Lesson 4: Structure of Atom | State Board · Class 11 · Chemistry

Subatomic particles, atomic number and mass number

Link to Std 9–10 : In Std 9–10 you learned that an atom has protons, neutrons and electrons; here we fix their numbers and the identity of an atom (Z, A).

From daily life

Every student has a roll number; every element has an atomic number, and changing it changes the element.

Simple definition

The atomic number (Z) is the number of protons. The mass number (A) is protons + neutrons. Neutrons N = A − Z. In a neutral atom electrons = protons.

In the figure : Nucleus has p and n; electrons are in the shells around it.

Rule 1: Identity of an atom — An atom is written ᴬ_Z X with A above and Z below. N = A − Z.

Easy : ²³Na: Z = 11, A = 23 ⇒ p = 11, e = 11, n = 12.
Medium : ³⁵Cl: Z = 17 ⇒ n = 35 − 17 = 18.
Exam-type : ²³⁸U: Z = 92 ⇒ n = 238 − 92 = 146.

Rule 2: Ions: number of electrons — Cation: electrons = Z − charge; anion: electrons = Z + charge. The proton number never changes.

Easy : Na⁺: 11 − 1 = 10 e⁻.
Medium : Fe³⁺: 26 − 3 = 23 e⁻.
Exam-type : S²⁻: 16 + 2 = 18 e⁻; Cl⁻, K⁺, Ar and Ca²⁺ all have 18 e⁻ (isoelectronic).

Rule 3: Properties of the particles — Electron −1.602 × 10⁻¹⁹ C, proton +1.602 × 10⁻¹⁹ C, neutron neutral. Almost all the mass is in the nucleus.

Easy : Proton ≈ neutron ≈ 1 u; electron ≈ 1/1836 u.
Medium : So A ≈ the mass of the atom in u.
Exam-type : The nucleus is about 10⁵ times smaller than the atom yet holds almost all the mass.

Derivation: finding p, n, e

Step 1: Read Z ⇒ protons = Z.
Step 2: Neutrons = A − Z.
Step 3: If neutral, e = Z; if an ion, e = Z ∓ charge.

Formulas, units and dimensional formulae

Particle	Charge	Mass (kg)	Mass (u)
Electron	−1.602 × 10⁻¹⁹ C	9.109 × 10⁻³¹	0.00055
Proton	+1.602 × 10⁻¹⁹ C	1.673 × 10⁻²⁷	1.0073
Neutron	0	1.675 × 10⁻²⁷	1.0087

CET level (MCQ)

CET 1) Neutrons in ²³⁸U? 1) 92 2) 146 3) 238 4) 330 — Answer: 2) 146. 238 − 92.
CET 2) Which pair has equal electrons? 1) Na⁺, Mg²⁺ 2) Na, Mg 3) Cl, Cl⁻ 4) O, O²⁻ — Answer: 1). Both have 10 e⁻.

Real-life use: Medical imaging, carbon dating and nuclear power all use the structure of the nucleus.

Remember

Z = p; A = p + n; N = A − Z; neutral atom e = Z.

Common mistake

Thinking protons change in an ion; only electrons change.

CET trap

Subtract e⁻ for cations, add for anions; options hide the opposite (Fe³⁺ = 23, not 29).

Shortcut

Memory clue: "Z = roll number (protons); A = total weight; N = A − Z".

Spot the difference

Particle	Location	Charge
Electron	Shells	−1
Proton	Nucleus	+1
Neutron	Nucleus	0

Practice questions

Easy

1What is the charge of an electron?
2What is mass number?
3What is Z for Na?

Medium

4Give p, n, e in ²³Na.
5How many electrons in Fe³⁺ (Z = 26)?
6Electrons in S²⁻ (Z = 16)?

Challenging

7Neutrons in ²³⁸U (Z = 92)?
8What do Cl⁻, K⁺, Ar and Ca²⁺ have in common?

Show answers (try them yourself first)

1−1.602 × 10⁻¹⁹ C.
2Protons + neutrons.
311.
4p = 11, n = 12, e = 11.
523.
618.
7146.
8All have 18 electrons (isoelectronic).

Isotopes, isobars, isotones and average atomic mass

From daily life

Two people with the same name can have different weights; atoms of one element can also differ in mass.

Simple definition

Isotopes: same Z, different A. Isobars: same A, different Z. Isotones: same N.

In the figure : Three isotopes of hydrogen: same p, different n.

Rule 1: Isotopes — Chemical properties are alike (same e⁻); physical properties differ slightly (mass).

Easy : ¹H, ²H, ³H: n = 0, 1, 2.
Medium : ¹²C, ¹³C, ¹⁴C: Z = 6, n = 6, 7, 8.
Exam-type : Heavy water D₂O has mass 2 × 2 + 16 = 20 u.

Rule 2: Isobars and isotones — Isobars: equal A, unequal Z. Isotones: equal N = A − Z.

Easy : ⁴⁰Ar and ⁴⁰Ca are isobars (A = 40).
Medium : ¹³C and ¹⁴N are isotones (both n = 7).
Exam-type : ³⁰Si and ³¹P are isotones (16 n each).

Rule 3: Average atomic mass — Average = Σ(mass × abundance) ÷ 100.

Easy : Cl: (35 × 75 + 37 × 25) ÷ 100 = 35.5.
Medium : Mg: 24 × 0.79 + 25 × 0.10 + 26 × 0.11 = 24.32.
Exam-type : Cu: ⁶³Cu 70 %, ⁶⁵Cu 30 % ⇒ 63.6.

Derivation: average mass of Mg

Step 1: Take 100 atoms: 79 of ²⁴Mg, 10 of ²⁵Mg, 11 of ²⁶Mg.
Step 2: Total mass = 79 × 24 + 10 × 25 + 11 × 26 = 1896 + 250 + 286 = 2432 u.
Step 3: Average = 2432 ÷ 100 = 24.32 u.

CET level (MCQ)

CET 1) ¹³C and ¹⁴N are? 1) isotopes 2) isobars 3) isotones 4) isoelectronic — Answer: 3). Both have 7 neutrons.
CET 2) Cu: ⁶³Cu 70 %, ⁶⁵Cu 30 % ⇒ average? 1) 63.6 2) 64 3) 63.3 4) 64.4 — Answer: 1) 63.6.

Real-life use: ¹⁴C carbon dating gives the age of old bones and wood; ¹³¹I is used in thyroid treatment.

Remember

Isotopes equal Z; isobars equal A; isotones equal N.

Common mistake

Believing isotopes react differently; their chemistry is the same.

CET trap

Forgetting to divide by 100; using A instead of A − Z for neutrons.

Shortcut

Memory clue: "iso-TOPE = Z; iso-BAR = A; iso-TONE = N".

Spot the difference

Type	What is equal	Example
Isotopes	Z	¹H, ²H
Isobars	A	⁴⁰Ar, ⁴⁰Ca
Isotones	N	¹³C, ¹⁴N

Practice questions

Easy

1What are isotopes?
2Give an example of isobars.
3How many neutrons in tritium?

Medium

4Find the average mass of Cl (75 % ³⁵Cl, 25 % ³⁷Cl).
5Molecular mass of D₂O?
6Neutrons in ¹²C, ¹³C, ¹⁴C?

Challenging

7Average mass of Mg (79 %, 10 %, 11 %).
8What type are ³⁰Si and ³¹P?

Show answers (try them yourself first)

1Atoms with the same Z but different A.
2⁴⁰Ar and ⁴⁰Ca.
32.
435.5 u.
520 u.
66, 7, 8.
724.32 u.
8Isotones (16 n).

Limits of Rutherford's model, electromagnetic radiation and Planck's theory

From daily life

Radio, mobile signals and sunlight are all electromagnetic waves; '98.3 MHz' on a radio is the frequency of the wave.

Simple definition

Rutherford's model: nucleus in the centre, electrons orbiting. Limits: (1) an orbiting electron would lose energy and spiral into the nucleus; (2) it cannot explain line spectra. Planck: energy is absorbed or emitted in quanta; E = hν.

In the figure : Wavelength, amplitude and the spectrum: radio → γ.

Rule 1: Wave properties — c = ν λ, c = 3 × 10⁸ m/s. Wavenumber = 1 ÷ λ.

Easy : λ = 500 nm ⇒ ν = 3 × 10⁸ ÷ 5 × 10⁻⁷ = 6 × 10¹⁴ Hz.
Medium : Wavenumber = 1 ÷ 500 nm = 2 × 10⁶ m⁻¹.
Exam-type : ν = 5 × 10¹⁴ Hz ⇒ λ = 600 nm.

Rule 2: Planck's equation — E = h ν = h c ÷ λ, h = 6.626 × 10⁻³⁴ J s.

Easy : ν = 6 × 10¹⁴ Hz ⇒ E = 3.98 × 10⁻¹⁹ J.
Medium : Per mole of photons: 3.976 × 10⁻¹⁹ × 6.022 × 10²³ = 239.4 kJ/mol.
Exam-type : A 100 W source at 500 nm emits 100 ÷ 3.976 × 10⁻¹⁹ = 2.5 × 10²⁰ photons per second.

Rule 3: Order in the spectrum — As λ increases ν and E decrease; E ∝ ν ∝ 1 ÷ λ.

Easy : Red light (long λ) has less E; blue has more.
Medium : 400 nm and 800 nm: E ratio is 2 : 1.
Exam-type : Highest ν: γ-rays; lowest: radio.

Derivation: E = h c ÷ λ

Step 1: Planck: E = h ν.
Step 2: Wave: c = ν λ ⇒ ν = c ÷ λ.
Step 3: Substitute: E = h c ÷ λ.

Formulas, units and dimensional formulae

Quantity	Formula	Unit
Frequency	ν = c ÷ λ	Hz = s⁻¹
Wavelength	λ = c ÷ ν	m, nm
Wavenumber	ν̃ = 1 ÷ λ	m⁻¹
Energy	E = h ν	J

CET level (MCQ)

CET 1) Ratio of photon energies at 400 nm and 800 nm? 1) 1 : 2 2) 2 : 1 3) 1 : 1 4) 4 : 1 — Answer: 2). E ∝ 1 ÷ λ.
CET 2) Highest ν? 1) radio 2) visible 3) X-rays 4) IR — Answer: 3).

Real-life use: Microwave ovens, X-ray scans and solar panels all depend on photon energy E = hν.

Remember

c = νλ; E = hν = hc ÷ λ; h = 6.626 × 10⁻³⁴ J s; c = 3 × 10⁸ m/s.

Common mistake

Not converting nm to m (1 nm = 10⁻⁹ m); mixing energy per photon with energy per mole.

CET trap

Options differ by factors of 10³ or 10⁹; convert everything to SI (m, Hz, J).

Shortcut

Memory clue: "short wave = high energy" (X-rays are dangerous, radio is safe).

Spot the difference

Radiation	λ	ν and E
Radio	Long	Low
Visible	Medium	Medium
γ-rays	Very short	Highest

Practice questions

Easy

1Name the letters in c = νλ.
2Write Planck's equation.
3State one limitation of Rutherford's model.

Medium

4Find ν for λ = 600 nm.
5Wavenumber for 400 nm?
6Energy of a photon with ν = 5 × 10¹⁴ Hz.

Challenging

7Energy of 500 nm photons per mole in kJ/mol.
8How many photons per second from a 2 W source at 600 nm?

Show answers (try them yourself first)

1c = speed, ν = frequency, λ = wavelength.
2E = hν.
3The electron would spiral in / it cannot explain line spectra.
45 × 10¹⁴ Hz.
52.5 × 10⁶ m⁻¹.
63.31 × 10⁻¹⁹ J.
7239.4 kJ/mol.
86.04 × 10¹⁸.

Hydrogen spectrum and the Rydberg equation

From daily life

White light through a prism makes a continuous rainbow, but excited hydrogen gives only fixed coloured lines — an atom's fingerprint.

Simple definition

Hydrogen's emission spectrum has groups (series) of lines. Rydberg equation: 1 ÷ λ = R_H (1 ÷ n₁² − 1 ÷ n₂²), R_H = 1.097 × 10⁷ m⁻¹ (= 109677 cm⁻¹), n₂ > n₁.

In the figure : Lyman (n₁ = 1, UV), Balmer (n₁ = 2, visible), Paschen (n₁ = 3, IR).

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Hydrogen spectrum and the Rydberg equation

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Rule 1: Finding the wavelength — n₁ is the lower level, n₂ the upper. 1 ÷ λ = R_H (1 ÷ n₁² − 1 ÷ n₂²).

Easy : Balmer 3 → 2: λ = 656.3 nm (red).
Medium : Balmer 4 → 2: 1 ÷ λ = 1.097 × 10⁷ × 3 ÷ 16 ⇒ λ = 486.2 nm.
Exam-type : Lyman 2 → 1: λ = 121.5 nm.

Rule 2: Series limit — Take n₂ → ∞ for the shortest λ: 1 ÷ λ = R_H ÷ n₁².

Easy : Lyman limit: λ = 91.2 nm.
Medium : Balmer limit: λ = 4 ÷ 1.097 × 10⁷ = 364.6 nm.
Exam-type : First Paschen line 4 → 3: λ = 1875 nm (IR).

Rule 3: Number of lines — All lines when an electron falls from level n down is n(n − 1) ÷ 2.

Easy : n = 3: 3 lines.
Medium : n = 4: 6 lines.
Exam-type : n = 5: 10 lines.

Derivation: the red Balmer line

Step 1: n₁ = 2, n₂ = 3.
Step 2: 1 ÷ λ = 1.097 × 10⁷ × (1 ÷ 4 − 1 ÷ 9) = 1.097 × 10⁷ × 0.1389 = 1.524 × 10⁶ m⁻¹.
Step 3: λ = 1 ÷ 1.524 × 10⁶ = 6.563 × 10⁻⁷ m = 656.3 nm.

Formulas, units and dimensional formulae

Series	n₁	n₂	Region
Lyman	1	2, 3, 4…	UV
Balmer	2	3, 4, 5…	Visible
Paschen	3	4, 5, 6…	IR
Brackett	4	5, 6…	IR
Pfund	5	6, 7…	IR

CET level (MCQ)

CET 1) The 486 nm line comes from? 1) 3 → 2 2) 4 → 2 3) 5 → 2 4) 2 → 1 — Answer: 2).
CET 2) Maximum lines from n = 5 to the ground? 1) 4 2) 6 3) 10 4) 15 — Answer: 3) 10. 5 × 4 ÷ 2.

Real-life use: The composition of stars is known from spectra; forensic labs use spectra too.

Remember

Lyman UV (n₁ = 1), Balmer visible (2), Paschen IR (3); n₂ > n₁.

Common mistake

Swapping n₁ and n₂ gives a negative λ; R_H is in m⁻¹, so λ comes in m.

CET trap

Only the first four Balmer lines are visible; 'visible region' means n₁ = 2.

Shortcut

Memory clue: "Lyman = 1 (UV), Balmer = 2 (visible), Paschen = 3 (IR)".

Spot the difference

Point	Emission	Absorption
Electron	Down	Up
Spectrum	Bright lines	Dark lines

Practice questions

Easy

1Which series lies in the visible region?
2Write the value of R_H.
3What value of n₂ gives the series limit?

Medium

4λ of the first Lyman line?
5λ at the Balmer limit?
6λ of the first Paschen line?

Challenging

7Number of lines from n = 5 to the ground?
8Which transition gives 486 nm?

Show answers (try them yourself first)

1Balmer.
21.097 × 10⁷ m⁻¹.
3∞.
4121.5 nm.
5364.6 nm.
61875 nm.
710.
8n = 4 → 2.

Bohr model of the atom

From daily life

A lift stops only at floors, never between them; in Bohr's atom an electron lives only in fixed orbits.

Simple definition

Bohr's postulates: (1) an electron moves in fixed circular orbits without losing energy; (2) angular momentum m v r = n h ÷ 2π; (3) moving between orbits absorbs or emits ΔE = hν.

In the figure : Orbit radii 1 : 4 : 9; energies −13.6, −3.4, −1.51 eV (not to scale).

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Bohr model of the atom

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Rule 1: Orbit radius — rₙ = 52.9 n² ÷ Z pm.

Easy : H, n = 2: r = 52.9 × 4 = 211.6 pm.
Medium : H, n = 3: 476.1 pm.
Exam-type : Li²⁺, n = 2: 52.9 × 4 ÷ 3 = 70.5 pm.

Rule 2: Orbit energy — Eₙ = −13.6 Z² ÷ n² eV = −2.18 × 10⁻¹⁸ Z² ÷ n² J.

Easy : H: E₁ = −13.6, E₂ = −3.4, E₃ = −1.51 eV.
Medium : He⁺, n = 1: −13.6 × 4 = −54.4 eV.
Exam-type : Li²⁺, n = 2: −13.6 × 9 ÷ 4 = −30.6 eV.

Rule 3: Transition and ionisation energy — ΔE = E_upper − E_lower; ionisation energy = E_∞ − E₁ = 13.6 eV (H).

Easy : 1 → 2: 10.2 eV.
Medium : 1 → 3: 13.6 − 1.51 = 12.09 eV.
Exam-type : H ionisation: 13.6 eV = 1312 kJ/mol; He⁺: 54.4 eV.

Derivation: the formula for rₙ

Step 1: Centripetal force = Coulomb force: m v² ÷ r = k Z e² ÷ r².
Step 2: Bohr: m v r = n h ÷ 2π ⇒ v = n h ÷ (2π m r).
Step 3: Substitute v: m r (n h ÷ 2π m r)² = k Z e² ⇒ r = n² h² ÷ (4π² m k Z e²).
Step 4: Put in the values: r = 52.9 n² ÷ Z pm.

Formulas, units and dimensional formulae

Quantity	Formula	Unit
Radius	rₙ = 52.9 n² ÷ Z	pm
Energy	Eₙ = −13.6 Z² ÷ n²	eV
Speed	vₙ = 2.18 × 10⁶ Z ÷ n	m s⁻¹

CET level (MCQ)

CET 1) Radius of the 3rd orbit of H? 1) 52.9 pm 2) 211.6 pm 3) 476.1 pm 4) 158.7 pm — Answer: 3). 52.9 × 9.
CET 2) Energy of n = 2 in Li²⁺? 1) −3.4 eV 2) −30.6 eV 3) −13.6 eV 4) −54.4 eV — Answer: 2). −13.6 × 9 ÷ 4.

Real-life use: The colours of neon signs and fireworks come from electron jumps between levels.

Remember

r ∝ n² ÷ Z; E ∝ −Z² ÷ n²; larger n means higher (less negative) energy.

Common mistake

Thinking a larger negative number is higher energy; E = 0 at n = ∞ is the highest.

CET trap

Bohr's formulas apply only to one-electron species (H, He⁺, Li²⁺), not neutral He or Li.

Shortcut

Memory clue: "radius goes as n squared; energy as 1 over n squared".

Spot the difference

n	Radius (H)	Energy (H)
1	52.9 pm	−13.6 eV
2	211.6 pm	−3.4 eV
3	476.1 pm	−1.51 eV

Practice questions

Easy

1State Bohr's second postulate.
2Radius of n = 1 in H?
3Ionisation energy of H in eV?

Medium

4Energy of n = 1 in He⁺.
5Energy for n = 1 → 3 in H.
6Radius of n = 2 in Li²⁺.

Challenging

7H ionisation energy in kJ/mol (1 eV = 1.602 × 10⁻¹⁹ J).
8λ for the 3 → 2 transition?

Show answers (try them yourself first)

1m v r = n h ÷ 2π.
252.9 pm.
313.6 eV.
4−54.4 eV.
512.09 eV.
670.5 pm.
7≈ 1312 kJ/mol.
8656.3 nm.

de Broglie relation, uncertainty principle and quantum numbers

From daily life

Shining light to see a tiny particle disturbs it, while a cricket ball is not moved by a torch. So at the tiny scale position and speed cannot both be exact.

Simple definition

de Broglie: a particle has wavelength λ = h ÷ (m v). Heisenberg: Δx · Δp ≥ h ÷ 4π. The quantum numbers n, l, mₗ, mₛ describe the state of an electron.

In the figure : The electron standing wave: 2πr = nλ.

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de Broglie relation, uncertainty principle and quantum numbers

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Rule 1: de Broglie wavelength — λ = h ÷ (m v); the λ of a large object is negligible.

Easy : Electron at v = 10⁶ m/s: λ = 7.27 × 10⁻¹⁰ m.
Medium : A 0.15 kg ball at 40 m/s: λ = 1.1 × 10⁻³⁴ m.
Exam-type : Double v ⇒ half λ.

Rule 2: Uncertainty principle — Δx · m Δv ≥ h ÷ 4π.

Easy : Δx = 10⁻¹⁰ m ⇒ electron Δv ≥ 5.79 × 10⁵ m/s.
Medium : A smaller Δx gives a larger Δv.
Exam-type : For a ball the effect is negligible.

Rule 3: Quantum numbers — n = 1, 2, 3…; l = 0 to n − 1; mₗ = −l to +l; mₛ = ±½.

Easy : n = 3 ⇒ l = 0, 1, 2 (3s, 3p, 3d).
Medium : l = 2 ⇒ mₗ = −2, −1, 0, +1, +2 (5 orbitals).
Exam-type : A shell n has n² orbitals and 2n² electrons: n = 3 ⇒ 9 and 18.

Derivation: Bohr's condition from 2πr = nλ

Step 1: For a standing wave a whole number of waves must fit the circumference: 2πr = nλ.
Step 2: Put λ = h ÷ m v: 2πr = n h ÷ m v.
Step 3: m v r = n h ÷ 2π — Bohr's second postulate.

Formulas, units and dimensional formulae

Quantum number	Symbol	Values	Decides
Principal	n	1, 2, 3…	shell, size, energy
Azimuthal	l	0 … n − 1	subshell, shape
Magnetic	mₗ	−l … +l	orbital orientation
Spin	mₛ	+½, −½	electron spin

CET level (MCQ)

CET 1) Which set is impossible? 1) n = 2, l = 1 2) n = 3, l = 2 3) n = 2, l = 2 4) n = 4, l = 3 — Answer: 3). l ≤ n − 1.
CET 2) Maximum electrons in the n = 4 shell? 1) 16 2) 18 3) 32 4) 8 — Answer: 3). 2 × 16.

Real-life use: An electron microscope uses the wave nature of electrons to see viruses.

Remember

λ = h ÷ mv; Δx·Δp ≥ h ÷ 4π; l ≤ n − 1; a shell holds 2n² electrons.

Common mistake

mₗ has 2l + 1 values, determined by l, not n.

CET trap

For 'impossible set' questions look for l = n or |mₗ| > l. Subshell capacity: s 2, p 6, d 10, f 14.

Shortcut

Memory clue: "n = shell, l = shape, mₗ = direction, mₛ = spin".

Spot the difference

Subshell	l	Orbitals	Electrons
s	0	1	2
p	1	3	6
d	2	5	10
f	3	7	14

Practice questions

Easy

1Whose formula is λ = h ÷ mv?
2Values of l for n = 3?
3Values of mₛ?

Medium

4λ of an electron at 10⁶ m/s.
5λ of a 0.15 kg ball at 40 m/s.
6Orbitals in the n = 3 shell?

Challenging

7Maximum electrons in n = 4?
8Δv for an electron with Δx = 10⁻¹⁰ m?

Show answers (try them yourself first)

1de Broglie's.
20, 1, 2.
3+½ and −½.
47.27 × 10⁻¹⁰ m.
51.1 × 10⁻³⁴ m.
69.
732.
8≥ 5.79 × 10⁵ m/s.

Orbitals, nodes and electronic configuration

From daily life

In a cinema the front rows fill first and people take separate seats before sharing; electrons fill low-energy orbitals first and singly before pairing.

Simple definition

An orbital is the region where an electron is most likely to be found. A node is where the probability is zero. Radial nodes = n − l − 1, angular nodes = l, total = n − 1.

In the figure : s is spherical; p dumbbell; d four-lobed.

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Orbitals, nodes and electronic configuration

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In the figure : Filling order by the n + l rule; Pauli and Hund rules.

Rule 1: Number of nodes — Radial = n − l − 1; angular = l; total = n − 1.

Easy : 2s: radial 1, angular 0, total 1.
Medium : 3p: radial 1, angular 1, total 2.
Exam-type : 4f: radial 0, angular 3, total 3; 4s: radial 3.

Rule 2: The three filling rules — Aufbau: lowest energy first (smaller n + l, if equal smaller n). Pauli: at most 2 e⁻ per orbital with opposite spins. Hund: fill equal-energy orbitals singly first with parallel spin.

Easy : N (Z = 7): 1s² 2s² 2p³ — 3 unpaired e⁻.
Medium : O (Z = 8): 1s² 2s² 2p⁴ — 2 unpaired.
Exam-type : 4s fills before 3d (4 + 0 < 3 + 2 by n + l).

Rule 3: Configurations and exceptions — In Cr and Cu half- or fully-filled d subshells are extra stable. Cations lose electrons first from the outer (4s) shell.

Easy : Fe: [Ar] 3d⁶ 4s² — 4 unpaired e⁻.
Medium : Cr: [Ar] 3d⁵ 4s¹ (exception); Cu: [Ar] 3d¹⁰ 4s¹.
Exam-type : Fe³⁺: [Ar] 3d⁵ — 5 unpaired; Mn²⁺ μ = √35 = 5.92 BM (μ = √(n(n+2))).

Derivation: configuration of Fe³⁺

Step 1: Fe (Z = 26): [Ar] 3d⁶ 4s².
Step 2: The two 4s electrons go first: [Ar] 3d⁶ (Fe²⁺).
Step 3: Then one 3d electron: [Ar] 3d⁵ (Fe³⁺) — half-filled and stable.

Formulas, units and dimensional formulae

Orbital	n	l	Radial nodes	Angular nodes
1s	1	0	0	0
2s	2	0	1	0
2p	2	1	0	1
3s	3	0	2	0
3p	3	1	1	1
3d	3	2	0	2

CET level (MCQ)

CET 1) Total nodes in a 3p orbital? 1) 0 2) 1 3) 2 4) 3 — Answer: 3) 2. n − 1 = 2.
CET 2) Configuration of Cr? 1) 3d⁴4s² 2) 3d⁵4s¹ 3) 3d⁶ 4) 3d⁴4s¹ — Answer: 2).

Real-life use: The colours of lamps and the magnetism of materials depend on electronic configuration.

Remember

Nodes: n − 1 (radial n − l − 1, angular l); 4s before 3d; Cr and Cu are exceptions.

Common mistake

Removing 3d electrons first when forming an ion; 4s electrons go first.

CET trap

Forgetting Hund's rule when counting unpaired electrons; forgetting the Cr and Cu (4s¹) exceptions.

Shortcut

Memory clue: "rows first (n + l), singles on benches (Hund), only two per bench (Pauli)".

Spot the difference

Rule	Says
Aufbau	Lowest energy first
Pauli	Max 2 e⁻ per orbital
Hund	Singly first

Practice questions

Easy

1What is an orbital?
2State Pauli's rule.
3Total nodes in 3p?

Medium

4Write the configuration of Fe.
5Why is Cr's configuration an exception?
6Unpaired electrons in Fe³⁺?

Challenging

7Find the magnetic moment of Mn²⁺.
8Radial nodes in 4s?

Show answers (try them yourself first)

1The region of highest probability of finding the electron.
2At most 2 e⁻ per orbital with opposite spins.
32.
4[Ar] 3d⁶ 4s².
5Half-filled 3d⁵ is extra stable.
65.
7√35 = 5.92 BM.
83.

Quick revision of the whole lesson

Concept	Key formula / fact
Basics	N = A − Z; e⁻ = Z ∓ charge
Wave	c = νλ; E = hν = hc ÷ λ
Rydberg	1 ÷ λ = R_H (1 ÷ n₁² − 1 ÷ n₂²)
Bohr	r = 52.9 n² ÷ Z pm; E = −13.6 Z² ÷ n² eV
de Broglie	λ = h ÷ mv
Uncertainty	Δx·Δp ≥ h ÷ 4π
Nodes	total n − 1; radial n − l − 1
Filling	Aufbau, Pauli, Hund

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🎯 Most important points for the exam

Z, A, N and electrons in ions.
Isotopes/isobars/isotones and average mass.
E = hν and nm → m conversion.
Balmer, Lyman, Paschen and number of lines.
Bohr r and E; one-electron species only.
Quantum number rules and configurations (Cr, Cu exceptions).

Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.