AP Chemistry Quiz: Valence Electrons And Ionic Compounds

What this quiz covers

This quiz focuses on Valence Electrons And Ionic Compounds, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\text{K}^{2+}$
2. $\text{K}^{19+}$
3. $\text{K}^-$
4. $\text{K}^+$ (correct answer)
5. $\text{K}^{3+}$

Explanation: This question tests understanding of valence electrons and ionic compounds. Potassium (Z=19) is in Group 1 with electron configuration [Ar]4s¹, meaning it has 1 valence electron in the fourth shell. To achieve the stable argon configuration (18 electrons), potassium loses its single valence electron to form K⁺. The K²⁺ option (A) is incorrect because losing 2 electrons would remove an electron from the filled 3p orbital, requiring much more energy and creating an unstable configuration that doesn't match any noble gas. Group 1 elements always form +1 cations by losing their single valence electron.

Question 2

1. $+9$
2. $0$
3. $+1$
4. $-9$
5. $-1$ (correct answer)

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons influence ion formation by showing how many electrons are gained or lost for stability. Element P with atomic number 9 is fluorine in Group 17, with 7 valence electrons, and it gains 1 electron to form a -1 ion. This achieves the stable configuration of neon. A tempting distractor is +1, but it is incorrect because halogens are nonmetals that gain electrons, not lose them. Main-group ions form to achieve noble-gas configurations by gaining or losing the fewest electrons possible.

Question 3

1. $\text{Q}^{0}$
2. $\text{Q}^{2+}$
3. $\text{Q}^{+1}$
4. $\text{Q}^{2-}$ (correct answer)
5. $\text{Q}^{6-}$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons determine ion formation by indicating whether an atom will gain or lose electrons for stability. Group 16 elements have 6 valence electrons and typically gain 2 electrons to complete an octet, forming anions with a -2 charge. This results in a noble-gas-like configuration, making Q^{2-} the most stable monatomic ion for element Q. A tempting distractor is Q^{2+}, but it is incorrect because nonmetals in Group 16 gain electrons rather than lose them to form negative ions. Main-group ions form to achieve noble-gas configurations by gaining or losing the fewest electrons possible.

Question 4

1. $\text{TW}$
2. $\text{T}_2\text{W}$ (correct answer)
3. $\text{TW}_2$
4. $\text{T}_2\text{W}_3$
5. $\text{T}_6\text{W}$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons determine the charges of ions, which in turn dictate the formulas of ionic compounds. Element T in Group 1 has 1 valence electron and forms a +1 ion, while W in Group 16 has 6 valence electrons and forms a -2 ion. To balance charges, two T ions combine with one W ion, resulting in the formula T_2W. A tempting distractor is TW, but it is incorrect because it does not account for the -2 charge of W requiring two +1 ions for neutrality. Main-group ions form to achieve noble-gas configurations, and compound formulas balance total positive and negative charges.

Question 5

1. $\text{M}^{+1}$
2. $\text{M}^{2+}$ (correct answer)
3. $\text{M}^{12+}$
4. $\text{M}^{2-}$
5. $\text{M}^{0}$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons play a key role in predicting ion charges as atoms lose or gain them to attain stable configurations. Element M with atomic number 12 is in Group 2 and has 2 valence electrons, which it loses to form a +2 ion, matching the noble gas neon's configuration. This +2 charge is common for alkaline earth metals due to their metallic nature. A tempting distractor is M^{2-}, but it is incorrect because metals lose electrons to form positive ions, not gain them. Main-group ions form to achieve noble-gas configurations by losing or gaining the fewest electrons possible.

Question 6

1. 2 (correct answer)
2. 12
3. 10
4. 8
5. 6

Explanation: This question assesses the skill of valence electrons and ionic compounds. Magnesium, in Group 2, has 2 valence electrons in its neutral atom, located in the s orbital. These valence electrons determine that magnesium loses 2 electrons to form a 2+ ion, achieving a noble-gas configuration like neon. The charge of the ion is positive because metals tend to lose electrons from their valence shell. A tempting distractor is 12, but that's the atomic number, not the valence electrons, which are only the outermost electrons. Valence electrons are key to predicting reactivity and bonding. Main-group ions form to achieve noble-gas configurations.

Question 7

1. 14
2. 2
3. 4 (correct answer)
4. 6
5. 8

Explanation: This question assesses the skill of valence electrons and ionic compounds. Silicon, in Group 14, has 4 valence electrons, which could be lost or shared, but for ions, it might form 4+ by losing them to neon's configuration. Valence electrons predict bonding behavior, including potential ion formation. The group number indicates valence electrons for main-group elements. A tempting distractor is 14, but that's the atomic number, not valence electrons. Silicon's configuration is [Ne] 3s2 3p2. Main-group ions form to achieve noble-gas configurations.

Question 8

1. tendency of both atoms to share their valence electrons to achieve a stable octet configuration.
2. attraction between the nuclei of the two atoms for the shared pair of valence electrons.
3. transfer of one or more valence electrons from the metal to the nonmetal, forming oppositely charged ions that attract. (correct answer)
4. delocalization of valence electrons from both atoms into a 'sea' of electrons that surrounds the resulting cations.

Explanation: Correct: This statement accurately describes the fundamental process of ionic bond formation. The metal atom loses its valence electrons (low ionization energy) and the nonmetal atom gains them (high electron affinity), creating cations and anions which are then held together by electrostatic attraction. A: Incorrect. Sharing of electrons describes covalent bonding, not ionic bonding. B: Incorrect. This also describes covalent bonding, focusing on the attraction for shared electrons. D: Incorrect. The delocalization of electrons into a 'sea' describes metallic bonding, not ionic bonding.

Question 9

1. $SrI$
2. $Sr_2I$
3. $SrI_2$ (correct answer)
4. $SrI_3$

Explanation: Correct: Elements in the same column of the periodic table tend to form analogous compounds. Strontium (Sr) is in the same group (Group 2) as calcium (Ca), so it forms a $Sr^{2+}$ ion. Iodine (I) is in the same group (Group 17) as fluorine (F), so it forms an $I^-$ ion. Therefore, the formula is $SrI_2$. A: Incorrect. This formula would imply a +1 charge for strontium, which is not typical for a Group 2 element. B: Incorrect. This formula would imply a -2 charge for iodine and a +1 charge for strontium, which are both incorrect. D: Incorrect. This formula would imply a +3 charge for strontium, which is incorrect.

Question 10

1. N most commonly forms $\text{N}^{8+}$ because it has 8 valence electrons.
2. N does not commonly form a monatomic ion because it already has a stable valence shell. (correct answer)
3. N most commonly forms $\text{N}^+$ because it is a nonmetal.
4. N most commonly forms $\text{N}^{2-}$ to complete an octet.
5. N has 18 valence electrons and therefore forms $\text{N}^{18-}$.

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons determine an atom's stability and likelihood of forming ions. Noble gases in Group 18, like element N, have 8 valence electrons, completing their octet and making them stable without needing to gain or lose electrons. Thus, they do not commonly form monatomic ions under typical conditions. A tempting distractor is that N forms N^{2-} to complete an octet, but it is incorrect because noble gases already have a complete octet. Main-group ions form to achieve noble-gas configurations, but noble gases are already stable and rarely ionize.

Question 11

1. 13
2. 3 (correct answer)
3. 10
4. 5
5. 8

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons are located in the highest energy level and influence an atom's reactivity and ion formation. For element Z with atomic number 13, which is in Group 13, the electron configuration ends with 3 valence electrons in the s and p orbitals. These 3 valence electrons can be lost to form a +3 ion, aligning with the group's tendency to achieve noble-gas stability. A tempting distractor is 13, but it is incorrect as it represents the total electrons, not just the valence ones. Main-group ions form to achieve noble-gas configurations by losing or gaining the fewest electrons possible.

Question 12

1. $+1$
2. $+7$
3. $-1$ (correct answer)
4. $0$
5. $-7$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons are the outermost electrons in an atom that participate in chemical bonding and determine the ion's charge. For elements in Group 17, like element X with atomic number 17, there are 7 valence electrons, and these elements tend to gain 1 electron to achieve a stable octet configuration similar to noble gases. This gain of one electron results in a monatomic ion with a charge of -1. A tempting distractor is +7, but it is incorrect because losing 7 electrons would require too much energy for a nonmetal, and nonmetals typically gain electrons rather than lose them. Main-group ions form to achieve noble-gas configurations by gaining or losing the fewest electrons possible.

Question 13

1. $\text{S}^{3-}$
2. $\text{S}^{13+}$
3. $\text{S}^{3+}$ (correct answer)
4. $\text{S}^{+1}$
5. $\text{S}^{0}$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Valence electrons help predict ion charges based on an element's position in the periodic table. Group 13 elements like S have 3 valence electrons and lose them to form +3 ions, attaining a noble-gas configuration. This is typical for metals in this group, such as aluminum. A tempting distractor is S^{3-}, but it is incorrect because Group 13 elements are metals that lose electrons, not gain them like nonmetals. Main-group ions form to achieve noble-gas configurations by losing or gaining the fewest electrons possible.

Question 14

1. $\text{Cl}^{2-}$
2. $\text{Cl}^+$
3. $\text{Cl}^-$ (correct answer)
4. $\text{Cl}^{3+}$
5. $\text{Cl}^{7-}$

Explanation: This question tests understanding of valence electrons and ionic compounds. Chlorine (Z=17) has an electron configuration of 1s²2s²2p⁶3s²3p⁵, giving it 7 valence electrons in the third shell. To achieve a noble-gas configuration like argon (18 electrons), chlorine needs to gain 1 electron, forming Cl⁻. The Cl²⁻ option (A) is incorrect because gaining 2 electrons would give chlorine 19 electrons total, exceeding argon's configuration and creating an unstable ion with excess negative charge. When forming ionic compounds, main-group elements gain or lose the fewest electrons possible to achieve the nearest noble-gas configuration.

Question 15

1. 8
2. 5 (correct answer)
3. 3
4. 15
5. 10

Explanation: This question assesses the skill of valence electrons and ionic compounds. Phosphorus, in Group 15, has 5 valence electrons and can gain 3 to form a 3- ion, achieving argon's configuration. Valence electrons determine the potential ion charge by the number needed for an octet. This is key for nonmetals in ionic compounds. A tempting distractor is 3, but that's the ion charge magnitude, not the valence electron count of 5. Phosphorus's configuration is [Ne] 3s2 3p3. Main-group ions form to achieve noble-gas configurations.

Question 16

1. $1+$
2. $3-$
3. $1-$ (correct answer)
4. $2+$
5. $0$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Chlorine, in Group 17, has 7 valence electrons in its neutral atom. To achieve a stable noble-gas configuration like argon, it gains 1 electron, resulting in a 1- charge. This ion formation is driven by the octet rule, where nonmetals in Group 17 typically form anions with a 1- charge. A tempting distractor is 3-, but that's incorrect for chlorine as it only needs to gain 1 electron, not 3, which would apply to Group 15 elements. The electron configuration of chlorine is [Ne] 3s2 3p5, so gaining one electron fills the 3p subshell. Main-group ions form to achieve noble-gas configurations.

Question 17

1. Gain 2 electrons (correct answer)
2. Lose 6 electrons
3. Lose 2 electrons
4. Gain 6 electrons
5. Gain 1 electron

Explanation: This question assesses the skill of valence electrons and ionic compounds. Sulfur, in Group 16, has 6 valence electrons and typically gains 2 to reach an octet, forming a 2- ion. This gain allows it to match the electron configuration of argon. The ion charge is determined by the number of electrons needed to fill the valence shell. A tempting distractor is lose 2 electrons, but that's incorrect for nonmetals like sulfur, which gain rather than lose to achieve stability. Sulfur's configuration is [Ne] 3s2 3p4, needing 2 electrons. Main-group ions form to achieve noble-gas configurations.

Question 18

1. 19
2. 1 (correct answer)
3. 8
4. 2
5. 18

Explanation: This question assesses the skill of valence electrons and ionic compounds. Potassium, in Group 1, has 1 valence electron in its 4s orbital. This valence electron is lost to form a 1+ ion, achieving argon's configuration. The ion charge is determined by the single valence electron in alkali metals. A tempting distractor is 19, but that's the atomic number, not the valence electrons, which are only the outermost one. Potassium's configuration is [Ar] 4s1, confirming 1 valence electron. Main-group ions form to achieve noble-gas configurations.

Question 19

1. $3+$
2. $2+$ (correct answer)
3. $1+$
4. $2-$
5. $1-$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Barium, in Group 2, has 2 valence electrons and loses them to form a 2+ ion with xenon's configuration. The positive charge results from electron loss in metals. Valence electrons directly predict the ion charge for Group 2 elements. A tempting distractor is 3+, but that's for Group 13, not Group 2 which loses only 2. Barium's configuration is [Xe] 6s2. Main-group ions form to achieve noble-gas configurations.

Question 20

1. $2-$
2. $3-$
3. $1+$
4. $1-$ (correct answer)
5. $2+$

Explanation: This question assesses the skill of valence electrons and ionic compounds. Bromine, in Group 17, has 7 valence electrons and gains 1 to form a 1- ion with krypton's configuration. This corrects the student's claim of 2-, as Group 17 needs only 1 electron. The charge is determined by valence electrons to reach octet. A tempting distractor is 2-, but that's for Group 16, not 17. Bromine's configuration is [Ar] 4s2 3d10 4p5. Main-group ions form to achieve noble-gas configurations.