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!

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!

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!

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!

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!

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!

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!

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!

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 aluminum with 13 electrons, start filling: 1s² (2), 2s² (4 total), 2p⁶ (10 total), 3s² (12 total), and the last electron goes into 3p¹, making the configuration 1s² 2s² 2p⁶ 3s² 3p¹ with 3 valence electrons in shell 3 (2 in 3s + 1 in 3p). Choice A correctly constructs the electron configuration following Aufbau filling order and properly accounts for total 13 electrons. Choice B fails by placing only one electron in 3s and two in 3p, which violates the order of filling 3s fully before 3p; remember to fill subshells in sequence! 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! Keep practicing, you're doing great!

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 with 20 electrons, fill up to 1s² 2s² 2p⁶ 3s² 3p⁶ (18 total), then the last two go into 4s², giving 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² with 2 valence electrons in shell 4. Choice A correctly constructs the electron configuration following Aufbau by filling 4s before 3d and accounting for total 20 electrons. Choice B incorrectly fills 3d before 4s, but remember the order: 4s comes before 3d in filling sequence for these elements! 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! Impressive handling of higher atomic numbers—keep shining!