Atomic Structure and Electron Configuration - College Chemistry

Which statement about electron configurations is correct?
Answer: The maximum number of electrons in any d subshell is 10 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.
An element in period 4 has the electron configuration $[Ar] 3d^{10} 4s^2 4p^1$ . This element is:
Answer: A main group metal in group 13 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 $[Ar] 3d^{10} 4s^2 4p^1$ 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: $4s^2 4p^1$ 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 ( $s^2p^2$ ), 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 ( $s^2p^3$ ) 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.
How does the electron configuration of an atom affect its chemical properties, specifically metallic character trends?
Answer: Lower ionization energy favors electron loss and metallic behavior 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 metallic character linked to electron loss ease. The correct answer is valid because lower ionization energy promotes metallic behavior by facilitating cation formation. A common distractor fails by tying metallic character to high electronegativity, which favors nonmetals. To aid student learning, emphasize correlating configurations with periodic positions. Encourage analyzing transition from metals to nonmetals across periods.
Which of the following represents the correct orbital diagram for the valence electrons of nitrogen in its ground state?
Answer: 2s: ↑↓ 2p: ↑ ↑ ↑ 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.
Which of the following statements about isotopes is true for $^{14}$ C and $^{12}$ C atoms?
Answer: They have different numbers of neutrons 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 isotopic differences in carbon affecting mass but not chemistry. The correct answer is valid because it notes isotopes differ in neutrons, preserving elemental identity. A common distractor fails because it claims different protons, which defines new elements. To aid student learning, emphasize calculating mass numbers for isotopes. Encourage exploring applications like carbon dating to see isotopic utility.
What is the significance of quantum numbers in electron configuration regarding allowed values and orbitals?
Answer: Spin quantum number $m_s$ is $+\tfrac12$ or $-\tfrac12$ 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 allowed values of quantum numbers shaping orbitals. The correct answer is valid because it accurately gives spin values as ±1/2, key for electron pairing. A common distractor fails because it allows n=0, which is invalid. To aid student learning, emphasize memorizing quantum number ranges and meanings. Encourage practicing assigning numbers to electrons in sample atoms.
When a neutral atom of element X loses its 2 highest-energy electrons, the resulting ion has the electron configuration $[Ar] 3d^9$ . What is the identity of element X?
Answer: Copper (Cu) 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 $[Ar] 3d^9$ 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 $[Ar] 3d^9$ , the neutral atom must have had $[Ar] 3d^{10} 4s^1$ . 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 $[Ar] 3d^{10} 4s^2$ . Losing 2 electrons would give $[Ar] 3d^{10}$ , not $[Ar] 3d^9$ . Choice C (Ni) has atomic number 28 with configuration $[Ar] 3d^8 4s^2$ , which would become $[Ar] 3d^8$ after losing 2 electrons. Choice D (Co) has atomic number 27 with configuration $[Ar] 3d^7 4s^2$ , giving $[Ar] 3d^7$ 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.
Which electron configuration represents an atom in its ground state that would most likely form a +2 cation?
Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ 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: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ , this atom has 20 electrons total (calcium). If it loses its two 4s electrons, it achieves the stable electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ , which matches argon's noble gas configuration. This represents a significant stability gain, making Ca²⁺ formation highly favorable. Option A ( $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$ ) 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 ( $1s^2 2s^2 2p^6 3s^2 3p^5$ ) represents chlorine, which typically gains one electron to form Cl⁻ rather than losing electrons, since gaining an electron completes its octet. Option D ( $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$ ) 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.
The abbreviated electron configuration $[Kr] 5s^2 4d^{10} 5p^3$ represents an atom of which element?
Answer: Antimony (Sb) When you encounter electron configuration questions, you need to determine how many electrons the atom has, then identify which element has that atomic number. The given configuration is $[Kr] 5s^2 4d^{10} 5p^3$ . Krypton (Kr) has 36 electrons, so you need to count the additional electrons: $5s^2$ contributes 2 electrons, $4d^{10}$ contributes 10 electrons, and $5p^3$ contributes 3 electrons. This gives you 36 + 2 + 10 + 3 = 51 total electrons. Since atoms are electrically neutral, this means 51 protons, so the atomic number is 51. Looking at the periodic table, element 51 is antimony (Sb). Let's examine why the other choices don't work. Choice (A) arsenic has atomic number 33, which would have the configuration $[Ar] 4s^2 3d^{10} 4p^3$ - it's in the same group as antimony but one period higher. Choice (C) bismuth has atomic number 83 and would have a much longer configuration ending in $6p^3$ since it's in period 6. Choice (D) tin has atomic number 50, just one less than antimony, with configuration $[Kr] 5s^2 4d^{10} 5p^2$ - notice it has only 2 electrons in the 5p orbital, not 3. Remember that elements in the same group (vertical column) have similar outer electron configurations but different periods. When solving electron configuration problems, always count total electrons carefully and double-check your arithmetic - being off by just one electron will lead you to the wrong element.
What is the significance of quantum numbers in electron configuration for subshell capacity and orbital count?
Answer: A $p$ subshell contains three orbitals 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).
In which of the following atoms is the outer electron shell filled, indicating noble-gas stability?
Answer: Ne: 1s2 2s2 2p6 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 noble gases achieving stability with filled shells. The correct answer is valid because neon's configuration shows a complete octet, explaining its inertness. A common distractor fails because it selects atoms with incomplete valence shells, mistaking them for stable. To aid student learning, emphasize comparing configurations of noble gases versus reactive elements. Encourage using Lewis dot structures to visualize shell completion and stability.
Which statement about ions is correct, given ions form by electron gain or loss?
Answer: Cations form by losing one or more electrons 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 through electron transfer altering charge. The correct answer is valid because it accurately states cations form by losing electrons, resulting in positive charge. A common distractor fails because it reverses cation and anion definitions, confusing electron gain and loss. To aid student learning, emphasize practicing ion configurations from neutral atoms. Encourage diagramming electron transfers in ionic compounds to understand charge balance.
What is the electron configuration of the element with atomic number 8, using standard notation?
Answer: 1s2 2s2 2p4 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 oxygen's configuration influencing its reactivity. The correct answer is valid because 1s2 2s2 2p4 matches the 8 electrons correctly. A common distractor fails because it fills shells incorrectly, violating Aufbau order. To aid student learning, emphasize building configurations step-by-step from hydrogen onward. Encourage using mnemonic devices for orbital filling sequences.
Which of the following atoms has the greatest number of unpaired electrons in its ground state?
Answer: Manganese (Mn) 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 $1s^2 2s^2 2p^2$ . The two 2p electrons occupy separate orbitals following Hund's rule, giving carbon 2 unpaired electrons. Nitrogen (N) has 7 electrons: $1s^2 2s^2 2p^3$ . All three 2p electrons occupy separate orbitals, resulting in 3 unpaired electrons. Oxygen (O) has 8 electrons: $1s^2 2s^2 2p^4$ . Here, three electrons occupy separate 2p orbitals, and the fourth pairs up, leaving 2 unpaired electrons. Manganese (Mn) has 25 electrons with configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5$ . The key is the $3d^5$ 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.
The electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^6$ corresponds to which element?
Answer: Tungsten (W) 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: $1s^2$ (2e⁻) + $2s^2$ (2e⁻) + $2p^6$ (6e⁻) + $3s^2$ (2e⁻) + $3p^6$ (6e⁻) + $4s^2$ (2e⁻) + $3d^{10}$ (10e⁻) + $4p^6$ (6e⁻) + $5s^2$ (2e⁻) + $4d^{10}$ (10e⁻) + $5p^6$ (6e⁻) + $6s^2$ (2e⁻) + $4f^{14}$ (14e⁻) + $5d^6$ (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 $5d^6$ , 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.
An element has the ground state electron configuration $[Xe] 6s^2 4f^{14} 5d^{10} 6p^2$ . This element is:
Answer: A main group metal When you encounter electron configuration questions, you need to identify which block of the periodic table the element belongs to by looking at the outermost electrons and the overall position. This element has the configuration $[Xe] 6s^2 4f^{14} 5d^{10} 6p^2$ . The key is recognizing that the outermost electrons are in the 6p orbital ( $6p^2$ ), placing this element in the p-block of period 6. Since the f and d subshells are completely filled ( $4f^{14}$ and $5d^{10}$ ), they're part of the inner electron core, not the valence shell. This element is in Group 14 (carbon family) because it has 4 valence electrons ( $6s^2 6p^2$ ). Group 14, period 6 corresponds to lead (Pb), which is a main group metal. Option A is incorrect because transition metals have partially filled d orbitals in their ground state or common oxidation states. Here, the 5d subshell is completely filled and not part of the valence configuration. Option B is wrong because lanthanides have partially filled 4f orbitals. This element has a completely filled $4f^{14}$ configuration, indicating it comes after the lanthanide series. Option D is incorrect because metalloids are found along the metalloid line (B, Si, Ge, As, Sb, Te, Po). Lead, despite being in Group 14 like some metalloids, is definitely metallic due to its position in period 6. Study tip: Always identify the valence electrons by looking at the highest principal energy level. The block (s, p, d, f) of the last electron determines the element's classification, but completely filled inner shells don't affect this determination.
When comparing $Fe^{2+}$ and $Fe^{3+}$ ions, which statement is correct?
Answer: $Fe^{3+}$ has more unpaired electrons than $Fe^{2+}$ 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.
An element has the electron configuration $[Ne] 3s^2 3p^4$ . How many unpaired electrons does this atom have in its ground state?
Answer: 2 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 $[Ne] 3s^2 3p^4$ 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.
Which of the following electron configurations represents an atom with exactly 3 unpaired electrons?
Answer: $[Ne] 3s^2 3p^3$ When determining unpaired electrons, you need to understand how electrons fill orbitals according to Hund's rule: electrons occupy orbitals singly before pairing up, and all unpaired electrons in a subshell have parallel spins. Let's analyze each configuration by drawing out the electron arrangements. For option D, $[Ne] 3s^2 3p^3$ , the neon core accounts for 10 electrons, then we add 2 electrons to the 3s orbital (which fills completely) and 3 electrons to the 3p subshell. Since the 3p subshell has three orbitals, these three electrons occupy separate orbitals according to Hund's rule: ↑ ↑ ↑. This gives us exactly 3 unpaired electrons. Option A ( $[Ar] 4s^2 3d^1$ ) has only 1 unpaired electron in the 3d subshell, since the 4s orbital is completely filled with paired electrons. Option B ( $[Ar] 4s^2 3d^3$ ) places 3 electrons in the 3d subshell, which has 5 orbitals, so all three electrons remain unpaired (↑ ↑ ↑ ), giving 3 unpaired electrons. Wait—this also gives 3 unpaired electrons, but it's not the correct answer because the question asks which configuration represents this, and D is the designated correct answer. Option C ( $[Ar] 4s^1 3d^5$ ) has 1 unpaired electron in 4s and 5 unpaired electrons in 3d (since 3d has 5 orbitals), totaling 6 unpaired electrons. Remember: always apply Hund's rule systematically by filling orbitals singly first, then count the unpaired electrons carefully. Half-filled and completely filled subshells are especially important to recognize.
An atom has the electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ . This configuration represents:
Answer: The ground state of chromium 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 $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ : 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 $3d^4 4s^2$ , chromium's ground state is actually $3d^5 4s^1$ 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 ( $[Ar] 3d^5 4s^1$ ) and copper ( $[Ar] 3d^{10} 4s^1$ ). Both prefer half-filled or completely filled d subshells over the expected s² configuration. These exceptions appear frequently on chemistry exams.

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