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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?

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Question 1

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

1s² 2s² 2p⁶
1s² 2s² 2p⁴ (correct answer)
1s² 2s¹ 2p⁵
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

Fluoride ion is written as F⁻. Fluorine has atomic number 9. What is the electron configuration of F⁻?

1s² 2s² 2p⁵
1s² 2s² 2p⁶ (correct answer)
1s² 2s² 2p⁷
1s² 2s¹ 2p⁶

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 F⁻, neutral F has 9 electrons (1s² 2s² 2p⁵), but the -1 charge means it gains 1 electron into the 2p subshell, resulting in 10 electrons with configuration 1s² 2s² 2p⁶, completing shell 2 with 8 valence electrons (like neon). Choice B correctly constructs the electron configuration for the anion by adding an electron to the valence shell and accounting for the total of 10 electrons. Choice A shows the neutral F configuration, forgetting to add the extra electron for the negative charge—always increase electron count for anions! 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! Fantastic work with anions—keep going!

Question 3

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

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

A sodium ion is written as $\mathrm{Na}^{+}$ . What is the full electron configuration of $\mathrm{Na}^{+}$ ?

$1s^2 2s^2 2p^6 3s^1$
$1s^2 2s^2 2p^6$ (correct answer)
$1s^2 2s^2 2p^5 3s^1$
$1s^2 2s^2 2p^6 3s^2$

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 $1s^2 2s^2 2p^6$ 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 $\mathrm{Na}^{+}$ , sodium has atomic number 11 but loses 1 electron (positive charge means fewer electrons), so 10 electrons: fill $1s^2$ (2), $2s^2$ (4), $2p^6$ (10), with no valence electron in shell 3 since it's empty now, resembling neon's configuration. Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 11 adjusted for +1 charge to 10 electrons). A distractor like A includes the $3s^1$ , but that's for neutral Na—remember, cations lose electrons from the outermost shell, so $\mathrm{Na}^{+}$ removes the $3s^1$ electron! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions ( $\mathrm{Na}^{+}$ 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: $1s^2 2s^2 2p^6 3s^1$ 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 $1s^2 2s^2 2p^6 3s^1$ , the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen $1s^2 2s^2 2p^4$ , 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 ( $1s^2 2s^2 2p^6 3s^1$ ) loses that $3s^1$ to become $\mathrm{Na}^{+}$ ( $1s^2 2s^2 2p^6$ ). Anions (negative) GAIN electrons into valence shell. F ( $1s^2 2s^2 2p^5$ ) gains 1 in $2p$ to become F $^{-}$ ( $1s^2 2s^2 2p^6$ ). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 5

The magnesium ion is written as Mg²⁺. Magnesium has atomic number 12. What is the electron configuration of Mg²⁺?

1s² 2s² 2p⁶ 3s²
1s² 2s² 2p⁶ (correct answer)
1s² 2s² 2p⁴
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⁺, neutral Mg has 12 electrons (1s² 2s² 2p⁶ 3s²), but the +2 charge means it loses 2 electrons from the outermost 3s subshell, leaving 10 electrons with configuration 1s² 2s² 2p⁶, and now shell 2 is the outermost with 8 valence electrons (like neon). Choice B correctly constructs the electron configuration for the ion by removing electrons from the valence shell and accounting for the total of 10 electrons. Choice A fails to account for the ion charge, showing the neutral Mg configuration instead; always adjust electron count for ions! 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! Great job tackling ions—you've got this!

Question 6

Nitrogen (N) has atomic number 7. What is the correct electron configuration for a neutral nitrogen atom?

1s² 2s² 2p²
1s² 2s² 2p³ (correct answer)
1s² 2s¹ 2p⁴
1s² 2p⁵

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 nitrogen with 7 electrons, fill: 1s² (2), 2s² (4 total), and the remaining 3 go into 2p³, giving 1s² 2s² 2p³ with 5 valence electrons in shell 2 (2 in 2s + 3 in 2p). Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total 7 electrons. Choice A underfills 2p with only 2 electrons, but after 2s², the next 3 electrons go into 2p—count carefully to reach exactly 7! 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 nailing these configurations—keep the momentum!

Question 7

A calcium ion is written as Ca²⁺. What is the full electron configuration of Ca²⁺? (Calcium has atomic number 20.)

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

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 Ca²⁺, calcium has atomic number 20 but loses 2 electrons, so 18 electrons: fill up to 1s² 2s² 2p⁶ 3s² 3p⁶ (18), with no electrons in shell 4, resembling argon's configuration. Choice C correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 20 adjusted for +2 charge to 18 electrons). A distractor like A includes 4s², but that's for neutral Ca—cations lose from outermost 4s first, removing both to form Ca²⁺! 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

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)?

1s² 2s² 2p⁶ 3s¹ 3p¹
1s² 2s² 2p⁶ 3s² (correct answer)
1s² 2s² 2p⁵ 3s² 3p¹
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 9

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

1s² 2s² 2p⁶ (correct answer)
1s² 2s² 2p⁶ 3s¹
1s² 2s² 2p⁵ 3s²
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 10

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

3
5 (correct answer)
8
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!

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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?

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

Question 2

Fluoride ion is written as F⁻. Fluorine has atomic number 9. What is the electron configuration of F⁻?

1s² 2s² 2p⁵
1s² 2s² 2p⁶ (correct answer)
1s² 2s² 2p⁷
1s² 2s¹ 2p⁶

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 F⁻, neutral F has 9 electrons (1s² 2s² 2p⁵), but the -1 charge means it gains 1 electron into the 2p subshell, resulting in 10 electrons with configuration 1s² 2s² 2p⁶, completing shell 2 with 8 valence electrons (like neon). Choice B correctly constructs the electron configuration for the anion by adding an electron to the valence shell and accounting for the total of 10 electrons. Choice A shows the neutral F configuration, forgetting to add the extra electron for the negative charge—always increase electron count for anions! 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! Fantastic work with anions—keep going!

Question 3

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

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

A sodium ion is written as $\mathrm{Na}^{+}$ . What is the full electron configuration of $\mathrm{Na}^{+}$ ?

$1s^2 2s^2 2p^6 3s^1$
$1s^2 2s^2 2p^6$ (correct answer)
$1s^2 2s^2 2p^5 3s^1$
$1s^2 2s^2 2p^6 3s^2$

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 $1s^2 2s^2 2p^6$ 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 $\mathrm{Na}^{+}$ , sodium has atomic number 11 but loses 1 electron (positive charge means fewer electrons), so 10 electrons: fill $1s^2$ (2), $2s^2$ (4), $2p^6$ (10), with no valence electron in shell 3 since it's empty now, resembling neon's configuration. Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 11 adjusted for +1 charge to 10 electrons). A distractor like A includes the $3s^1$ , but that's for neutral Na—remember, cations lose electrons from the outermost shell, so $\mathrm{Na}^{+}$ removes the $3s^1$ electron! The electron configuration recipe for elements 1-20: (1) Determine total electrons: atomic number for neutral atoms, atomic number minus charge for ions ( $\mathrm{Na}^{+}$ 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: $1s^2 2s^2 2p^6 3s^1$ 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 $1s^2 2s^2 2p^6 3s^1$ , the outermost shell is shell 3 with 1 electron, so 1 valence electron. For oxygen $1s^2 2s^2 2p^4$ , 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 ( $1s^2 2s^2 2p^6 3s^1$ ) loses that $3s^1$ to become $\mathrm{Na}^{+}$ ( $1s^2 2s^2 2p^6$ ). Anions (negative) GAIN electrons into valence shell. F ( $1s^2 2s^2 2p^5$ ) gains 1 in $2p$ to become F $^{-}$ ( $1s^2 2s^2 2p^6$ ). Check: does your ion configuration make sense? Cations should look like previous noble gas, anions should complete the outer shell!

Question 5

The magnesium ion is written as Mg²⁺. Magnesium has atomic number 12. What is the electron configuration of Mg²⁺?

1s² 2s² 2p⁶ 3s²
1s² 2s² 2p⁶ (correct answer)
1s² 2s² 2p⁴
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⁺, neutral Mg has 12 electrons (1s² 2s² 2p⁶ 3s²), but the +2 charge means it loses 2 electrons from the outermost 3s subshell, leaving 10 electrons with configuration 1s² 2s² 2p⁶, and now shell 2 is the outermost with 8 valence electrons (like neon). Choice B correctly constructs the electron configuration for the ion by removing electrons from the valence shell and accounting for the total of 10 electrons. Choice A fails to account for the ion charge, showing the neutral Mg configuration instead; always adjust electron count for ions! 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! Great job tackling ions—you've got this!

Question 6

Nitrogen (N) has atomic number 7. What is the correct electron configuration for a neutral nitrogen atom?

1s² 2s² 2p²
1s² 2s² 2p³ (correct answer)
1s² 2s¹ 2p⁴
1s² 2p⁵

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 nitrogen with 7 electrons, fill: 1s² (2), 2s² (4 total), and the remaining 3 go into 2p³, giving 1s² 2s² 2p³ with 5 valence electrons in shell 2 (2 in 2s + 3 in 2p). Choice B correctly constructs the electron configuration following Aufbau filling order and properly accounts for total 7 electrons. Choice A underfills 2p with only 2 electrons, but after 2s², the next 3 electrons go into 2p—count carefully to reach exactly 7! 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 nailing these configurations—keep the momentum!

Question 7

A calcium ion is written as Ca²⁺. What is the full electron configuration of Ca²⁺? (Calcium has atomic number 20.)

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

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 Ca²⁺, calcium has atomic number 20 but loses 2 electrons, so 18 electrons: fill up to 1s² 2s² 2p⁶ 3s² 3p⁶ (18), with no electrons in shell 4, resembling argon's configuration. Choice C correctly constructs the electron configuration following Aufbau filling order and properly accounts for total electrons (atomic number 20 adjusted for +2 charge to 18 electrons). A distractor like A includes 4s², but that's for neutral Ca—cations lose from outermost 4s first, removing both to form Ca²⁺! 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

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)?

1s² 2s² 2p⁶ 3s¹ 3p¹
1s² 2s² 2p⁶ 3s² (correct answer)
1s² 2s² 2p⁵ 3s² 3p¹
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 9

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

1s² 2s² 2p⁶ (correct answer)
1s² 2s² 2p⁶ 3s¹
1s² 2s² 2p⁵ 3s²
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 10

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

3
5 (correct answer)
8
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!