Valence Electrons and Ionic Compounds - AP Chemistry

What is the most probable chemical formula for the ionic compound formed between aluminum (Al) and sulfur (S)?
Answer: $Al_2S_3$ Correct: Aluminum (Al) is in Group 13 and typically forms a +3 ion ( $Al^{3+}$ ). Sulfur (S) is in Group 16 and typically forms a -2 ion ( $S^{2-}$ ). To form a neutral compound, two $Al^{3+}$ ions (total charge +6) are needed to balance three $S^{2-}$ ions (total charge -6), resulting in the formula $Al_2S_3$ . A: Incorrect. This formula would result if both ions had charges of equal magnitude, which is not the case. C: Incorrect. This formula incorrectly assumes aluminum forms a +2 ion and sulfur forms a -3 ion. D: Incorrect. This formula does not result in a neutral compound based on the typical charges of aluminum and sulfur.
Element A is fluorine, $Z = 9$ , in Group 17. Which ion is most likely formed by A?
Answer: $\mathrm{A^-}$ This question assesses the skill of valence electrons and ionic compounds. Fluorine, in Group 17, has 7 valence electrons and gains 1 to form a 1- ion with neon's configuration. This gain fills the valence shell to 8 electrons. The negative charge is typical for halogens. A tempting distractor is A3-, but that's for Group 15 elements needing 3 electrons, not fluorine needing 1. Fluorine's configuration is [He] 2s2 2p5. Main-group ions form to achieve noble-gas configurations.
Element V is nitrogen, $Z = 7$ , in Group 15. What is the most likely ionic charge for V in an ionic compound with a metal?
Answer: $3-$ This question assesses the skill of valence electrons and ionic compounds. Nitrogen, in Group 15, has 5 valence electrons and gains 3 to achieve an octet, forming a 3- ion. This results in a noble-gas configuration like neon. The charge is negative due to electron gain in nonmetals. A tempting distractor is 3+, but that's for metals losing 3 electrons, not nitrogen gaining them. Nitrogen's configuration is [He] 2s2 2p3, needing 3 more electrons. Main-group ions form to achieve noble-gas configurations.
The formation of an ionic bond between a metal atom and a nonmetal atom is primarily driven by the
Answer: transfer of one or more valence electrons from the metal to the nonmetal, forming oppositely charged ions that attract. 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.
Which of the following compounds is LEAST likely to be formed based on periodic trends of ionic bonding?
Answer: $ArS$ Correct: Argon (Ar) is a noble gas (Group 18) with a complete valence shell. It is chemically inert and does not readily form ions or chemical bonds. An ionic compound between Argon and Sulfur is extremely unlikely to form. A: Incorrect. Rubidium (Group 1) forms $Rb^+$ and Chlorine (Group 17) forms $Cl^-$ , creating the stable ionic compound $RbCl$ . B: Incorrect. Barium (Group 2) forms $Ba^{2+}$ and Oxygen (Group 16) forms $O^{2-}$ , creating the stable ionic compound $BaO$ . C: Incorrect. Indium (Group 13) typically forms $In^{3+}$ and Oxygen forms $O^{2-}$ , creating the ionic compound $In_2O_3$ .
Element P has atomic number 9. What is the most likely charge of the monatomic ion formed by P?
Answer: $-1$ 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.
Element B is phosphorus, $Z = 15$ , in Group 15. How many valence electrons are in a neutral atom of B?
Answer: 5 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.
Which of the following describes the number of valence electrons and the typical ionic charge for Barium (Ba)?
Answer: 2 valence electrons, +2 charge Correct: Barium (Ba) is in Group 2 of the periodic table. Elements in this group, the alkaline earth metals, have 2 valence electrons and readily lose them to form stable cations with a +2 charge. A: Incorrect. This describes elements in Group 1. C: Incorrect. This describes elements in Group 16. D: Incorrect. This describes elements in Group 17.
Element E is in Group 13 and has atomic number $Z = 5$ (boron). If E were to form a simple ion based only on achieving a noble-gas configuration, what charge would be predicted?
Answer: $\mathrm{E^{3+}}$ This question assesses the skill of valence electrons and ionic compounds. Boron, in Group 13, has 3 valence electrons and would lose them to form a 3+ ion with helium's configuration. Though boron often forms covalent bonds, the prediction is based on valence electron loss. The charge matches the group number minus 10 for Group 13. A tempting distractor is E3-, but that's gaining electrons, incorrect for metals like boron. Boron's configuration is [He] 2s2 2p1. Main-group ions form to achieve noble-gas configurations.
Element R is in Group 13 and has atomic number $Z=31$ (gallium, Ga). In ionic compounds, which charge is Ga most likely to have based on its valence electrons?
Answer: $\text{Ga}^{3+}$ This question tests understanding of valence electrons and ionic compounds. Gallium (Z=31) is in Group 13 with electron configuration [Ar]3d¹⁰4s²4p¹, giving it 3 valence electrons in the fourth shell (4s²4p¹). To achieve a stable configuration, gallium loses all 3 valence electrons to form Ga³⁺, resulting in a filled 3d¹⁰ subshell. The Ga⁺ option (B) is incorrect because losing only 1 electron would leave gallium with 2 unpaired electrons in the 4s orbital, creating an unstable configuration. Group 13 elements typically form +3 cations by losing all their valence electrons to achieve a stable electron configuration.
Element S is sodium, $Z = 11$ , a Group 1 element. When S forms a stable ion, which charge is most likely?
Answer: $1+$ This question assesses the skill of valence electrons and ionic compounds. Sodium, in Group 1, has 1 valence electron in its s orbital. It loses this electron to form a 1+ ion, achieving the noble-gas configuration of neon. The positive charge arises because metals lose valence electrons to stabilize. A tempting distractor is 1-, but that's for Group 17 elements gaining an electron, not Group 1 losing one. Sodium's electron configuration is [Ne] 3s1, confirming the loss of one electron. Main-group ions form to achieve noble-gas configurations.
Sulfur is in Group 16 and has atomic number $Z=16$ . Which ion is sulfur most likely to form to achieve a noble-gas electron configuration?
Answer: $\text{S}^{2-}$ This question tests understanding of valence electrons and ionic compounds. Sulfur (Z=16) has an electron configuration of 1s²2s²2p⁶3s²3p⁴, giving it 6 valence electrons in the third shell. To achieve the argon configuration (18 electrons), sulfur gains 2 electrons to fill its 3p orbital, forming S²⁻. The S⁶⁺ option (B) is incorrect because losing 6 electrons would require enormous energy and would strip sulfur down to its neon core, creating an extremely unstable ion that never forms in typical ionic compounds. Nonmetals in Groups 16-17 gain electrons to complete their valence shell rather than losing many electrons.
Element Z is oxygen, $Z=8$ . How many valence electrons does a neutral oxygen atom have?
Answer: 6 This question tests understanding of valence electrons and ionic compounds. Oxygen (Z=8) has an electron configuration of 1s²2s²2p⁴, placing it in Group 16 of the periodic table. The six electrons in the second shell (2s²2p⁴) are its valence electrons, as they occupy the outermost energy level. The answer 8 (C) is incorrect because it represents the total number of electrons in a neutral oxygen atom, not just the valence electrons—remember that valence electrons are only those in the highest principal energy level. For main-group elements, the group number (16) minus 10 gives the number of valence electrons (6).
Element X is in Period 3 and reacts with chlorine to form the compound $XCl_2$ . Element X is most likely
Answer: Magnesium (Mg) Correct: Chlorine, a halogen, typically forms a chloride ion with a -1 charge ( $Cl^-$ ). For the compound to have the formula $XCl_2$ , element X must form an ion with a +2 charge ( $X^{2+}$ ). Magnesium (Mg) is in Period 3 and Group 2, and it characteristically loses its two valence electrons to form a $Mg^{2+}$ ion. A: Incorrect. Sodium (Na) is in Group 1 and forms $Na^+$ , which would result in the formula NaCl. C: Incorrect. Aluminum (Al) is in Group 13 and forms $Al^{3+}$ , which would result in the formula $AlCl_3$ . D: Incorrect. Silicon (Si) is a metalloid and forms covalent bonds with chlorine, not an ionic compound with a simple +2 charge.
An atom of magnesium has atomic number $Z=12$ . How many valence electrons does a neutral Mg atom have?
Answer: 2 This question tests understanding of valence electrons and ionic compounds. Magnesium (Z=12) has an electron configuration of 1s²2s²2p⁶3s², placing it in Group 2 of the periodic table. The two electrons in the 3s orbital are its valence electrons, as they are in the outermost shell. The answer 12 (A) is incorrect because it confuses the total number of electrons with valence electrons—valence electrons are only those in the outermost shell, not all electrons in the atom. To determine valence electrons for main-group elements, count the electrons in the highest principal energy level (n=3 for Mg).
An element Y is in Group 2 of the periodic table (alkaline earth metals). Which of the following is the most likely charge of the stable ion formed by Y?
Answer: $+2$ This question assesses the skill of valence electrons and ionic compounds. Valence electrons dictate how atoms form ions by losing or gaining electrons to reach a stable electron configuration. Elements in Group 2, such as alkaline earth metals, have 2 valence electrons and lose them to form cations with a +2 charge, achieving the electron configuration of the nearest noble gas. This loss of two electrons is energetically favorable for metals in this group. A tempting distractor is +1, but it is incorrect because Group 2 elements lose both valence electrons, not just one, to achieve stability. Main-group ions form to achieve noble-gas configurations by losing or gaining the fewest electrons possible.
Element X is chlorine, $Z=17$ . Based on its position in the periodic table, which ion is chlorine most likely to form in order to achieve a noble-gas electron configuration?
Answer: $\text{Cl}^-$ 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.
Which statement best describes why elements in Group 1 (alkali metals) are highly reactive in forming ionic compounds?
Answer: They have a single valence electron that is easily removed to form a stable cation with a +1 charge. Correct: The high reactivity of alkali metals is due to their single valence electron. This electron has a relatively low ionization energy, meaning it can be easily lost to form a stable +1 cation with a noble gas electron configuration. This process drives the formation of ionic bonds. A: Incorrect. Alkali metals have very low electronegativities; they tend to lose electrons, not attract them. C: Incorrect. This statement describes noble gases (Group 18), not alkali metals. D: Incorrect. Alkali metals have only one valence electron and they lose it to form ionic bonds, they do not share two electrons.
Calcium has atomic number $Z=20$ and is in Group 2. Which ion is calcium most likely to form in ionic compounds?
Answer: $\text{Ca}^{2+}$ This question tests understanding of valence electrons and ionic compounds. Calcium (Z=20) is in Group 2 with electron configuration [Ar]4s², meaning it has 2 valence electrons in the fourth shell. To achieve the stable argon configuration (18 electrons), calcium loses both valence electrons to form Ca²⁺. The Ca⁺ option (A) is incorrect because losing only 1 electron would leave calcium with an unpaired electron in the 4s orbital, creating an unstable configuration that doesn't match any noble gas. Group 2 elements (alkaline earth metals) always form +2 cations by losing both valence electrons.
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?
Answer: N does not commonly form a monatomic ion because it already has a stable valence shell. 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.

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