--- title: "Lewis Structures, Resonance and VSEPR" book: "University Chemistry — Year 1" subject: chemistry language: en chapter: 3 exercises: 12 source: https://one-course.com/books/chemistry/2/en/chapter/3-lewis-structures-resonance-and-vsepr license: CC-BY-NC-SA-4.0 credit: "One Chemistry Book, One Course (one-course.com)" --- # Chapter 3 — Lewis Structures, Resonance and VSEPR The ozone molecule, $\ce{O3}$, is three oxygen atoms in a bent chain. Draw it with the rules of Book 1 and one bond comes out double, the other single: one bond should be short, the other long. Yet microwave spectroscopy finds the two $\ce{O-O}$ bonds of ozone exactly equal, at $127.8\,\mathrm{pm}$, between the double bond of $\ce{O2}$ ($120.8\,\mathrm{pm}$) and the single bond of hydrogen peroxide ($147.5\,\mathrm{pm}$). No single drawing is right; two drawings together are. This chapter makes the Lewis model precise — [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge), octets that are incomplete or exceeded, resonance between several structures — and then uses it to predict the shapes of molecules and the [dipole moments](#def-b1-lewis-resonance-vsepr-dipole) that follow from them. **You already know.** Book 1 (grade 10) described the [covalent bond](#def-b1-lewis-resonance-vsepr-lewis) as a shared pair of electrons, drew [Lewis structures](#def-b1-lewis-resonance-vsepr-lewis) with bonding and [lone pairs](#def-b1-lewis-resonance-vsepr-lewis), and predicted four shapes (linear, bent, trigonal, tetrahedral). [Electronegativity](https://one-course.com/books/chemistry/2/en/chapter/2-periodicity#def-b1-periodicity-electronegativity) and its differences were made quantitative in [Chapter 2](https://one-course.com/books/chemistry/2/en/chapter/2-periodicity#ch-b1-periodicity); the [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) of an atom are read from its configuration ([Definition 1.11](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration)). ## 3.1 Lewis structures and formal charges **Definition 3.1 (Covalent bond, Lewis structure).** A *covalent bond* is a pair of electrons shared by two atoms, the *bonding pair*; two or three pairs make a double or a triple bond. A pair of [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) that belongs to a single atom is a *lone pair*. A *Lewis structure* shows every valence electron of a molecule or ion, as bonds (lines between atoms) and lone pairs (lines or pairs of dots on an atom). By the *octet rule*, the atoms of [period](https://one-course.com/books/chemistry/2/en/chapter/2-periodicity#def-b1-periodicity-table) 2 (C, N, O, F) are surrounded in a stable structure by four pairs, eight electrons; by the *duet rule*, hydrogen by one pair. **Definition 3.2 (Formal charge).** The *formal charge* of an atom in a [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) is $$ q_F = N_v - N_{\text{lone}} - \tfrac12 N_{\text{bond}}, $$ where $N_v$ is the number of [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) of the free atom, $N_{\text{lone}}$ the number of electrons in its [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) and $N_{\text{bond}}$ the number of electrons in the bonds it forms. It is written $\oplus$ or $\ominus$ next to the atom when it is not zero. **Proposition 3.3 (Formal charges add up to the charge).** The sum of the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) of the atoms of a [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) equals the total charge of the molecule or ion. **Proof.** Summing $q_F$ over the atoms gives $\sum N_v - (\text{electrons in lone pairs} + \text{electrons in bonds})$, because each bond’s electrons are counted half on each of its two atoms. The bracket is the total number of [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) drawn in the structure, which equals $\sum N_v$ minus the charge $z$ of the species. Hence $\sum q_F = z$. ∎ **Method 3.4 (Drawing a Lewis structure).** To draw the [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) of a molecule or ion: 1. count the [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) $N = \sum N_v - z$ and the pairs $N/2$ ; 2. draw the skeleton: the least electronegative atom (not H) is usually central; hydrogen and fluorine are always terminal; 3. complete the octets of the terminal atoms with [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) , then place the remaining pairs on the central atom; 4. if the central atom lacks an octet, turn [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) of its neighbours into multiple bonds; 5. compute the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) ; among possible structures prefer the one with the fewest [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) , and with negative charges on the most electronegative atoms. **Example 3.5 (Nitric acid).** $\ce{HNO3}$: $N = 1 + 5 + 3 \times 6 = 24$ electrons, 12 pairs. Skeleton: nitrogen central, bonded to three oxygens, one of which carries the hydrogen. Completing octets with single bonds leaves nitrogen with six electrons; one oxygen [lone pair](#def-b1-lewis-resonance-vsepr-lewis) becomes an $\ce{N=O}$ bond. [Formal charges](#def-b1-lewis-resonance-vsepr-formal-charge): nitrogen $5 - 0 - 4 = +1$; the singly bonded oxygen without hydrogen $6 - 6 - 1 = -1$; the others 0. The total is 0, as [Proposition 3.3](#prop-b1-lewis-resonance-vsepr-formal-sum) requires. ![Lewis structures with lone pairs and formal charges.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-0b74cee4071c.svg) *[Lewis structures](#def-b1-lewis-resonance-vsepr-lewis) with [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) and [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge).* ## 3.2 Beyond the octet **Definition 3.6 (Electron-deficient and hypervalent molecules).** A molecule in which an atom has fewer than eight [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) is *electron-deficient*: in $\ce{BF3}$ boron has six. A molecule in which an atom of [period](https://one-course.com/books/chemistry/2/en/chapter/2-periodicity#def-b1-periodicity-table) 3 or beyond is surrounded by more than four pairs is *hypervalent*: in $\ce{PCl5}$ phosphorus forms five bonds, in $\ce{SF6}$ sulfur six. Molecules with an odd number of electrons, such as $\ce{NO}$ and $\ce{NO2}$, cannot satisfy the [octet rule](#def-b1-lewis-resonance-vsepr-lewis) on every atom: one electron is unpaired. **Remark 3.7 (Why period 3 and not period 2).** Nitrogen forms $\ce{NF3}$ but never $\ce{NF5}$, while phosphorus forms both $\ce{PF3}$ and $\ce{PF5}$. A period-2 atom is too small to bond to more than four neighbours; a period-3 atom is large enough to hold five or six. Drawings of $\ce{SO4^{2-}}$ or $\ce{H2SO4}$ with sulfur double bonds (12 electrons around S) or with single bonds and [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) (an octet) are both found; the second reflects better the charge distribution, and both give the right geometry. **Definition 3.8 (Lewis acid, Lewis base, dative bond).** A *Lewis acid* is a species able to accept a pair of electrons into a vacant place of its valence shell ($\ce{BF3}$, $\ce{AlCl3}$, $\ce{H+}$, metal cations); a *Lewis base* is a species with a [lone pair](#def-b1-lewis-resonance-vsepr-lewis) it can share ($\ce{NH3}$, $\ce{H2O}$, $\ce{F-}$). When the base gives the pair to the acid, the bond formed is a *dative bond*: a [covalent bond](#def-b1-lewis-resonance-vsepr-lewis) whose two electrons both came from one partner. **Example 3.9 (Ammonia and boron trifluoride).** $\ce{NH3 + BF3 -> H3NBF3}$. In the adduct, boron completes its octet; nitrogen, which now shares its [lone pair](#def-b1-lewis-resonance-vsepr-lewis), carries the [formal charge](#def-b1-lewis-resonance-vsepr-formal-charge) $+1$ and boron $-1$. Once formed, the $\ce{N-B}$ bond is an ordinary [covalent bond](#def-b1-lewis-resonance-vsepr-lewis). ![A dative bond: the lone pair of ammonia (a Lewis base) fills the vacant place of boron trifluoride (a Lewis acid). Lone pairs of fluorine are not drawn.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-30c7b4ef18d7.svg) *A [dative bond](#def-b1-lewis-resonance-vsepr-lewis-acid): the [lone pair](#def-b1-lewis-resonance-vsepr-lewis) of ammonia (a [Lewis base](#def-b1-lewis-resonance-vsepr-lewis-acid)) fills the vacant place of boron trifluoride (a [Lewis acid](#def-b1-lewis-resonance-vsepr-lewis-acid)). [Lone pairs](#def-b1-lewis-resonance-vsepr-lewis) of fluorine are not drawn.* ## 3.3 Resonance **Definition 3.10 (Resonance structures, resonance hybrid).** When the electrons of a molecule can be placed in several [Lewis structures](#def-b1-lewis-resonance-vsepr-lewis) that differ only by the position of $\pi$ bonds and [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) — the nuclei staying in place — each is a *resonance structure* (or mesomeric form), linked to the others by a double-headed arrow $\leftrightarrow$. The real molecule is none of them: it is the *resonance hybrid*, a single structure whose electron distribution is a weighted average of theirs. **Remark 3.11 (An arrow, not an equilibrium).** The arrow $\leftrightarrow$ does not describe a reaction: ozone does not switch from one structure to the other. The hybrid is one molecule, described by several drawings because one drawing is not enough. Never use $\rightleftharpoons$ between [resonance structures](#def-b1-lewis-resonance-vsepr-resonance). **Method 3.12 (Writing and weighting resonance structures).** To find the [resonance structures](#def-b1-lewis-resonance-vsepr-resonance) of a species: 1. start from one [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) ; look for a [lone pair](#def-b1-lewis-resonance-vsepr-lewis) or a $\pi$ bond next to a $\pi$ bond, a vacant place or a positive charge; 2. move pairs with curly arrows (a [lone pair](#def-b1-lewis-resonance-vsepr-lewis) becomes a $\pi$ bond, a $\pi$ bond becomes a [lone pair](#def-b1-lewis-resonance-vsepr-lewis) ), never moving a nucleus and never exceeding the octet of a period-2 atom; 3. recompute the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) ; 4. weight the structures: the most important ones obey the [octet rule](#def-b1-lewis-resonance-vsepr-lewis) , have the fewest [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) , and put negative charges on the most electronegative atoms; equivalent structures weigh the same. **Example 3.13 (Ozone, nitrate, benzene).** Ozone has two equivalent structures: each $\ce{O-O}$ bond is double in one and single in the other, so both have a bond order $1.5$ and the same length, $127.8\,\mathrm{pm}$, between $\ce{O=O}$ and $\ce{O-O}$. The nitrate ion $\ce{NO3-}$ has three equivalent structures; each $\ce{N-O}$ bond has order $4/3$. Benzene, $\ce{C6H6}$, has two Kekulé structures; its six $\ce{C-C}$ bonds are equal, $139.7\,\mathrm{pm}$, between the single bond of ethane ($153.6\,\mathrm{pm}$) and the double bond of ethene ($133.9\,\mathrm{pm}$). ![Resonance in ozone (one lone pair of the right-hand oxygen becomes a π bond while the π bond on the left becomes a lone pair) and in benzene, and one of the three equivalent structures of the nitrate ion, whose charge is shared by the three oxygens.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-7776f7e8d348.svg) *Resonance in ozone (one [lone pair](#def-b1-lewis-resonance-vsepr-lewis) of the right-hand oxygen becomes a $\pi$ bond while the $\pi$ bond on the left becomes a [lone pair](#def-b1-lewis-resonance-vsepr-lewis)) and in benzene, and one of the three equivalent structures of the nitrate ion, whose charge is shared by the three oxygens.* **Definition 3.14 (σ\sigmaσ bond, π\piπ bond).** A single bond is a *$\sigma$ bond*: its electron pair lies along the axis joining the two nuclei, from the head-on overlap of two orbitals. The second and third bonds of a multiple bond are *$\pi$ bonds*: their pairs lie on either side of the axis, from the side-on overlap of two p orbitals perpendicular to it. A double bond is one $\sigma$ and one $\pi$ bond; a triple bond one $\sigma$ and two $\pi$. ![The two kinds of overlap. A bond allows rotation about its axis; a π bond does not, which is why a double bond is rigid ().](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-04b8b6103fd2.svg) *The two kinds of overlap. A $\sigma$ bond allows rotation about its axis; a $\pi$ bond does not, which is why a double bond is rigid ([Chapter 16](https://one-course.com/books/chemistry/2/en/chapter/16-stereochemistry-in-depth#ch-b1-stereochemistry-in-depth)).* ## 3.4 The VSEPR model **Definition 3.15 (VSEPR model, steric number).** In the *VSEPR model* (valence-shell electron-pair repulsion), the pairs around a central atom A repel one another and take the arrangement that keeps them farthest apart. A molecule is written $\ce{AX_nE_m}$, where $n$ is the number of atoms bonded to A (a multiple bond counts once) and $m$ the number of [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) of A. The *steric number* $n + m$ fixes the arrangement of the pairs: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral. The shape of the molecule is the arrangement of its atoms only. **Proposition 3.16 (VSEPR geometries).** The [VSEPR model](#def-b1-lewis-resonance-vsepr-vsepr) predicts the shapes and ideal angles of the table below; measured angles are close. | type | arrangement of pairs | shape | example | measured angle | | --- | --- | --- | --- | --- | | $\ce{AX2}$ | linear | linear | $\ce{CO2}$ | $180^\circ$ | | $\ce{AX3}$ | trigonal planar | trigonal planar | $\ce{BF3}$ | $120^\circ$ | | $\ce{AX2E}$ | trigonal planar | bent | $\ce{SO2}$ | $119.5^\circ$ | | $\ce{AX4}$ | tetrahedral | tetrahedral | $\ce{CH4}$ | $109.5^\circ$ | | $\ce{AX3E}$ | tetrahedral | trigonal pyramidal | $\ce{NH3}$ | $106.7^\circ$ | | $\ce{AX2E2}$ | tetrahedral | bent | $\ce{H2O}$ | $104.5^\circ$ | | $\ce{AX5}$ | trigonal bipyramidal | trigonal bipyramidal | $\ce{PCl5}$ | $90^\circ$, $120^\circ$ | | $\ce{AX4E}$ | trigonal bipyramidal | seesaw | $\ce{SF4}$ | $101.6^\circ$, $173.1^\circ$ | | $\ce{AX3E2}$ | trigonal bipyramidal | T-shaped | $\ce{ClF3}$ | $87.5^\circ$ | | $\ce{AX2E3}$ | trigonal bipyramidal | linear | $\ce{XeF2}$ | $180^\circ$ | | $\ce{AX6}$ | octahedral | octahedral | $\ce{SF6}$ | $90^\circ$ | | $\ce{AX5E}$ | octahedral | square pyramidal | $\ce{BrF5}$ | — | | $\ce{AX4E2}$ | octahedral | square planar | $\ce{XeF4}$ | — | **Proof.** *Admitted at this level.* ∎ **Proposition 3.17 (Lone pairs close the angles).** A [lone pair](#def-b1-lewis-resonance-vsepr-lewis), held by one nucleus only, spreads more than a [bonding pair](#def-b1-lewis-resonance-vsepr-lewis) and repels more strongly: repulsions decrease in the order [lone pair](#def-b1-lewis-resonance-vsepr-lewis)/[lone pair](#def-b1-lewis-resonance-vsepr-lewis) $>$ [lone pair](#def-b1-lewis-resonance-vsepr-lewis)/[bonding pair](#def-b1-lewis-resonance-vsepr-lewis) $>$ [bonding pair](#def-b1-lewis-resonance-vsepr-lewis)/[bonding pair](#def-b1-lewis-resonance-vsepr-lewis). Hence the angles between bonds shrink as [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) are added ($109.5^\circ$ in $\ce{CH4}$, $106.7^\circ$ in $\ce{NH3}$, $104.5^\circ$ in $\ce{H2O}$), and in a trigonal bipyramid [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) take the equatorial positions, where they have only two neighbours at $90^\circ$. **Proof.** *Admitted at this level.* ∎ ![Shapes predicted by VSEPR. A wedge points towards the reader, a hashed wedge away; plain bonds lie in the plane of the page. Lone pairs of the central atom are shown as pairs of dots; XeF4 is drawn face on, its two lone pairs pointing towards and away from the reader.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-5777d91880f4.svg) *Shapes predicted by VSEPR. A wedge points towards the reader, a hashed wedge away; plain bonds lie in the plane of the page. [Lone pairs](#def-b1-lewis-resonance-vsepr-lewis) of the central atom are shown as pairs of dots; $\ce{XeF4}$ is drawn face on, its two [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) pointing towards and away from the reader.* **Method 3.18 (Predicting a shape).** To predict the shape around a central atom: 1. draw the [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) ( [Method 3.4](#met-b1-lewis-resonance-vsepr-lewis) ); 2. count the atoms bonded to the central atom ( $n$ ) and its [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) ( $m$ ); write $\ce{AX_nE_m}$ ; 3. read the arrangement from $n + m$ and the shape from the positions of the $n$ atoms; 4. correct the ideal angles: [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) and multiple bonds take more room than single bonds. **Definition 3.19 (Hybridisation).** The *hybridisation* of an atom is a description of its bonds by orbitals pointing along the directions predicted by VSEPR: an atom of [steric number](#def-b1-lewis-resonance-vsepr-vsepr) 4 is said to be sp$^3$ (four equivalent orbitals at $109.5^\circ$), of [steric number](#def-b1-lewis-resonance-vsepr-vsepr) 3 sp$^2$ (three orbitals at $120^\circ$ in a plane, one p orbital left perpendicular to it), of [steric number](#def-b1-lewis-resonance-vsepr-vsepr) 2 sp (two orbitals at $180^\circ$, two p orbitals left). The p orbitals left over form the $\pi$ bonds. **Example 3.20 (The carbons of organic chemistry).** In ethane each carbon is sp$^3$, tetrahedral; in ethene sp$^2$, the six atoms in one plane with angles near $120^\circ$ (measured $\ce{H-C-H}$ $117.6^\circ$); in ethyne sp, the four atoms on one line. The bond lengths shrink from 153.6 to 133.9 to $120.3\,\mathrm{pm}$ as $\pi$ bonds are added. **Remark 3.21 (A description, not a cause).** [Hybridisation](#def-b1-lewis-resonance-vsepr-hybridisation) describes a geometry already known; it does not explain it, and it fails for the energies of electrons, which molecular orbital theory, in the Year 2 volume, accounts for. It remains the everyday language of organic chemistry: an “sp$^2$ carbon” means a planar carbon with a p orbital free for a $\pi$ bond. ## 3.5 Polar molecules: dipole moments **Definition 3.22 (Bond dipole, dipole moment).** In a bond between atoms of different electronegativities the electrons lean towards the more electronegative atom, which bears a partial charge $-\delta e$ while its partner bears $+\delta e$ ($0 < \delta < 1$). The *bond dipole* is the vector $\vec\mu = \delta e\, \vec d$, of norm $\delta e d$, pointing from the negative to the positive partial charge, $\vec d$ being the vector between them. The *dipole moment* of a molecule is the vector sum of its bond dipoles (and of the contributions of its [lone pairs](#def-b1-lewis-resonance-vsepr-lewis)); it is expressed in coulomb metres or in debyes, $1\,\mathrm{D} = 3.336 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$. A molecule with a non-zero dipole moment is a *polar molecule*. **Remark 3.23 (Direction convention).** Physics draws the dipole vector from the negative to the positive charge, the convention used here. Older chemistry books draw an arrow with a crossed tail pointing towards the negative end; only the direction differs. **Proposition 3.24 (Dipole of a symmetric molecule).** A molecule $\ce{AX_n}$ without [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) on A (linear $\ce{AX2}$, trigonal $\ce{AX3}$, tetrahedral $\ce{AX4}$, …) has a zero [dipole moment](#def-b1-lewis-resonance-vsepr-dipole), whatever the polarity of its bonds. A bent $\ce{AX2}$ molecule of angle $\theta$ whose bonds have the dipole $\mu_b$ has the [dipole moment](#def-b1-lewis-resonance-vsepr-dipole) $$ \mu = 2\mu_b \cos\frac{\theta}{2} . $$ **Proof.** In the symmetric shapes the bond vectors are equal in norm and their directions are arranged symmetrically around A, so they add up to zero (for $\ce{AX2}$ linear they are opposite; for $\ce{AX3}$ at $120^\circ$ three vectors of the same norm sum to zero; for $\ce{AX4}$ the sum is invariant under the rotations of the tetrahedron, so it is zero). For the bent molecule, each bond vector makes the angle $\theta/2$ with the bisector; the components perpendicular to the bisector cancel and those along it add: $2\mu_b\cos(\theta/2)$. ∎ ![Bond dipoles (orange, from the partial negative to the partial positive charge) and their sum (red). In bent water they add; in linear carbon dioxide they cancel.](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-99f818b3708e.svg) *[Bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) (orange, from the partial negative to the partial positive charge) and their sum (red). In bent water they add; in linear carbon dioxide they cancel.* **Example 3.25 (Polar and non-polar molecules).** $\ce{CO2}$, $\ce{BF3}$, $\ce{CH4}$ and $\ce{CCl4}$ have zero [dipole moments](#def-b1-lewis-resonance-vsepr-dipole). $\ce{H2O}$ ($1.86\,\mathrm{D}$), $\ce{NH3}$ ($1.48\,\mathrm{D}$) and $\ce{SO2}$ ($1.63\,\mathrm{D}$) are polar. Along $\ce{CH3Cl}$, $\ce{CH2Cl2}$, $\ce{CHCl3}$ the moment falls, 1.87, 1.62, $1.04\,\mathrm{D}$, to vanish in $\ce{CCl4}$: the $\ce{C-Cl}$ [bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) increasingly cancel. Among the hydrogen halides it follows the [electronegativity](https://one-course.com/books/chemistry/2/en/chapter/2-periodicity#def-b1-periodicity-electronegativity) difference: HF 1.83, HCl 1.09, HBr 0.83, HI $0.45\,\mathrm{D}$. ## 3.6 Exercises **Exercise 3.1 ★.** Draw the [Lewis structures](#def-b1-lewis-resonance-vsepr-lewis) of $\ce{H2O}$, $\ce{NH3}$, $\ce{CO2}$, $\ce{HCN}$, $\ce{CH2O}$ (methanal) and $\ce{NH4+}$, and give the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge). **Solution of Exercise 3.1.** $\ce{H2O}$: two $\ce{O-H}$ bonds, two [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) on O. $\ce{NH3}$: three $\ce{N-H}$ bonds, one [lone pair](#def-b1-lewis-resonance-vsepr-lewis) on N. $\ce{CO2}$: $\ce{O=C=O}$, two [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) on each O. $\ce{HCN}$: $\ce{H-C#N}$, one [lone pair](#def-b1-lewis-resonance-vsepr-lewis) on N. $\ce{CH2O}$: carbon bonded to two H and double-bonded to O, two [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) on O. All [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) are zero in these five. $\ce{NH4+}$: four $\ce{N-H}$ bonds, no [lone pair](#def-b1-lewis-resonance-vsepr-lewis); nitrogen $5 - 0 - 4 = +1$, the charge of the ion. **Exercise 3.2 ★.** Give the type $\ce{AX_nE_m}$, the shape and the approximate angle of $\ce{H2S}$, $\ce{PCl3}$, $\ce{BF3}$, $\ce{SiH4}$ and $\ce{CO3^{2-}}$ (central atom first in each formula). **Solution of Exercise 3.2.** $\ce{H2S}$: $\ce{AX2E2}$, bent, below $109.5^\circ$ (measured $92^\circ$). $\ce{PCl3}$: $\ce{AX3E}$, trigonal pyramidal, about $100^\circ$. $\ce{BF3}$: $\ce{AX3}$, trigonal planar, $120^\circ$. $\ce{SiH4}$: $\ce{AX4}$, tetrahedral, $109.5^\circ$. $\ce{CO3^{2-}}$: $\ce{AX3}$ (the double bond counts once), trigonal planar, $120^\circ$. **Exercise 3.3 ★.** Which of these molecules are polar: $\ce{CCl4}$, $\ce{CH2Cl2}$, $\ce{BF3}$, $\ce{NH3}$, $\ce{CO2}$, $\ce{SO2}$? Justify from the shapes. **Solution of Exercise 3.3.** Non-polar: $\ce{CCl4}$ (tetrahedral $\ce{AX4}$), $\ce{BF3}$ ($\ce{AX3}$), $\ce{CO2}$ (linear $\ce{AX2}$): the [bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) cancel. Polar: $\ce{CH2Cl2}$ (two different kinds of bonds on a tetrahedron), $\ce{NH3}$ (pyramidal), $\ce{SO2}$ (bent). **Exercise 3.4 ★.** Identify the [Lewis acid](#def-b1-lewis-resonance-vsepr-lewis-acid) and the [Lewis base](#def-b1-lewis-resonance-vsepr-lewis-acid) in $\ce{H+ + H2O -> H3O+}$ and in $\ce{Cu^{2+} + 4NH3 -> [Cu(NH3)4]^{2+}}$. Which bonds are dative? **Solution of Exercise 3.4.** $\ce{H+}$ is the [Lewis acid](#def-b1-lewis-resonance-vsepr-lewis-acid) and $\ce{H2O}$ the base: the new $\ce{O-H}$ bond of $\ce{H3O+}$ is dative when formed (afterwards the three $\ce{O-H}$ bonds are identical). $\ce{Cu^{2+}}$ is the acid and each $\ce{NH3}$ a base: the four $\ce{Cu-N}$ bonds are dative. **Exercise 3.5 ★★.** Write two [resonance structures](#def-b1-lewis-resonance-vsepr-resonance) of the ethanoate ion $\ce{CH3COO-}$. What do they predict for the two $\ce{C-O}$ bond lengths? Compare with methanoic acid $\ce{HCOOH}$, whose two $\ce{C-O}$ bonds measure 120.2 and $134.3\,\mathrm{pm}$. **Solution of Exercise 3.5.** ![](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-cc5311659b07.svg) $\leftrightarrow$ ![](https://one-course.com/images/onecourse/chapters/chemistry-2/b1-lewis-resonance-vsepr/fig-a0195b978dd4.svg). The two structures are equivalent: both $\ce{C-O}$ bonds have order 1.5 and the same length, between the double and the single bond. In methanoic acid the two oxygens differ (one carries the H), the structures are not equivalent, and the bonds keep distinct lengths, 120.2 ($\ce{C=O}$) and $134.3\,\mathrm{pm}$ ($\ce{C-OH}$). **Exercise 3.6 ★★.** Draw the [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) of the sulfate ion $\ce{SO4^{2-}}$ with four single $\ce{S-O}$ bonds, compute the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge), and predict the shape. How many [resonance structures](#def-b1-lewis-resonance-vsepr-resonance) with two $\ce{S=O}$ bonds and no charge on sulfur can be written? **Solution of Exercise 3.6.** $N = 6 + 4 \times 6 + 2 = 32$ electrons. With four single bonds and three [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) on each oxygen: sulfur $6 - 0 - 4 = +2$, each oxygen $6 - 6 - 1 = -1$; total $+2 - 4 = -2$. Shape $\ce{AX4}$, tetrahedral. With two $\ce{S=O}$ bonds, choosing which two of the four oxygens are double-bonded gives $\binom{4}{2} = 6$ structures. **Exercise 3.7 ★★.** Use VSEPR to predict the shapes of $\ce{SF4}$, $\ce{ClF3}$, $\ce{XeF2}$ and $\ce{XeF4}$. Explain why the [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) of a trigonal bipyramid sit in the equatorial plane. **Solution of Exercise 3.7.** $\ce{SF4}$: $\ce{AX4E}$, seesaw. $\ce{ClF3}$: $\ce{AX3E2}$, T-shaped. $\ce{XeF2}$: $\ce{AX2E3}$, linear. $\ce{XeF4}$: $\ce{AX4E2}$, square planar. In a trigonal bipyramid an axial position has three neighbours at $90^\circ$, an equatorial position only two: the bulky [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) go where they meet the fewest $90^\circ$ repulsions. **Exercise 3.8 ★★.** The angles $\ce{H-X-H}$ are $104.5^\circ$ in $\ce{H2O}$ and $92.1^\circ$ in $\ce{H2S}$; $106.7^\circ$ in $\ce{NH3}$ and $93.3^\circ$ in $\ce{PH3}$. Propose an explanation in terms of the size and [electronegativity](https://one-course.com/books/chemistry/2/en/chapter/2-periodicity#def-b1-periodicity-electronegativity) of the central atom. **Solution of Exercise 3.8.** Sulfur and phosphorus are larger and less electronegative than oxygen and nitrogen: the [bonding pairs](#def-b1-lewis-resonance-vsepr-lewis) lie farther from the central atom and are pulled towards hydrogen, so they repel one another less, and the [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) squeeze the bonds towards $90^\circ$. **Exercise 3.9 ★★.** Give the [hybridisation](#def-b1-lewis-resonance-vsepr-hybridisation) of each carbon and of the nitrogen in $\ce{CH3-C#N}$ (ethanenitrile) and in $\ce{CH2=CH-CH3}$, and the number of $\sigma$ and $\pi$ bonds in each molecule. **Solution of Exercise 3.9.** $\ce{CH3-C#N}$: the $\ce{CH3}$ carbon is sp$^3$, the nitrile carbon sp, the nitrogen sp; 5 $\sigma$ bonds (three $\ce{C-H}$, $\ce{C-C}$, one in $\ce{C#N}$) and 2 $\pi$ bonds. $\ce{CH2=CH-CH3}$: the two alkene carbons are sp$^2$, the methyl carbon sp$^3$; 8 $\sigma$ bonds (six $\ce{C-H}$, two $\ce{C-C}$) and 1 $\pi$ bond. **Exercise 3.10 ★★★.** The [dipole moment](#def-b1-lewis-resonance-vsepr-dipole) of $\ce{NF3}$ is only $0.24\,\mathrm{D}$, against $1.48\,\mathrm{D}$ for $\ce{NH3}$, although the $\ce{N-F}$ bond is more polar than the $\ce{N-H}$ bond. Using the [bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) and the [lone pair](#def-b1-lewis-resonance-vsepr-lewis) of nitrogen, explain the difference. **Solution of Exercise 3.10.** Both molecules are pyramidal with a [lone pair](#def-b1-lewis-resonance-vsepr-lewis) on nitrogen. The [lone pair](#def-b1-lewis-resonance-vsepr-lewis) gives a contribution pointing (negative to positive) from the [lone pair](#def-b1-lewis-resonance-vsepr-lewis) towards the nucleus. In $\ce{NH3}$, nitrogen is the negative end of each bond: the [bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) point from N towards the hydrogens, in the same direction as the lone-pair contribution, and they add. In $\ce{NF3}$, fluorine is the negative end: the [bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) point from the fluorines towards N, against the lone-pair contribution, and the two nearly cancel. **Exercise 3.11 ★★★.** The $\ce{SO2}$ molecule has an angle of $119.5^\circ$ and a [dipole moment](#def-b1-lewis-resonance-vsepr-dipole) of $1.63\,\mathrm{D}$. Compute the dipole of one $\ce{S-O}$ bond; with the bond length $143.2\,\mathrm{pm}$, deduce the partial charge $\delta$ on each oxygen. **Solution of Exercise 3.11.** $\mu_b = \mu/(2\cos(\theta/2)) = 1.63/(2\cos 59.75^\circ) = 1.62\,\mathrm{D} = 5.40 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$. Then $\delta = \mu_b/(e\,d) = 5.40 \times 10^{-30}/(1.602 \times 10^{-19} \times 143.2 \times 10^{-12}) = 0.24$. **Exercise 3.12 ★★★.** Draw the most important [Lewis structures](#def-b1-lewis-resonance-vsepr-lewis) of dinitrogen monoxide $\ce{N2O}$ (linear, $\ce{N-N-O}$) and compute the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge). The measured bond lengths are $\ce{N-N}$ $112.8\,\mathrm{pm}$ and $\ce{N-O}$ $118.4\,\mathrm{pm}$; in $\ce{N2}$ the triple bond measures $109.8\,\mathrm{pm}$. Which structure weighs more? **Solution of Exercise 3.12.** $\ce{N=N=O}$ with [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) $-1$, $+1$, $0$; $\ce{N#N-O}$ with $0$, $+1$, $-1$. The $\ce{N-N}$ bond ($112.8\,\mathrm{pm}$) is close to the triple bond of $\ce{N2}$ ($109.8\,\mathrm{pm}$): the structure $\ce{N#N-O}$ weighs more, and it places the negative charge on oxygen, the more electronegative atom. The $\ce{N-O}$ bond is however shorter than a single bond: the other structure contributes too. ## 3.7 Problem: The Molecules of a Lightning Strike **Problem 3.1.** Weekend problem — the nitrogen and oxygen species made when lightning heats air: Lewis structures, resonance, shapes, and the partial charges of water A lightning bolt heats air to thousands of degrees: $\ce{N2}$ and $\ce{O2}$ give $\ce{NO}$, then $\ce{NO2}$, $\ce{N2O}$, ozone and finally nitric acid in rain. Data: $1\,\mathrm{D} = 3.336 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$, $e = 1.602 \times 10^{-19}\,\mathrm{C}$. Measured values are given where needed. **Part I — [Lewis structures](#def-b1-lewis-resonance-vsepr-lewis).** 1. Count the [valence electrons](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-configuration) of $\ce{N2}$ , $\ce{NO}$ , $\ce{NO2}$ , $\ce{N2O}$ , $\ce{O3}$ and $\ce{HNO3}$ . 2. Draw the [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) of $\ce{N2}$ . 3. Draw a [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) of $\ce{NO}$ . Why can the [octet rule](#def-b1-lewis-resonance-vsepr-lewis) not be satisfied on both atoms? 4. Draw two [Lewis structures](#def-b1-lewis-resonance-vsepr-lewis) of $\ce{N2O}$ , with $\ce{N=N=O}$ and with $\ce{N#N-O}$ , and give the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) . 5. Draw a [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) of ozone and give the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) . 6. Draw a [Lewis structure](#def-b1-lewis-resonance-vsepr-lewis) of $\ce{HNO3}$ and give the [formal charges](#def-b1-lewis-resonance-vsepr-formal-charge) . 7. Which of the six species have an [unpaired electron](https://one-course.com/books/chemistry/2/en/chapter/1-quantum-numbers-and-electron-configurations#def-b1-quantum-numbers-unpaired) ? **Part II — Resonance and bond lengths.** 8. Write the two [resonance structures](#def-b1-lewis-resonance-vsepr-resonance) of ozone. What bond order do they predict? 9. The $\ce{O-O}$ bond measures $127.8\,\mathrm{pm}$ in ozone, $120.8\,\mathrm{pm}$ in $\ce{O2}$ and $147.5\,\mathrm{pm}$ in $\ce{H2O2}$ . Is the prediction confirmed? 10. Write the three [resonance structures](#def-b1-lewis-resonance-vsepr-resonance) of $\ce{NO3-}$ and give the order of each $\ce{N-O}$ bond. 11. In $\ce{HNO3}$ the three $\ce{N-O}$ bonds measure 140.6, 121.1 and $119.9\,\mathrm{pm}$ . Assign them and explain using resonance. 12. Explain why the angle $\ce{O-N-O}$ of $\ce{NO2}$ ( $134^\circ$ ) is larger than $120^\circ$ . 13. Which of the two structures of question 4 places the negative [formal charge](#def-b1-lewis-resonance-vsepr-formal-charge) on the more electronegative atom? **Part III — Shapes and polarity.** 14. Give the type $\ce{AX_nE_m}$ of the central atom of $\ce{O3}$ , $\ce{N2O}$ , $\ce{NO3-}$ and the nitrogen of $\ce{HNO3}$ , and their shapes. 15. The ozone angle is $116.8^\circ$ . Explain why it is below $120^\circ$ . 16. Ozone has a [dipole moment](#def-b1-lewis-resonance-vsepr-dipole) of $0.53\,\mathrm{D}$ , $\ce{N2O}$ $0.16\,\mathrm{D}$ , $\ce{N2}$ none. Explain. 17. Why is $\ce{CO2}$ non-polar while $\ce{SO2}$ is polar? 18. Predict the angle of $\ce{NH3}$ relative to $109.5^\circ$ , then compare with the measured $106.7^\circ$ . 19. Same question for water ( $104.5^\circ$ ). **Part IV — The partial charges of water.** The [dipole moment](#def-b1-lewis-resonance-vsepr-dipole) of water is $1.857\,\mathrm{D}$, its angle $104.48^\circ$ and its $\ce{O-H}$ bond length $95.8\,\mathrm{pm}$. 20. Express the [dipole moment](#def-b1-lewis-resonance-vsepr-dipole) of water as a function of the [bond dipole](#def-b1-lewis-resonance-vsepr-dipole) $\mu_{OH}$ and the angle. 21. Compute $\mu_{OH}$ in debyes. 22. Convert it into coulomb metres. 23. If the [bond dipole](#def-b1-lewis-resonance-vsepr-dipole) were made of two full charges $\pm e$ at the two nuclei, what would it be? 24. Deduce the fractional ionic character $\delta$ of the $\ce{O-H}$ bond, the ratio of the measured [bond dipole](#def-b1-lewis-resonance-vsepr-dipole) to that of full charges. **Solution of Problem 3.1.** **1.** $\ce{N2}$ 10, $\ce{NO}$ 11, $\ce{NO2}$ 17, $\ce{N2O}$ 16, $\ce{O3}$ 18, $\ce{HNO3}$ 24. **2.** $\ce{N#N}$ with one [lone pair](#def-b1-lewis-resonance-vsepr-lewis) on each nitrogen. **3.** $\ce{N=O}$ with two [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) on oxygen, one [lone pair](#def-b1-lewis-resonance-vsepr-lewis) and one single electron on nitrogen. With 11 electrons, an odd number, one atom always has an odd count: nitrogen has 7 electrons around it. **4.** $\ce{N=N=O}$: $-1$, $+1$, $0$; $\ce{N#N-O}$: $0$, $+1$, $-1$. **5.** $\ce{O=O-O}$: central oxygen $+1$ (one [lone pair](#def-b1-lewis-resonance-vsepr-lewis), three bonds), the singly bonded terminal oxygen $-1$, the other 0. **6.** As in [Example 3.5](#ex-b1-lewis-resonance-vsepr-hno3): nitrogen $+1$, the oxygen bearing a single bond and no hydrogen $-1$, the others 0. **7.** $\ce{NO}$ and $\ce{NO2}$ (11 and 17 electrons). **8.** Two equivalent structures, the double bond on either side: order 1.5 for each bond. **9.** Yes: $127.8\,\mathrm{pm}$ lies between the double bond ($120.8\,\mathrm{pm}$) and the single bond ($147.5\,\mathrm{pm}$). **10.** The double bond on each of the three oxygens in turn; each $\ce{N-O}$ bond is double in one structure of three: order $4/3$. **11.** The long bond, $140.6\,\mathrm{pm}$, is $\ce{N-OH}$, a single bond. The two oxygens without hydrogen share one $\pi$ bond through two equivalent [resonance structures](#def-b1-lewis-resonance-vsepr-resonance): two bonds of order 1.5, 121.1 and $119.9\,\mathrm{pm}$. **12.** Nitrogen carries a single electron rather than a [lone pair](#def-b1-lewis-resonance-vsepr-lewis): it repels the [bonding pairs](#def-b1-lewis-resonance-vsepr-lewis) less than a pair would, and the bonds open beyond $120^\circ$. **13.** $\ce{N#N-O}$, with the negative charge on oxygen. **14.** $\ce{O3}$: $\ce{AX2E}$, bent. $\ce{N2O}$: $\ce{AX2}$ (central N), linear. $\ce{NO3-}$: $\ce{AX3}$, trigonal planar. Nitrogen of $\ce{HNO3}$: $\ce{AX3}$, trigonal planar. **15.** The [lone pair](#def-b1-lewis-resonance-vsepr-lewis) of the central oxygen repels the [bonding pairs](#def-b1-lewis-resonance-vsepr-lewis) more than they repel each other. **16.** Ozone is bent and its central atom differs from the end atoms (positive [formal charge](#def-b1-lewis-resonance-vsepr-formal-charge) in the middle, negative charge shared by the ends): a small dipole along the bisector. $\ce{N2O}$ is linear but its two ends are different atoms: a small dipole. $\ce{N2}$ is symmetric: none. **17.** $\ce{CO2}$ is linear ($\ce{AX2}$): its [bond dipoles](#def-b1-lewis-resonance-vsepr-dipole) cancel. $\ce{SO2}$ is bent ($\ce{AX2E}$): they add. **18.** $\ce{AX3E}$: the [lone pair](#def-b1-lewis-resonance-vsepr-lewis) closes the angles below $109.5^\circ$; measured $106.7^\circ$. **19.** $\ce{AX2E2}$: two [lone pairs](#def-b1-lewis-resonance-vsepr-lewis) close it further; $104.5^\circ$. **20.** $\mu = 2\mu_{OH}\cos(\theta/2)$ ([Proposition 3.24](#prop-b1-lewis-resonance-vsepr-dipole-sum)). **21.** $\mu_{OH} = 1.857/(2\cos 52.24^\circ) = 1.857/1.2246 = 1.52\,\mathrm{D}$. **22.** $1.516 \times 3.336 \times 10^{-30} = 5.06 \times 10^{-30}\,\mathrm{C}\,\mathrm{m}$. **23.** $e\,d = 1.602 \times 10^{-19} \times 95.8 \times 10^{-12} = 1.535 \times 10^{-29}\,\mathrm{C}\,\mathrm{m}$, that is $4.60\,\mathrm{D}$. **24.** $\delta = 5.06 \times 10^{-30}/1.535 \times 10^{-29} = \textbf{0.33}$: the $\ce{O-H}$ bond of water carries a third of the charge a fully ionic bond would carry.