Question 1:

To determine the electronic configurations of the Al³⁺ and S²⁻ ions:

1. Aluminum (Al) atom has an atomic number of 13, so its electron configuration is:

   • Neutral Al: 1s² 2s² 2p⁶ 3s² 3p¹ or [Ne] 3s² 3p¹ (which is 2,8,3).

   • Al³⁺ ion loses 3 electrons (from the 3s and 3p orbitals), so it becomes:
     1s² 2s² 2p⁶ or [Ne] (which is 2,8).

2. Sulfur (S) atom has an atomic number of 16, so its electron configuration is:

   • Neutral S: 1s² 2s² 2p⁶ 3s² 3p⁴ or [Ne] 3s² 3p⁴ (which is 2,8,6).

   • S²⁻ ion gains 2 electrons (added to the 3p orbital), so it becomes:
     1s² 2s² 2p⁶ 3s² 3p⁶ or [Ar] (which is 2,8,8).

Now, comparing with the options:

• A: Al³⁺ (2,8,3) and S²⁻ (2,8,6) → Incorrect for both.

• B: Al³⁺ (2,8) and S²⁻ (2,8,8) → Correct.

• C: Al³⁺ (2,8,6) and S²⁻ (2,8,4) → Incorrect for both.

• D: Al³⁺ (2,8,8) and S²⁻ (2,8) → Incorrect for both.

Answer: B (Al³⁺: 2,8 and S²⁻: 2,8,8)

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 2:

The number of orbitals in a sub-shell is determined by the magnetic quantum number (ml m l​ ). For a given sub-shell with angular momentum quantum number l l, the number of orbitals is 2l+1 2l + 1.

• For an f sub-shell, l=3 l = 3.

• Therefore, the number of orbitals = 2×3+1=7 2 × 3 + 1 = 7.

So, there are 7 orbitals in an f sub-shell.

Answer: B. 7

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 3:

To determine which list of species all have the electronic configuration 1s²2s²2p⁶3s²3p⁶ (which is the same as the electron configuration of argon, [Ar]), we need to check the number of electrons in each species and see if it matches argon (18 electrons).

Let's evaluate each option:

Option A: K⁺, Ar, P³⁻

• K⁺ (potassium ion): Atomic number of K is 19, so K⁺ has 19 - 1 = 18 electrons. ✅

• Ar (argon): Atomic number 18, so 18 electrons. ✅

• P³⁻ (phosphide ion): Atomic number of P is 15, so P³⁻ has 15 + 3 = 18 electrons. ✅
  All have 18 electrons and the configuration [Ar].

Option B: Ne, S²⁻, Cl⁻

• Ne (neon): Atomic number 10, so 10 electrons (configuration is 1s22s22p6 1s 22s 22p 6, not [Ar]). ❌

• S²⁻ (sulfide ion): Atomic number of S is 16, so S²⁻ has 16 + 2 = 18 electrons. ✅

• Cl⁻ (chloride ion): Atomic number of Cl is 17, so Cl⁻ has 17 + 1 = 18 electrons. ✅
  But Ne does not have 18 electrons, so not all species have the same configuration.

Option C: Ar, Mg²⁺, Ca²⁺

• Ar (argon): 18 electrons. ✅

• Mg²⁺ (magnesium ion): Atomic number of Mg is 12, so Mg²⁺ has 12 - 2 = 10 electrons (configuration is [Ne], not [Ar]). ❌

• Ca²⁺ (calcium ion): Atomic number of Ca is 20, so Ca²⁺ has 20 - 2 = 18 electrons. ✅
  But Mg²⁺ does not have 18 electrons.

Option D: Mg²⁺, Na⁺, O²⁻

• Mg²⁺: As above, has 10 electrons ([Ne]). ❌

• Na⁺ (sodium ion): Atomic number of Na is 11, so Na⁺ has 11 - 1 = 10 electrons ([Ne]). ❌

• O²⁻ (oxide ion): Atomic number of O is 8, so O²⁻ has 8 + 2 = 10 electrons ([Ne]). ❌
  None have 18 electrons; all have 10 electrons.

Only Option A has all species with 18 electrons and the configuration 1s²2s²2p⁶3s²3p⁶

Answer: A. K⁺, Ar, P³⁻

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 4:

To determine which element has a ground state electronic configuration ending in np⁴, we need to look for an element whose configuration has np⁴ as the last part. This means the element is in the p-block and has four electrons in its p subshell.

Let's analyze each option:

A. Iodine (I) - Atomic number 53

• Electron configuration: [Kr]5s²4d¹⁰5p⁵ [Kr]5s 24d 105p 5

• Ends with 5p⁵, not np⁴.

B. Chromium (Cr) - Atomic number 24

• Electron configuration: [Ar]4s¹³d⁵ (exception to the rule)

• Does not have a p subshell as the last part; it ends in d.

C. Selenium (Se) - Atomic number 34

• Electron configuration: [Ar]4s²3d¹⁰4p⁴

• Ends with 4p⁴.

D. Silicon (Si) - Atomic number 14

• Electron configuration: [Ne]3s²3p²

• Ends with 3p², not np⁴.

Only selenium (Se) has the configuration ending in np⁴.

Answer: C. Selenium, Se

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 5:

To determine the electronic configuration of the iron(III) ion (Fe³⁺), let's first recall the electron configuration of neutral iron (Fe).

• Iron (Fe) has an atomic number of 26.

• The ground state electron configuration of neutral Fe is: [Ar]4s²3d⁶

Now, the iron(III) ion (Fe³⁺) has lost 3 electrons. When forming ions, electrons are removed first from the 4s orbital and then from the 3d orbital.

• Remove 2 electrons from the 4s orbital: becomes [Ar]3d⁶

• Remove 1 more electron from the 3d orbital: becomes [Ar]3d⁵

So, the electron configuration of Fe³⁺ is [Ar]3d⁵.

Answer: B. [Ar] 3d⁵

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 6:

To determine the correct "electron in boxes" diagram for the phosphorus (P) atom, we need to know its electron configuration and how the electrons are arranged in the orbitals.

• Phosphorus has an atomic number of 15.

• Its electron configuration is: 1s²2s²2p⁶3s²3p³ or [Ne]3s²3p³.

• The [Ne] core is 1s²2s²2p⁶, which is fully filled.

• The valence electrons are in the 3s and 3p subshells:

  • The 3s orbital has 2 electrons (paired, with opposite spins).

  • The 3p orbitals have 3 electrons. According to Hund's rule, these electrons will occupy the three 3p orbitals singly with parallel spins (all up arrows or all down arrows) before pairing.

So, the diagram should show:

• 3s: two electrons (one up arrow and one down arrow) in the same box.

• 3p: three electrons, each in a separate box, all with the same spin (e.g., all up arrows).

Now, let's evaluate the options:

Option A:

• 3s: "up down half arrows" – this likely means two electrons with opposite spins (correct).

• 3p: "up down half arrows" (which implies paired electrons in one orbital) and "up half arrow" (one electron in another). This would be for 3 electrons but incorrectly paired in one orbital instead of spreading out.

Option B:

• 3s: "up down half arrows" – correct for paired electrons.

• 3p: "up half arrow", "up half arrow", "up half arrow" – three unpaired electrons with parallel spins (correct).

Option C:

• 3s: "up down half arrows" – correct.

• 3p: "up half arrow", "down half arrow", "up half arrow" – three electrons but with mixed spins (violates Hund's rule).

Option D:

• 3s: "up half arrow" – only one electron? Incorrect, should be two.

• 3p: "up down half arrows" (paired), "down half arrow", "up half arrow" – incorrect for 3s and mixed pairing in 3p.

Only Option B correctly shows:

• 3s: two paired electrons (up and down).

• 3p: three unpaired electrons with parallel spins (all up).

Answer: B. [Ne] up down half arrows (3s) | up half arrow, up half arrow, up half arrow (3p)

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 7:

To determine the total number of electrons in p subshells in a ground state atom of germanium (Ge):

1. Atomic number of germanium (Ge) is 32.

2. The electron configuration of germanium in the ground state is:

   • 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p²

   • Alternatively, written as: [Ar]4s²3d¹⁰4p²

3. Identify all electrons in p subshells:

   • 2p⁶: 6 electrons

   • 3p⁶: 6 electrons

   • 4p²: 2 electrons

4. Total electrons in p subshells:

   • 6+6+2=14

Answer: C. 14

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 8:

To determine in which block of the Periodic Table zinc (Zn) is found:

• Zinc has an atomic number of 30.

• Its electron configuration is [Ar]4s²3d¹⁰

• The last electron enters the d subshell (specifically, the 3d orbital).

• Elements where the last electron occupies a d orbital are classified as d-block elements.

Therefore, zinc is in the d-block.

Answer: C. d-block

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 9:

To determine which species has the least number of unpaired electrons, we need to examine the electron configuration of each and count the unpaired electrons.

Option A: Cu⁺ (Copper ion)

• Atomic number of Cu is 29.

• Neutral Cu: [Ar]4s¹3d¹⁰ (exception to the rule).

• Cu⁺ loses one electron (from the 4s orbital), so configuration: [Ar]3d¹⁰.

• All electrons are paired in the d subshell (d¹⁰ is fully filled).

• Unpaired electrons: 0.

Option B: Mn²⁺ (Manganese ion)

• Atomic number of Mn is 25.

• Neutral Mn: [Ar]4s²3d⁵

• Mn²⁺ loses two electrons (from 4s first), so configuration: [Ar]3d⁵.

• In a half-filled d subshell (d⁵), all five electrons are unpaired (Hund's rule).

• Unpaired electrons: 5.

Option C: P (Phosphorus atom)

• Atomic number of P is 15.

• Configuration: [Ne]3s²3p³.

• In the 3p subshell, three electrons are unpaired (each in a separate orbital with parallel spins).

• Unpaired electrons: 3.

Option D: Al (Aluminum atom)

• Atomic number of Al is 13.

• Configuration: [Ne]3s²3p¹.

• The 3p electron is unpaired.

• Unpaired electrons: 1.

Comparing unpaired electrons:

• Cu⁺: 0

• Mn²⁺: 5

• P: 3

• Al: 1

Cu⁺ has the least number of unpaired electrons (0).

Answer: A. Cu⁺

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 10:

To determine the correct "electron in boxes" diagram for the chromium (Cr) atom, we need to recall its electron configuration. Chromium has an atomic number of 24, and it is an exception to the standard Aufbau principle.

• The expected configuration might be [Ar]4s²3d⁴, but the actual ground state configuration is [Ar]4s¹3d⁵.

• This is because a half-filled d subshell (d⁵) is more stable.

• So, the valence electrons are: 4s¹ and 3d⁵.

According to Hund's rule:

• The 4s orbital has 1 electron (unpaired, represented as an up arrow).

• The 3d subshell has 5 orbitals. Each orbital gets one electron with parallel spins (all up arrows). This gives a half-filled d subshell with all unpaired electrons.

Now, let's evaluate the options:

Option A:

• 4s: "up down half arrows" (meaning two electrons with opposite spins) ❌ Incorrect, as Cr has only one 4s electron.

• 3d: "up down half arrows" (paired electrons in one orbital) and then more? This suggests pairing, which is not correct for Cr.

Option B:

• 4s: "up down half arrows" (two electrons) ❌ Incorrect.

• 3d: "up half arrow" repeated five times? This would be correct for the 3d part (all unpaired), but the 4s is wrong.

Option C:

• 4s: "up half arrow" ✅ Correct (one unpaired electron).

• 3d: "up down half arrows" (paired in one orbital), "up down half arrows" (paired in another), and "up half arrow" (unpaired in one). This implies only one unpaired in 3d, but Cr has five unpaired in 3d. ❌ Incorrect.

Option D:

• 4s: "up half arrow" ✅ Correct (one unpaired electron).

• 3d: Five "up half arrow" ✅ Correct (each of the five 3d orbitals has one electron with parallel spins).

So, Option D correctly shows:

• 4s: one electron (up arrow)

• 3d: five electrons, each in a separate orbital with the same spin (all up arrows).

Answer: D. [Ar] up half arrow (4s) | up half arrow, up half arrow, up half arrow, up half arrow, up half arrow (3d)

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.