Electronic Configurations

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The Electromagnetic Spectrum

The electromagnetic spectrum covers all electromagnetic radiation, divided into bands or regions by frequency, wavelength and energy.
Gamma rays, X-rays and UV are high frequency, high energy and can damage cells, making them hazardous to health.
All electromagnetic waves travel at the same speed in a vacuum: the speed of light, c = 3.00 × 10⁸ m s⁻¹.
Frequency and wavelength are inversely related: higher frequency means shorter wavelength, lower frequency means longer wavelength.
The key relationship is c = fλ, so frequency can be calculated from wavelength, or vice versa.
A continuous spectrum contains all colours of visible light, blending smoothly with no gaps, produced when white light is refracted through a prism or water droplets.
A line spectrum only shows certain fixed frequencies, telling us that emitted light from atoms is quantised (quanta means 'little packet').
Electrons can only possess certain amounts of energy — they cannot have any energy value.

Emission Spectra

Electrons move in energy levels around the nucleus; when they absorb energy they move to higher levels, and they release energy when returning to lower levels.
The frequency of the emitted energy matches the frequency of the energy absorbed.
If emitted energy is in the visible region, passing it through a diffraction grating produces a line emission spectrum.
Each line is a specific energy value, suggesting electrons can only possess a limited choice of allowed energies.
In the hydrogen spectrum, lines get closer together towards the blue end — this is called convergence, converging towards the higher energy end.
The maximum energy corresponds to the ionisation energy of the electron.
The visible lines were first observed by Johannes Balmer and correspond to electron jumps from higher levels down to n = 2.
The line emission spectrum of hydrogen provided evidence of energy levels; Niels Bohr applied Quantum Theory in 1913, proposing electrons exist only in fixed energy levels.

Electron Jumps in the Hydrogen Spectrum

Transitions from n = ∞ to n = 3 are in the infrared region and have low energy.
Transitions from n = ∞ to n = 2 are in the visible region and have medium energy.
Transitions from n = ∞ to n = 1 are in the ultraviolet region and have high energy.
Emission in the visible region occurs for an electron jumping from any higher energy level to n = 2.
The full hydrogen spectrum shows sets or families of lines corresponding to electrons jumping from higher levels to lower levels.

Energy Levels, Sublevels & Orbitals

Electrons are arranged around the nucleus in principal energy levels or principal quantum shells, numbered by principal quantum numbers (n).
The lower the principal quantum number, the closer the shell is to the nucleus; the higher the n, the greater the energy of the electron in that shell.
Each principal quantum number holds a fixed maximum number of electrons: n = 1 up to 2, n = 2 up to 8, n = 3 up to 18, n = 4 up to 32.
The mathematical relationship is 2n²; for example, in the third shell n = 3 and the number of electrons is 2 × (3²) = 18.
Principal quantum shells are split into subshells given the letters s, p and d; elements with more than 57 electrons also have an f subshell.
The energy of electrons in subshells increases in the order s < p < d < f.
The order of subshells overlaps for higher principal quantum shells.

Orbitals and Their Shapes

Subshells contain one or more atomic orbitals; orbitals exist at specific energy levels and electrons can only be found at these levels, not in between.
Each atomic orbital can be occupied by a maximum of two electrons.
The s orbitals are spherical in shape, and their size increases with increasing shell number (e.g. the 3s orbital is bigger than the 1s orbital).
The p orbitals are dumbbell-shaped; every shell has three p orbitals except the first one (n = 1).
The p orbitals occupy the x, y and z axes and point at right angles to each other, so are oriented perpendicular to one another.
The lobes of the p orbitals become larger and longer with increasing shell number.
The shape of the d orbitals is not required for IB Chemistry.

Subshells and Orbitals: Numbers and Energies

Each subshell holds a specific number of orbitals: s has 1, p has 3 (labelled px, py and pz), d has 5, and f has 7.
Each orbital holds a maximum of 2 electrons, so the maximum electrons per subshell are: s = 2, p = 6, d = 10, f = 14.
In the ground state, orbitals in the same subshell have the same energy and are said to be degenerate (e.g. the px and py orbitals have the same energy).
The ground state is the most stable electronic configuration of an atom, with the lowest amount of energy.
This is achieved by filling the subshells of lowest energy first (1s) — this is called the Aufbau Principle.
The subshells increase in energy as s < p < d < f, but the 3d orbital has slightly higher energy than the 4s orbital, so 4s is filled before 3d.
The order of subshells in terms of increasing energy does not follow a regular pattern at n = 3 and higher.

Writing Electron Configurations

The electron configuration gives information about the number of electrons in each shell, subshell and orbital of an atom; subshells are filled in order of increasing energy.
Electrons can be imagined as small spinning charges rotating clockwise or anticlockwise around their own axis.
Electrons with the same spin repel each other — this is spin–pair repulsion.
Hund's rule: electrons fill degenerate orbitals singly first, with parallel spins, to minimise repulsion; they pair only after all orbitals are singly occupied, with opposite spins.
In a p subshell, electrons fill px, py and pz orbitals singly before pairing.
Pauli Exclusion principle: an orbital holds two electrons with opposite spins only; electrons pair only when no empty orbital of the same energy is available.
Pairing costs less energy than jumping to a higher orbital.
There are two formats: full configuration lists all electrons from 1s onward; shorthand configuration uses the symbol of the nearest noble gas in brackets (e.g. [Ar]) to represent inner electrons.

Electron shells and the Periodic Table

Electron Configurations of Ions and the Periodic Table Blocks

Anions (negative) form by adding electrons to the outer shell; cations (positive) form by removing electrons from the outer shell.
Transition metals fill the 4s before 3d when neutral, but lose electrons from 4s first, not 3d, when forming ions.
In the Periodic Table, elements are grouped into blocks based on their valence subshell: s-block (valence electrons in an s orbital), p-block (p orbital), d-block (d orbital), f-block (f orbital).
Example — Fe (atomic number 26): full configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶; shorthand [Ar] 4s² 3d⁶.
Example — Fe²⁺ has 24 electrons; electrons are removed from the 4s orbital before the 3d: full configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶; shorthand [Ar] 3d⁶.
Exceptions to the Aufbau principle: Cr is [Ar] 3d⁵ 4s¹ (not [Ar] 3d⁴ 4s²) and Cu is [Ar] 3d¹⁰ 4s¹ (not [Ar] 3d⁹ 4s²), because promoting an electron from 4s to 3d achieves a half full or full d-subshell, which is energetically favourable.

Ionisation Energy from an Emission Spectrum (HL)

Energy (e.g. heat or electricity) can excite electrons to higher levels (n = 2 to ∞); when electrons return to lower levels, they emit energy as electromagnetic radiation.
A fall to n = 1 releases ultraviolet radiation — this is the Lyman series.
The amount of energy released corresponds to the difference between the energy levels.
The line spectrum supports Bohr's model, which proposes electrons occupy discrete energy levels; electrons require a specific amount of energy to move between levels, like steps on a ladder.
Limitations of Bohr's model: assumes electrons have fixed positions, assumes all energy levels are spherical, and only accurately explains the hydrogen spectrum — it fails for atoms with more than one electron.
As spectral lines are produced, they become closer together at higher energy levels; the point where lines appear to merge is the limit of convergence.
The limit of convergence corresponds to the energy required to remove an electron from the atom — the first ionisation energy.
In the Lyman series, the convergence limit represents an electron falling from n = ∞ to n = 1; for hydrogen this corresponds to a wavelength of 91.16 nm (91.16 × 10⁻⁹ m).

Calculating First Ionisation Energy from the Convergence Limit

The first ionisation energy (IE₁) of an atom can be calculated using the frequency (or wavelength) of the convergence limit.
Use the equations ΔE = hf and c = fλ; rearrange c = fλ to f = c ÷ λ if frequency is not given.
Constants: h = Planck's constant (6.63 × 10⁻³⁴ J s), c = speed of light (3.00 × 10⁸ m s⁻¹), f = frequency (s⁻¹), λ = wavelength (m).
Step 1: calculate the frequency of the convergence limit, converting wavelength into m (nm to m = × 10⁻⁹).
Step 2: substitute into ΔE = hf to give the energy change per atom in J atom⁻¹.
Step 3: calculate the energy per mole by multiplying by Avogadro's constant (NA = 6.02 × 10²³).
Step 4: convert J mol⁻¹ to kJ mol⁻¹ by dividing by 1000.
Example — sodium: frequency 1.24 × 10¹⁵ s⁻¹ gives IE₁ = 8.22 × 10⁻¹⁹ J atom⁻¹, then × 6.02 × 10²³ = 494 916 J mol⁻¹ = 495 kJ mol⁻¹.

Successive Ionisation Energies (HL)

Successive ionisation energies of an element increase, because once the outer electron is removed the atom becomes a positive ion, making further electron removal more difficult.
As more electrons are removed: shielding decreases, the proton-to-electron ratio increases, and attraction between the nucleus and remaining electrons increases.
The increase is not constant — it depends on the electronic configuration.
Large jumps in ionisation energy show a change in shell, while smaller jumps show changes within a subshell.
Successive ionisation energy data can predict or confirm the electronic configuration, identify the number of outer electrons, and determine the group number in the Periodic Table.
Example — calcium: IE₁ = 590, IE₂ = 1150, IE₃ = 4940, IE₄ = 6480 kJ mol⁻¹; the first electron is easy to remove due to spin-pair repulsion in the 4s orbital, the second is harder (no spin-pair repulsion), and the third is much harder as it comes from the 3p subshell, which is closer to the nucleus.
Example — sodium: a large increase between the first and second ionisation energies shows the first electron removed is from the valence shell, so Na is in Group 1; the jump corresponds to removing an electron from the 3s to the full 2p subshell.
Example — aluminium: a large increase between the third and fourth ionisation energies shows the first three electrons (3p and 3s) are easier to remove, so Al belongs to Group 13 (Group III).

Ionisation Energy Trends and Periodicity

Ionisation energy shows periodicity — a repeating trend across a period in the Periodic Table.
Group 1 metals have low first ionisation energies; noble gases have very high values due to stable configurations.
Four key factors affect first ionisation energy: nuclear charge (more protons = stronger attraction), distance of outer electrons from the nucleus, shielding by inner electrons, and spin–pair repulsion between paired electrons.
Across a period, first ionisation energy increases because nuclear charge increases (pulling electrons closer), atomic radius decreases, and shielding stays roughly constant since electrons are added to the same shell.
Down a group, first ionisation energy decreases due to greater atomic radius and more shielding.
There is a slight decrease in IE between beryllium and boron because boron's fifth electron is in the 2p subshell, which is further from the nucleus than beryllium's 2s subshell (Be: 1s² 2s², IE₁ = 900 kJ mol⁻¹; B: 1s² 2s² 2p¹, IE₁ = 800 kJ mol⁻¹).
There is a slight decrease in IE between nitrogen and oxygen due to spin-pair repulsion in the 2p orbital of oxygen (N: 1s² 2s² 2px¹ 2py¹ 2pz¹, IE₁ = 1400 kJ mol⁻¹; O: 1s² 2s² 2px² 2py¹ 2pz¹, IE₁ = 1310 kJ mol⁻¹).
There is a large decrease in ionisation energy between the last element in one period and the first element in the next period, because of increased distance between the nucleus and outer electrons (a new shell is added) and increased shielding by inner electrons; these outweigh the increased nuclear charge.

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1.Which region of the electromagnetic spectrum has the highest frequency?
Easy
- ARadio waves
- BInfrared
- CVisible light
- DGamma rays
2.What is the speed of all electromagnetic waves in a vacuum?
Easy
- A3.00 × 10⁸ m s^-1
- B3.00 × 10⁶ m s^-1
- C6.63 × 10^-34 m s^-1
- D1.00 × 10⁸ m s^-1
3.A continuous spectrum contains all frequencies of visible light, whereas a line spectrum contains only certain specific frequencies.
Easy
True or false?
4.Which of the following correctly describes the relationship between frequency and wavelength for electromagnetic radiation?
Easy
- AThey are directly proportional
- BThey are inversely proportional
- CThey are independent of each other
- DThey are equal in magnitude
5.What is the maximum number of electrons that can occupy the n = 3 energy level?
Easy
- A8
- B18
- C32
- D2
6.Which of the following gives the correct order of subshell energies from lowest to highest?
Easy
- As < p < d < f
- Bs < d < p < f
- Cf < d < p < s
- Dp < s < d < f
7.Which of the following statements about atomic orbitals are correct? (select all that apply)
Medium
- AAn s orbital is spherical in shape.
- BA p orbital is dumbbell-shaped.
- CEach orbital can hold a maximum of two electrons.
- DThe d orbitals are spherical.
- EAll p orbitals in the same subshell have different energies.
8.Match each subshell with the number of orbitals it contains.
Medium
- s
- p
- d
- f
- 1
- 3
- 5
- 7

Soal lainnya

9.Which electron transition in the hydrogen atom would emit a photon in the ultraviolet region?
Medium
- An = 3 to n = 2
- Bn = 2 to n = 1
- Cn = 4 to n = 3
- Dn = 5 to n = 4
10.The successive ionisation energies of an element are: 590, 1150, 4940, 6480 kJ mol^-1. Which group is this element in?
Medium
- AGroup 1
- BGroup 2
- CGroup 3
- DGroup 4
11.Which of the following statements about the electromagnetic spectrum are true? (select all that apply)
Medium
- AGamma rays have higher energy than X-rays.
- BRadio waves have shorter wavelengths than visible light.
- CAll electromagnetic waves travel at the same speed in a vacuum.
- DEnergy is directly proportional to frequency.
- EWavelength increases as frequency increases.
12.The 4s orbital is filled before the 3d orbital because 4s has lower energy than 3d.
Easy
True or false?
13.What is the shape of a p orbital?
Easy
- ASpherical
- BDumbbell-shaped
- CTetrahedral
- DPlanar
14.Place the following regions of the electromagnetic spectrum in order of increasing wavelength.
Medium
- Gamma rays
- Ultraviolet
- Visible light
- Infrared
- Radio waves
15.Match each region of the electromagnetic spectrum with the correct description of its energy.
Medium
- Gamma rays
- Visible light
- Radio waves
- High energy
- Medium energy
- Low energy
16.Which of the following statements about a continuous spectrum and a line spectrum are correct? (select all that apply)
Medium
- AA continuous spectrum contains all frequencies of visible light.
- BA line spectrum contains only certain specific frequencies.
- CA line spectrum is produced when white light passes through a prism.
- DA continuous spectrum shows that energy levels in atoms are quantised.
- EA line spectrum provides evidence for discrete energy levels in atoms.
17.Place the following electron transitions in a hydrogen atom in order of increasing energy of the emitted photon.
Medium
- n = 4 to n = 3
- n = 3 to n = 2
- n = 2 to n = 1
18.Which statement correctly describes the relationship between the frequency and energy of electromagnetic radiation?
Medium
- AHigher frequency radiation has higher energy.
- BHigher frequency radiation has lower energy.
- CEnergy is independent of frequency.
- DEnergy is inversely proportional to frequency.
19.Which of the following correctly lists the regions of the electromagnetic spectrum in order of increasing wavelength?
Medium
- AGamma rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves
- BRadio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays
- CGamma rays, ultraviolet, X-rays, visible, infrared, microwaves, radio waves
- DRadio waves, infrared, microwaves, visible, ultraviolet, X-rays, gamma rays
20.Place the following subshells in order of increasing energy according to the Aufbau principle.
Medium
- 1s
- 2s
- 2p
- 3s
- 3p
- 4s
- 3d
21.Which of the following is the correct full electron configuration for a chromium atom in its ground state?
Medium
- A1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
- B1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
- C1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s⁰
- D1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s³
22.Which statements about Hund's rule and the Pauli exclusion principle are correct? (select all that apply)
Medium
- AElectrons fill degenerate orbitals singly first, with parallel spins.
- BElectrons pair only after all orbitals in a subshell are singly occupied.
- CAn orbital can hold two electrons with opposite spins.
- DElectrons with the same spin attract each other.
- EPairing electrons in the same orbital costs less energy than jumping to a higher orbital.
23.The first six successive ionisation energies of an element are: 577, 1820, 2740, 11600, 14800, 18400 kJ mol⁻¹. Which group is this element in?
Medium
- AGroup 1
- BGroup 2
- CGroup 13
- DGroup 14
24.The convergence limit for a hydrogen atom has a frequency of 3.29 × 10¹⁵ Hz. Calculate the first ionisation energy of hydrogen in kJ mol⁻¹. Use h = 6.63 × 10⁻³⁴ J s and N_A = 6.02 × 10²³ mol⁻¹.
Hard
- A1310 kJ mol⁻¹
- B2.18 × 10⁻¹⁸ kJ mol⁻¹
- C1313 kJ mol⁻¹
- D1.31 × 10⁶ kJ mol⁻¹
25.Which of the following statements about successive ionisation energies are correct? (select all that apply)
Hard
- ASuccessive ionisation energies generally increase because the ion becomes more positively charged.
- BA large jump in ionisation energy indicates removal of an electron from an inner shell.
- CSmaller jumps in ionisation energy indicate changes within the same subshell.
- DSuccessive ionisation energies decrease as more electrons are removed.
- EThe number of valence electrons can be deduced from the position of the largest jump.
26.Which equation correctly represents the second ionisation energy of aluminium?
Medium
- AAl(g) → Al⁺(g) + e⁻
- BAl⁺(g) → Al²⁺(g) + e⁻
- CAl(g) → Al²⁺(g) + 2e⁻
- DAl²⁺(g) → Al³⁺(g) + e⁻
27.Which factors affect the first ionisation energy of an element? (select all that apply)
Medium
- ANuclear charge
- BDistance of outer electrons from the nucleus
- CShielding by inner electrons
- DSpin-pair repulsion between paired electrons
- EThe mass number of the element
28.Place the following elements in order of increasing first ionisation energy.
Medium
- Na
- Mg
- Al
- Si
29.Which of the following is the correct condensed electron configuration for a chromium(III) ion, Cr³⁺?
Medium
- A[Ar] 3d³
- B[Ar] 3d⁵ 4s¹
- C[Ar] 3d⁴ 4s²
- D[Ar] 3d⁶
30.State what group of the Periodic Table element Y belongs to.
Medium
- AGroup 1
- BGroup 2
- CGroup 13
- DGroup 3
31.Which element in Period 2 has the highest first ionisation energy?
Medium
- ALithium
- BBeryllium
- CNeon
- DFluorine
32.Which equation correctly represents the third ionisation energy of beryllium?
Medium
- ABe(g) → Be⁺(g) + e⁻
- BBe⁺(g) → Be²⁺(g) + e⁻
- CBe²⁺(g) → Be³⁺(g) + e⁻
- DBe(g) → Be³⁺(g) + 3e⁻
33.The successive ionisation energies of element X are shown below. How many shells does element X have?
Medium
- A1
- B2
- C3
- D4
34.The successive ionisation energies of element X are shown below. Which group is element X in?
Medium
- AGroup 1
- BGroup 2
- CGroup 13
- DGroup 14
35.What is the general trend in first ionisation energies across Period 3?
Medium
- AThey increase from left to right.
- BThey decrease from left to right.
- CThey remain constant.
- DThey increase then decrease.
36.Which of the following explains why the first ionisation energy of aluminium is lower than that of magnesium?
Medium
- AAluminium has a smaller atomic radius.
- BAluminium has more shielding.
- CAluminium's outer electron is in a higher energy 3p subshell.
- DAluminium has a lower nuclear charge.
37.Which of the following explains why the first ionisation energy of sulfur is lower than that of phosphorus?
Hard
- ASulfur has a larger atomic radius.
- BSulfur has more shielding.
- CSulfur has spin-pair repulsion in its 3p orbital.
- DSulfur has a lower nuclear charge.
38.The successive ionisation energies of vanadium are shown. From which sub-levels are the first four electrons lost?
Medium
- A4s, 4s, 3d, 3d
- B3d, 3d, 3d, 3d
- C4s, 3d, 3d, 3d
- D3d, 4s, 4s, 4s
39.Why is there a large increase in ionisation energy between the fifth and sixth electrons removed from vanadium?
Hard
- AThe sixth electron is removed from a 4s orbital.
- BThe sixth electron is removed from a 3d orbital.
- CThe sixth electron is removed from a 3p orbital.
- DThe sixth electron is removed from a 4p orbital.
40.The first six ionisation energies of an element are: 578, 1816, 2744, 11576, 14829, 18375 kJ mol⁻¹. What causes the large increase between the third and fourth ionisation energies?
Medium
- ARemoving an electron from a 3p orbital.
- BRemoving an electron from a 3s orbital.
- CRemoving an electron from a 2p orbital.
- DRemoving an electron from a 2s orbital.
41.Complete the following sentences about electron configurations.
Easy
The ____ principle states that electrons fill the lowest energy orbitals first. ____ rule states that electrons fill degenerate orbitals singly before pairing. The ____ exclusion principle states that an orbital can hold a maximum of two electrons with opposite spins.
42.The frequency of the convergence limit for a hydrogen atom is 3.29 × 10¹⁵ Hz. Calculate the first ionisation energy of hydrogen in kJ mol^-1. Use h = 6.63 × 10^-34 J s and NA = 6.02 × 10²³ mol^-1. Give your answer to 3 significant figures.
Medium
43.Complete the electron configuration for a chromium atom in its ground state.
Medium
The full electron configuration of chromium is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d^____ 4s^____.
44.Calculate the wavelength, in nm, of electromagnetic radiation with a frequency of 5.00 × 10¹⁴ Hz. Use c = 3.00 × 10⁸ m s^-1. Give your answer to 3 significant figures.
Medium
45.Complete the sentence about the electromagnetic spectrum.
Easy
All electromagnetic waves travel at the same speed in a vacuum, which is the speed of ____.
46.Calculate the frequency, in Hz, of electromagnetic radiation with a wavelength of 5.00 × 10⁻⁷ m. Use c = 3.00 × 10⁸ m s⁻¹. Give your answer in scientific notation to 3 significant figures.
Medium
47.Complete the statement about the ionisation energy data.
Medium
The large increase between the ____ and ____ ionisation energies indicates a change in shell.
48.Calculate the ionisation energy, in J, for a single atom of hydrogen.
Medium
49.Calculate the wavelength, in m, for the electron transition corresponding to this frequency. Use c = 3.00 × 10⁸ m s⁻¹.
Medium
50.Explain why the first ionisation energy of aluminium is lower than that of magnesium.
Medium
51.Explain how the successive ionisation energies of an element can be used to determine its group in the Periodic Table.
Hard
52.The frequency of the convergence limit for a hydrogen atom is 3.29 × 10¹⁵ Hz. Calculate the first ionisation energy of hydrogen in kJ mol^-1. Use h = 6.63 × 10^-34 J s and NA = 6.02 × 10²³ mol^-1. Give your answer to 3 significant figures.
Medium
- A1182 kJ mol^-1
- B1313 kJ mol^-1
- C1575 kJ mol^-1
- D1444 kJ mol^-1
53.Calculate the wavelength, in nm, of electromagnetic radiation with a frequency of 5.00 × 10¹⁴ Hz. Use c = 3.00 × 10⁸ m s^-1. Give your answer to 3 significant figures.
Medium
- A660 nm
- B540 nm
- C720 nm
- D600 nm
54.Calculate the frequency, in Hz, of electromagnetic radiation with a wavelength of 5.00 × 10⁻⁷ m. Use c = 3.00 × 10⁸ m s⁻¹. Give your answer in scientific notation to 3 significant figures.
Medium
- A600000000000000 Hz
- B660000000000000 Hz
- C540000000000000 Hz
- D720000000000000 Hz
55.Calculate the ionisation energy, in J, for a single atom of hydrogen.
Medium
- A2 J
- B-1 J
- C2.19e-18 J
- D1 J
56.Calculate the wavelength, in m, for the electron transition corresponding to this frequency. Use c = 3.00 × 10⁸ m s⁻¹.
Medium
- A9.09e-8 m
- B2 m
- C1 m
- D-1 m

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