Sign up
← Back to Learn by Concept

Chemistry · Learn by Concept

Chemistry Help: Model Electron Configuration

Review real example questions for Model Electron Configuration in Chemistry.

Question 1 / 10

0 of 10 answered

Oxygen (O) has atomic number 8. What is the electron configuration of a neutral oxygen atom?

All questions

Question 1

Oxygen (O) has atomic number 8. What is the electron configuration of a neutral oxygen atom?

  1. 1s² 2s² 2p⁶
  2. 1s² 2s² 2p⁴ (correct answer)
  3. 1s² 2s¹ 2p⁵
  4. 1s² 2s² 2p³ 3s¹

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For oxygen with 8 electrons, fill: 1s² (2), 2s² (4 total), and the remaining 4 go into 2p⁴, giving 1s² 2s² 2p⁴ with 6 valence electrons in shell 2 (2 in 2s + 4 in 2p). Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total 8 electrons. Choice A shows 10 electrons, which would be for neon, not oxygen—always double-check the total electron count matches the atomic number! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell! You're making excellent progress—stay confident!

Question 2

An oxide ion is written as O²⁻. What is the full electron configuration of O²⁻?

  1. 1s² 2s² 2p⁴
  2. 1s² 2s² 2p⁶ (correct answer)
  3. 1s² 2s² 2p⁷
  4. 1s² 2s² 2p⁶ 3s²

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, then 3d, etc. The filling order for the first 20 elements goes: 1s, 2s, 2p, 3s, 3p, 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For O²⁻, oxygen has atomic number 8 but gains 2 electrons (negative charge means more electrons), so 10 electrons: fill 1s² (2), 2s² (4), 2p⁶ (10), with 8 valence electrons in shell 2, matching neon's stable configuration. Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 8 adjusted for -2 charge to 10 electrons). A distractor like A is for neutral oxygen—remember, anions gain electrons to fill the valence shell, so O²⁻ adds 2 to 2p⁴ making 2p⁶! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 3

Using the Aufbau filling order (1s, 2s, 2p, 3s, 3p, 4s), what is the electron configuration of the neutral magnesium atom (Mg, atomic number 12)?

  1. 1s² 2s² 2p⁶ 3s¹ 3p¹
  2. 1s² 2s² 2p⁶ 3s² (correct answer)
  3. 1s² 2s² 2p⁵ 3s² 3p¹
  4. 1s² 2s² 2p⁶ 3s² 3p²

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For magnesium (Mg, atomic number 12), we fill: 1s² (2 electrons), 2s² (4 total), 2p⁶ (10 total), 3s² (12 total), so the configuration is 1s² 2s² 2p⁶ 3s², with 2 valence electrons in the 3s subshell. Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 12). For example, choice A incorrectly places electrons in 3p before fully filling 3s, which violates the filling order—keep practicing to spot these! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 4

What is the electron configuration of the sodium ion, Na⁺ (sodium has atomic number 11)?

  1. 1s² 2s² 2p⁶ (correct answer)
  2. 1s² 2s² 2p⁶ 3s¹
  3. 1s² 2s² 2p⁵ 3s²
  4. 1s² 2s² 2p⁶ 3s²

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For Na⁺ (sodium atomic number 11, but ion has 10 electrons), start with neutral Na 1s² 2s² 2p⁶ 3s¹ and remove 1 electron from the valence 3s, resulting in 1s² 2s² 2p⁶, with no valence electrons in shell 3 anymore (it matches neon's configuration). Choice A correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons adjusted for ion charge (11 - 1 = 10). For example, choice B keeps the 3s¹, which would be neutral Na, not the ion—great job spotting that cations lose electrons! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 5

How many electrons are in the outermost shell of phosphorus (P), atomic number 15?

  1. 3
  2. 5 (correct answer)
  3. 8
  4. 15

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, then 3d, etc. The filling order for the first 20 elements goes: 1s, 2s, 2p, 3s, 3p, 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For phosphorus with atomic number 15, configuration is 1s² 2s² 2p⁶ 3s² 3p³, so outermost shell 3 has 2 (3s) + 3 (3p) = 5 electrons, matching group 15. Choice B correctly counts 5 electrons in the outermost shell. A distractor like A might only count 3p, but include the entire shell—add 3s and 3p for the total in shell 3! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 6

How many valence electrons does chlorine (Cl), atomic number 17, have?

  1. 2
  2. 5
  3. 7 (correct answer)
  4. 17

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, then 3d, etc. The filling order for the first 20 elements goes: 1s, 2s, 2p, 3s, 3p, 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For chlorine with atomic number 17, the configuration is 1s² 2s² 2p⁶ 3s² 3p⁵, so the outermost shell 3 has 2 (in 3s) + 5 (in 3p) = 7 valence electrons, which matches its group 17 position. Choice C correctly identifies 7 valence electrons following the configuration and valence counting. A distractor like B might count only the 3p electrons, but remember to include all electrons in the highest shell, so add 3s and 3p together for valence! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 7

Which full electron configuration correctly represents fluorine (F), atomic number 9?

  1. 1s² 2s² 2p⁵ (correct answer)
  2. 1s² 2s² 2p⁶
  3. 1s² 2s¹ 2p⁶
  4. 1s² 2s² 2p⁷

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, then 3d, etc. The filling order for the first 20 elements goes: 1s, 2s, 2p, 3s, 3p, 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For fluorine with atomic number 9, fill: 1s² (2), 2s² (4), 2p⁵ (9), so configuration 1s² 2s² 2p⁵, with 7 valence electrons in shell 2 (2+5). Choice A correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 9). A distractor like B has 2p⁶, which is 10 electrons for neon—fluorine has only 9, so stop at 2p⁵ after adding exactly 9! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 8

What is the full electron configuration of magnesium (Mg), atomic number 12?

  1. 1s² 2s² 2p⁶ 3s¹
  2. 1s² 2s² 2p⁶ 3s² (correct answer)
  3. 1s² 2s² 2p⁶ 3p²
  4. 1s² 2s² 2p⁸ 3s²

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, then 3d, etc. The filling order for the first 20 elements goes: 1s, 2s, 2p, 3s, 3p, 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For magnesium with atomic number 12, fill: 1s² (2), 2s² (4), 2p⁶ (10), 3s² (12), so configuration 1s² 2s² 2p⁶ 3s², with 2 valence electrons in shell 3. Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 12). A distractor like D uses 2p⁸, but 2p can hold only 6 electrons max—always respect subshell capacities: s=2, p=6! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 9

Magnesium has atomic number 12. When magnesium forms Mg²⁺, which electron configuration represents Mg²⁺?

  1. 1s² 2s² 2p⁶ 3s²
  2. 1s² 2s² 2p⁶ (correct answer)
  3. 1s² 2s² 2p⁴ 3s²
  4. 1s² 2s² 2p⁶ 3s¹

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For Mg²⁺ (magnesium atomic number 12, ion has 10 electrons), start with neutral Mg 1s² 2s² 2p⁶ 3s² and remove 2 electrons from valence 3s, resulting in 1s² 2s² 2p⁶ (matching neon). Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons adjusted for ion charge (12 - 2 = 10). For example, choice A is neutral Mg, not accounting for electron loss—cations lose from the outer shell first! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 10

Calcium has atomic number 20. What is the electron configuration of a neutral calcium atom (Ca)?

  1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² (correct answer)
  2. 1s² 2s² 2p⁶ 3s² 3p⁸
  3. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d²
  4. 1s² 2s² 2p⁶ 3s² 3p⁵ 4s³

Explanation: This question tests your ability to construct electron configurations showing how electrons are distributed in shells and subshells around the nucleus, following the Aufbau principle (filling order), Pauli exclusion principle (max 2 per orbital), and recognizing valence electrons. Electron configuration describes where electrons are located using notation like 1s² 2s² 2p⁶ where the number indicates the shell (1, 2, 3...), the letter indicates the subshell type (s, p, d), and the superscript shows how many electrons are in that subshell. Electrons fill in a specific order from lowest to highest energy: 1s (holds 2), then 2s (holds 2), then 2p (holds 6), then 3s (holds 2), then 3p (holds 6), then 4s, and you keep adding electrons until you've placed all of them (total electrons = atomic number for neutral atoms). Valence electrons are the electrons in the outermost shell—these are the ones involved in bonding and chemical reactions! For calcium (Ca, atomic number 20), fill up to 3p⁶ (18 electrons), then add 4s² (20 total), so 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², with 2 valence electrons in shell 4. Choice A correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 20). For example, choice C incorrectly places electrons in 3d before 4s, but remember 4s fills before 3d for these elements—keep that order in mind! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions (Na⁺ has 11 - 1 = 10 electrons). (2) Fill in order: 1s (add 2 electrons), 2s (add 2 more), 2p (add 6 more), 3s (add 2 more), 3p (add 6 more), 4s (add 2 more). Stop when you've placed all electrons. (3) Write configuration: 1s² 2s² 2p⁶ 3s¹ for sodium (11 total: 2+2+6+1=11). Check your total matches atomic number! (4) Identify valence: the outermost shell (highest n) electrons. For sodium 1s² 2s² 2p⁶ 3s¹, the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen 1s² 2s² 2p⁴, outermost is shell 2 with 2+4=6 electrons, so 6 valence electrons. Quick valence shortcut for main group elements: group number often equals valence electrons! Group 1 = 1 valence, group 2 = 2 valence, group 13 = 3 valence, group 14 = 4 valence, etc. For ions, remember: cations (positive) LOSE electrons from outermost shell first. Na (1s² 2s² 2p⁶ 3s¹) loses that 3s¹ to become Na⁺ (1s² 2s² 2p⁶). Anions (negative) GAIN electrons into valence shell. F (1s² 2s² 2p⁵) gains 1 in 2p to become F⁻ (1s² 2s² 2p⁶). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!