All questions
Question 1
Which of the following statements about valence electrons and ionic compound formation is correct?
- Metals gain valence electrons to achieve noble gas configurations when forming ionic compounds
- The number of valence electrons equals the group number for all main group elements
- Nonmetals in Groups 15-17 typically gain electrons equal to (18 minus their group number) to achieve stability (correct answer)
- Transition metals always lose all their valence electrons when forming ions
- Elements in the same group form ions with identical charges in all ionic compounds
Explanation: When analyzing ionic compound formation, you need to understand how atoms achieve stable electron configurations by gaining or losing valence electrons to resemble the nearest noble gas.
Option C correctly describes how nonmetals in Groups 15-17 achieve stability. These elements have 5, 6, or 7 valence electrons respectively, and they gain electrons to reach 8 (the stable octet). The formula "18 minus group number" works perfectly: Group 15 elements gain 18-15=3 electrons, Group 16 gains 18-16=2 electrons, and Group 17 gains 18-17=1 electron. For example, chlorine (Group 17) gains 1 electron to form Cl⁻.
Option A is backwards—metals lose valence electrons, not gain them, to achieve noble gas configurations. Sodium loses its one valence electron to form Na⁺, resembling neon's electron configuration.
Option B contains a subtle error. While this rule works for Groups 1-2 and 13-18, it fails for transition metals (Groups 3-12), where the relationship between group number and valence electrons is more complex due to d-orbital involvement.
Option D overgeneralizes transition metal behavior. These metals can form multiple ions with different charges (like Fe²⁺ and Fe³⁺) and don't always lose all possible valence electrons. Their variable oxidation states make them versatile but unpredictable.
Remember this pattern: main group metals lose electrons, nonmetals gain electrons, and both aim for the nearest noble gas configuration. Focus on the 18-minus rule for Groups 15-17—it's a reliable shortcut for predicting common ionic charges.
Question 2
Which of the following pairs of ions would form an ionic compound with the formula AB2?
- Ca2+ and Cl− (correct answer)
- Mg2+ and O2−
- Al3+ and F−
- Na+ and S2−
- K+ and Br−
Explanation: When you encounter questions about ionic compound formulas, you need to apply the principle of charge neutrality: the total positive charge must equal the total negative charge in the compound.
To determine which ion pair forms AB2, you need to find a combination where one cation (A) pairs with two anions (B) to achieve electrical neutrality.
Let's examine each option systematically. For choice A, Ca2+ has a +2 charge and Cl− has a -1 charge. To balance the +2 charge from calcium, you need two chloride ions (each contributing -1), giving you CaCl2. This matches the AB2 formula perfectly.
Choice B pairs Mg2+ (+2 charge) with O2− (-2 charge). Since the charges are equal in magnitude, you only need one of each ion, forming MgO with an AB formula, not AB2.
Choice C combines Al3+ (+3 charge) with F− (-1 charge). To balance the +3 charge, you need three fluoride ions, creating AlF3 with an AB3 formula.
Choice D involves Na+ (+1 charge) and S2− (-2 charge). To balance the -2 charge from sulfur, you need two sodium ions, forming Na2S with an A2B formula.
Study tip: When predicting ionic formulas, use the "crisscross method" - the absolute value of each ion's charge becomes the subscript of the other ion, then simplify if needed. This helps you quickly identify the correct formula pattern. Question 3
Which of the following electron configurations represents an atom that would most likely form a −2 ion?
- 1s22s22p63s2
- 1s22s22p63s23p4 (correct answer)
- 1s22s22p63s23p64s1
- 1s22s22p63s23p64s2
- 1s22s22p63s23p5
Explanation: When you encounter questions about ion formation, focus on how atoms achieve stable electron configurations by gaining or losing electrons to reach the nearest noble gas configuration.
To form a −2 ion, an atom must gain two electrons. This typically happens when an atom needs exactly two more electrons to complete its outermost shell and achieve a stable octet (8 electrons in the outer shell).
Option B (1s22s22p63s23p4) represents sulfur, which has 6 electrons in its outer shell (3s² 3p⁴). By gaining 2 electrons, sulfur achieves the stable configuration 3s23p6, completing its octet. This makes sulfur highly likely to form a −2 ion (sulfide, S²⁻).
Option A represents magnesium, which has a complete outer shell (3s²) and would more likely lose 2 electrons to form a +2 ion, not gain electrons. Option C represents potassium, which has 1 electron in its outermost shell (4s¹) and would preferentially lose this electron to form a +1 ion rather than gain 7 electrons. Option D represents calcium, which like magnesium would lose its 2 outer electrons (4s²) to form a +2 ion.
Remember this pattern: atoms in Group 16 (chalcogens) like oxygen, sulfur, and selenium typically form −2 ions because they need exactly 2 electrons to complete their octet. Look for electron configurations with 6 valence electrons when predicting −2 ion formation. Question 4
Element Z has 5 valence electrons and forms an ionic compound with magnesium. What is the most likely formula for this compound?
- MgZ
- Mg2Z
- MgZ2
- Mg3Z2 (correct answer)
- Mg2Z3
Explanation: When you encounter questions about ionic compound formation, focus on how electrons transfer between metals and nonmetals to achieve stable electron configurations. Elements gain or lose electrons to reach the nearest noble gas configuration.
Element Z has 5 valence electrons, meaning it needs 3 more electrons to complete its octet (8 valence electrons). Therefore, Z will gain 3 electrons, forming a Z3− ion. Magnesium, being in Group 2, has 2 valence electrons that it readily loses to form Mg2+ ions.
To create a neutral ionic compound, the total positive charge must equal the total negative charge. With Mg2+ and Z3− ions, you need to find the least common multiple of their charges: LCM(2,3) = 6. This requires 3 magnesium ions (3×2+=6+) and 2 Z ions (2×3−=6−), giving the formula Mg3Z2.
Option A (MgZ) incorrectly assumes Z forms a 1− ion, which would leave Z with only 6 valence electrons. Option B (Mg2Z) wrongly treats Z as having a 4− charge, requiring Z to gain 7 electrons. Option C (MgZ2) mistakenly assumes Z forms a 1− ion and creates an unbalanced compound with unequal charges.
Remember this pattern: always determine each element's most likely ionic charge based on valence electrons, then balance the charges using the smallest whole number ratio. Group numbers are your best guide for predicting ionic charges. Question 5
An element in Group 13 forms an ionic compound with an element in Group 16. If the Group 13 element has 13 total electrons in its neutral state, what is the simplest formula for the ionic compound?
- AB
- A2B
- AB2
- A2B3 (correct answer)
- A3B2
Explanation: When you encounter ionic compound formation questions, you need to determine the charges of the ions based on their group positions, then balance those charges to achieve electrical neutrality.
The Group 13 element with 13 electrons is aluminum (Al). Group 13 elements lose 3 electrons to achieve a stable electron configuration, forming Al3+ ions. Group 16 elements gain 2 electrons to fill their outer shell, forming ions with a 2- charge (like O2− or S2−).
To balance the charges in the ionic compound, you need the total positive charge to equal the total negative charge. With Al3+ and a Group 16 element forming X2−, you need to find the lowest common multiple of 3 and 2, which is 6. This requires 2 aluminum ions (2×3+=6+) and 3 Group 16 ions (3×2−=6−), giving the formula Al2X3.
Answer choice A (AB) assumes both elements have 1+ and 1- charges, which doesn't match their group positions. Choice B (A2B) would work if the Group 13 element had a 1+ charge and Group 16 had a 2- charge, but Group 13 elements form 3+ ions. Choice C (AB2) incorrectly assumes the Group 13 element has a 2+ charge while the Group 16 element has a 1- charge.
The correct answer is D (A2B3).
Study tip: Always determine ion charges from group numbers first (Groups 1, 2, 13 lose electrons; Groups 15, 16, 17 gain electrons), then use the crisscross method to balance charges in the simplest whole-number ratio. Question 6
Which of the following statements about the formation of ionic compounds is most accurate?
- The total number of valence electrons is conserved during ionic compound formation, but some electrons are redistributed between atoms (correct answer)
- Valence electrons are destroyed when metals and nonmetals combine to form ionic compounds
- New valence electrons are created to fill the octets of both metals and nonmetals in ionic compounds
- Only the valence electrons of nonmetals are involved in ionic compound formation
- Valence electrons are converted to core electrons during ionic compound formation
Explanation: When analyzing ionic compound formation, you're dealing with electron transfer and conservation principles. The key insight is that electrons aren't created or destroyed—they're simply redistributed between atoms as metals lose electrons and nonmetals gain them.
Answer A correctly captures this fundamental principle. During ionic bonding, the total number of valence electrons remains constant throughout the process. Metals donate their valence electrons to achieve stable electron configurations (usually resembling the nearest noble gas), while nonmetals accept these electrons to complete their own octets. For example, when sodium (1 valence electron) combines with chlorine (7 valence electrons), sodium loses its electron to chlorine, creating Na⁺ and Cl⁻ ions. The total count remains 8 valence electrons—just redistributed.
Answer B is incorrect because electrons are never destroyed in chemical reactions. This would violate the law of conservation of mass and charge. Answer C is wrong because no new electrons are created—nature conserves electrons, and atoms achieve stability by sharing or transferring existing electrons. Answer D incorrectly suggests only nonmetal electrons participate, when in reality, the process specifically involves metals losing their valence electrons to nonmetals.
Remember this pattern: in ionic bonding questions, always think "transfer and conserve." Electrons move from metals to nonmetals, but the total electron count stays the same. Watch for answer choices that suggest electrons are created or destroyed—these violate fundamental conservation laws and are typically incorrect.
Question 7
A hypothetical element Q has 4 valence electrons. If Q were to form an ionic compound with fluorine, which of the following would be the most likely outcome?
- Q would lose 4 electrons to form Q4+
- Q would gain 4 electrons to form Q4−
- Q would not form a stable ionic compound with fluorine (correct answer)
- Q would form covalent bonds with fluorine instead
- Q would both lose and gain electrons simultaneously
Explanation: When you encounter questions about ionic compound formation, you need to consider both the energy required to remove electrons and the stability of the resulting compound. Elements with 4 valence electrons occupy a unique position on the periodic table that makes ionic bonding challenging.
For element Q to form an ionic compound, it would need to either lose or gain electrons to achieve a stable electron configuration. However, both pathways present significant energy barriers. Losing all 4 valence electrons (option A) would require an enormous amount of energy because removing electrons becomes progressively more difficult as you remove each one. The fourth ionization energy would be prohibitively high, making Q4+ formation energetically unfavorable under normal conditions.
Gaining 4 electrons to form Q4− (option B) is equally problematic. Adding electrons to an already neutral atom becomes increasingly difficult due to electron-electron repulsion, and a Q4− ion would be extremely unstable.
Option D suggests covalent bonding, which is actually what elements with 4 valence electrons typically do. However, the question specifically asks about ionic compound formation, making this answer incorrect in context.
The correct answer is C because elements with 4 valence electrons (like carbon and silicon) rarely form stable ionic compounds due to these extreme energy requirements. Instead, they achieve stability through electron sharing in covalent bonds.
Study tip: Remember that elements in the middle of the periodic table (groups 13-15) generally avoid ionic bonding due to high ionization energies and electron affinities. Focus on groups 1-2 and 16-17 for classic ionic compound formation. Question 8
An element X has the electron configuration [Ne]3s23p1 in its neutral state. When X forms an ionic compound with oxygen, what is the most likely formula?
- XO
- X2O
- XO2
- X2O3 (correct answer)
- X3O2
Explanation: When you encounter electron configuration problems involving ionic compounds, you need to determine the charges of the ions that will form, then balance those charges to find the correct formula.
Element X has the configuration [Ne]3s23p1, meaning it has 3 valence electrons. To achieve a stable electron configuration, X will lose these 3 electrons to form X3+, matching the stable neon configuration. Oxygen, with 6 valence electrons, gains 2 electrons to form O2− and achieve the stable octet.
To write the ionic compound formula, you need to balance the positive and negative charges. With X3+ and O2−, you need the least common multiple of 3 and 2, which is 6. This requires 2 aluminum ions (2×3+=6+) and 3 oxide ions (3×2−=6−), giving the formula X2O3.
Looking at the wrong answers: A) XO would require both ions to have the same charge magnitude, which doesn't match X's +3 charge. B) X2O suggests X has a +1 charge, but X has 3 valence electrons to lose. C) XO2 implies X has a +4 charge, but again, X only has 3 valence electrons in its outer shell.
Study tip: For ionic compound formulas, always determine each element's most likely charge from its valence electrons first, then use the "crisscross method" – the charge number becomes the subscript for the other element. Remember that metals typically lose electrons to match the nearest noble gas configuration. Question 9
In the ionic compound Al2(SO4)3, what is the total number of valence electrons that were transferred from aluminum atoms to form the aluminum ions present?
- 3
- 6 (correct answer)
- 9
- 12
- 18
Explanation: When analyzing ionic compounds, you need to understand how electrons are transferred from metals to nonmetals to form ions. This question specifically asks about the electrons lost by aluminum atoms during ion formation.
In Al2(SO4)3, aluminum forms Al3+ ions by losing electrons. Since aluminum is in Group 13, it has 3 valence electrons that it loses to achieve a stable electron configuration. The compound contains 2 aluminum atoms (indicated by the subscript 2), so the total electrons transferred from aluminum atoms is: 2 atoms × 3 electrons per atom = 6 electrons.
Looking at the wrong answers: Choice A (3) represents the electrons lost by only one aluminum atom, ignoring that there are two aluminum atoms in the formula. Choice C (9) incorrectly adds the electrons from aluminum (6) plus the charge on sulfate ions (3 × -2 = -6), but the question only asks about aluminum's transferred electrons. Choice D (12) might result from confusing the total number of electrons involved in the entire compound or miscounting the aluminum atoms and their electron loss.
The correct answer is B (6 electrons).
Study tip: When counting transferred electrons in ionic compounds, always multiply the number of electrons lost or gained per ion by the number of that type of ion present in the formula unit. Pay careful attention to subscripts that tell you how many of each ion are present. Question 10
Element X is in Period 4 and Group 2. Element Y is in Period 3 and Group 17. When these elements form an ionic compound, what is the total number of electrons in the formula unit?
- 35
- 37
- 54 (correct answer)
- 57
- 75
Explanation: When you encounter questions about ionic compounds, you need to identify the elements, predict their ionic charges based on their periodic table positions, determine the compound's formula, and then count the total electrons.
Element X (Period 4, Group 2) is calcium (Ca), which loses 2 electrons to form Ca²⁺. Element Y (Period 3, Group 17) is chlorine (Cl), which gains 1 electron to form Cl⁻. To balance charges in the ionic compound, you need one Ca²⁺ and two Cl⁻ ions, giving the formula CaCl₂.
Now count the electrons in this formula unit. Calcium has 20 protons, so Ca²⁺ has 18 electrons (lost 2). Each chlorine has 17 protons, so each Cl⁻ has 18 electrons (gained 1). The total electrons in CaCl₂ = 18 + 18 + 18 = 54 electrons, making C correct.
Looking at the wrong answers: A (35) might come from incorrectly adding the atomic numbers of Ca and Cl (20 + 17 = 37, then subtracting 2), but this ignores the second chlorine. B (37) could result from simply adding atomic numbers without considering the formula. D (57) might occur if you correctly identified 54 electrons but mistakenly added the net charge of the compound.
The key strategy here is systematic thinking: identify elements → determine charges → write correct formula → count electrons in all ions. Remember that in ionic compounds, you're counting electrons in the ions, not neutral atoms, and the formula must reflect charge balance.
Question 11
Which of the following electron configurations represents an atom that would require the least energy to remove all of its valence electrons?
- 1s22s22p63s1 (correct answer)
- 1s22s22p63s2
- 1s22s22p63s23p1
- 1s22s22p63s23p3
- 1s22s22p63s23p5
Explanation: When you encounter questions about removing valence electrons, focus on two key factors: how many valence electrons need to be removed and how tightly they're held by the nucleus.
Valence electrons are those in the outermost shell. For these atoms, you need to identify which electrons are valence electrons and consider the energy required to remove them all. The fewer valence electrons an atom has, and the more loosely they're held, the less total energy removal requires.
Option A (1s22s22p63s1) represents sodium with just one valence electron in the 3s orbital. This single electron experiences significant shielding from the filled inner shells and is relatively far from the nucleus, making it easy to remove. Since there's only one valence electron, the total energy required is minimal.
Option B (1s22s22p63s2) represents magnesium with two 3s valence electrons. While these are also in the third shell, you need to remove two electrons, requiring more total energy than option A.
Option C (1s22s22p63s23p1) represents aluminum with three valence electrons (two 3s and one 3p). Removing all three requires significantly more energy than removing just one.
Option D (1s22s22p63s23p3) represents phosphorus with five valence electrons total. This requires the most energy since you're removing the greatest number of electrons.
Study tip: For valence electron removal questions, count the outermost shell electrons first. Atoms with fewer valence electrons (especially Group 1 metals) will always require less total energy for complete valence electron removal. Question 12
In the ionic compound Ca3N2, what is the ratio of the number of valence electrons lost by all calcium atoms to the number of valence electrons gained by all nitrogen atoms?
- 1:1 (correct answer)
- 2:3
- 3:2
- 6:6
- 9:6
Explanation: This question tests your understanding of electron transfer in ionic compound formation. When metals and nonmetals form ionic compounds, electrons are transferred from metal atoms (which become cations) to nonmetal atoms (which become anions).
Let's analyze Ca3N2 systematically. Calcium is in Group 2, so each Ca atom has 2 valence electrons and forms Ca2+ ions by losing both electrons. Nitrogen is in Group 15 with 5 valence electrons, and it forms N3− ions by gaining 3 electrons to achieve a stable octet.
In Ca3N2, you have 3 calcium atoms and 2 nitrogen atoms. The total valence electrons lost by all calcium atoms is 3×2=6 electrons. The total valence electrons gained by all nitrogen atoms is 2×3=6 electrons. The ratio is 6:6, which simplifies to 1:1.
Answer A (1:1) is correct because electron transfer must be balanced - the total electrons lost must equal the total electrons gained.
Answer B (2:3) incorrectly uses the electrons per individual atom rather than the total for all atoms in the formula.
Answer C (3:2) mistakenly uses the subscripts from the chemical formula instead of calculating actual electron transfer.
Answer D (6:6) shows the correct calculation but fails to simplify the ratio to its lowest terms.
Remember: In any ionic compound, the total electrons lost always equals the total electrons gained. Always multiply the charge per ion by the number of atoms present, then simplify your ratio. Question 13
Consider the ionic compound Li3P. If this compound were to dissociate completely in water, what would be the total concentration of ions if the compound concentration is 0.2 M?
- 0.2 M
- 0.4 M
- 0.6 M
- 0.8 M (correct answer)
- 1.0 M
Explanation: When you encounter ionic dissociation problems, you need to determine how many ions are produced when the compound breaks apart in water. The key is understanding the compound's formula and applying stoichiometry.
Lithium phosphide (Li3P) dissociates according to this equation:
Li3P→3Li++P3−
Each formula unit of Li3P produces 4 total ions (3 lithium ions plus 1 phosphide ion). With a starting concentration of 0.2 M Li3P, you can calculate the total ion concentration:
0.2 M Li3P × 4 ions per formula unit = 0.8 M total ions
Therefore, answer D (0.8 M) is correct.
Let's examine why the other answers are wrong. Answer A (0.2 M) assumes no dissociation occurred, which ignores the fact that ionic compounds completely dissociate in water. Answer B (0.4 M) suggests only 2 ions are produced per formula unit, which would be true for a compound like NaCl but not Li3P. Answer C (0.6 M) might result from only counting the lithium ions (3 × 0.2 M = 0.6 M) while forgetting about the phosphide ion.
The pattern to remember: always count all ions produced during dissociation. Look at the subscripts in the formula to determine how many of each ion type is formed, then multiply by the molarity and sum everything up. Practice identifying the dissociation equation first, then the math becomes straightforward. Question 14
In which of the following ionic compounds would you expect the metal to have lost exactly 2 valence electrons per atom?
- NaCl
- AlF3
- CaBr2 (correct answer)
- Li2O
- FeCl3
Explanation: When analyzing ionic compounds, you need to determine how many electrons each metal atom loses by examining the charges that balance the compound. The key is using the known charges of nonmetals to work backwards and find the metal's oxidation state.
For CaBr2, bromine (a halogen) always forms Br− ions with a -1 charge. Since there are two bromide ions contributing a total charge of -2, the calcium must have a +2 charge to balance the compound. This means each calcium atom lost exactly 2 electrons to form Ca2+, making C correct.
Let's examine why the other options don't work. In option A, NaCl, chlorine forms Cl− (charge -1), so sodium must be Na+ (charge +1), meaning sodium lost only 1 electron. In option B, AlF3, fluorine forms F− ions, and with three of them (total charge -3), aluminum must be Al3+, so aluminum lost 3 electrons. In option D, Li2O, oxygen forms O2− (charge -2), and since there are two lithium atoms to balance this, each lithium is Li+ (charge +1), so each lithium atom lost only 1 electron.
Remember this strategy: Always use the predictable charges of nonmetals (halogens = -1, oxygen family = -2) to determine the metal's charge. Count the total negative charge, then figure out how that charge is distributed among the metal atoms to find electrons lost per metal atom. Question 15
Which of the following statements correctly describes the relationship between valence electrons and ionic charge for main group elements?
- Elements with 1-3 valence electrons typically gain electrons equal to their group number
- Elements with 5-7 valence electrons typically lose electrons to achieve a stable octet configuration
- The ionic charge magnitude equals the number of valence electrons for all main group elements
- Elements with 4 valence electrons typically form ions with charges of +4 or -4
- Elements with 1-3 valence electrons typically form positive ions with charge equal to their number of valence electrons (correct answer)
Explanation: When analyzing ionic charge patterns for main group elements, you need to consider how atoms achieve stable electron configurations by gaining or losing electrons to reach the nearest noble gas configuration.
Elements typically follow the octet rule by achieving eight valence electrons (or two for hydrogen and helium). Elements with 1-3 valence electrons find it energetically favorable to lose electrons, while elements with 5-7 valence electrons prefer to gain electrons. The resulting ionic charge reflects how many electrons were transferred.
Looking at the incorrect options: Choice A reverses the actual behavior - elements with 1-3 valence electrons typically lose electrons (not gain them equal to their group number) to form positive ions. Choice B also reverses reality - elements with 5-7 valence electrons typically gain electrons (not lose them) to complete their octet, forming negative ions. Choice C oversimplifies the relationship - while there's a connection between valence electrons and ionic charge, the magnitude doesn't simply equal the number of valence electrons for all elements. Choice D is unrealistic because forming ions with ±4 charges requires enormous energy; carbon group elements typically share electrons through covalent bonding rather than forming highly charged ions.
Since all the given statements contain significant errors about fundamental ionic behavior patterns, none accurately describes the relationship between valence electrons and ionic charge.
Remember this key pattern: metals (left side of periodic table) lose electrons to form positive ions, while nonmetals (right side) gain electrons to form negative ions, both seeking the stability of a complete outer shell.
Question 16
A metal M in Group 2 combines with a nonmetal Y in Group 17 to form an ionic compound. If the metal atom has 20 electrons in its neutral state, what is the most likely formula for the ionic compound formed?
- MY
- MY2 (correct answer)
- M2Y
- M2Y3
- MY3
Explanation: When you encounter ionic compound formation questions, focus on electron configuration and charge balance. The key is determining how many electrons each atom gains or loses to achieve a stable octet.
Since the metal M has 20 electrons in its neutral state, it's calcium (Ca). As a Group 2 metal, calcium loses 2 electrons to form Ca2+, achieving the stable electron configuration of argon. The nonmetal Y from Group 17 (halogens) gains 1 electron to form Y−, completing its octet.
For the compound to be electrically neutral, the total positive charge must equal the total negative charge. With one Ca2+ ion carrying a +2 charge, you need two Y− ions (each with a -1 charge) to balance it: (+2) + 2(-1) = 0. This gives the formula MY2.
Choice A (MY) is incorrect because one +2 cation and one -1 anion would give a net charge of +1, violating electrical neutrality. Choice C (M2Y) fails because two +2 cations and one -1 anion would result in a net charge of +3. Choice D (M2Y3) is wrong because the charges don't balance: 2(+2) + 3(-1) = +1.
Study tip: For ionic compounds, always use the "crisscross method" or ensure charge balance. Group 2 metals form +2 ions, Group 17 nonmetals form -1 ions. The subscripts in the formula must make the total charge zero, so you'll often see patterns like M2X (Group 1 + Group 16) or MX2 (Group 2 + Group 17). Question 17
Two elements, R and S, form an ionic compound with the formula R2S3. If element R loses 3 valence electrons per atom when forming this compound, how many valence electrons does element S gain per atom?
- 1
- 2 (correct answer)
- 3
- 4
- 6
Explanation: When you encounter ionic compound formulas, remember that the total positive and negative charges must balance to create a neutral compound. This principle of charge neutrality is your key to solving these problems.
In the compound R2S3, you have 2 atoms of element R and 3 atoms of element S. Since R loses 3 electrons per atom, each R atom becomes R3+. With 2 R atoms, the total positive charge is 2×(+3)=+6.
For the compound to be electrically neutral, the 3 S atoms must collectively provide a -6 charge to balance the +6 from the R atoms. Since you have 3 S atoms sharing this -6 charge equally, each S atom must gain 36=2 electrons, making each one S2−.
Looking at the wrong answers: A) suggests S gains only 1 electron, which would give a total negative charge of only -3, leaving the compound with a net +3 charge. C) proposes S gains 3 electrons each, creating a total negative charge of -9, which would exceed the +6 positive charge. D) suggests 4 electrons per S atom, resulting in a -12 charge, creating an even greater imbalance.
Only B) gives you the charge balance: 2(+3)+3(−2)=+6−6=0.
Study tip: Always check that total positive charges equal total negative charges in ionic compounds. Set up the equation: (number of cations × charge per cation) + (number of anions × charge per anion) = 0. Question 18
An unknown element X forms a compound with the formula X2O3. If element X has 13 protons, how many valence electrons does a neutral atom of X possess?
- 2
- 3 (correct answer)
- 5
- 8
- 13
Explanation: When you encounter a question asking about valence electrons based on a chemical formula, you need to connect three key pieces: the element's identity, its oxidation state in the compound, and electron configuration principles.
Element X has 13 protons, making it aluminum (Al). In the compound X2O3, you can determine aluminum's oxidation state by working with oxygen's known -2 charge. Since the compound is neutral, the total positive charge must balance the total negative charge. With three oxygen atoms contributing -6 charge total, the two aluminum atoms must contribute +6 charge total, meaning each aluminum has a +3 oxidation state.
Aluminum's electron configuration is 1s22s22p63s23p1, giving it 3 valence electrons in its outermost shell (3s² 3p¹). When aluminum forms Al2O3, it loses these 3 valence electrons to achieve the +3 oxidation state, confirming that aluminum has 3 valence electrons. The answer is B.
Choice A (2 electrons) might tempt you if you only counted the 3s electrons, but valence electrons include all outermost shell electrons. Choice C (5 electrons) could result from mistakenly adding the shell number (3) to the valence electrons (3). Choice D (8 electrons) represents a common misconception of thinking about the octet rule rather than the actual valence electrons present.
Remember: valence electrons equal the group number for main group elements. Aluminum is in Group 13, so it has 3 valence electrons, which you can verify through its chemical behavior in compounds.