Sub-atomic particles, early atomic models, atomic number and mass number
Link to Std 9–10 : You learned that an atom has electrons, protons and neutrons; now we see how each was discovered and how the picture of the atom changed.
A tube light glows when electricity passes through the gas inside it. Experiments with glowing gas tubes at low pressure were the first window into the atom.
The electron (charge −1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg) was found from cathode rays. The proton (+1.602 × 10⁻¹⁹ C, 1.673 × 10⁻²⁷ kg) was found from canal rays. The neutron (no charge, 1.675 × 10⁻²⁷ kg) was found by Chadwick in 1932. Atomic number Z = number of protons; mass number A = protons + neutrons.
In the figure : A tiny dense nucleus (protons and neutrons) with electrons in the space around it; the nucleus is about 10⁻¹⁵ m across, the atom about 10⁻¹⁰ m.
Rule 1: Cathode rays — Rays from the negative plate of a gas discharge tube are streams of electrons; they travel in straight lines, are bent by electric and magnetic fields towards the positive side, and are the same for every gas. J. J. Thomson measured their charge-to-mass ratio, 1.76 × 10⁸ C/g.
- Easy : Cathode rays are bent towards the positive plate, so they carry negative charge.
- Medium : Millikan's oil drop experiment gave the charge of the electron; every drop carried a whole-number multiple of 1.602 × 10⁻¹⁹ C.
- Exam-type : A drop carrying 4.806 × 10⁻¹⁹ C holds 4.806 ÷ 1.602 = 3 extra electrons.
Rule 2: Models of the atom — Thomson (plum pudding): positive sphere with electrons embedded like seeds in a watermelon. Rutherford (α-scattering): when α-particles hit a thin gold foil, most passed straight, some turned through small angles and a very few bounced back. So most of the atom is empty space and the positive charge and mass sit in a tiny nucleus.
- Easy : Most α-particles passed undeflected, so most of the atom is empty.
- Medium : A few were deflected strongly, so the nucleus is positively charged.
- Exam-type : Very few bounced back, so the nucleus is very small and very dense.
Rule 3: Notation, isotopes and isobars — Write an atom with A on the top left and Z on the bottom left of the symbol, for example ³⁵₁₇Cl; neutrons = A − Z. Isotopes: same Z, different A (¹H, ²H, ³H). Isobars: same A, different Z (⁴⁰Ar, ⁴⁰Ca). Isotones: same number of neutrons (¹⁴C and ¹⁶O have 8 neutrons each).
- Easy : ³⁵Cl has Z = 17, A = 35, so neutrons = 18.
- Medium : Al³⁺ (Z = 13) has 13 − 3 = 10 electrons.
- Exam-type : ³⁷Cl⁻ has 17 protons, 20 neutrons and 18 electrons.
The number of protons decides the element; the number of neutrons only changes the isotope.
Subtracting electrons from A. Neutrons = A − Z, never A − electrons.
Memory clue: "Z = P, A = P + N" (Z is the proton count; A is protons plus neutrons).
Spot the difference
| Point | Isotopes | Isobars |
|---|---|---|
| Same | Atomic number | Mass number |
| Different | Mass number | Atomic number |
| Example | ¹²C, ¹⁴C | ⁴⁰Ar, ⁴⁰Ca |
Practice questions
- 1Name the scientist who discovered the neutron.
- 2How many neutrons are there in ²³Na (Z = 11)?
- 3Which particle is lightest: electron, proton or neutron?
- 4How many electrons are there in the Mg²⁺ ion (Z = 12)?
- 5Which of ¹⁴C, ¹⁴N, ¹²C are isobars?
- 6Which observation of the α-scattering experiment shows that the nucleus is small?
- 7An ion has mass number 56, atomic number 26 and charge +3. Give its protons, neutrons and electrons.
- 8An oil drop has a charge of 8.01 × 10⁻¹⁹ C. How many electron charges does it hold?
Show answers (try them yourself first)
- 1James Chadwick (1932).
- 223 − 11 = 12.
- 3Electron.
- 412 − 2 = 10.
- 5¹⁴C and ¹⁴N (same A = 14).
- 6Very few α-particles bounced back.
- 7Protons 26; neutrons 56 − 26 = 30; electrons 26 − 3 = 23.
- 88.01 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 5.
Electromagnetic radiation, Planck's quantum theory and the photoelectric effect
Link to Std 9–10 : You know light travels as a wave; now we see that light also comes in packets, called photons.
Radio, microwave oven, mobile phone, sunlight and X-ray machine all use electromagnetic radiation. They differ only in wavelength and frequency.
Electromagnetic radiation is a wave of electric and magnetic fields. Wavelength λ is the crest-to-crest distance; frequency ν is the number of waves passing a point per second. In a vacuum c = ν × λ, where c = 3.0 × 10⁸ m/s.
In the figure : The electromagnetic spectrum from radio waves (long λ) to gamma rays (short λ); visible light is a thin band of about 400–750 nm.
Rule 1: Wave relations — c = ν λ and wavenumber ν̄ = 1 ÷ λ (in m⁻¹).
- Easy : λ = 500 nm = 5.00 × 10⁻⁷ m gives ν = 3.0 × 10⁸ ÷ 5.00 × 10⁻⁷ = 6.0 × 10¹⁴ Hz.
- Medium : The same light has wavenumber 1 ÷ 5.00 × 10⁻⁷ = 2.0 × 10⁶ m⁻¹.
- Exam-type : A wave of frequency 1.5 × 10¹⁵ Hz has λ = 3.0 × 10⁸ ÷ 1.5 × 10¹⁵ = 2.0 × 10⁻⁷ m = 200 nm.
Rule 2: Planck's quantum theory — Energy is emitted or absorbed in packets called quanta: E = h ν = h c ÷ λ with h = 6.626 × 10⁻³⁴ J s. This explained black body radiation, which wave theory could not.
- Easy : For 500 nm light: E = 6.626 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J per photon.
- Medium : Shorter wavelength means higher frequency and higher energy; violet photons carry more energy than red photons.
- Exam-type : 1 J of 500 nm light contains 1 ÷ 3.98 × 10⁻¹⁹ = 2.5 × 10¹⁸ photons.
Rule 3: Photoelectric effect — Electrons are emitted from a metal only when the light frequency is at least the threshold frequency ν₀. Kinetic energy of the electron = hν − hν₀ (hν₀ is the work function W₀).
- Easy : Below the threshold frequency no electrons are emitted, however bright the light.
- Medium : A metal with W₀ = 2.5 × 10⁻¹⁹ J has ν₀ = 2.5 × 10⁻¹⁹ ÷ 6.626 × 10⁻³⁴ = 3.77 × 10¹⁴ Hz.
- Exam-type : Light of λ = 4.0 × 10⁻⁷ m has E = 4.97 × 10⁻¹⁹ J, so the electron leaves with KE = 4.97 × 10⁻¹⁹ − 2.5 × 10⁻¹⁹ = 2.47 × 10⁻¹⁹ J.
Brighter light means more photons, not more energy per photon. Energy per photon depends only on frequency.
Writing wavelength in nm or cm directly into c = νλ. Convert to metres first.
Memory clue: "Short wave, high energy" (X-rays are more dangerous than radio waves).
Spot the difference
| Point | Wave nature | Particle nature |
|---|---|---|
| Explains | Interference, diffraction | Black body, photoelectric effect |
| Key quantity | λ, ν | Photon energy hν |
Practice questions
- 1State the relation between c, ν and λ.
- 2Which has more energy per photon, red or blue light?
- 3What is the SI unit of frequency?
- 4Find the frequency of light of wavelength 600 nm.
- 5Find the wavenumber for λ = 400 nm.
- 6Why does bright red light fail to eject electrons from a metal whose threshold lies in the blue?
- 7Find the energy of one photon of wavelength 600 nm (h = 6.626 × 10⁻³⁴ J s).
- 8A metal has W₀ = 3.0 × 10⁻¹⁹ J. Does light of 5.0 × 10¹⁴ Hz eject electrons? If yes, find the KE.
Show answers (try them yourself first)
- 1c = νλ.
- 2Blue light.
- 3Hertz (Hz = s⁻¹).
- 43.0 × 10⁸ ÷ 6.0 × 10⁻⁷ = 5.0 × 10¹⁴ Hz.
- 51 ÷ 4.0 × 10⁻⁷ = 2.5 × 10⁶ m⁻¹.
- 6Each red photon has less energy than the work function; more photons do not add up.
- 7E = 6.626 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.31 × 10⁻¹⁹ J.
- 8E = 6.626 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.31 × 10⁻¹⁹ J, which is more than 3.0 × 10⁻¹⁹ J, so yes; KE = 3.1 × 10⁻²⁰ J.
Hydrogen spectrum and Bohr's model of the atom
Link to Std 9–10 : You saw that a prism splits white light into a continuous spectrum; the light from hydrogen gas gives only a few bright lines.
Sodium street lamps glow yellow and fireworks show different colours for different metals. Each element sends out its own set of colours.
An emission line spectrum has bright lines at definite wavelengths. Bohr proposed that the electron in hydrogen moves only in certain fixed orbits (energy levels n = 1, 2, 3 ...) without radiating, and emits or absorbs a photon only when it jumps between orbits.
In the figure : Energy levels of hydrogen: transitions down to n = 1 make the Lyman series (ultraviolet), down to n = 2 the Balmer series (visible), down to n = 3 the Paschen series (infrared).
Rule 1: Rydberg equation — 1 ÷ λ = R_H (1 ÷ n₁² − 1 ÷ n₂²) where R_H = 1.097 × 10⁷ m⁻¹ and n₂ > n₁.
- Easy : For n₂ = 3, n₁ = 2 (Balmer): 1 ÷ λ = 1.097 × 10⁷ × (1/4 − 1/9), so λ = 656 nm (red).
- Medium : For n₂ = 2, n₁ = 1 (Lyman): λ = 121.5 nm (ultraviolet).
- Exam-type : The number of different lines when an electron falls from level n is n(n − 1) ÷ 2; from n = 4 it is 6.
Rule 2: Bohr's results for hydrogen-like species — Radius rₙ = 52.9 × n² ÷ Z pm and energy Eₙ = −2.18 × 10⁻¹⁸ × Z² ÷ n² J.
- Easy : H, n = 1: r = 52.9 pm; n = 3: r = 52.9 × 9 = 476 pm.
- Medium : H, n = 2: E = −2.18 × 10⁻¹⁸ ÷ 4 = −5.45 × 10⁻¹⁹ J.
- Exam-type : Li²⁺ (Z = 3), n = 1: E = −2.18 × 10⁻¹⁸ × 9 = −1.96 × 10⁻¹⁷ J.
Rule 3: Energy of a transition — ΔE = E(final) − E(initial) = h ν.
- Easy : 3 → 2 in H: ΔE = E₂ − E₃ = −5.45 × 10⁻¹⁹ − (−2.42 × 10⁻¹⁹) = −3.03 × 10⁻¹⁹ J, so 3.03 × 10⁻¹⁹ J is emitted.
- Medium : Ionising H from n = 2 needs 5.45 × 10⁻¹⁹ J.
- Exam-type : Ionisation from n = 1 needs 2.18 × 10⁻¹⁸ J per atom, which is 1312 kJ/mol.
Rule 4: Limitations of Bohr's model — It fails for atoms with more than one electron, cannot explain the splitting of lines in a magnetic field (Zeeman) or an electric field (Stark), and treats the electron as a particle with a known path, which the uncertainty principle forbids.
Hydrogen spectrum and Bohr's model of the atom
In the app you can move this figure with your finger.
Open in the appHydrogen spectrum and Bohr's model of the atom
In the app you can move this figure with your finger.
Open in the appEnergy values are negative because the electron is bound; n = ∞ (free electron) is taken as zero.
Taking n₁ larger than n₂ in the Rydberg equation, which gives a negative wavelength.
Memory clue: "Lyman – Ultraviolet, Balmer – Blue/visible, Paschen – infra", first letters L, B, P in order of n₁ = 1, 2, 3.
Spot the difference
| Point | Lyman series | Balmer series |
|---|---|---|
| Ends at | n = 1 | n = 2 |
| Region | Ultraviolet | Visible |
Practice questions
- 1Which series lies in the visible region?
- 2What is the radius of the first Bohr orbit of hydrogen?
- 3How many lines can arise when an electron falls from n = 3?
- 4Energy of the electron in hydrogen for n = 3.
- 5Compute ΔE for the transition n = 4 → n = 2 using Eₙ = −2.18 × 10⁻¹⁸ ÷ n² J.
- 6State two limitations of Bohr's model.
- 7Find the wavelength of the Balmer line for n₂ = 4 (R_H = 1.097 × 10⁷ m⁻¹).
- 8Radius of the n = 2 orbit of Li²⁺ (Z = 3).
Show answers (try them yourself first)
- 1Balmer series.
- 252.9 pm.
- 33 lines (3→2, 3→1, 2→1).
- 4E₃ = −2.18 × 10⁻¹⁸ ÷ 9 = −2.42 × 10⁻¹⁹ J.
- 5E₂ − E₄ = −2.18 × 10⁻¹⁸ × (1/4 − 1/16) = −4.09 × 10⁻¹⁹ J, so 4.09 × 10⁻¹⁹ J is released.
- 6It fails for multi-electron atoms; it cannot explain Zeeman and Stark effects.
- 71 ÷ λ = 1.097 × 10⁷ × (1/4 − 1/16) = 2.056 × 10⁶ m⁻¹, so λ = 486 nm.
- 8r = 52.9 × 4 ÷ 3 = 70.5 pm.
Dual nature of matter and the uncertainty principle
Link to Std 9–10 : Light behaves as both wave and particle; de Broglie asked whether matter does the same.
A cricket ball flies along a clear path and you can see where it is and how fast it moves. An electron inside an atom does not behave like that ball.
de Broglie proposed that every moving particle has a wavelength λ = h ÷ (m v). The Heisenberg uncertainty principle says that position and momentum of a small particle cannot both be known exactly: Δx × Δp ≥ h ÷ 4π.
In the figure : The wavelength of a cricket ball is far too small to see; that of an electron is comparable to the size of an atom.
Rule 1: de Broglie wavelength — λ = h ÷ (m v) (m = mass in kg, v = speed in m/s, h = 6.626 × 10⁻³⁴ J s).
- Easy : A 0.1 kg ball at 10 m/s: λ = 6.626 × 10⁻³⁴ ÷ 1.0 = 6.6 × 10⁻³⁴ m (not detectable).
- Medium : An electron (9.109 × 10⁻³¹ kg) at 3.0 × 10⁶ m/s: λ = 2.42 × 10⁻¹⁰ m (comparable with atomic size).
- Exam-type : Doubling the speed halves the wavelength.
Rule 2: Uncertainty principle — Δx × Δ(mv) ≥ h ÷ 4π.
- Easy : The more exactly the position is known, the less exactly the speed is known.
- Medium : If Δx = 1 × 10⁻¹⁰ m for an electron, Δv ≥ 6.626 × 10⁻³⁴ ÷ (4π × 9.109 × 10⁻³¹ × 10⁻¹⁰) = 5.8 × 10⁵ m/s.
- Exam-type : This is why Bohr's fixed orbits are replaced by orbitals, regions of high probability.
Dual nature of matter and the uncertainty principle
In the app you can move this figure with your finger.
Open in the appWave nature of matter matters only for tiny particles; for large bodies the wavelength is negligible.
Using mass in grams. Use kg with h in J s.
Memory clue: "λ = h over p" (p = momentum = mv).
Spot the difference
| Point | Bohr orbit | Orbital |
|---|---|---|
| Idea | Fixed circular path | Region of probability |
| Consistent with uncertainty principle? | No | Yes |
Practice questions
- 1Write the de Broglie relation.
- 2Who showed experimentally that electrons show diffraction?
- 3Does a moving truck have a wavelength?
- 4Find λ of a 0.2 kg ball moving at 5.0 m/s.
- 5Why is the uncertainty principle important only for sub-atomic particles?
- 6What happens to λ when speed becomes three times?
- 7Find λ of an electron moving at 6.0 × 10⁶ m/s (m = 9.109 × 10⁻³¹ kg).
- 8Find the minimum Δv of an electron located within Δx = 2 × 10⁻¹⁰ m.
Show answers (try them yourself first)
- 1λ = h ÷ (m v).
- 2Davisson and Germer (1927).
- 3Yes, but it is far too small to detect.
- 46.626 × 10⁻³⁴ ÷ (0.2 × 5.0) = 6.6 × 10⁻³⁴ m.
- 5For large bodies Δx and Δp are negligible compared with their values.
- 6It becomes one-third.
- 76.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 6.0 × 10⁶) = 1.21 × 10⁻¹⁰ m.
- 8Half of the earlier 5.8 × 10⁵: 2.9 × 10⁵ m/s.
Quantum mechanical model, quantum numbers and orbitals
Link to Std 9–10 : You drew electrons in shells; in Std 11 the shells are broken into sub-shells and orbitals.
Think of a stadium. The country is the atom, the stand is the shell, the row is the sub-shell, and the seat is the orbital. A seat holds at most 2 spectators (electrons).
In the quantum mechanical model, an electron is described by a wave function ψ obtained from the Schrödinger equation. An orbital is the region around the nucleus where the probability of finding the electron is high; it is described by the quantum numbers n, l and mₗ, and the electron's spin by mₛ.
In the figure : Shapes: s is a sphere, p is a dumbbell along x, y or z, d has four lobes.
Rule 1: Quantum numbers — n (principal) = 1, 2, 3 ...; l (azimuthal) = 0 to n − 1 (s, p, d, f for l = 0, 1, 2, 3); mₗ (magnetic) = −l to +l; mₛ (spin) = +½ or −½.
- Easy : n = 2 allows l = 0 (2s) and l = 1 (2p).
- Medium : For l = 2 (d), mₗ = −2, −1, 0, +1, +2, so 5 d orbitals.
- Exam-type : n = 4 has 4 sub-shells (4s, 4p, 4d, 4f) with 16 orbitals and a capacity of 32 electrons.
Rule 2: Counting — Orbitals in a shell = n²; electrons in a shell = 2n²; orbitals in a sub-shell = 2l + 1.
- Easy : n = 3: 9 orbitals, 18 electrons.
- Medium : 4f: 7 orbitals, 14 electrons.
- Exam-type : A set (n = 2, l = 2) is not allowed, since l must be less than n.
Rule 3: Nodes — Radial nodes = n − l − 1; angular nodes = l; total nodes = n − 1.
- Easy : 1s has 0 nodes.
- Medium : 2p has 0 radial and 1 angular node.
- Exam-type : 3p has 1 radial and 1 angular node, total 2.
Rule 4: Energy of orbitals — In a hydrogen atom, energy depends only on n. In a multi-electron atom, a lower value of (n + l) means lower energy, and for equal (n + l) the lower n has lower energy.
- Easy : 2s < 2p because l is smaller.
- Medium : 4s (n + l = 4) is lower than 3d (n + l = 5).
- Exam-type : 4p and 5s both have n + l = 5; 4p has the lower n, so 4p lies below 5s.
An orbital can have at most 2 electrons, with opposite spins.
Believing that 3d is filled before 4s. 4s fills first, and in cations the 4s electrons are also lost first.
Memory clue: "s-p-d-f = sharp, principal, diffuse, fundamental" with 1, 3, 5, 7 orbitals.
Spot the difference
| Point | Orbit (Bohr) | Orbital (quantum model) |
|---|---|---|
| Meaning | Definite path | Probability region |
| Number | n per shell | n² per shell |
Practice questions
- 1Give the value of l for a p sub-shell.
- 2How many orbitals are in the 3p sub-shell?
- 3What is the maximum number of electrons in the L shell (n = 2)?
- 4Give the possible values of mₗ for l = 1.
- 5Which sub-shell has lower energy, 3d or 4s?
- 6Why is (n = 1, l = 1) not possible?
- 7How many radial and angular nodes does a 4d orbital have?
- 8Write one valid set of four quantum numbers for the last electron of an atom with configuration 2p¹.
Show answers (try them yourself first)
- 1l = 1.
- 23 orbitals.
- 32 × 2² = 8 electrons.
- 4−1, 0, +1.
- 54s (n + l = 4 against 5).
- 6l can run only from 0 to n − 1 = 0, so l = 1 is not allowed when n = 1.
- 7Radial = 4 − 2 − 1 = 1; angular = 2.
- 8n = 2, l = 1, mₗ = −1, mₛ = +½ (other choices of mₗ and mₛ are also correct).
Electronic configuration: Aufbau principle, Pauli's exclusion principle and Hund's rule
Link to Std 9–10 : You wrote electrons as 2, 8, 8; now we write the exact sub-shell arrangement, such as 1s² 2s² 2p⁶.
Passengers in a bus first take empty seats one by one; only when every row has one passenger do they sit together. Electrons in equal-energy orbitals behave in the same way.
The Aufbau principle says orbitals are filled in order of increasing energy. Pauli's exclusion principle: no two electrons in an atom have the same four quantum numbers (so an orbital holds at most 2 electrons of opposite spin). Hund's rule: electrons occupy degenerate orbitals singly with parallel spins before pairing.
In the figure : Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Rule 1: Writing configurations — Fill orbitals in the above order and write electrons as superscripts.
- Easy : N (Z = 7): 1s² 2s² 2p³, three unpaired electrons.
- Medium : O (Z = 8): 1s² 2s² 2p⁴, two unpaired electrons.
- Exam-type : Fe (Z = 26): [Ar] 3d⁶ 4s², four unpaired electrons.
Rule 2: Cations and anions — For anions add electrons by the same order. For cations remove electrons from the outermost shell first (4s before 3d).
- Easy : Mg²⁺: 1s² 2s² 2p⁶.
- Medium : Fe³⁺: [Ar] 3d⁵ (the two 4s electrons and one 3d electron are lost).
- Exam-type : Cl⁻ has the same configuration as Ar: 1s² 2s² 2p⁶ 3s² 3p⁶.
Rule 3: Extra stability of half-filled and fully filled sub-shells — Cr is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) and Cu is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²), because symmetry and exchange energy make d⁵ and d¹⁰ extra stable.
- Easy : Cr has 6 unpaired electrons (five 3d and one 4s).
- Medium : Cu has 1 unpaired electron.
- Exam-type : Mn²⁺ is [Ar] 3d⁵ and has 5 unpaired electrons.
Electronic configuration: Aufbau principle, Pauli's exclusion principle and Hund's rule
In the app you can move this figure with your finger.
Open in the appUnpaired electrons make an atom or ion paramagnetic; none unpaired means diamagnetic.
Removing 3d electrons before 4s while making a transition metal cation.
Memory clue: "Cr and Cu cheat", they borrow from 4s to make 3d half or fully filled.
Spot the difference
| Point | Pauli's principle | Hund's rule |
|---|---|---|
| Concerns | Electrons in one orbital | Electrons in several equal orbitals |
| Statement | Opposite spins in an orbital | Fill singly first |
Practice questions
- 1Write the configuration of carbon (Z = 6).
- 2How many unpaired electrons does phosphorus (Z = 15) have?
- 3State Pauli's exclusion principle.
- 4Write the configuration of Ca²⁺ (Z = 20).
- 5How many unpaired electrons are in Zn (Z = 30)?
- 6Why is Cu [Ar] 3d¹⁰ 4s¹ rather than [Ar] 3d⁹ 4s²?
- 7Write the configuration of Ni (Z = 28) and Ni²⁺, and state the unpaired electrons in Ni²⁺.
- 8Which of Sc³⁺, Ti⁴⁺, V⁵⁺ (Z = 21, 22, 23) are isoelectronic with Ar?
Show answers (try them yourself first)
- 11s² 2s² 2p².
- 2Three (3p³).
- 3No two electrons in an atom can have the same set of four quantum numbers.
- 41s² 2s² 2p⁶ 3s² 3p⁶.
- 5None: Zn is [Ar] 3d¹⁰ 4s².
- 6A fully filled d¹⁰ sub-shell is extra stable.
- 7Ni: [Ar] 3d⁸ 4s²; Ni²⁺: [Ar] 3d⁸, with 2 unpaired electrons.
- 8All three have 18 electrons, so all are isoelectronic with Ar.