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AP Chemistry Help: Valence Electrons And Ionic Compounds

Review real example questions for Valence Electrons And Ionic Compounds in AP Chemistry.

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Potassium is an element in Group 1 with atomic number $Z=19$ . Which ion is potassium most likely to form?

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Question 1

Potassium is an element in Group 1 with atomic number $Z=19$ . Which ion is potassium most likely to form?

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

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

Element P has atomic number 9. What is the most likely charge of the monatomic ion formed by P?

$+9$
$0$
$+1$
$-9$
$-1$ (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

A neutral atom of element Q is in Group 16 of the periodic table. Which monatomic ion is Q most likely to form to achieve a noble-gas electron configuration?

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

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

Element T is in Group 1 of the periodic table. Which formula represents the most likely ionic compound formed between T and an element W in Group 16?

$\text{TW}$
$\text{T}_2\text{W}$ (correct answer)
$\text{TW}_2$
$\text{T}_2\text{W}_3$
$\text{T}_6\text{W}$

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

Element M has atomic number 12. Which of the following best predicts the charge on the common ion formed by M?

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

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

A neutral atom of element Y has atomic number $Z = 12$ (magnesium), a Group 2 element. How many valence electrons does a neutral Y atom have?

2 (correct answer)
12
10
8
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

Element G is silicon, $Z = 14$ , in Group 14. A student uses the group number to predict valence electrons. How many valence electrons does a neutral atom of G have?

14
2
4 (correct answer)
6
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

The formation of an ionic bond between a metal atom and a nonmetal atom is primarily driven by the

tendency of both atoms to share their valence electrons to achieve a stable octet configuration.
attraction between the nuclei of the two atoms for the shared pair of valence electrons.
transfer of one or more valence electrons from the metal to the nonmetal, forming oppositely charged ions that attract. (correct answer)
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

The compound calcium fluoride has the chemical formula $CaF_2$ . Based on periodic trends, what is the most likely chemical formula for the compound formed between strontium (Sr) and iodine (I)?

$SrI$
$Sr_2I$
$SrI_2$ (correct answer)
$SrI_3$

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

Element N is a noble gas in Group 18. Based on valence electrons and stability, which statement best describes the tendency of N to form a monatomic ion under typical conditions?

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