College Chemistry Quiz: Atomic Structure And Electron Configuration
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Atomic Structure And Electron ConfigurationQuestion 1 of 20

How does the electron configuration of an atom affect its chemical properties in forming common ions?

Atoms gain neutrons to reach noble-gas configuration
Atoms transfer valence electrons to reach lower energy
Atoms change proton number to complete octets
Atoms rearrange nuclei to satisfy Hund's rule
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College Chemistry Quiz

College Chemistry Quiz: Atomic Structure And Electron Configuration

Practice Atomic Structure And Electron Configuration in College 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 Atomic Structure And Electron Configuration, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

All questions

Question 1

How does the electron configuration of an atom affect its chemical properties in forming common ions?

  1. Atoms gain neutrons to reach noble-gas configuration
  2. Atoms transfer valence electrons to reach lower energy (correct answer)
  3. Atoms change proton number to complete octets
  4. Atoms rearrange nuclei to satisfy Hund's rule
Explanation: This question tests college-level chemistry skills in understanding atomic structure and electron configuration. Atomic structure describes the arrangement of subatomic particles within an atom, including protons, neutrons, and electrons. Electron configuration is determined by quantum numbers and impacts an element's chemical properties. In this passage, specific attention is given to ion formation driven by achieving stable configurations. The correct answer is valid because it describes electron transfer for energy minimization and octet completion. A common distractor fails because it involves proton changes, impossible in chemical reactions. To aid student learning, emphasize predicting common ions from group positions. Encourage analyzing ionic compounds to link configurations to stability.

Question 2

How does the electron configuration of an atom affect its chemical properties, especially bonding behavior?

  1. Only neutrons control bonding and reactivity
  2. Valence electrons largely determine bonding patterns (correct answer)
  3. Core electrons primarily determine typical oxidation states
  4. Proton number changes during ordinary reactions
Explanation: This question tests college-level chemistry skills in understanding atomic structure and electron configuration. Atomic structure describes the arrangement of subatomic particles within an atom, including protons, neutrons, and electrons. Electron configuration is determined by quantum numbers and impacts an element's chemical properties. In this passage, specific attention is given to how valence electrons dictate bonding and reactivity. The correct answer is valid because it accurately highlights that valence electrons determine bonding patterns, linking configuration to chemistry. A common distractor fails because it wrongly attributes bonding control to neutrons, ignoring electron roles. To aid student learning, emphasize identifying valence electrons from configurations for various elements. Encourage exploring reaction examples to see how configurations predict compound formation.

Question 3

What is the significance of quantum numbers in electron configuration for subshell capacity and orbital count?

  1. An ss subshell holds up to 6 electrons
  2. A pp subshell contains three orbitals (correct answer)
  3. A dd subshell contains two orbitals
  4. An ff subshell holds up to 10 electrons
Explanation: This question tests college-level chemistry skills in understanding atomic structure and electron configuration. Atomic structure describes the arrangement of subatomic particles within an atom, including protons, neutrons, and electrons. Electron configuration is determined by quantum numbers and impacts an element's chemical properties. In this passage, specific attention is given to subshell structures defined by azimuthal quantum number. The correct answer is valid because p subshells have three orbitals, each holding two electrons. A common distractor fails by misstating s subshell capacity as 6. To aid student learning, emphasize orbital counts: s=1, p=3, d=5, f=7. Encourage calculating maximum electrons per subshell using 2(2l+1).

Question 4

In which of the following atoms is the outer electron shell filled, based on valence configuration?

  1. He: 1s2 (correct answer)
  2. Li: 1s2 2s1
  3. B: 1s2 2s2 2p1
  4. O: 1s2 2s2 2p4
Explanation: This question tests college-level chemistry skills in understanding atomic structure and electron configuration. Atomic structure describes the arrangement of subatomic particles within an atom, including protons, neutrons, and electrons. Electron configuration is determined by quantum numbers and impacts an element's chemical properties. In this passage, specific attention is given to helium's unique stability as a noble gas. The correct answer is valid because 1s2 fills its only shell, explaining inertness. A common distractor fails by selecting elements with unfilled valence orbitals. To aid student learning, emphasize valence shell concepts for light elements. Encourage contrasting helium with other group 18 elements.

Question 5

An atom has the electron configuration 1s22s22p63s23p63d54s11s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1. This configuration represents:

  1. The ground state of chromium (correct answer)
  2. An excited state of manganese
  3. The ground state of iron
  4. An excited state of chromium
  5. The Cr5+Cr^{5+} ion
Explanation: When you encounter electron configuration questions, you need to identify both the element and whether it's in its ground state or an excited state. Start by counting the total electrons to determine the element, then check if the configuration follows the expected filling order. Counting the electrons in 1s22s22p63s23p63d54s11s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1: 2 + 2 + 6 + 2 + 6 + 5 + 1 = 24 electrons. Since atoms are neutral, this means 24 protons, identifying the element as chromium (atomic number 24). Now, is this the ground state? Chromium is one of the exceptions to normal electron filling rules. Instead of filling 3d44s23d^4 4s^2, chromium's ground state is actually 3d54s13d^5 4s^1 because having a half-filled d subshell provides extra stability. This matches the given configuration exactly. Looking at the wrong answers: B) suggests this is an excited manganese (25 electrons), but we only have 24 electrons. C) claims this is ground state iron (26 electrons), but again, we only have 24 electrons. D) suggests this is excited chromium, but we've established this is actually chromium's stable ground state configuration. The answer is A - this represents the ground state of chromium. Study tip: Memorize the two main electron configuration exceptions: chromium ([Ar]3d54s1[Ar] 3d^5 4s^1) and copper ([Ar]3d104s1[Ar] 3d^{10} 4s^1). Both prefer half-filled or completely filled d subshells over the expected s² configuration. These exceptions appear frequently on chemistry exams.

Question 6

An atom of element X has 17 protons and 20 neutrons. After losing 3 electrons, what is the mass number and charge of the resulting ion?

  1. Mass number = 17, charge = +3
  2. Mass number = 37, charge = +3 (correct answer)
  3. Mass number = 20, charge = +3
  4. Mass number = 37, charge = -3
  5. Mass number = 34, charge = +3
Explanation: When you encounter atomic structure problems, focus on three key numbers: protons (atomic number), neutrons, and electrons. The mass number equals protons plus neutrons, while charge depends on the difference between protons and electrons. Let's work through this systematically. Element X has 17 protons and 20 neutrons. The mass number is 17+20=3717 + 20 = 37. This number never changes unless the nucleus itself is altered—losing or gaining electrons doesn't affect mass number. For the charge, start with a neutral atom: 17 protons (+17 charge) balanced by 17 electrons (-17 charge) gives zero net charge. When the atom loses 3 electrons, it has 17 protons but only 14 electrons. The net charge becomes 17−14=+317 - 14 = +3. Therefore, the ion has mass number 37 and charge +3, making B correct. Here's why the other answers fail: A gives mass number 17, which incorrectly uses only the number of protons instead of protons plus neutrons. C gives mass number 20, which incorrectly uses only the number of neutrons. D correctly calculates the mass number as 37 but shows charge as -3, which represents the opposite error—this would be the charge if the atom gained 3 electrons rather than lost them. Remember this pattern: mass number always equals protons plus neutrons (electrons have negligible mass), and positive ions form when atoms lose electrons. The more electrons lost, the more positive the charge becomes.

Question 7

Which electron configuration represents an atom in its ground state that would most likely form a +2 cation?

  1. 1s22s22p63s23p64s11s^2 2s^2 2p^6 3s^2 3p^6 4s^1
  2. 1s22s22p63s23p64s21s^2 2s^2 2p^6 3s^2 3p^6 4s^2 (correct answer)
  3. 1s22s22p63s23p51s^2 2s^2 2p^6 3s^2 3p^5
  4. 1s22s22p63s23p63d104s11s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1
  5. 1s22s22p63s23p63d54s21s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2
Explanation: When determining which atom is most likely to form a +2 cation, you need to consider which electron configuration would become more stable by losing exactly two electrons. Atoms form cations when removing electrons leads to a more stable, lower-energy configuration—typically achieving a noble gas electron arrangement. Looking at option B: 1s22s22p63s23p64s21s^2 2s^2 2p^6 3s^2 3p^6 4s^2, this atom has 20 electrons total (calcium). If it loses its two 4s electrons, it achieves the stable electron configuration 1s22s22p63s23p61s^2 2s^2 2p^6 3s^2 3p^6, which matches argon's noble gas configuration. This represents a significant stability gain, making Ca²⁺ formation highly favorable. Option A (1s22s22p63s23p64s11s^2 2s^2 2p^6 3s^2 3p^6 4s^1) represents potassium, which would more likely lose just one electron to form K⁺ and achieve the same argon configuration. Losing two electrons would be energetically unfavorable. Option C (1s22s22p63s23p51s^2 2s^2 2p^6 3s^2 3p^5) represents chlorine, which typically gains one electron to form Cl⁻ rather than losing electrons, since gaining an electron completes its octet. Option D (1s22s22p63s23p63d104s11s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1) represents copper, which can form Cu⁺ by losing one electron, but Cu²⁺ formation requires removing an electron from the stable, filled 3d subshell—less favorable than the straightforward 4s² loss in calcium. Remember: atoms in Groups 1 and 2 readily form +1 and +2 cations respectively because losing their outermost s electrons achieves noble gas configurations with minimal energy cost.

Question 8

An element has the electron configuration [Ne]3s23p4[Ne] 3s^2 3p^4. How many unpaired electrons does this atom have in its ground state?

  1. 0
  2. 1
  3. 2 (correct answer)
  4. 4
  5. 6
Explanation: When you encounter electron configurations, you need to determine how electrons fill orbitals and whether they pair up or remain unpaired. This requires understanding Hund's rule and orbital diagrams. The given configuration [Ne]3s23p4[Ne] 3s^2 3p^4 tells you this atom has 2 electrons in the 3s orbital and 4 electrons in the 3p subshell. The 3s orbital is completely filled with 2 paired electrons, so focus on the 3p subshell. The 3p subshell contains three orbitals (3p_x, 3p_y, 3p_z), each capable of holding 2 electrons. According to Hund's rule, electrons first occupy each orbital singly before pairing up. With 4 electrons in 3p: the first three electrons occupy each orbital singly (all unpaired), then the fourth electron pairs with one of them. This gives you two orbitals with one electron each (unpaired) and one orbital with two paired electrons, resulting in 2 unpaired electrons total. Choice A (0 unpaired electrons) would only be correct if all orbitals were completely filled or empty. Choice B (1 unpaired electron) suggests only one orbital contains a single electron, which ignores Hund's rule. Choice D (4 unpaired electrons) incorrectly assumes all four 3p electrons remain unpaired, but this violates the fact that you only have three 3p orbitals available. The correct answer is C. Study tip: Always draw orbital diagrams when counting unpaired electrons. Remember Hund's rule: electrons prefer to occupy orbitals singly before pairing, which helps you visualize the electron arrangement accurately.

Question 9

Which ion has the same electron configuration as the noble gas argon?

  1. Ca+Ca^+
  2. K2+K^{2+}
  3. Cl2−Cl^{2-}
  4. S2−S^{2-} (correct answer)
  5. Mg2+Mg^{2+}
Explanation: When you encounter questions about electron configurations and noble gas equivalents, you're working with the concept of isoelectronic species—atoms or ions that have the same number of electrons. Noble gases are stable because they have complete outer electron shells, so ions often gain or lose electrons to achieve this same configuration. Argon has 18 electrons (atomic number 18). To find which ion matches this, you need to calculate the electron count for each option by considering how many electrons are gained or lost during ion formation. For option D, sulfur (SS, atomic number 16) normally has 16 electrons. When it forms S2−S^{2-}, it gains 2 electrons, giving it 16 + 2 = 18 electrons—exactly matching argon's configuration. Now let's check why the other options don't work: Option A gives calcium (atomic number 20) with a +1 charge, meaning 20 - 1 = 19 electrons. Option B shows potassium (atomic number 19) with a +2 charge, yielding 19 - 2 = 17 electrons. Option C presents chlorine (atomic number 17) with a -2 charge, resulting in 17 + 2 = 19 electrons. Only S2−S^{2-} achieves the 18-electron configuration of argon. Study tip: For isoelectronic problems, always write out the calculation: starting electrons (atomic number) ± charge = final electron count. Then compare to your target noble gas. Remember that gaining electrons means adding to the atomic number, while losing electrons means subtracting.

Question 10

Which of the following electron configurations is impossible for any atom or ion?

  1. 1s22s22p63s11s^2 2s^2 2p^6 3s^1
  2. 1s22s22p63s23p63d101s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}
  3. 1s22s12p61s^2 2s^1 2p^6 (correct answer)
  4. 1s22s22p63s23p64s21s^2 2s^2 2p^6 3s^2 3p^6 4s^2
  5. 1s22s22p51s^2 2s^2 2p^5
Explanation: When evaluating electron configurations, you need to check whether they follow the fundamental rules of electron filling: the Aufbau principle (electrons fill lower energy orbitals first), Hund's rule (electrons occupy orbitals singly before pairing), and the Pauli exclusion principle (maximum two electrons per orbital with opposite spins). The key insight here is recognizing violations of the Aufbau principle. Electrons must fill orbitals in order of increasing energy: 1s, then 2s, then 2p, then 3s, and so on. Choice C (1s22s12p61s^2 2s^1 2p^6) is impossible because it shows the 2p subshell completely filled with 6 electrons while the 2s subshell only has 1 electron. Since 2s orbitals are lower in energy than 2p orbitals, the 2s subshell must be completely filled (2s22s^2) before any electrons can occupy 2p orbitals. This configuration violates the Aufbau principle and cannot exist. Choice A represents sodium (Na) and follows proper filling order. Choice B could represent a neutral zinc atom (Zn) or various transition metal ions, with electrons properly filling through the 3d subshell. Choice D represents calcium (Ca) with correct sequential filling through 4s. Remember that while atoms can lose or gain electrons to form ions, and excited states can temporarily promote electrons to higher orbitals, the fundamental energy ordering still applies. Any configuration that shows higher-energy orbitals filled while lower-energy orbitals remain unfilled violates basic quantum mechanical principles and is impossible.

Question 11

An element in period 4 has the electron configuration [Ar]3d104s24p1[Ar] 3d^{10} 4s^2 4p^1. This element is:

  1. A transition metal in group 3
  2. A main group metal in group 13 (correct answer)
  3. A metalloid in group 14
  4. A nonmetal in group 15
  5. A noble gas in group 18
Explanation: When you encounter electron configuration questions, you need to determine both the group number and the element type by analyzing the valence electrons and their location. The given configuration [Ar]3d104s24p1[Ar] 3d^{10} 4s^2 4p^1 shows a completely filled 3d subshell (10 electrons) followed by filled 4s and partially filled 4p orbitals. The key insight is that since the 3d subshell is completely filled, this isn't a transition metal actively using d electrons for bonding. Instead, the valence electrons are in the 4s and 4p orbitals: 4s24p14s^2 4p^1 gives us 3 valence electrons total. For main group elements, the group number equals the number of valence electrons. With 3 valence electrons, this element belongs to group 13. The electron configuration pattern (filled d-block plus s and p electrons) and the metallic character typical of group 13 elements like aluminum confirm this is a main group metal. Looking at the wrong answers: A is incorrect because transition metals have partially filled d orbitals, but here the 3d subshell is completely filled. C is wrong because group 14 elements have 4 valence electrons (s2p2s^2p^2), not 3, and while some group 13 elements show metalloid properties, the question asks for classification. D is incorrect because group 15 elements have 5 valence electrons (s2p3s^2p^3) and are typically nonmetals or metalloids. Study tip: Count valence electrons in the outermost s and p orbitals for main group elements—ignore completely filled d subshells when determining group number.

Question 12

Which of the following represents the correct orbital diagram for the valence electrons of nitrogen in its ground state?

  1. 2s: ↑↓ 2p: ↑↓ ↑ ↑
  2. 2s: ↑↓ 2p: ↑ ↑ ↑ (correct answer)
  3. 2s: ↑↓ 2p: ↑↓ ↑↓ ↑
  4. 2s: ↑ 2p: ↑ ↑ ↑ ↑
  5. 2s: ↑↓ 2p: ↑↓ ↑↓ ↑↓
Explanation: When you encounter orbital diagram questions, you need to apply three fundamental rules: the Aufbau principle (fill lowest energy orbitals first), Hund's rule (place one electron in each orbital before pairing), and the Pauli exclusion principle (maximum two electrons per orbital with opposite spins). Nitrogen has 7 electrons total, with 5 valence electrons in the second shell (2s² 2p³). First, fill the 2s orbital completely with two paired electrons (↑↓). Then distribute the remaining three electrons among the three 2p orbitals. According to Hund's rule, you must place one electron in each 2p orbital before pairing any electrons. This gives you 2s: ↑↓ and 2p: ↑ ↑ ↑, which matches answer B. Answer A violates Hund's rule by pairing electrons in the first 2p orbital while the other 2p orbitals remain empty. This higher-energy configuration wouldn't occur in the ground state. Answer C shows six electrons in the 2p subshell (2p⁶), which would be correct for neon, not nitrogen. This represents a common mistake of confusing electron counts between elements. Answer D shows an unpaired electron in the 2s orbital, violating the Aufbau principle since the 2s orbital should be completely filled before any electrons enter the 2p orbitals. Remember this pattern: always completely fill lower energy orbitals first, then follow Hund's rule by placing one electron in each orbital of the same energy before pairing. This systematic approach will help you correctly diagram any element's electron configuration.

Question 13

The electron configuration 1s22s22p63s23p63d104s24p64d105s11s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^1 corresponds to which element?

  1. Silver (Ag) (correct answer)
  2. Cadmium (Cd)
  3. Indium (In)
  4. Tin (Sn)
  5. Strontium (Sr)
Explanation: When you encounter electron configuration problems, you need to count the total number of electrons and match it to an element's atomic number, since neutral atoms have equal numbers of protons and electrons. Let's count the electrons in this configuration: 1s22s22p63s23p63d104s24p64d105s11s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^1 Adding up all the superscripts: 2 + 2 + 6 + 2 + 6 + 10 + 2 + 6 + 10 + 1 = 47 electrons An element with 47 electrons has atomic number 47, which is silver (Ag). This configuration shows silver's ground state with its characteristic single electron in the 5s orbital after completely filling the 4d subshell. Let's examine why the other options don't work: Option B (Cadmium) has atomic number 48, so it would have 48 electrons and end in 5s25s^2, not 5s15s^1. Option C (Indium) has atomic number 49, requiring 49 electrons, which would continue past 5s15s^1 into the 5p orbital (5s25p15s^2 5p^1). Option D (Tin) has atomic number 50, needing 50 electrons and ending in 5s25p25s^2 5p^2. The key strategy here is always count electrons systematically and remember that the electron configuration directly corresponds to atomic number for neutral atoms. Also, be familiar with transition metals like silver that have unexpected configurations due to d-orbital stability—silver actually has the configuration [Kr]4d105s1[Kr] 4d^{10} 5s^1 rather than the expected [Kr]4d95s2[Kr] 4d^9 5s^2 because a filled d subshell provides extra stability.

Question 14

Which of the following atoms has the greatest number of unpaired electrons in its ground state?

  1. Carbon (C)
  2. Nitrogen (N)
  3. Oxygen (O)
  4. Manganese (Mn) (correct answer)
  5. Iron (Fe)
Explanation: This question tests your understanding of electron configurations and Hund's rule, which states that electrons occupy orbitals singly before pairing up. To find unpaired electrons, you need to write the electron configuration for each atom's ground state. Let's work through each option systematically. Carbon (C) has 6 electrons with configuration 1s22s22p21s^2 2s^2 2p^2. The two 2p electrons occupy separate orbitals following Hund's rule, giving carbon 2 unpaired electrons. Nitrogen (N) has 7 electrons: 1s22s22p31s^2 2s^2 2p^3. All three 2p electrons occupy separate orbitals, resulting in 3 unpaired electrons. Oxygen (O) has 8 electrons: 1s22s22p41s^2 2s^2 2p^4. Here, three electrons occupy separate 2p orbitals, and the fourth pairs up, leaving 2 unpaired electrons. Manganese (Mn) has 25 electrons with configuration 1s22s22p63s23p64s23d51s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5. The key is the 3d53d^5 portion - all five d orbitals contain one electron each before any pairing occurs, giving manganese 5 unpaired electrons. Answer A (Carbon) has only 2 unpaired electrons in its 2p orbitals. Answer B (Nitrogen) has 3 unpaired electrons, all in 2p orbitals. Answer C (Oxygen) has 2 unpaired electrons since one pair forms in the 2p subshell. Answer D (Manganese) has 5 unpaired electrons in its half-filled d subshell, making it correct. When comparing unpaired electrons, always pay special attention to transition metals with their d orbitals - they often have more unpaired electrons than main group elements due to the availability of five d orbitals.

Question 15

Which statement about electron configurations is correct?

  1. The 3d orbitals are filled before the 4s orbital
  2. Atoms always have the same number of protons and electrons
  3. All atoms follow the aufbau principle without exception
  4. The maximum number of electrons in any d subshell is 10 (correct answer)
  5. Hund's rule only applies to p orbitals
Explanation: Understanding electron configurations requires knowing the fundamental rules that govern how electrons fill atomic orbitals. These rules determine the electronic structure of atoms and explain their chemical behavior. The maximum number of electrons that can occupy any d subshell is indeed 10, making answer D correct. Each d subshell contains five d orbitals, and since each orbital can hold a maximum of two electrons (with opposite spins according to the Pauli exclusion principle), the total capacity is 5 × 2 = 10 electrons. Let's examine why the other options are incorrect. Answer A reverses the actual filling order - the 4s orbital is filled before the 3d orbitals due to its lower energy, which is why we see configurations like [Ar] 4s² 3d⁶ for iron. Answer B is false because ions exist everywhere in chemistry; atoms gain or lose electrons to form charged species, creating an imbalance between protons and electrons. Answer C overstates the aufbau principle's universality - while it works for most ground-state atoms, there are notable exceptions like chromium and copper, which have unexpected configurations (Cr: [Ar] 4s¹ 3d⁵ instead of [Ar] 4s² 3d⁴) due to the stability of half-filled and fully-filled subshells. When studying electron configurations, remember that the maximum electron capacity follows the formula 2n², where n is the principal quantum number, but for subshells specifically: s holds 2, p holds 6, d holds 10, and f holds 14 electrons. These numbers are fundamental and have no exceptions.

Question 16

How many electrons can have the quantum numbers n = 3 and l = 1 in a single atom?

  1. 2
  2. 3
  3. 6 (correct answer)
  4. 8
  5. 10
Explanation: When you encounter quantum number problems, you're working with the electron configuration rules that govern how electrons occupy atomic orbitals. The key is understanding what each quantum number tells you about electron capacity. Given n = 3 and l = 1, you're looking at the 3p subshell. The principal quantum number n = 3 indicates the third energy level, while the angular momentum quantum number l = 1 corresponds to a p subshell (since l = 0 is s, l = 1 is p, l = 2 is d, etc.). Every p subshell contains exactly three orbitals, corresponding to the three possible values of the magnetic quantum number: ml = -1, 0, +1. Each orbital can hold a maximum of two electrons (one with spin up, one with spin down). Therefore, the total electron capacity is 3 orbitals × 2 electrons per orbital = 6 electrons. Looking at the wrong answers: A) 2 represents the maximum number of electrons in a single orbital, but ignores that p subshells have three orbitals. B) 3 counts only the number of orbitals without considering that each holds two electrons. D) 8 might come from confusing this with d subshells, which hold 10 electrons, or from some other miscalculation. Remember this pattern: s subshells hold 2 electrons, p subshells hold 6, d subshells hold 10, and f subshells hold 14. When given specific n and l values, identify the subshell type first, then recall its electron capacity.

Question 17

The electron configuration 1s22s22p63s23p64s23d104p65s24d105p66s24f145d61s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^6 corresponds to which element?

  1. Tungsten (W) (correct answer)
  2. Osmium (Os)
  3. Platinum (Pt)
  4. Gold (Au)
  5. Mercury (Hg)
Explanation: When you encounter electron configuration problems, you need to count the total number of electrons to determine the atomic number, which directly identifies the element. Let's count the electrons systematically: 1s21s^2 (2e⁻) + 2s22s^2 (2e⁻) + 2p62p^6 (6e⁻) + 3s23s^2 (2e⁻) + 3p63p^6 (6e⁻) + 4s24s^2 (2e⁻) + 3d103d^{10} (10e⁻) + 4p64p^6 (6e⁻) + 5s25s^2 (2e⁻) + 4d104d^{10} (10e⁻) + 5p65p^6 (6e⁻) + 6s26s^2 (2e⁻) + 4f144f^{14} (14e⁻) + 5d65d^6 (6e⁻) = 74 total electrons. Since atoms are electrically neutral, 74 electrons means 74 protons, giving us atomic number 74. Looking at the periodic table, element 74 is tungsten (W), making answer A correct. Let's check why the other options are wrong: B) Osmium (Os) has atomic number 76, which would require 76 electrons total. C) Platinum (Pt) has atomic number 78, needing 78 electrons. D) Gold (Au) has atomic number 79, requiring 79 electrons. Each of these would need additional electrons beyond what's shown in the given configuration. Notice that this configuration ends in 5d65d^6, which confirms we're dealing with a transition metal in the third transition series (6th period), where the 5d orbitals are being filled. Study tip: Always count electrons systematically by adding up all the superscripts. The total equals the atomic number for neutral atoms. Practice recognizing that elements in the same period will have similar electron configurations, differing mainly in how many d or f electrons they contain.

Question 18

An atom in its ground state has 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, 6 electrons in the 2p orbitals, and 3 electrons in the 3s orbital. This description:

  1. Represents sodium in an excited state
  2. Represents aluminum in its ground state
  3. Violates the Pauli exclusion principle
  4. Violates the aufbau principle
  5. Is impossible because 3s can only hold 2 electrons (correct answer)
Explanation: When analyzing electron configurations, you need to check whether the arrangement follows fundamental quantum mechanical principles and represents a valid atomic state. Let's examine this configuration: 1s² 2s² 2p⁶ 3s³. First, count the total electrons: 2 + 2 + 6 + 3 = 13 electrons, which corresponds to aluminum (atomic number 13). However, there's a critical problem with the 3s orbital containing 3 electrons. The Pauli exclusion principle states that each orbital can hold a maximum of 2 electrons with opposite spins. Since any s orbital (including 3s) can only accommodate 2 electrons maximum, having 3 electrons in the 3s orbital directly violates this fundamental principle. This makes the described configuration physically impossible. Looking at the wrong answers: (A) is incorrect because sodium has 11 electrons, not 13, and this isn't an excited state anyway due to the Pauli violation. (B) is wrong because aluminum's correct ground state configuration is 1s² 2s² 2p⁶ 3s², not 3s³. (D) is incorrect because the aufbau principle (filling orbitals from lowest to highest energy) is actually followed here - the violation is with orbital capacity, not filling order. The answer is (C) - this configuration violates the Pauli exclusion principle. Study tip: When checking electron configurations, always verify that no orbital exceeds its electron capacity: s orbitals hold 2 maximum, p orbitals hold 6 maximum, d orbitals hold 10 maximum, and f orbitals hold 14 maximum. This is often tested on chemistry exams.

Question 19

When a neutral atom of element X loses its 2 highest-energy electrons, the resulting ion has the electron configuration [Ar]3d9[Ar] 3d^9. What is the identity of element X?

  1. Copper (Cu) (correct answer)
  2. Zinc (Zn)
  3. Nickel (Ni)
  4. Cobalt (Co)
  5. Iron (Fe)
Explanation: When you encounter electron configuration questions involving ion formation, you need to work backwards from the given ionic configuration to determine the neutral atom's identity. The ion has configuration [Ar]3d9[Ar] 3d^9 after losing its 2 highest-energy electrons. To find the neutral atom, you must add these 2 electrons back. The key insight is understanding electron filling order: the 4s orbital fills before 3d orbitals, but when atoms ionize, they lose 4s electrons first because they become higher in energy than 3d electrons in transition metals. Since the ion lost 2 electrons and has [Ar]3d9[Ar] 3d^9, the neutral atom must have had [Ar]3d104s1[Ar] 3d^{10} 4s^1. This gives us 18 (Ar) + 10 (3d) + 1 (4s) = 29 electrons, identifying element 29, copper (Cu). Choice A is correct. Let's examine why the other options fail: Choice B (Zn) has atomic number 30 with configuration [Ar]3d104s2[Ar] 3d^{10} 4s^2. Losing 2 electrons would give [Ar]3d10[Ar] 3d^{10}, not [Ar]3d9[Ar] 3d^9. Choice C (Ni) has atomic number 28 with configuration [Ar]3d84s2[Ar] 3d^8 4s^2, which would become [Ar]3d8[Ar] 3d^8 after losing 2 electrons. Choice D (Co) has atomic number 27 with configuration [Ar]3d74s2[Ar] 3d^7 4s^2, giving [Ar]3d7[Ar] 3d^7 as a +2 ion. Remember: transition metals lose 4s electrons before 3d electrons during ionization, even though 4s fills first. Always work backwards by adding electrons to the highest-energy orbitals available.

Question 20

When comparing Fe2+Fe^{2+} and Fe3+Fe^{3+} ions, which statement is correct?

  1. Fe2+Fe^{2+} has more unpaired electrons than Fe3+Fe^{3+}
  2. Fe3+Fe^{3+} has more unpaired electrons than Fe2+Fe^{2+} (correct answer)
  3. Both ions have the same number of unpaired electrons
  4. Fe2+Fe^{2+} has a half-filled d subshell
  5. Fe3+Fe^{3+} has a completely filled d subshell
Explanation: When you encounter questions about transition metal ions and unpaired electrons, you need to determine the electron configurations and apply Hund's rule to count unpaired electrons. Iron (Fe) has the electron configuration [Ar] 3d⁶ 4s². When Fe forms cations, electrons are removed first from the 4s orbital, then from the 3d orbitals. For Fe²⁺, two electrons are removed (both 4s electrons), giving [Ar] 3d⁶. For Fe³⁺, three electrons are removed (two 4s and one 3d), giving [Ar] 3d⁵. Now apply Hund's rule: electrons fill orbitals singly before pairing up. In Fe²⁺ (3d⁶), the five d orbitals each get one electron first, then one orbital gets a second electron, resulting in 4 unpaired electrons. In Fe³⁺ (3d⁵), each of the five d orbitals gets exactly one electron, resulting in 5 unpaired electrons. Since Fe³⁺ has 5 unpaired electrons compared to Fe²⁺'s 4 unpaired electrons, answer B is correct. Answer A is wrong because Fe²⁺ actually has fewer unpaired electrons (4) than Fe³⁺ (5). Answer C is incorrect since the ions have different numbers of unpaired electrons (4 vs. 5). Answer D is wrong because Fe²⁺ has 6 d electrons, not the 5 needed for a half-filled d subshell—that describes Fe³⁺. Study tip: Remember that d⁵ and d¹⁰ configurations are especially stable due to half-filled and fully-filled subshells. Always remove 4s electrons before 3d electrons when forming transition metal cations.