IB Chemistry Quiz: Apply Electron Configurations
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Apply Electron ConfigurationsQuestion 1 of 20

The first four successive ionization energies of an element are 738, 1451, 7733, and 10540 kJ mol⁻¹. To which group in the periodic table does this element belong?

Group 1
Group 2
Group 13
Group 14
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Electron Configurations

Practice Apply Electron Configurations in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Electron Configurations, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The first four successive ionization energies of an element are 738, 1451, 7733, and 10540 kJ mol⁻¹. To which group in the periodic table does this element belong?

  1. Group 1
  2. Group 2 (correct answer)
  3. Group 13
  4. Group 14
Explanation: A large increase in successive ionization energies indicates the removal of an electron from a stable, core electron shell. The values show a relatively small increase from the first to the second ionization energy (738 to 1451), but a very large jump to the third ionization energy (1451 to 7733). This indicates that the first two electrons are valence electrons, and the third electron is being removed from a stable inner shell. An element with two valence electrons belongs to Group 2.

Question 2

How many unpaired electrons are present in a ground-state gaseous Fe²⁺ ion?

  1. 6
  2. 5
  3. 4 (correct answer)
  4. 2
Explanation: First, write the configuration for a neutral iron atom (Fe, Z=26): [Ar] 4s²3d⁶. To form the Fe²⁺ ion, two electrons are removed from the orbital with the highest principal quantum number, which is the 4s orbital. The resulting configuration for Fe²⁺ is [Ar] 3d⁶. Now, distribute the six 3d electrons into the five d-orbitals according to Hund's rule: five electrons will occupy the five orbitals singly with parallel spins, and the sixth electron will pair up in one of the orbitals. This leaves four unpaired electrons.

Question 3

An ion with a charge of 2- has the ground-state electron configuration 1s²2s²2p⁶3s²3p⁶. What is the identity of the element?

  1. Argon (Ar)
  2. Sulfur (S) (correct answer)
  3. Calcium (Ca)
  4. Scandium (Sc)
Explanation: The ion has a total of 18 electrons (2+2+6+2+6). Since the ion has a 2- charge, it has gained two electrons. Therefore, the neutral atom must have had 18 - 2 = 16 electrons. The element with atomic number 16 is Sulfur (S). Argon is a noble gas with 18 electrons. Calcium has 20 electrons and forms a Ca²⁺ ion with 18 electrons. Scandium has 21 electrons.

Question 4

The lines in the visible region of the hydrogen emission spectrum (the Balmer series) are produced by electrons transitioning to the n=2 energy level. Which transition produces the line with the shortest wavelength?

  1. n=3 to n=2
  2. n=4 to n=2
  3. n=5 to n=2
  4. n=∞ to n=2 (correct answer)
Explanation: Wavelength is inversely proportional to energy (E = hc/λ). The shortest wavelength corresponds to the highest energy transition. The energy levels of the hydrogen atom converge at higher n values. Therefore, the largest possible energy drop into the n=2 level will be from the highest possible energy level, which is represented by n=∞ (the convergence limit, corresponding to ionization). This transition from n=∞ to n=2 releases the most energy among all possible transitions into n=2, thus producing the photon with the shortest wavelength.

Question 5

Element Y has the electron configuration [Kr]5s24d105p4[Kr] 5s^2 4d^{10} 5p^4. When Y forms a compound with oxygen, it typically exhibits a +6 oxidation state. What is the electron configuration of Y in this oxidation state?

  1. [Kr]4d10[Kr] 4d^{10} with all d electrons remaining as core electrons (correct answer)
  2. [Kr]5s24d8[Kr] 5s^2 4d^8 with electrons removed only from p and d orbitals
  3. [Kr]4d4[Kr] 4d^4 with electrons removed from s, p, and d orbitals
  4. [Kr]5s14d9[Kr] 5s^1 4d^9 with partial electron removal from multiple orbitals
Explanation: Element Y is tellurium (Te) with configuration [Kr] 5s² 4d¹⁰ 5p⁴. In the +6 oxidation state, 6 electrons are removed: 2 from 5s, 4 from 5p, and 0 from 4d (since d electrons are more tightly held). This gives [Kr] 4d¹⁰. The 4d electrons remain as they are considered core electrons in this context. Option B incorrectly removes d electrons while keeping s electrons. Option C removes too many electrons including d electrons. Option D shows an impossible partial removal pattern.

Question 6

Two isoelectronic species, Mn2+Mn^{2+} and Fe3+Fe^{3+}, both have 23 electrons. However, they exhibit different chemical properties. What best explains this difference despite having identical electron configurations?

  1. Mn2+Mn^{2+} has different d orbital splitting patterns due to its position in the periodic table
  2. Fe3+Fe^{3+} has more unpaired electrons than Mn2+Mn^{2+}, making it more paramagnetic and chemically reactive
  3. Mn2+Mn^{2+} has larger ionic radius due to lower nuclear charge, affecting its coordination chemistry and bond strengths (correct answer)
  4. Fe3+Fe^{3+} exhibits stronger crystal field effects because of its higher oxidation state and electron density
Explanation: When you encounter isoelectronic species with different chemical properties, focus on how nuclear charge affects ionic size and bonding behavior, not just electron configuration. Both Mn2+Mn^{2+} and Fe3+Fe^{3+} have identical electron configurations ([Ar] 3d⁵), but iron has one more proton in its nucleus (26 vs 25). This higher nuclear charge in Fe3+Fe^{3+} pulls the electron cloud more tightly, creating a smaller ionic radius. The larger Mn2+Mn^{2+} ion affects coordination chemistry significantly—it can accommodate different numbers of ligands, forms longer and weaker bonds, and exhibits different geometric preferences compared to the more compact Fe3+Fe^{3+}. This size difference is the primary driver of their distinct chemical behaviors. Option A incorrectly suggests different d orbital splitting patterns, but isoelectronic ions in similar environments have identical d⁵ configurations and orbital energies. Option B is wrong because both ions have exactly five unpaired d electrons, making them equally paramagnetic. Option D confuses cause and effect—while Fe3+Fe^{3+} does experience stronger crystal field effects, this results from its smaller size and higher charge density, not higher oxidation state alone. Remember this key principle: isoelectronic species differ chemically primarily due to size effects from nuclear charge differences. When comparing ions with identical electron counts, always consider how the number of protons affects ionic radius and subsequent bonding interactions.

Question 7

When copper (Cu) forms Cu2+Cu^{2+} ions, electrons are removed to give the configuration [Ar]3d9[Ar] 3d^9. Why are electrons removed from the 4s orbital before the 3d orbitals, despite 4s being lower in energy during filling?

  1. The 4s orbital becomes higher in energy than 3d orbitals once the atom is ionized due to decreased shielding effects
  2. The 4s electrons are removed first because they are in the outermost shell and experience less nuclear attraction (correct answer)
  3. Removing 4s electrons first maintains the stability of the half-filled 3d subshell in the resulting ion
  4. The 4s orbital contracts more than 3d orbitals when electrons are removed, making 4s electrons easier to remove
Explanation: In neutral Cu, 4s is indeed lower in energy than 3d during filling. However, 4s electrons are in the outermost shell (n=4) while 3d electrons are in a lower shell (n=3). The 4s electrons experience less effective nuclear charge due to their greater distance from the nucleus and are therefore easier to remove during ionization. Option A incorrectly suggests orbital energy reversal upon ionization. Option C is wrong because Cu²⁺ has d⁹, not d⁵ configuration. Option D incorrectly describes orbital contraction effects.

Question 8

An element X has the electron configuration [Ar]4s23d104p5[Ar] 4s^2 3d^{10} 4p^5. When this element forms its most stable monatomic ion, what is the electron configuration of the resulting ion and its magnetic properties?

  1. [Ar]4s23d104p6[Ar] 4s^2 3d^{10} 4p^6; diamagnetic due to all paired electrons (correct answer)
  2. [Ar]4s23d104p4[Ar] 4s^2 3d^{10} 4p^4; paramagnetic due to two unpaired electrons
  3. [Ar]3d104p6[Ar] 3d^{10} 4p^6; diamagnetic due to complete p subshell
  4. [Ar]4s13d104p5[Ar] 4s^1 3d^{10} 4p^5; paramagnetic due to unpaired s electron
Explanation: Element X is bromine (Br) with electron configuration [Ar] 4s² 3d¹⁰ 4p⁵. Bromine most commonly forms Br⁻ by gaining one electron to achieve a noble gas configuration. The resulting ion has configuration [Ar] 4s² 3d¹⁰ 4p⁶, which is diamagnetic because all electrons are paired. Option B shows electron loss instead of gain. Option C incorrectly removes the 4s electrons. Option D shows an impossible configuration with only one 4s electron.

Question 9

An unknown element has the following properties: it forms a stable +2 ion with electron configuration [Ar]3d6[Ar] 3d^6, and its neutral atom is paramagnetic. What is the most likely electron configuration of the neutral atom?

  1. [Ar]4s23d8[Ar] 4s^2 3d^8 because two electrons are removed from the d subshell to form the +2 ion
  2. [Ar]4s13d7[Ar] 4s^1 3d^7 because this configuration maximizes unpaired electrons for paramagnetism
  3. [Ar]4s23d6[Ar] 4s^2 3d^6 because the ion configuration directly indicates the neutral atom configuration (correct answer)
  4. [Ar]3d8[Ar] 3d^8 because the neutral atom has the same d electron count as the ion plus the charge
Explanation: When you encounter questions about electron configurations and ion formation, remember that electrons are removed from the highest energy orbitals first when forming cations, and this follows specific rules for transition metals. The key insight is understanding electron removal order. For transition metals, when forming cations, electrons are removed from the 4s orbital before the 3d orbitals, even though 3d fills after 4s in neutral atoms. Since the ion has configuration [Ar]3d6[Ar] 3d^6 with a +2 charge, two electrons were removed from the neutral atom. These two electrons came from the 4s orbital, meaning the neutral atom originally had [Ar]4s23d6[Ar] 4s^2 3d^6. You can verify this: the neutral atom [Ar]4s23d6[Ar] 4s^2 3d^6 has four unpaired electrons in the d orbitals, making it paramagnetic as stated. Option A incorrectly assumes electrons are removed from d orbitals to form the +2 ion. This violates the fundamental rule that 4s electrons are removed first in transition metal cations. Option B gives a configuration that would lose one 4s and one 3d electron to reach [Ar]3d6[Ar] 3d^6, but this doesn't follow the systematic removal pattern where both 4s electrons are removed before any d electrons. Option D completely omits the 4s electrons, which is impossible for a neutral transition metal atom where 4s fills before 3d. Study tip: Always remember "4s electrons leave first" when transition metals form cations. This is one of the most tested concepts in electron configuration problems on the IB exam.

Question 10

Consider the isoelectronic series: S2−S^{2-}, Cl−Cl^-, ArAr, K+K^+, and Ca2+Ca^{2+}. All have 18 electrons with configuration [Ne]3s23p6[Ne] 3s^2 3p^6. Which statement best explains the trend in their ionic/atomic radii?

  1. Radius increases with increasing atomic number because more electron shells are added to accommodate the electrons
  2. Radius decreases with increasing positive charge because cations are always smaller than anions with the same electron count
  3. Radius remains constant because all species have identical electron configurations and the same number of electron shells
  4. Radius decreases with increasing nuclear charge because the same electron configuration experiences stronger nuclear attraction (correct answer)
Explanation: When you encounter isoelectronic species on the IB Chemistry exam, focus on how nuclear charge affects electron-cloud attraction. Isoelectronic means these species have identical electron configurations, so differences in size must come from nuclear effects. In this series, all species have 18 electrons arranged as [Ne]3s23p6[Ne] 3s^2 3p^6, but their nuclear charges differ: S (16 protons), Cl (17), Ar (18), K (19), and Ca (20). As nuclear charge increases while electron count stays constant, the nucleus pulls the electron cloud more tightly inward, decreasing the radius. This creates the trend: S2−>Cl−>Ar>K+>Ca2+S^{2-} > Cl^- > Ar > K^+ > Ca^{2+}. Answer D correctly identifies that increasing nuclear charge pulls the same electron configuration closer to the nucleus. Answer A is wrong because no additional electron shells are added—all species use the same three shells (1s, 2s/2p, 3s/3p). Answer B oversimplifies by focusing only on charge type rather than the underlying nuclear attraction; while cations are generally smaller than anions, the key factor here is nuclear charge, not just ionic charge. Answer C incorrectly assumes that identical electron configurations guarantee identical sizes, ignoring how nuclear charge affects electron-nucleus attraction. Study tip: For isoelectronic series questions, always compare nuclear charge first. More protons = stronger pull on the same electron configuration = smaller radius. This pattern appears frequently on IB exams, so practice identifying isoelectronic species and ranking them by nuclear charge.

Question 11

How does the number of occupied p-orbitals in a Cl⁻ ion compare to the number in a K⁺ ion?

  1. Cl⁻ has more occupied p-orbitals than K⁺.
  2. K⁺ has more occupied p-orbitals than Cl⁻.
  3. Both ions have the same number of occupied p-orbitals. (correct answer)
  4. Neither ion has any occupied p-orbitals.
Explanation: First, determine the electron configuration of each ion. Chlorine (Cl, Z=17) is [Ne] 3s²3p⁵. Cl⁻ gains one electron to become [Ne] 3s²3p⁶. Potassium (K, Z=19) is [Ar] 4s¹. K⁺ loses one electron to become [Ar]. The full configuration of Ar is 1s²2s²2p⁶3s²3p⁶. Both Cl⁻ and K⁺ are isoelectronic with Ar. Both have filled 2p and 3p sub-levels. The 2p sub-level has 3 orbitals and the 3p sub-level has 3 orbitals, so both ions have a total of 6 occupied p-orbitals.

Question 12

An element in Period 4 is a p-block element that typically forms a 2- ion. What is the element's ground-state electron configuration?

  1. [Ar] 4s²3d¹⁰4p⁴ (correct answer)
  2. [Ar] 4s²3d¹⁰4p²
  3. [Ar] 4s²4p⁶
  4. [Ar] 4s²3d⁴
Explanation: An element that forms a 2- ion typically has 6 valence electrons and belongs to Group 16. A Period 4, Group 16 element will have its valence electrons in the n=4 shell. The electron configuration will fill up to Argon, then the 4s, 3d, and finally the 4p sub-levels. For 6 valence electrons, the configuration will be 4s²4p⁴. The full configuration is therefore [Ar] 4s²3d¹⁰4p⁴. This element is Selenium (Se).

Question 13

According to the Aufbau principle, when an electron is added to an oxygen atom (1s²2s²2p⁴) to form an O⁻ ion, into which orbital is the electron placed?

  1. A 2s orbital
  2. A 2p orbital (correct answer)
  3. A 3s orbital
  4. A 3p orbital
Explanation: The Aufbau principle states that electrons fill orbitals starting at the lowest available energy states before filling higher states. In an oxygen atom, the 2p sub-level is the highest-energy occupied sub-level, and it is not full (it contains 4 out of a possible 6 electrons). Therefore, the added electron will go into the 2p sub-level, as it is the lowest-energy available orbital.

Question 14

What is the correct ground-state electron configuration for a chromium(III) ion, Cr³⁺?

  1. [Ar] 4s²3d¹
  2. [Ar] 4s¹3d²
  3. [Ar] 3d³ (correct answer)
  4. [Ar] 3d⁴
Explanation: First, determine the configuration of a neutral chromium atom (Cr, Z=24). Chromium is an exception to the Aufbau principle, with the configuration [Ar] 4s¹3d⁵. To form an ion, electrons are removed from the highest principal energy level first. So, the 4s electron is removed first, followed by electrons from the 3d sub-level. To form Cr³⁺, one 4s electron and two 3d electrons are removed, leaving the configuration [Ar] 3d³.

Question 15

An element has a ground-state electron configuration of [Kr] 5s²4d¹⁰5p⁴. In which group and period of the periodic table is this element located?

  1. Period 5, Group 14
  2. Period 5, Group 16 (correct answer)
  3. Period 4, Group 16
  4. Period 5, Group 4
Explanation: The highest principal energy level (n) of the valence electrons determines the period. Here, the valence shell is n=5, so the element is in Period 5. For a p-block element, the group number is 10 + (number of s electrons + number of p electrons in the valence shell). The number of valence electrons is 2 (from 5s) + 4 (from 5p) = 6. The group number is 10 + 6 = 16. The element is Tellurium (Te).

Question 16

What is the ground-state electron configuration of a nitride ion, N³⁻?

  1. 1s²2s²2p³
  2. 1s²2s²2p⁶ (correct answer)
  3. 1s²2s²
  4. 1s²2s²2p⁶3s²3p³
Explanation: A neutral nitrogen atom (N, Z=7) has the configuration 1s²2s²2p³. To form the nitride ion, N³⁻, the atom gains three electrons. These electrons are added to the outermost partially filled sub-level, the 2p sub-level, until it is full. This results in the configuration 1s²2s²2p⁶, which is isoelectronic with Neon (Ne).

Question 17

What is the maximum number of electrons that can occupy the n=4 principal energy level?

  1. 8
  2. 16
  3. 18
  4. 32 (correct answer)
Explanation: The maximum number of electrons in a principal energy level 'n' is given by the formula 2n². For n=4, the maximum number of electrons is 2(4)² = 2(16) = 32. Alternatively, the n=4 level contains the 4s (2 electrons), 4p (6 electrons), 4d (10 electrons), and 4f (14 electrons) sub-levels. The total is 2 + 6 + 10 + 14 = 32 electrons.

Question 18

The emission spectrum of an element provides direct evidence for which of the following?

  1. The existence of isotopes with different numbers of neutrons.
  2. The existence of quantized electron energy levels in atoms. (correct answer)
  3. The spherical shape of the s-orbitals.
  4. The relative abundance of elements in the universe.
Explanation: An emission spectrum consists of discrete lines of specific frequencies (or wavelengths), not a continuous spectrum. This indicates that electrons can only release specific, discrete amounts of energy when they transition between energy levels. This is direct evidence that the energy levels themselves are quantized, meaning electrons can only exist in specific, fixed energy states within an atom.

Question 19

The ground-state electron configuration of a neutral silicon atom is 1s²2s²2p⁶3s²3p². Which configuration represents a silicon atom in an excited state?

  1. 1s²2s²2p⁶3s²3p¹
  2. 1s²2s²2p⁶3s²3p²3d¹
  3. 1s²2s²2p⁶3s¹3p³ (correct answer)
  4. 1s²2s²2p⁶3s²
Explanation: An atom in an excited state has the same number of electrons as its ground state, but at least one electron has been promoted to a higher energy orbital. Neutral silicon has 14 electrons. Choice C, 1s²2s²2p⁶3s¹3p³, also has 14 electrons, but one electron has been promoted from the 3s orbital to the 3p orbital, which is a higher energy state. Choice A is an ion (Si⁺). Choice B has 15 electrons, so it is a different element (P). Choice D is an ion (Si²⁺).

Question 20

Which ion has a ground-state electron configuration with a completely filled 3d sub-level and an empty 4s sub-level?

  1. Sc³⁺
  2. Ti²⁺
  3. V³⁺
  4. Zn²⁺ (correct answer)
Explanation: We need to find an ion with the configuration [Ar] 3d¹⁰. Let's examine the options. Sc (Z=21) is [Ar] 4s²3d¹. Sc³⁺ is [Ar]. Ti (Z=22) is [Ar] 4s²3d². Ti²⁺ is [Ar] 3d². V (Z=23) is [Ar] 4s²3d³. V³⁺ is [Ar] 3d². Zn (Z=30) is [Ar] 4s²3d¹⁰. To form Zn²⁺, the two 4s electrons are removed, leaving the configuration [Ar] 3d¹⁰. This matches the description.