on the gas. Neutron: Chadwick 1932, Be + $\alpha$. Proton and neutron are about 1 u each; a proton is about 1836 times an electron's mass.

Models. Thomson: plum pudding, explains neutrality only. Rutherford 1911: $\alpha$-scattering, tiny dense nucleus ($10^{-15}$ m) in a $10^{-10}$ m atom; fails on stability and spectra. Bohr 1913: quantised orbits, $mvr = nh/2\pi$; explains H and hydrogen-like spectra; fails on fine structure, multi-electron atoms, Zeeman and Stark effects. Quantum mechanical: orbitals, $\lvert\psi\rvert^2$.

Numbers on the symbol. $Z$ = protons; $A = p + n$; neutrons $= A - Z$; electrons $= Z - \text{charge}$. Isotopes same $Z$; isobars same $A$; isoelectronic same electron count ($\mathrm{O^{2-}}$, $\mathrm{F^-}$, Ne, $\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$ all 10).

Radiation. $c = \nu\lambda$; $\bar{\nu} = 1/\lambda$; $c = 3.0 \times 10^8\ \mathrm{m\ s^{-1}}$. Order by rising $\lambda$: $\gamma$, X, UV, visible (400 to 750 nm), IR, microwave, radio. $E = h\nu = hc/\lambda$, $h = 6.626 \times 10^{-34}$ J s; $E(\text{eV}) = 1240/\lambda(\text{nm})$. Photoelectric: $h\nu = W_0 + KE$; threshold frequency $\nu_0 = W_0/h$; KE depends on $\nu$, number on intensity, no time lag.

Spectra. $\bar{\nu} = 109{,}677\,(1/n_1^2 - 1/n_2^2)\ \mathrm{cm^{-1}}$. Lyman ($n_1 = 1$, UV), Balmer (2, visible: 656.3, 486.1, 434.0, 410.2 nm), Paschen (3), Brackett (4), Pfund (5) — all IR. First line $n_2 = n_1 + 1$; limit $n_2 = \infty$. Lines from level $n$: $n(n-1)/2$.

Bohr. $r_n = 52.9\,n^2/Z$ pm; $E_n = -2.18 \times 10^{-18}\,Z^2/n^2$ J $= -13.6\,Z^2/n^2$ eV; $v_n = 2.18 \times 10^6\,Z/n\ \mathrm{m\ s^{-1}}$; $\Delta E = 13.6\,Z^2(1/n_1^2 - 1/n_2^2)$ eV; $IE(\mathrm{H}) = 13.6$ eV $= 1312\ \mathrm{kJ\ mol^{-1}}$; $\mathrm{He^+}$: $\times 4$ on energy, $\div 2$ on radius.

Waves and uncertainty. $\lambda = h/mv = h/\sqrt{2mKE}$; electron through $V$ volts: $12.27/\sqrt{V}\ \text{\AA}$; $2\pi r = n\lambda$. $\Delta x \cdot \Delta p \geq h/4\pi$; $\Delta x \cdot \Delta v \geq h/4\pi m$. Both together forbid orbits and give orbitals (90% probability surface).

Quantum numbers. $n = 1, 2, 3, \ldots$ (size, energy; $n^2$ orbitals, $2n^2$ electrons); $l = 0$ to $n - 1$ ($s, p, d, f$; shape; $\sqrt{l(l+1)}\,h/2\pi$); $m_l = -l$ to $+l$ ($2l + 1$ orbitals; orientation); $m_s = \pm\frac{1}{2}$. Capacity: $s$ 2, $p$ 6, $d$ 10, $f$ 14. Nodes: radial $n - l - 1$, angular $l$, total $n - 1$. $s$ sphere, $p$ dumbbell, $d$ cloverleaf ($d_{z^2}$ with a ring).

Filling. Order: $1s\ 2s\ 2p\ 3s\ 3p\ 4s\ 3d\ 4p\ 5s\ 4d\ 5p\ 6s\ 4f\ 5d\ 6p\ 7s\ 5f\ 6d\ 7p$ — lower $(n + l)$ first, then lower $n$. Hydrogen: energy depends on $n$ only. Aufbau, Pauli (no two electrons with the same four quantum numbers), Hund (singly occupy first, parallel spins). $\mathrm{Cr}\ [\mathrm{Ar}]\,3d^5 4s^1$; $\mathrm{Cu}\ [\mathrm{Ar}]\,3d^{10} 4s^1$; $\mathrm{Fe^{2+}}\ 3d^6$; $\mathrm{Fe^{3+}}\ 3d^5$; cations lose $4s$ before $3d$. $\mu = \sqrt{n(n+2)}$ BM: 1.73, 2.83, 3.87, 4.90, 5.92 for $n = 1$ to 5.

That is the whole chapter.