AP Chemistry Unit 2: Molecular and Ionic Bonding — Free Review Games.

Sodium (a metal) transfers an electron to chlorine (a nonmetal), forming Na⁺ and Cl⁻ ions held together by electrostatic attraction (ionic bond).

Oxygen forms two covalent bonds with hydrogen atoms and retains two lone pairs, giving it an octet of electrons.

Four electron domains with no lone pairs arrange in a tetrahedral geometry (109.5° bond angles) to minimize electron repulsion.

CO₂ is linear with two equal C=O polar bonds pointing in opposite directions, so the dipole moments cancel out, making the molecule nonpolar.

The strong Coulombic attractions between cations and anions in a crystal lattice require large amounts of thermal energy to disrupt, resulting in high melting points.

Four electron domains give tetrahedral electron geometry, but with one lone pair the molecular shape is trigonal pyramidal (like NH₃).

Formal charge = valence electrons - nonbonding electrons - (½ × bonding electrons). It helps determine the most favorable Lewis structure.

Lone pair electrons occupy more space than bonding pairs in VSEPR theory, pushing the bonding pairs closer together and reducing the bond angle below the ideal tetrahedral value.

Resonance structures represent different possible electron arrangements. The actual molecule is a resonance hybrid with delocalized electrons, not a mixture of discrete structures.

In metallic bonding, valence electrons are delocalized across the entire metal lattice, creating a 'sea' of electrons that holds the metal cations together.

XeF₄ has 6 electron domains (4 bonding, 2 lone pairs). The lone pairs are opposite each other (trans), creating square planar molecular geometry. The symmetrical arrangement makes it nonpolar.

While oxygen is more electronegative, the lone pair on carbon in CO creates electron density that partially counteracts the bond dipole, making CO's dipole moment surprisingly small.

The triple bond structure with C(-1) and O(+1) gives complete octets and the lowest magnitude of formal charges overall, making it the most stable structure despite the counterintuitive charge placement.

MgO has the highest lattice energy because Mg²⁺ and O²⁻ have higher charges than the singly charged ions in the other compounds, and both ions are relatively small.

sp³d² hybridization involves 6 hybrid orbitals (one s, three p, two d), creating 6 electron domains arranged in octahedral geometry.

An electronegativity difference greater than approximately 1.7 is the conventional threshold used to classify a bond as predominantly ionic, meaning one atom has essentially transferred its electron to the other. Choice C is incorrect because a difference of 3.0 is rarely achieved; the maximum possible electronegativity difference is about 3.3 (Cs-F), and many clearly ionic compounds like NaCl have differences around 2.1, well below 3.0.

Nitrogen has 5 valence electrons. In NH₃, it forms 3 single bonds with hydrogen atoms, using 3 electrons for bonding. The remaining 2 electrons form 1 lone pair on nitrogen. Choice A is wrong because nitrogen would need to form 5 bonds to have no lone pairs, which exceeds its typical bonding capacity. Choice C would require nitrogen to have only 3 valence electrons assigned to bonding, leaving 4 for lone pairs, which does not match nitrogen's electron count.

With only 2 electron domains and no lone pairs, VSEPR predicts maximum separation at 180 degrees, producing a linear geometry. BeCl₂ and CO₂ are classic examples. Choice A (bent) requires 2 bonding pairs plus lone pairs on the central atom, as in water, where lone pair repulsion compresses the ideal 180 degree angle down to about 104.5 degrees.

A polar covalent bond involves unequal sharing of electrons between two atoms of differing electronegativity. The more electronegative atom attracts the shared electrons more strongly, creating partial negative and positive charges (delta minus and delta plus). Choice A describes ionic bonding, where electron transfer is essentially complete. Choice D describes metallic bonding, which is an entirely different bonding model.

In the solid state, ionic compounds do not conduct electricity because the ions are locked in a rigid crystal lattice and cannot move freely to carry charge. Electrical conductivity occurs only when the compound is melted or dissolved in water, freeing the ions. Choice A incorrectly describes molten or aqueous ionic compounds. Choice C is wrong because ionic compounds typically have high melting points due to the strong electrostatic forces between oppositely charged ions.

The octet rule holds that atoms tend to gain, lose, or share electrons until they have 8 valence electrons, mimicking the electron configuration of the nearest noble gas. This drives bond formation in most second-period and third-period main-group elements. Choice C is wrong because the number of bonds varies by element: carbon forms 4, nitrogen typically forms 3, oxygen typically forms 2, and so on.

Boron has only 3 valence electrons and commonly forms 3 covalent bonds in stable neutral compounds like BF₃ and BCl₃, giving it only 6 electrons in its valence shell. Unlike carbon, nitrogen, and oxygen, boron frequently forms stable molecules with an incomplete octet without carrying a formal charge. Carbon always forms 4 bonds to achieve an octet, and nitrogen and oxygen achieve 8 electrons in typical stable neutral compounds.

PCl₃ has 3 bonding pairs and 1 lone pair on phosphorus (4 total electron domains), giving a trigonal pyramidal molecular geometry with bond angles compressed below the tetrahedral ideal of 109.5 degrees. BCl₃ has only 3 bonding pairs and no lone pairs on boron (incomplete octet), giving a perfectly trigonal planar geometry with 120 degree angles. Choice C is incorrect because boron in BCl₃ has no lone pair; its incomplete octet is precisely why the angle is 120 degrees.

Percent ionic character increases with increasing electronegativity difference between the bonded atoms. Fluorine is the most electronegative element (EN approximately 4.0), giving the H-F bond the largest electronegativity difference with hydrogen (EN approximately 2.1), about 1.9. As you descend the halogen group from F to I, electronegativity decreases, so the H-X bond becomes progressively less polar. HI has the smallest electronegativity difference and therefore the least ionic character.

Phosphorus in PCl₅ forms 5 covalent bonds, placing 10 electrons in its valence shell, exceeding the octet. This is possible because phosphorus is in the third period and has access to d orbitals. CCl₄ has exactly 8 electrons around carbon (4 bonds). NH₃ has 8 electrons around nitrogen (3 bonds plus 1 lone pair). BF₃ has only 6 electrons around boron, which is an incomplete octet rather than an expanded one.

Bond length decreases as bond order increases because greater shared electron density pulls the two nuclei closer together. The triple bond (C≡C, approximately 120 pm) is shorter than the double bond (C=C, approximately 134 pm), which is shorter than the single bond (C-C, approximately 154 pm). Choice B presents the exact reverse order, listing bonds from longest to shortest rather than shortest to longest.

Each double bond consists of exactly one sigma bond (end-to-end orbital overlap forming the bond axis) and one pi bond (side-by-side overlap of p orbitals). CO₂ has two double bonds, giving 2 sigma bonds and 2 pi bonds in total. Choice C incorrectly assigns 2 pi bonds per double bond; a double bond has only 1 pi bond. Choice B is wrong because pi bonds are present in every double and triple bond.

CCl₄ has a tetrahedral geometry with four identical C-Cl bond dipoles pointing symmetrically outward from the central carbon. Because of this perfect symmetry, the four bond dipoles cancel vectorially, resulting in a net dipole moment of zero and a nonpolar molecule. H₂O and NH₃ both have lone pairs that create asymmetric electron distributions, producing net dipoles. SO₂ has a bent shape with a lone pair on sulfur, so its two S=O dipoles do not cancel.

According to Coulomb's law, lattice energy is proportional to the product of the ionic charges divided by interionic distance. MgO has doubly charged ions (Mg²⁺ and O²⁻) versus singly charged ions in NaCl (Na⁺ and Cl⁻). The fourfold increase in the charge product (2 x 2 = 4 versus 1 x 1 = 1) dramatically increases lattice energy. Choice A is factually wrong; Mg²⁺ has a smaller ionic radius than Na⁺, which would further increase lattice energy rather than decrease it.

The nitrite ion cannot be fully described by a single Lewis structure; two resonance structures are needed, each placing the double bond on a different oxygen. The actual molecule is a resonance hybrid with pi electron density delocalized equally over both N-O bonds, giving each a bond order of 1.5 and identical intermediate lengths. Choice C describes a localized structure, which would predict two unequal bond lengths, contradicting the experimental evidence of equivalent bonds.

SO₂ is polar because its bent geometry (caused by the lone pair on sulfur) means the two S=O bond dipoles do not point in opposite directions and cannot cancel. Instead, they combine to produce a net molecular dipole moment. Choice A incorrectly describes SO₂ as if it were linear like CO₂; in a linear AX₂ molecule, equal and opposite dipoles do cancel, but the bent geometry of SO₂ prevents this.

C-F has an electronegativity difference of about 1.5 (C = 2.5, F = 4.0), while C-O has a difference of about 1.0 (O = 3.5), so C-F bonds are indeed more polar. Choice A is wrong because oxygen (EN approximately 3.5) is more electronegative than nitrogen (EN approximately 3.0), making O-H more polar than N-H, not less. Choice D is wrong because bond polarity depends on electronegativity difference, not atomic size; fluorine is far more electronegative than chlorine.

With 5 total electron domains (3 bonding and 2 lone pairs), the electron geometry is trigonal bipyramidal. To minimize lone pair-lone pair repulsion, both lone pairs occupy equatorial positions, leaving 1 equatorial bond and 2 axial bonds. This arrangement produces a T-shaped molecular geometry, as seen in ClF₃. Choice C (seesaw) is wrong because it requires 4 bonding pairs and 1 lone pair. Choice D (square planar) requires 4 bonding pairs and 2 lone pairs across 6 electron domains total.

Iodine in IF₅ has 7 valence electrons, forms 5 bonds to F, and retains 1 lone pair, giving 6 total electron domains. The electron geometry is octahedral. Placing the lone pair in one of the six positions leaves the 5 F atoms in the remaining positions, producing a square pyramidal molecular geometry. The lone pair generates an asymmetric arrangement so the I-F bond dipoles do not cancel, making IF₅ polar. Choice C incorrectly calls the molecular geometry octahedral, which is only the electron geometry.

Ozone requires two resonance structures because a single Lewis structure would imply one shorter double bond and one longer single bond. Experimentally, both O-O bonds in ozone are identical in length (approximately 128 pm), lying between the length of a typical O-O single bond and an O=O double bond. The real molecule is a resonance hybrid with pi electrons delocalized over both bonds, giving a bond order of 1.5. Choice A is wrong precisely because the single-structure prediction of unequal bond lengths contradicts experimental observation.

Oxygen in H₂O undergoes sp³ hybridization; the four electron domains adopt a near-tetrahedral arrangement and the two lone pairs compress the H-O-H angle from 109.5 to 104.5 degrees. Sulfur in H₂S, being a larger third-period atom with more diffuse electron density, hybridizes minimally and uses essentially pure p orbitals (which are oriented at 90 degrees) for bonding, resulting in a bond angle close to 90 degrees. Choice A is misleading because greater electronegativity in oxygen would draw bonding electrons away from the H atoms, but this effect is much smaller than the hybridization difference and cannot explain a roughly 13 degree gap.

Lattice energy increases as interionic distance decreases, according to Coulomb's law. Li⁺ has the smallest ionic radius among the three alkali metal cations, so LiF has the shortest Li-F distance and the strongest electrostatic attraction, giving it the highest lattice energy. As cation size increases from Li to Na to K, interionic distance increases and lattice energy decreases accordingly. Choice C is wrong because equal charges do not guarantee equal lattice energies; the interionic distances differ substantially, producing meaningful differences in lattice energy.

Using delta H = (bonds broken) minus (bonds formed): bonds broken = H-H + F-F = 436 + 158 = 594 kJ/mol absorbed; bonds formed = 2 x H-F = 2 x 569 = 1138 kJ/mol released. Delta H = 594 minus 1138 = negative 544 kJ/mol, which is strongly exothermic. Choice A (138 kJ endothermic) is a common error that arises from subtracting only one H-F bond energy instead of two, forgetting that two moles of HF are produced per mole of reaction.

Xenon in XeF₂ has 8 valence electrons. After forming 2 bonds to F, it retains 3 lone pairs, giving 5 total electron domains. VSEPR predicts trigonal bipyramidal electron geometry. To minimize lone pair-lone pair repulsion, all 3 lone pairs occupy the equatorial positions of the trigonal bipyramid, placing both F atoms in the axial positions. With the two F atoms directly opposite each other, the molecular geometry is linear. Choice A (tetrahedral/bent) describes molecules with 4 electron domains and 2 lone pairs, such as H₂O.

A core principle of Lewis structure evaluation is that the best representation minimizes formal charges on all atoms and places negative formal charges on more electronegative atoms. Structure B gives sulfur a formal charge of 0 compared to +2 in Structure A, which much better reflects the actual electron distribution. Sulfur is a third-period element with available d orbitals that allow octet expansion, so the double bonds in Structure B are chemically allowed. Choice A is wrong because the octet rule can be legitimately exceeded for third-period and heavier elements. Choice C is a subtle distractor: while all four oxygens are indeed equivalent in the full resonance hybrid of SO₄²⁻ (requiring 6 equivalent resonance structures with two double bonds each), that equivalence does not mean Structure A and Structure B contribute equally.

Ionic bonds form when one atom (typically a metal) transfers one or more electrons to another atom (typically a nonmetal), creating oppositely charged ions. The electrostatic attraction between these ions constitutes the ionic bond. Choice A describes a nonpolar covalent bond, choice C describes a polar covalent bond, and choice D describes a coordinate (dative) covalent bond.

Phosphorus is in Group 15 of the periodic table, so it has 5 valence electrons (electron configuration ends in 3s²3p³). These 5 electrons are available for bonding or lone pairs in a Lewis structure. Choice A (3) is a common error — it reflects the number of unpaired p electrons that form bonds in simple phosphorus compounds, not the total valence electron count.

With 2 bonding pairs and no lone pairs, the two bonded atoms repel each other to opposite sides of the central atom, producing a 180 degree bond angle and linear molecular geometry. Bent or V-shaped geometries require lone pairs that compress the bond angle below 180 degrees (as in water). Trigonal planar requires 3 electron domains.

Ionic bonds typically form between a metal and a nonmetal when the electronegativity difference is large (generally greater than 1.7). Potassium is a Group 1 alkali metal that readily loses one electron, and bromine is a Group 17 halogen that readily gains one, forming K+ and Br-. The other three pairs are all nonmetal-nonmetal combinations that would form covalent bonds.

To find total valence electrons, sum the contributions of each atom: nitrogen contributes 5 valence electrons, and each of the 3 hydrogen atoms contributes 1, giving 5 + 3(1) = 8 total. These 8 electrons appear as 3 N-H bonding pairs (6 electrons) and 1 lone pair on nitrogen (2 electrons). Choice A (6) would be incorrect because it omits the lone pair on nitrogen.

A lone pair (also called a nonbonding pair) consists of two electrons that belong entirely to one atom and are not shared with any neighboring atom. Lone pairs influence molecular geometry because they occupy space and exert repulsive forces on bonding pairs. Choice A describes a bonding pair, and choice C describes a coordinate covalent (dative) bond — not a lone pair.

Fluorine has the highest electronegativity of all elements, approximately 4.0 on the Pauling scale. Electronegativity generally increases moving across a period (left to right) and up a group (bottom to top), placing fluorine — in the upper right of the periodic table — at the maximum. Oxygen is second (about 3.5), followed by chlorine (about 3.2) and nitrogen (about 3.0).

Sigma bonds form from direct, head-on overlap of orbitals along the internuclear axis, producing the greatest electron density between the nuclei and making sigma bonds the stronger type. Pi bonds form from lateral, side-by-side overlap of parallel p orbitals above and below the internuclear axis. Every single bond is a sigma bond; a double bond contains one sigma and one pi bond. Choice D is incorrect — sigma bonds can form from s-p or p-p overlap as well.

The ammonium ion has total valence electrons = 5 (N) + 4(1) (H) - 1 (positive charge) = 8. All 8 electrons form 4 N-H bonding pairs, leaving no lone pairs on nitrogen. In neutral NH₃, nitrogen has 3 bonds and 1 lone pair; when that lone pair forms a coordinate covalent bond with H+, NH₄+ results with tetrahedral geometry and no lone pairs on nitrogen. Choice A describes neutral NH₃, not the ion.

Boron in BF₃ forms 3 covalent bonds, placing only 6 electrons around it — 2 short of a full octet. This electron deficiency makes BF₃ a strong Lewis acid. In contrast, PCl₅ (10 electrons on P), SF₆ (12 electrons on S), and XeF₂ (10 electrons on Xe including 3 lone pairs) all have expanded octets — they are exceptions to the octet rule for the opposite reason: too many electrons, not too few.

In BF₃, the three polar B-F bond dipoles each point from boron toward the more electronegative fluorine. Because the molecule is trigonal planar with 120 degree angles, the vector sum of the three bond dipoles equals exactly zero — they cancel symmetrically. This illustrates a key principle: a molecule can have polar bonds yet be nonpolar overall if its geometry is sufficiently symmetric. The lone pairs on fluorine do not generate opposing dipoles that cancel; they are already included in the bond polarity analysis.

Formal charge = valence electrons - nonbonding electrons - (1/2 × bonding electrons). Nitrogen has 0 lone pairs and participates in a total of 8 bonding electrons: the double bond contributes 4 and the two single bonds contribute 4. FC(N) = 5 - 0 - (8/2) = 5 - 4 = +1. This positive formal charge on nitrogen is characteristic of nitrogen in highly oxidized states. The overall -1 charge of the ion is distributed across the three oxygen atoms.

As bond order increases from single to double to triple, more electron density accumulates between the nuclei, drawing the atoms closer together and resisting separation more strongly. For carbon-oxygen bonds: C-O single bond (approximately 143 pm, about 360 kJ/mol) is longest and weakest; C=O double bond (approximately 123 pm, about 745 kJ/mol) is intermediate; C≡O triple bond (approximately 113 pm, about 1072 kJ/mol) is shortest and strongest. Choice B inverts this relationship entirely.

The electron geometry and molecular geometry are the same only when there are no lone pairs on the central atom. BeCl₂ has 2 bonding pairs and 0 lone pairs, making both its electron geometry and molecular geometry linear. In contrast, H₂O has 2 lone pairs (tetrahedral electron geometry, bent molecular geometry), NH₃ has 1 lone pair (tetrahedral electron geometry, trigonal pyramidal molecular geometry), and SO₂ has 1 lone pair (trigonal planar electron geometry, bent molecular geometry).

The partial positive charge on hydrogen in an X-H bond is proportional to the electronegativity difference between hydrogen and atom X. The greater the electronegativity of X, the more electron density is pulled away from H, increasing δ+. Fluorine has the highest electronegativity (about 4.0), creating the largest electronegativity difference with hydrogen and the greatest δ+ on hydrogen in HF. Sulfur and phosphorus actually have lower electronegativities than hydrogen in some scales, resulting in very small or even reversed bond dipoles in H₂S and PH₃.

Ozone has two resonance structures: one shows a double bond on the left (O=O-O) and the other shows a double bond on the right (O-O=O). The actual molecule is a resonance hybrid, and the true bond order is the average: (2 + 1)/2 = 1.5. This is confirmed experimentally — both O-O bonds in ozone are equal in length (approximately 128 pm), intermediate between a single bond (about 148 pm) and a double bond (about 121 pm). Choosing answer A or B would mean treating only one resonance structure as real, which misrepresents the actual bonding.

Six total electron domains (5 bonding pairs + 1 lone pair) produce an octahedral electron geometry. When one of the six positions is occupied by a lone pair rather than a bonded atom, the molecular geometry — based only on atom positions — becomes square pyramidal: four atoms form a square base around the central atom, and one atom sits directly above the center. Choice A describes a molecule with 5 total electron domains (4 bonding + 1 lone pair), which gives see-saw molecular geometry. Choice C incorrectly identifies the molecular geometry as the same as the electron geometry.

When one P-O single bond becomes a double bond, the doubly-bonded oxygen loses one lone pair (going from 3 to 2 lone pairs) and gains one more bond. Its formal charge becomes FC = 6 - 4 - (4/2) = 0. Phosphorus now owns one additional half-bond, changing FC from 5 - 0 - 4 = +1 (with 4 single bonds) to FC = 5 - 0 - 5 = 0 (with 1 double + 3 single bonds). The three remaining singly-bonded oxygens retain FC = -1. The sum checks out: 0 + 0 + 3(-1) = -3, consistent with the -3 charge of the ion.

In a trigonal bipyramidal arrangement, 90-degree interactions are the most destabilizing. An axial position forms 90-degree angles with all three equatorial positions (three close interactions), while an equatorial position forms 90-degree angles with only the two axial positions (two close interactions). Since lone pairs exert greater repulsion than bonding pairs, they go where they create the fewest 90-degree interactions — equatorial positions. In ClF₃, both lone pairs occupy equatorial positions, placing the three fluorines in one equatorial and two axial positions, producing the T-shaped geometry.

Benzene is best described as a resonance hybrid of its two Kekule structures — not as a molecule that rapidly switches between them (choice A is a common misconception). The six pi electrons are delocalized over all six carbon atoms equally, giving each C-C bond a bond order of 1.5. This delocalization produces bonds of equal length (140 pm, intermediate between single and double) and confers extra thermodynamic stability known as resonance energy (approximately 150 kJ/mol compared to hypothetical cyclohexatriene). The ring geometry itself does not force bond lengths to converge.

Lattice energy is governed by Coulomb's law: it increases with greater ionic charges and decreases with greater interionic distance. Since all three compounds have the same ionic charges (+1 and -1) and the same anion (F-), the only variable is cation size. Li+ is the smallest alkali cation, giving LiF the shortest interionic distance and the highest lattice energy. Cs+ is the largest, giving CsF the longest interionic distance and lowest lattice energy. Choice B is incorrect — larger ions mean greater separation, which reduces, not increases, lattice energy. Ionization energy (choice D) relates to formation of the gaseous ion, not to the lattice energy itself.

Sulfur in SF₄ has 5 electron domains, producing a trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position because it minimizes repulsion: an equatorial lone pair has only two 90-degree interactions (with the two axial bonding pairs), while an axial lone pair would have three 90-degree interactions (with all three equatorial bonding pairs). The resulting molecular shape — 2 axial fluorines, 2 equatorial fluorines, and 1 equatorial lone pair — is called see-saw. Choice D is incorrect: though axial positions are farther apart from each other (180 degrees), an axial lone pair still generates more total 90-degree repulsions than an equatorial lone pair.

Using the formula FC = valence electrons - nonbonding electrons - (1/2 x bonding electrons): for carbon with 0 lone pairs and 3 single bonds (6 bonding electrons), FC(C) = 4 - 0 - 3 = +1. For each oxygen with 3 lone pairs (6 nonbonding electrons) and 1 single bond (2 bonding electrons), FC(O) = 6 - 6 - 1 = -1. The sum confirms the ion charge: +1 + 3(-1) = -2. Converting one C-O single bond to a double bond in the preferred Lewis structure reduces the formal charge on both that carbon and that oxygen to 0, lowering the overall charge separation.

In Structure I (O=C=O): FC(C) = 4 - 0 - 4 = 0, and FC(each O) = 6 - 4 - 2 = 0. All formal charges are zero. In Structure II (O≡C-O): the triply-bonded oxygen has FC = 6 - 2 - 3 = +1, and the singly-bonded oxygen has FC = 6 - 6 - 1 = -1, while carbon has FC = 0. The positive formal charge on oxygen in Structure II is especially unfavorable because it places a positive charge on the more electronegative atom. Structure I dominates overwhelmingly. These are not equivalent resonance structures — Structure II is a minor contributor at best.

With n double bonds and no lone pairs on chlorine, chlorine owns n×2 + (4-n)×1 = n + 4 electrons from bonding. FC(Cl) = 7 - 0 - (n + 4) = 3 - n. Setting this equal to 0 gives n = 3. With three Cl=O double bonds: each doubly-bonded oxygen has 2 lone pairs and 1 double bond, so FC = 6 - 4 - (4/2) = 0. The one remaining singly-bonded oxygen has 3 lone pairs and 1 single bond: FC = 6 - 6 - 1 = -1. The sum confirms the ion charge: 0 (Cl) + 3(0) + (-1) = -1. Choice D (4 double bonds) would give FC(Cl) = -1, which overshoots the target of 0.

Fluorine is the most electronegative element on the periodic table (EN = 4.0 on the Pauling scale). Electronegativity generally increases across a period and up a group, making fluorine — in the top-right corner of the periodic table — the maximum. Oxygen (3.5) is second, so it is a tempting distractor, but fluorine surpasses it.

Nitrogen is in Group 15 (VA), so it has 5 valence electrons. The number 3 is a common distractor because nitrogen typically forms 3 bonds, but bonding pairs are not the same as valence electrons. The number 7 is the valence electron count for halogens (Group 17), not nitrogen.

Water has 2 bonding pairs and 2 lone pairs around the central oxygen atom, giving it a tetrahedral electron geometry. Because molecular geometry describes only the positions of atoms (not lone pairs), the shape is bent (approximately 104.5°). Trigonal pyramidal is wrong — that describes molecules like NH₃, which have 3 bonding pairs and 1 lone pair.

As bond order increases, the atoms are pulled closer together, so bond length decreases. A C≡C triple bond (bond order 3) is shorter than a C=C double bond (bond order 2), which is shorter than a C–C single bond (bond order 1). Higher bond order also means greater bond energy, so triple bonds are both shorter and stronger.

A tetrahedral arrangement of four electron domains around a central atom produces bond angles of 109.5°. This is the geometry of methane (CH₄), for example. The 120° angle corresponds to trigonal planar geometry (three domains, no lone pairs), and 180° corresponds to linear geometry (two domains).

MgO is formed between a metal (Mg) and a nonmetal (O) with a large electronegativity difference, resulting in full electron transfer and an ionic bond. CO₂ and NH₃ are molecular compounds held together by covalent bonds. HCl is a polar covalent molecule — although it is highly polar, the electronegativity difference (~0.9) is below the ~1.7 threshold typically used to classify a bond as ionic.

Carbon contributes 4 valence electrons and each oxygen contributes 6, giving 4 + 6 + 6 = 16 total valence electrons. These are distributed as two C=O double bonds (8 electrons) and two lone pairs on each oxygen (8 electrons), accounting for all 16. The answer 14 is a common error from miscounting the oxygen contribution.

NH₃ has 3 bonding pairs and 1 lone pair around nitrogen, giving a tetrahedral electron geometry. Because molecular geometry counts only atom positions, the lone pair creates a trigonal pyramidal shape with bond angles of approximately 107°. Trigonal planar is wrong because that geometry has no lone pairs (e.g., BF₃). Tetrahedral describes the electron geometry, not the molecular geometry.

CO₂ is linear (180° bond angle), so the two C=O dipoles are equal in magnitude but point in exactly opposite directions. They cancel vectorially, producing a net dipole of zero. The C=O bond is actually quite polar (electronegativity difference ~1.0), eliminating choice B. Carbon has no lone pairs in the standard Lewis structure of CO₂, making choice C incorrect.

Formal charge = (valence electrons) – (nonbonding electrons) – ½(bonding electrons). Carbon has 4 valence electrons, 2 nonbonding electrons (one lone pair), and 6 bonding electrons (triple bond). So FC(C) = 4 – 2 – 3 = –1. The formal charge on oxygen in CO is +1, which seems counterintuitive given oxygen's higher electronegativity — this is why CO is considered an unusual molecule. Choosing 0 is wrong because the lone pair on carbon is often overlooked.

SO₃ has three equivalent resonance structures, each with the double bond on a different oxygen. The actual molecule is a resonance hybrid — a weighted average — so no single bond is purely single or double. Each bond has a bond order of 4/3, making all three bonds identical in length and energy. Resonance does not mean bonds 'rotate' (choice A); that is a misconception. Choice C involves an expanded octet on S, which is a separate concept and does not fully explain why all bonds are equal.

As a general rule, a bond is classified as predominantly ionic when the electronegativity difference exceeds approximately 1.7 (Pauling scale). A difference of 2.2 clearly exceeds this threshold and indicates nearly complete electron transfer. Differences of 0.1 and 0.5 indicate nonpolar or slightly polar covalent bonds, while 1.2 represents a polar covalent bond but falls short of the ionic threshold.

CHCl₃ (chloroform) is polar because it has three C–Cl bonds and one C–H bond arranged tetrahedrally. The C–Cl bond dipoles do not cancel the C–H dipole, so there is a net dipole moment. BF₃ is trigonal planar — its three B–F dipoles point outward symmetrically and cancel. CCl₄ is tetrahedral with four identical C–Cl bonds that cancel. SF₆ is octahedral with six identical S–F bonds that cancel.

All three molecules have a tetrahedral electron geometry. Lone pairs exert greater repulsion than bonding pairs, compressing bond angles. CH₄ has zero lone pairs: 109.5°. NH₃ has one lone pair: ~107°. H₂O has two lone pairs: ~104.5°. Therefore the order from smallest to largest is H₂O < NH₃ < CH₄. Choice A incorrectly reverses the trend.

BeCl₂ has only 2 bonds around beryllium, giving it just 4 electrons — well below the octet. Beryllium is a common electron-deficient exception. PCl₅ and SF₆ are expanded-octet molecules (10 and 12 electrons around the central atom, respectively) — they exceed the octet rather than fall short. XeF₂ also has an expanded octet (10 electrons around Xe, including 3 lone pairs).

Sulfur contributes 6 valence electrons and each of the four oxygen atoms contributes 6, for a subtotal of 6 + (4 × 6) = 30. The 2– charge means 2 additional electrons must be added: 30 + 2 = 32 total. A common error is forgetting to add the extra electrons from the negative charge, yielding 30 instead of 32.

As bond order increases (single → double → triple), the nuclei are pulled closer together by the greater electron density between them, so bond length decreases. For example, C–C is about 154 pm, C=C is about 134 pm, and C≡C is about 120 pm. Higher bond order also corresponds to greater bond energy, making choice D incorrect.

Lattice energy increases with higher ionic charges and smaller ionic radii. Mg²⁺ and O²⁻ each carry a charge of magnitude 2, so the electrostatic attraction is much stronger than for singly charged ions. The lattice energy is proportional to (q₊ × q₋)/r, so going from ±1 to ±2 roughly quadruples the charge product. MgO has a very high lattice energy (~3850 kJ/mol) compared to NaCl (~787 kJ/mol).

XeF₄ has 6 electron domains (4 bonds + 2 lone pairs), giving an octahedral electron geometry. To minimize lone pair–lone pair repulsion (which is greater than LP–BP repulsion), the two lone pairs are placed on opposite axial positions, 180° apart. This leaves the four fluorine atoms in a square plane, so the molecular geometry is square planar. Choice C (see-saw) would result if only one lone pair were present, as in SF₄.

Formal charge = valence electrons – nonbonding electrons – ½(bonding electrons). Sulfur: 6 – 2 – ½(6) = +1. Double-bonded oxygen (2 lone pairs, 4 bonding electrons): 6 – 4 – ½(4) = 0. Single-bonded oxygen (3 lone pairs, 2 bonding electrons): 6 – 6 – ½(2) = –1. The total is +1 + 0 – 1 = 0, consistent with a neutral molecule. Choice C reverses the oxygen formal charges, incorrectly placing –1 on the oxygen that forms a double bond.

In the structure :N≡N–O: (with one lone pair on terminal N and three lone pairs on O): Terminal N has 1 lone pair and a triple bond: FC = 5 – 2 – ½(6) = 0. Central N has no lone pairs, a triple bond to one N, and a single bond to O: FC = 5 – 0 – ½(8) = +1. Oxygen has 3 lone pairs and a single bond: FC = 6 – 6 – ½(2) = –1. The negative formal charge resides on the more electronegative oxygen, making this the most stable structure. Choice A corresponds to the alternative resonance structure :N=N=O, where the terminal N bears the negative charge — less stable because N is less electronegative than O.

PCl₅ has 5 bonding pairs around phosphorus (5 P–Cl bonds), giving 10 electrons — an expanded octet possible because phosphorus is in the third period and has available d orbitals. Five electron domains arrange to minimize repulsion in a trigonal bipyramidal geometry with bond angles of 90° and 120°. Square pyramidal (choice D) would require one lone pair occupying one of the six positions in an octahedral arrangement, as in BrF₅.

BCl₃ is trigonal planar (3 bonding pairs, no lone pairs on B). The three B–Cl bond dipoles point outward at 120° to each other and cancel exactly, producing a net dipole moment of zero. H₂O and NH₃ both have lone pairs that break symmetry, so their bond dipoles do not cancel and both are polar. CHCl₃ is tetrahedral but asymmetric (one H and three Cl), so its dipoles do not cancel.

The actual nitrite ion is a resonance hybrid of the two contributing structures. Both N–O bonds are identical in length and energy, with a bond order of (1+2)/2 = 1.5 — intermediate between a single (bond order 1) and double (bond order 2) bond. Resonance does not mean the bonds 'alternate' rapidly (choice A); the hybrid is a single real structure. Choice C is backwards: higher bond order (double bond character) means shorter and stronger, not longer and weaker. In the standard Lewis structures, nitrogen in NO₂⁻ carries a formal charge of 0 in the structure with a double bond and a lone pair, but +1 in the structure without a lone pair — they are not both zero.

Lattice energy is governed by Coulomb's law: it is proportional to the product of ionic charges divided by interionic distance. Going from ±1 ions (NaF) to ±2 ions (MgO) increases the charge product by a factor of 4, which dominates over any difference in ionic radius. MgO has a lattice energy of approximately 3850 kJ/mol versus about 923 kJ/mol for NaF. Choice A confuses electronegativity (a property related to covalent bonding) with lattice energy. Choice D incorrectly applies the properties of metallic bonding to ionic lattice energy.

Ionic compounds form between elements with large electronegativity differences, typically a metal and a nonmetal. Potassium is an alkali metal (low electronegativity) and bromine is a halogen (high electronegativity), producing a large difference that favors electron transfer and ionic bonding. The other pairs involve two nonmetals with similar electronegativities, which leads to electron sharing (covalent bonds) rather than electron transfer.

The Lewis structure of N₂ shows a triple bond: N≡N. Each nitrogen contributes 5 valence electrons; after forming a triple bond (3 shared pairs = 6 electrons) and one lone pair on each nitrogen, all electrons are accounted for. Bond order equals the number of shared electron pairs between two atoms, which is 3. A bond order of 4 does not exist for nitrogen under normal conditions.

Oxygen has 6 valence electrons. In H₂O, oxygen forms 2 covalent bonds with the two hydrogen atoms, using 2 electrons for bonding. The remaining 4 electrons exist as 2 lone pairs on oxygen. These lone pairs are critical for VSEPR analysis: they push the bonding pairs closer together, resulting in the bent molecular geometry and a bond angle of approximately 104.5°.

Lattice energy is the energy released when gaseous cations and anions come together to form a crystalline ionic solid. It is always exothermic (negative in sign) because forming the stable crystal lattice releases energy. Choice A describes ionization energy, and choice C describes the reverse process (sometimes called the lattice dissociation enthalpy), which requires energy input rather than releasing it.

A double bond consists of two shared pairs of electrons, for a total of 4 electrons shared between the two bonded atoms. A single bond contains 2 shared electrons (1 pair), and a triple bond contains 6 shared electrons (3 pairs). Choice A describes a single bond, and choice C describes a triple bond.

CO₂ has two C=O double bonds and no lone pairs on carbon, giving 2 electron domains. VSEPR predicts a linear arrangement to maximize separation, resulting in a 180° bond angle. H₂O is bent (2 bonds plus 2 lone pairs), NH₃ is trigonal pyramidal (3 bonds plus 1 lone pair), and CH₄ is tetrahedral (4 bonds, no lone pairs). Lone pairs on the central atom always result in nonlinear molecular geometries for these examples.

When a central atom has 4 bonding pairs and no lone pairs, the 4 electron domains arrange themselves as far apart as possible, producing tetrahedral geometry with bond angles of approximately 109.5°. Trigonal pyramidal also has 4 electron domains but includes 1 lone pair (reducing bond angles). Square planar requires 6 electron domains with 2 lone pairs. Trigonal planar involves only 3 electron domains.

Across a period, the number of protons (nuclear charge) increases while electrons are added to the same shell, so atomic radius decreases. The higher nuclear charge and smaller radius mean the nucleus attracts shared electrons more strongly, increasing electronegativity. Choice A has the correct observation about radius but draws the wrong conclusion. Choice D is incorrect because new electron shells are only added when moving down a group, not across a period.

Formal charge = (valence electrons of neutral atom) - (lone pair electrons) - (1/2)(bonding electrons). Nitrogen normally has 5 valence electrons. In NH₄⁺, nitrogen forms 4 N-H bonds and has no lone pairs. FC = 5 - 0 - (8/2) = 5 - 4 = +1. Choice B is wrong because nitrogen does not retain its neutral electron count here — it has donated electron density into 4 bonds without keeping any lone pairs, giving it a positive formal charge.

According to VSEPR theory, lone pairs occupy more space than bonding pairs because they are held by only one nucleus instead of two. The lone pair on nitrogen repels the three N-H bonding pairs more strongly than those bonding pairs repel each other, compressing the H-N-H angle below the ideal 109.5°. Choice A is a common misconception — electronegativity differences affect bond polarity but are not the VSEPR explanation for bond angle compression. Choice C is incomplete; it is specifically the lone pair that drives the change.

Higher bond order corresponds to shorter, stronger bonds. A triple bond (bond order 3) is the shortest, followed by a double bond (bond order 2), and a single bond (bond order 1) is the longest. This is because more shared electron pairs pull the two nuclei closer together. Typical values: C≡C ≈ 120 pm, C=C ≈ 134 pm, C-C ≈ 154 pm. Choices A and D reverse this trend.

Three equivalent resonance structures can be drawn for CO₃²⁻, each placing the double bond on a different oxygen. The actual structure is a resonance hybrid — a weighted average of all three contributors — giving each C-O bond an effective bond order of 4/3 ≈ 1.33. This delocalization explains the equal, intermediate bond lengths. Choice A is wrong because carbon is a second-period element that does not expand its octet. Choice C misapplies the concept of coordinate covalent bonds.

Lattice energy increases with higher ionic charges and smaller ionic radii. MgO consists of Mg²⁺ and O²⁻ ions, which carry charges of 2+ and 2-, and both ions are relatively small. The Coulombic attraction between these doubly charged, small ions is much stronger than for singly charged ions. NaF and KCl have only 1+ and 1- charges. CaS also has 2+ and 2- charges but Ca²⁺ and S²⁻ are significantly larger than Mg²⁺ and O²⁻, reducing the lattice energy.

Chlorine has 7 valence electrons; forming 3 bonds to fluorine uses 3 electrons, leaving 4 as 2 lone pairs. ClF₃ therefore has 5 electron domains (3 bonding pairs plus 2 lone pairs), giving a trigonal bipyramidal electron geometry. To minimize lone pair-lone pair repulsion, both lone pairs occupy the equatorial positions of the trigonal bipyramid, pushing the three fluorine atoms into axial and equatorial positions that form a T-shaped molecular geometry. Trigonal planar would require 3 bonding pairs and no lone pairs.

Ozone is a resonance hybrid of two equivalent structures, each showing one O=O double bond and one O-O single bond on opposite sides. The actual molecule does not alternate between these structures — it exists as a single hybrid in which both bonds are equivalent, with a bond order of 1.5. This gives bond lengths between those of a pure single bond (148 pm) and a pure double bond (121 pm), experimentally measured at approximately 128 pm. Choice A incorrectly treats one resonance structure as the actual structure.

Sulfur has 6 valence electrons. Forming 4 bonds to fluorine uses 4 electrons, leaving 2 as 1 lone pair. SF₄ therefore has 5 electron domains (4 bonding pairs plus 1 lone pair). VSEPR theory predicts 5 electron domains adopt a trigonal bipyramidal electron geometry to minimize repulsion. The molecular geometry (describing only atom positions) is see-saw, but the question asks for electron geometry, which counts all electron domains including lone pairs. Choice C is the molecular geometry, not the electron geometry.

Formal charge assigns each atom its own valence electrons plus one-half of the bonding electrons (since each bond is shared equally). The formula is: FC = (valence electrons of neutral atom) - (nonbonding/lone pair electrons) - (1/2)(bonding electrons). Choice A incorrectly subtracts all bonding electrons rather than half of them. Choices C and D add lone pair electrons instead of subtracting, which would give an incorrect result.

In Structure I, the formal charge on S is +2 and each O carries -1 (four single bonds: S FC = 6 - 0 - 4 = +2). In Structure II, the S formal charge drops to 0 and only the two singly-bonded O atoms carry -1 each (S FC = 6 - 0 - 6 = 0). Lower formal charges and negative charges on the more electronegative oxygen atoms make Structure II more stable. Choice A is tempting but wrong because sulfur is in period 3 and CAN expand its octet. Choice D incorrectly states that sulfur can never exceed an octet — period 3 and beyond elements regularly do so.

The central atom has 3 bonding pairs plus 1 lone pair, giving 4 electron domains total. Four electron domains adopt a tetrahedral electron geometry. Because one position is occupied by a lone pair rather than an atom, the molecular geometry (based only on atom positions) is trigonal pyramidal, as seen in NH₃. Choice C describes a molecule with 5 electron domains (such as ClF₃). Choice D is wrong because trigonal pyramidal is a molecular geometry, not an electron geometry.

Fluorine atoms are extremely small, so when two fluorine atoms bond, their lone pairs are in very close proximity and repel each other strongly. This lone pair-lone pair repulsion partially destabilizes the F-F sigma bond, lowering its dissociation energy below that of Cl₂. Chlorine atoms are larger, so lone pairs are farther apart and repel each other less. Choice A is factually wrong — both F₂ and Cl₂ have single bonds (bond order 1). Choice C is incorrect; fluorine is the most electronegative element.

Three resonance structures can be drawn for NO₃⁻, each placing the double bond on a different oxygen. The actual ion is a resonance hybrid — all three contribute equally — so the pi electron density is delocalized over all three N-O bonds. Each bond has an effective bond order of (1+1+2)/3 = 4/3 ≈ 1.33, which corresponds to an intermediate bond length. Choice C is a misconception; resonance structures do not describe dynamic interconversion but rather the single hybrid structure. Choice D is wrong because nitrogen is a period 2 element with no available d orbitals.

NH₃ has the largest bond angle at approximately 107°. Although both NH₃ and H₂O have a central atom with lone pairs, H₂O has two lone pairs versus one for NH₃, so water experiences greater lone pair repression of its bond angle. The period 3 hydrides (H₂S and PH₃) have much smaller angles near 90° because the larger, more diffuse 3p orbitals on sulfur and phosphorus have poor overlap with hydrogen 1s orbitals, and the bonding pairs are farther apart, reducing repulsion and allowing the angles to approach 90°.

Formal charge = (valence electrons) - (lone pair electrons) - (1/2)(bonding electrons). Phosphorus has 5 valence electrons. In the second structure, phosphorus forms 1 double bond and 3 single bonds, giving 5 total bonding pairs (10 bonding electrons) and no lone pairs. FC = 5 - 0 - (10/2) = 5 - 5 = 0. This is why the second structure is preferred — it reduces the formal charge on phosphorus from +1 to 0, making the overall charge distribution more stable, consistent with the rule of minimizing formal charges.

Xenon has 8 valence electrons; forming 2 bonds uses 2 electrons, leaving 6 as 3 lone pairs. XeF₂ has 5 electron domains (2 bonding plus 3 lone pairs), giving a trigonal bipyramidal electron geometry. The 3 lone pairs occupy the equatorial positions (where repulsion is minimized), and the 2 fluorine atoms occupy the axial positions, producing a linear molecular geometry. The two Xe-F bond dipoles point in exactly opposite directions and cancel, giving a net dipole of zero — so XeF₂ is nonpolar despite having polar bonds. Choice A would require only 2 electron domains on the central atom.

In Structure I (O=C=O), all formal charges are zero: FC on C = 4 - 0 - 4 = 0, and FC on each O = 6 - 4 - 2 = 0. In Structure II, the triply-bonded oxygen has FC = 6 - 2 - 3 = +1, and the singly-bonded oxygen has FC = 6 - 6 - 1 = -1, while carbon remains 0. The guiding principle in Lewis structures is to minimize formal charges; when all formal charges can be zero, that structure is strongly preferred. Choice C is wrong because these are not resonance structures — they are different connectivity arrangements, and Structure I is the correct one. Choice D is incorrect since both structures satisfy the octet rule on oxygen.

Ionic bonds form through the complete transfer of electrons from a metal to a nonmetal, producing cations and anions that attract each other electrostatically. Choice A describes a nonpolar covalent bond, Choice C describes a polar covalent bond, and Choice D describes a coordinate (dative) covalent bond.

Oxygen contributes 6 valence electrons and each hydrogen contributes 1, for a total of 6 + 1 + 1 = 8 electrons. Choice A (4) only counts the two hydrogen contributions, Choice B (6) only counts oxygen, and Choice D (10) overcounts by assuming each hydrogen contributes 2 electrons.

Four electron groups (2 bonding pairs and 2 lone pairs) give a tetrahedral electron geometry, but molecular geometry describes only atom positions. With two lone pairs occupying two of the four positions, the two bonded atoms form a bent (V-shaped) arrangement. Water is the classic example. Tetrahedral molecular geometry requires 4 bonding pairs and 0 lone pairs; trigonal pyramidal requires 3 bonding pairs and 1 lone pair.

When ionic compounds dissolve in water, they dissociate into mobile cations and anions that carry electric current. Ionic compounds are typically solids at room temperature with high melting points due to strong electrostatic forces, and many dissolve readily in polar solvents. Choices A, C, and D are all false for most ionic compounds.

N₂ contains a triple bond (N≡N) with a bond order of 3 — one sigma bond and two pi bonds. O₂ has a double bond (bond order 2), while F₂ and Cl₂ each have only a single bond (bond order 1) because each halogen atom has only one unpaired electron available for bonding.

Lone pairs (also called nonbonding pairs) belong to one atom and do not contribute to bonding between atoms. They do, however, influence molecular geometry according to VSEPR theory because they occupy space around the central atom. Choice A describes a bonding pair, Choice C describes a radical electron, and Choice D describes a coordinate (dative) covalent bond — which is a bonding pair, not a lone pair.

Bond strength increases with bond order because more electron density between the nuclei provides stronger electrostatic attraction. Single bonds (one shared pair) are weakest, double bonds (two shared pairs) are intermediate, and triple bonds (three shared pairs) are strongest and shortest. Choice B reverses the correct order, and Choice C incorrectly places the triple bond between single and double.

Formal charge = valence electrons minus nonbonding electrons minus one-half the bonding electrons. Nitrogen has 5 valence electrons. With one lone pair (2 nonbonding electrons) and 6 bonding electrons (1 double bond plus 1 single bond): FC(N) = 5 - 2 - (6/2) = 5 - 2 - 3 = 0. The overall ion charge of -1 is accounted for by formal charges on the two oxygen atoms. Choice B (+1) would result if nitrogen had no lone pair and formed three bonds with 8 total bonding electrons.

BF₃ has trigonal planar geometry with three equivalent B-F bonds oriented at 120° angles. The vector sum of the three bond dipoles equals zero because of the perfect three-fold symmetry. BF₃ has no lone pair on boron — it is an electron-deficient molecule with only 6 electrons around boron. Choice A is false because the B-F bond is actually quite polar. Choice D confuses identical substituents with dipole cancellation, which requires geometric symmetry to occur.

SO₂ is polar because its bent molecular geometry — caused by the lone pair on sulfur — prevents the two bond dipoles from canceling, producing a net dipole moment. CO₂ is linear and its two C=O dipoles cancel. CCl₄ is tetrahedral and its four C-Cl dipoles cancel. BF₃ is trigonal planar and its three B-F dipoles cancel. Molecular polarity requires both polar bonds and an asymmetric geometry that prevents cancellation.

The actual bonding in ozone is the average of its two resonance contributors. Bond order = (sum of bond orders across all resonance structures) divided by (number of resonance structures) = (1 + 2) / 2 = 1.5. This is confirmed experimentally: both O-O bonds in ozone are identical in length (128 pm), which is intermediate between a typical O-O single bond (148 pm) and a typical O=O double bond (121 pm). A bond order of exactly 1 or 2 would imply only one resonance structure adequately describes the molecule.

Lattice energy is proportional to (Q₊ × Q₋) / r, where Q represents ionic charges and r is the interionic distance. Replacing ions with smaller ones decreases r, which increases the ratio and therefore the lattice energy. Choices A and B both increase the interionic distance by substituting larger ions, decreasing lattice energy. Choice D reduces the charge product from (2 × charge of anion) to (1 × charge of anion), significantly lowering lattice energy.

Electron geometry is determined by the total number of electron groups — both bonding and lone pairs — regardless of bond order. Sulfur has 3 electron groups: one double bond counted as one group, one single bond, and one lone pair. Three electron groups produce trigonal planar electron geometry. The molecular geometry is bent because one of the three positions is occupied by a lone pair rather than an atom. Tetrahedral electron geometry (Choice C) would require 4 electron groups.

In H₃O⁺, oxygen forms three single bonds to three hydrogen atoms and retains one lone pair. Formal charge = valence electrons minus nonbonding electrons minus one-half the bonding electrons = 6 - 2 - (6/2) = 6 - 2 - 3 = +1. This +1 formal charge is consistent with the overall +1 charge of the ion. Choice B (0) applies to oxygen in neutral water, where oxygen has two bonds and two lone pairs: 6 - 4 - (4/2) = 0.

BeCl₂ has a central beryllium atom with only 4 electrons (two single bonds, no lone pairs), making it electron-deficient and an octet-rule exception in the deficient direction. PCl₅ (10 electrons around P), SF₆ (12 electrons around S), and XeF₂ (10 electrons around Xe) are exceptions in the other direction — their central atoms accommodate expanded octets with more than 8 electrons, which is possible for elements in period 3 and beyond.

Ionic character correlates with electronegativity difference. Cesium has the lowest electronegativity among metals (approximately 0.79) and fluorine has the highest electronegativity of all elements (3.98), giving a difference of approximately 3.19. Sodium and chlorine also form an ionic bond, but the electronegativity difference (approximately 2.23) is smaller. Carbon-oxygen and nitrogen-chlorine pairs have differences below 1.7 and form polar covalent, not ionic, bonds.

Phosphorus contributes 5 valence electrons, each of the four oxygen atoms contributes 6 (total = 24 from oxygen), and the 3- ionic charge adds 3 more electrons: 5 + 24 + 3 = 32 total valence electrons. Choice A (26) omits the ionic charge. Choice B (29) reflects an arithmetic error, likely adding only 1 for the charge. Choice D (35) overcounts, possibly using 7 electrons per oxygen instead of 6.

As bond order increases, more electron density accumulates between the nuclei, resulting in greater electrostatic attraction (stronger bond) and shorter internuclear distance. C=O double bonds (bond energy approximately 745 kJ/mol, length approximately 120 pm) are both stronger and shorter than C-O single bonds (bond energy approximately 358 kJ/mol, length approximately 143 pm). Choice D is wrong because pi bonds add net bonding character and increase bond strength — they do not destabilize the bond.

With 2 bonds and 2 lone pairs on Xe: FC(Xe) = 8 - 4 (nonbonding electrons) - (4/2) = +2. Each F in this structure has FC = 7 - 6 - 1 = 0, giving a total formal charge of +2 for a supposedly neutral molecule, which is invalid. In the accepted expanded-octet structure, Xe has 2 bonds and 3 lone pairs (10 electrons): FC(Xe) = 8 - 6 - (4/2) = 0. All formal charges equal zero, correctly representing a neutral molecule. Xenon is in period 5 and can accommodate an expanded octet. Minimizing formal charges is a key criterion in evaluating Lewis structures.

Lattice energy is proportional to (Q₊ × Q₋) / r. MgO contains Mg²⁺ and O²⁻ (charge product = 4), while NaCl contains Na⁺ and Cl⁻ (charge product = 1). Additionally, Mg²⁺ and O²⁻ are small ions with a short interionic distance. Both effects combine to give MgO a lattice energy of approximately 3795 kJ/mol versus NaCl's approximately 786 kJ/mol. Choice C is incorrect — electron count does not directly determine lattice energy. Choice D is incorrect because Li⁺ is smaller than Na⁺, not larger, so LiF actually has a higher lattice energy than NaF.

AX₄E₁ has 5 total electron groups, giving trigonal bipyramidal electron geometry. The lone pair preferentially occupies an equatorial position (120° environment) to minimize repulsion with neighboring pairs. The 4 bonded atoms then adopt a seesaw shape. Ideal bond angles are 90° (axial-equatorial) and 120° (equatorial-equatorial), but lone pair repulsion compresses these slightly below ideal values. SF₄ is a well-known example. Choice C describes AX₅E₁ (6 electron groups), and Choice A describes AX₄E₀.

For the doubly-bonded oxygen (2 lone pairs, 4 bonding electrons): FC = 6 - 4 - (4/2) = 0. For each singly-bonded oxygen (3 lone pairs, 2 bonding electrons): FC = 6 - 6 - (2/2) = -1. Nitrogen in this contributor has 0 lone pairs and 4 bonds (8 bonding electrons): FC(N) = 5 - 0 - (8/2) = +1. Summing all formal charges: +1 + 0 + (-1) + (-1) = -1, which matches the ion's overall charge. Choice B incorrectly places the negative formal charge on the atom with more bonds, which is the opposite of the correct result.

With 4 single bonds and 0 lone pairs on sulfur: FC(S) = 6 - 0 - (8/2) = +2. Each singly-bonded oxygen with 3 lone pairs has FC = 6 - 6 - (2/2) = -1. Total: +2 + 4(-1) = -2, which correctly matches the ion charge. After converting two S-O bonds to double bonds, sulfur has 2 double bonds and 2 single bonds (12 bonding electrons, 0 lone pairs): FC(S) = 6 - 0 - (12/2) = 0. The doubly-bonded oxygens now have FC = 6 - 4 - (4/2) = 0. This structure minimizes formal charges and is often considered a more accurate representation of the actual bonding in sulfate.

AX₂E₂ molecules always adopt bent molecular geometry because 4 total electron groups force a tetrahedral electron arrangement, and 2 lone pairs occupy 2 of those positions. In a bent molecule, the two bond dipoles point in directions that reinforce rather than cancel: both point toward the more electronegative atom, and their vector sum is nonzero. Even when both X atoms are identical — as in H₂O or OF₂ — the asymmetric lone-pair geometry ensures a net dipole remains. Choice C is incorrect because AX₂E₂ geometry cannot be linear; lone pairs prevent it.

In VSEPR theory, a double bond counts as one electron group. Three electron groups (one double bond, one single bond, one lone pair) produce trigonal planar electron geometry. With one lone pair occupying one of the three positions, only two atoms are bonded to the central atom, giving bent molecular geometry. The ideal trigonal planar angle is 120°, but lone pair repulsion exceeds bonding pair repulsion, compressing the actual bond angle slightly below 120°. SO₂ is a well-known example of this arrangement. Choice A describes a molecule with 4 electron groups (tetrahedral electron geometry).