4.1 Lewis Structures, Ionic Bonds and Bond Parameters
Why atoms bond to reach an octet, how to draw Lewis structures and choose between them with formal charges, where the octet rule fails, how ions build a crystal held by lattice enthalpy, and the numbers that describe every bond — length, angle, enthalpy, order and polarity — with resonance and Fajans' rules.
Table salt melts at $801\ ^{\circ}\mathrm{C}$ and conducts electricity when molten; methane boils at $-161\ ^{\circ}\mathrm{C}$ and never conducts at all. Both are held together by chemical bonds, yet the bonds are clearly of different kinds. In sodium chloride, sodium has handed an electron to chlorine and the resulting ions attract one another through the whole crystal. In methane, carbon and hydrogen share pairs of electrons, and each molecule is a small, separate unit. Why do atoms bond at all, why in these two different ways, and how can we describe a bond with numbers?
The first useful answer was given in 1916 by Walther Kössel and Gilbert N. Lewis: atoms bond so as to reach the very stable electron arrangement of a noble gas — usually eight outer electrons, an octet. This section turns that idea into a working tool: Lewis structures and formal charges for molecules and ions, the energetics of ionic bonding, and the bond parameters (length, angle, enthalpy, order) that are measured in the laboratory. Resonance repairs a gap in the Lewis picture, and dipole moments and Fajans' rules show that most real bonds lie somewhere between purely ionic and purely covalent.
The Octet Rule and Lewis Structures
Only the electrons of the outermost shell — the valence electrons — take part in bonding. Lewis showed them as dots around the symbol of the element; the inner electrons and the nucleus together form the kernel and are represented by the symbol itself.
📐 Rule 4.1 — The octet rule (Kössel–Lewis approach)
Atoms of the main-group elements combine by transferring valence electrons from one atom to another, or by sharing them, so that each atom ends up with eight electrons in its valence shell — the configuration of a noble gas. (Hydrogen, lithium and beryllium aim for the two-electron duplet of helium.) Transfer produces ions held by electrostatic attraction (an electrovalent or ionic bond, Kössel); sharing produces a covalent bond (Lewis and Langmuir).
📘 Definition 4.1 — Covalent bond and Lewis structure
- A covalent bond is formed when two atoms share a pair of electrons; each atom usually contributes one electron. Sharing one pair gives a single bond ($\mathrm{Cl{-}Cl}$), two pairs a double bond ($\mathrm{O{=}C{=}O}$), three pairs a triple bond ($\mathrm{N{\equiv}N}$). A shared pair is a bond pair; a pair belonging to one atom only is a lone pair.
- A Lewis structure (electron-dot structure) shows how the valence electrons of a molecule or polyatomic ion are arranged: shared pairs as lines (or pairs of dots) between atoms, lone pairs as dots on atoms. The total number of electrons shown equals the sum of the valence electrons of all atoms, plus one for each unit of negative charge, minus one for each unit of positive charge.
🧭 Problem-solving strategy — Writing a Lewis structure
- Count the total valence electrons; add the negative charge, subtract the positive charge. Example $\mathrm{CO_{3}^{2-}}$: $4+3(6)+2=24$ electrons, i.e. $12$ pairs.
- Skeleton. The least electronegative atom (never H or F) usually goes in the centre; hydrogen and fluorine are always terminal. Join each outer atom to the centre with a single bond.
- Complete the octets of the outer atoms with lone pairs, then put any electrons left over on the central atom.
- If the central atom still has fewer than eight electrons, move lone pairs from outer atoms into multiple bonds until it has an octet.
- Check: the electron total is right, every atom (except H) has an octet where possible, and the formal charges (below) are as small as possible.
✏️ Example 1 — Drawing Lewis structures
Write Lewis structures for (a) hydrogen cyanide, $\mathrm{HCN}$; (b) the carbonate ion, $\mathrm{CO_{3}^{2-}}$; (c) carbon monoxide, $\mathrm{CO}$; (d) the ammonium ion, $\mathrm{NH_{4}^{+}}$.
💡 Show solution▶
(a) Electrons: $1+4+5=10$ (5 pairs). Skeleton $\mathrm{H{-}C{-}N}$ uses 2 pairs; 3 lone pairs on N leave C with only 4 electrons. Move two N lone pairs into bonds: $\mathrm{H{-}C{\equiv}N\!:}$ — C has 8 electrons ($2+6$), N has $6+2=8$, H has 2. ✓ 10 electrons.
(b) 24 electrons. Three $\mathrm{C{-}O}$ single bonds (6) and three lone pairs on each O (18) use all 24, but C has only 6. Make one $\mathrm{C{=}O}$: C now has 8. The structure is one $\mathrm{C{=}O}$ (O with 2 lone pairs) and two $\mathrm{C{-}O}$ (each O with 3 lone pairs), in brackets with charge $2-$.
(c) $4+6=10$ electrons. Only a triple bond gives both atoms an octet: $\mathrm{:C{\equiv}O:}$ — 6 shared + one lone pair on each atom $=10$. ✓
(d) $5+4(1)-1=8$ electrons = four $\mathrm{N{-}H}$ bonds, no lone pairs: N has an octet, each H a duplet. The whole ion is placed in brackets with charge $+$.
A Lewis structure books every electron, so it can tell us how the electrons of a bond are "owned". Counting each bonding pair as shared equally and each lone pair as fully owned gives every atom a formal charge. It is a bookkeeping charge, not a real one — but it is the best guide to which of several possible Lewis structures is the most important.
📘 Definition 4.2 — Formal charge
The formal charge (FC) of an atom in a Lewis structure is
$$\text{FC}=V-L-\tfrac12B$$
where $V$ is the number of valence electrons of the free atom, $L$ the number of non-bonding (lone-pair) electrons on the atom in the structure, and $B$ the number of bonding electrons around it (two per bond). The formal charges of all the atoms add up to the overall charge of the molecule or ion.
📐 Rule 4.2 — Choosing the best Lewis structure
When more than one Lewis structure can be written, the one of lowest energy (the most important one) is generally the structure in which (i) the formal charges are as small as possible (zero where possible), and (ii) any negative formal charge sits on the more electronegative atom. Like charges on neighbouring atoms are unfavourable.
✏️ Example 2 — Formal charges and the best structure
(a) Find the formal charge on each oxygen in ozone, $\mathrm{O{=}O{-}O}$, with two lone pairs on the left O, one on the central O and three on the right O. (b) Find the formal charges in $\mathrm{:C{\equiv}O:}$. (c) For the cyanate ion $\mathrm{OCN^{-}}$ (C in the centre) compare $[\mathrm{O{=}C{=}N}]^{-}$, $[\mathrm{O{-}C{\equiv}N}]^{-}$ and $[\mathrm{O{\equiv}C{-}N}]^{-}$, each with octets on all atoms. Which is best?
💡 Show solution▶
(a) Left O: $6-4-\tfrac12(4)=0$. Centre O: $6-2-\tfrac12(6)=+1$. Right O: $6-6-\tfrac12(2)=-1$. Sum $=0$ ✓ (a neutral molecule).
(b) C: $4-2-\tfrac12(6)=-1$; O: $6-2-\tfrac12(6)=+1$. The octet structure puts the negative charge on the less electronegative atom — one reason CO has only a very small dipole moment ($0.11\ \mathrm{D}$).
(c) count $6+4+5+1=16$ electrons. $[\mathrm{O{=}C{=}N}]^{-}$ (2 lone pairs on O and on N): O $0$, C $0$, N $5-4-2=-1$. $[\mathrm{O{-}C{\equiv}N}]^{-}$ (3 lone pairs on O, 1 on N): O $6-6-1=-1$, C $0$, N $5-2-3=0$. $[\mathrm{O{\equiv}C{-}N}]^{-}$ (1 lone pair on O, 3 on N): O $6-2-3=+1$, C $0$, N $5-6-1=-2$.
(c) choose The third has large charges ($+1$, $-2$) — worst. The first two both have a single $-1$; Rule 4.2 prefers it on the more electronegative atom, so $[\mathrm{O{-}C{\equiv}N}]^{-}$ is best, with $[\mathrm{O{=}C{=}N}]^{-}$ also contributing (see resonance, below).
⚠️ Limitations of the octet rule
The octet rule is a guide, not a law. It fails in three kinds of molecules, and has other drawbacks:
- Incomplete octet — the central atom has fewer than eight electrons, usually because it has fewer than four valence electrons: $\mathrm{LiCl}$ (Li: 2), $\mathrm{BeH_{2}}$ (Be: 4), $\mathrm{BCl_{3}}$ and $\mathrm{AlCl_{3}}$ (B, Al: 6).
- Odd-electron molecules — the total number of valence electrons is odd, so not every atom can have an octet: $\mathrm{NO}$ ($5+6=11$), $\mathrm{NO_{2}}$ ($5+12=17$).
- Expanded octet — elements of period 3 and beyond have empty $3d$ orbitals in the valence shell and can hold more than eight electrons: $\mathrm{PF_{5}}$ (10 around P), $\mathrm{SF_{6}}$ (12 around S), $\mathrm{H_{2}SO_{4}}$ (12 around S in its usual Lewis structure). Period-2 elements can never do this (Section 3.3).
- Other drawbacks — noble gases are not completely inert: xenon and krypton form compounds such as $\mathrm{XeF_{2}}$, $\mathrm{KrF_{2}}$ and $\mathrm{XeOF_{2}}$. The rule says nothing about the shape of a molecule (Section 4.2) and does not explain why bonds form in energy terms or why some molecules are more stable than others (Sections 4.3–4.4).
✏️ Example 3 — Which octets are broken?
For each species, count the electrons around the central atom in the usual Lewis structure and say whether the octet is complete, incomplete or expanded, or whether the molecule is odd-electron: (a) $\mathrm{BeCl_{2}}$ (gas-phase molecule); (b) $\mathrm{BF_{3}}$; (c) $\mathrm{ClO_{2}}$; (d) $\mathrm{PCl_{5}}$; (e) $\mathrm{SF_{4}}$; (f) $\mathrm{IF_{7}}$; (g) $\mathrm{CCl_{4}}$.
💡 Show solution▶
(a), (b) $\mathrm{Cl{-}Be{-}Cl}$: two bond pairs, $4$ electrons — incomplete. $\mathrm{BF_{3}}$: three bond pairs, $6$ electrons — incomplete.
(c) $7+2(6)=19$ valence electrons, an odd number: an odd-electron molecule (chlorine dioxide, like $\mathrm{NO}$ and $\mathrm{NO_{2}}$, has an unpaired electron).
(d), (e) $\mathrm{PCl_{5}}$: five bond pairs, $10$ electrons — expanded. $\mathrm{SF_{4}}$: $6$ valence electrons of S, four used in bonds, one lone pair left: $4\times2+2=10$ — expanded.
(f), (g) $\mathrm{IF_{7}}$: seven bond pairs, $14$ electrons — expanded (iodine is in period 5). $\mathrm{CCl_{4}}$: four bond pairs, $8$ — a normal octet.
The Ionic Bond and Lattice Enthalpy
When an atom of very low ionisation enthalpy meets an atom of very negative electron gain enthalpy, electron transfer is the easier route to two noble-gas configurations: $\mathrm{Na}\ [\mathrm{Ne}]3s^{1}$ gives its electron to $\mathrm{Cl}\ [\mathrm{Ne}]3s^{2}3p^{5}$, producing $\mathrm{Na^{+}}\ [\mathrm{Ne}]$ and $\mathrm{Cl^{-}}\ [\mathrm{Ar}]$. The number of electrons an atom gains or loses is its electrovalence (Na $1$, Mg $2$, O $2$).
📘 Definition 4.3 — Ionic (electrovalent) bond
An ionic bond is the electrostatic attraction between oppositely charged ions formed by the complete transfer of one or more electrons from one atom to another. Ionic compounds do not consist of molecules: in solid sodium chloride every $\mathrm{Na^{+}}$ is surrounded by six $\mathrm{Cl^{-}}$ and every $\mathrm{Cl^{-}}$ by six $\mathrm{Na^{+}}$ in a regular three-dimensional crystal lattice; the formula $\mathrm{NaCl}$ gives only the ratio of ions. Ionic bonding is favoured by (i) a low ionisation enthalpy of the metal, (ii) a very negative electron gain enthalpy of the non-metal, and (iii) a large lattice enthalpy of the solid formed.
Condition (iii) is the decisive one. Turning gaseous atoms into gaseous ions almost always costs energy, because even the lowest ionisation enthalpy is larger than the most negative electron gain enthalpy. The energy is repaid — with plenty to spare — when the ions come together into a crystal.
📘 Definition 4.4 — Lattice enthalpy
The lattice enthalpy of an ionic solid is the energy required to separate one mole of the solid completely into its gaseous ions. For sodium chloride
$$\mathrm{NaCl(s)\rightarrow Na^{+}(g)+Cl^{-}(g)}\qquad\Delta_{\mathrm{lattice}}H=+788\ \mathrm{kJ\,mol^{-1}}$$
so $788\ \mathrm{kJ}$ is released when one mole of solid forms from the gaseous ions. Lattice enthalpy is larger when the ions carry larger charges and when they are smaller (closer together). It cannot be measured directly; Chapter 5 shows how to obtain it from a thermochemical cycle.
✏️ Example 4 — Why does sodium chloride form?
(a) For gaseous atoms, $\Delta_{i}H_{1}(\mathrm{Na})=495.8\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{eg}H(\mathrm{Cl})=-348.7\ \mathrm{kJ\,mol^{-1}}$. Find the enthalpy change for $\mathrm{Na(g)+Cl(g)\rightarrow Na^{+}(g)+Cl^{-}(g)}$. Is electron transfer between isolated atoms favourable? (b) Using the lattice enthalpy $788\ \mathrm{kJ\,mol^{-1}}$, find the enthalpy change for $\mathrm{Na(g)+Cl(g)\rightarrow NaCl(s)}$. (c) Without using numbers, arrange $\mathrm{KCl}$, $\mathrm{MgO}$, $\mathrm{NaCl}$ and $\mathrm{LiF}$ in order of increasing lattice enthalpy.
💡 Show solution▶
(a) $\Delta H=495.8+(-348.7)=+147.1\ \mathrm{kJ\,mol^{-1}}$. Positive: forming a pair of gaseous ions from gaseous atoms costs energy, so electron transfer alone is not favourable.
(b) Add the formation of the lattice from the gaseous ions, $-788\ \mathrm{kJ\,mol^{-1}}$: $\Delta H=147.1-788=-640.9\ \mathrm{kJ\,mol^{-1}}$. Strongly negative: the lattice enthalpy is what makes the ionic solid stable.
(c) MgO has doubly charged ions (charge product $2\times2=4$ against $1$), so it is far the largest. Among the $1{:}1$ salts, smaller ions give a larger value: $\mathrm{K^{+}}\gt\mathrm{Na^{+}}\gt\mathrm{Li^{+}}$ in size and $\mathrm{Cl^{-}}\gt\mathrm{F^{-}}$. Order: $\mathrm{KCl}\lt\mathrm{NaCl}\lt\mathrm{LiF}\lt\mathrm{MgO}$ (about $717$, $788$, $1037$ and $3800\ \mathrm{kJ\,mol^{-1}}$).
Check The higher lattice enthalpy of MgO shows up as a much higher melting point: about $2850\ ^{\circ}\mathrm{C}$ against $801\ ^{\circ}\mathrm{C}$ for NaCl.
Bond Parameters and Resonance
Every covalent bond can be described by a few measurable numbers. X-ray and electron diffraction and spectroscopy give distances and angles; thermochemistry gives the energy needed to break the bond.
📘 Definition 4.5 — Bond parameters
- Bond length — the equilibrium distance between the nuclei of two bonded atoms. It is roughly the sum of the covalent radii: $r_{\mathrm{C}}+r_{\mathrm{Cl}}=77+99=176\ \mathrm{pm}$ (measured $\mathrm{C{-}Cl}$ in $\mathrm{CCl_{4}}$: $177\ \mathrm{pm}$).
- Bond angle — the angle between the lines joining the nucleus of a central atom to the nuclei of two atoms bonded to it, e.g. $\angle\mathrm{HOH}=104.5^{\circ}$ in water. Bond angles reveal the shape of a molecule (Section 4.2).
- Bond enthalpy — the energy needed to break one mole of bonds of a given type between two atoms in the gaseous state: $\mathrm{H_{2}(g)\rightarrow2H(g)}$, $\Delta_{\mathrm{bond}}H^{\circ}=435.8\ \mathrm{kJ\,mol^{-1}}$. In polyatomic molecules the successive bonds of one type need different energies ($\mathrm{H_{2}O}$: $502$ then $427\ \mathrm{kJ\,mol^{-1}}$), so tables give the mean (average) bond enthalpy.
- Bond order — in the Lewis description, the number of bonds (shared pairs) between two atoms: $1$ in $\mathrm{H_{2}}$ and $\mathrm{F_{2}}$, $2$ in $\mathrm{O_{2}}$, $3$ in $\mathrm{N_{2}}$ and $\mathrm{CO}$.
| Bond | $\mathrm{H{-}H}$ | $\mathrm{F{-}F}$ | $\mathrm{Cl{-}Cl}$ | $\mathrm{C{-}H}$ | $\mathrm{O{-}H}$ | $\mathrm{C{-}C}$ | $\mathrm{C{=}C}$ | $\mathrm{C{\equiv}C}$ | $\mathrm{O{=}O}$ | $\mathrm{N{\equiv}N}$ |
|---|---|---|---|---|---|---|---|---|---|---|
| Length / pm | 74 | 142 | 198 | 109 | 96 | 154 | 134 | 120 | 121 | 110 |
| Bond enthalpy / $\mathrm{kJ\,mol^{-1}}$ | 436 | 155 | 243 | 413 | 464 | 348 | 614 | 839 | 498 | 946 |
C–H, O–H and the carbon–carbon values are mean bond enthalpies; tables in different books differ by a few $\mathrm{kJ\,mol^{-1}}$ and pm. Note the unusually weak $\mathrm{F{-}F}$ bond: the lone pairs on the two small fluorine atoms repel each other strongly.
📐 Rule 4.3 — Bond order, bond length and bond enthalpy
For bonds between the same two elements, the higher the bond order, the shorter and stronger the bond: $\mathrm{C{-}C}\ (154\ \mathrm{pm},\ 348)$, $\mathrm{C{=}C}\ (134,\ 614)$, $\mathrm{C{\equiv}C}\ (120\ \mathrm{pm},\ 839\ \mathrm{kJ\,mol^{-1}})$. Isoelectronic molecules and ions have the same bond order: $\mathrm{F_{2}}$ and $\mathrm{O_{2}^{2-}}$ ($14$ electrons) have bond order $1$; $\mathrm{N_{2}}$, $\mathrm{CO}$, $\mathrm{NO^{+}}$ and $\mathrm{CN^{-}}$ ($10$ electrons) have bond order $3$.
Some molecules cannot be described by a single Lewis structure. In ozone the structure $\mathrm{O{=}O{-}O}$ suggests one short double bond ($121\ \mathrm{pm}$) and one long single bond ($148\ \mathrm{pm}$), but experiment finds two identical bonds of $128\ \mathrm{pm}$, in between. The structure $\mathrm{O{-}O{=}O}$ is just as good, and neither alone is right.
📘 Definition 4.6 — Resonance
When a molecule or ion is described by two or more Lewis structures that differ only in the positions of electrons (not of nuclei), each one is a canonical (resonance) structure, and the real species is the resonance hybrid of them, written with $\leftrightarrow$: $\ \mathrm{O{=}O{-}O}\ \leftrightarrow\ \mathrm{O{-}O{=}O}$. The hybrid is more stable (lower in energy) than any single canonical structure; the difference is the resonance stabilisation energy. The canonical structures have no real existence: the molecule does not flip between them — it has one structure, the hybrid, which no single Lewis structure can draw. When equivalent canonical structures spread $n_{\mathrm{b}}$ bonds over $n_{\mathrm{p}}$ equivalent positions, the average bond order is $n_{\mathrm{b}}/n_{\mathrm{p}}$.
✏️ Example 5 — Resonance and bond order
(a) Draw the canonical structures of the carbonate ion and find the average $\mathrm{C{-}O}$ bond order. (b) The $\mathrm{C{-}O}$ bond lengths in $\mathrm{CO}$, $\mathrm{CO_{2}}$ and $\mathrm{CO_{3}^{2-}}$ are $113$, $115$ and $129\ \mathrm{pm}$ (not in that order of species). Match them, and place the three in order of increasing $\mathrm{C{-}O}$ bond enthalpy. (c) Find the mean $\mathrm{O{-}H}$ bond enthalpy in water from the two successive values $502$ and $427\ \mathrm{kJ\,mol^{-1}}$. (d) Which of $\mathrm{N_{2}}$, $\mathrm{O_{2}^{2-}}$, $\mathrm{CN^{-}}$, $\mathrm{NO^{+}}$, $\mathrm{F_{2}}$ have bond order $3$?
💡 Show solution▶
(a) From Example 1(b), one $\mathrm{C{=}O}$ and two $\mathrm{C{-}O^{-}}$. The double bond can be placed on any of the three oxygens: three equivalent canonical structures. Bonds: $4$ (one double + two single) spread over $3$ positions → bond order $4/3\approx1.33$. All three bonds are identical in the real ion.
(b) Bond orders: $\mathrm{CO}$ $3$, $\mathrm{CO_{2}}$ $2$ (each $\mathrm{C{=}O}$), $\mathrm{CO_{3}^{2-}}$ $1.33$. Higher order → shorter: $\mathrm{CO}\ 113$, $\mathrm{CO_{2}}\ 115$, $\mathrm{CO_{3}^{2-}}\ 129\ \mathrm{pm}$. Bond enthalpy rises with bond order: $\mathrm{CO_{3}^{2-}}\lt\mathrm{CO_{2}}\lt\mathrm{CO}$.
(c) $\tfrac12(502+427)=464.5\ \mathrm{kJ\,mol^{-1}}$.
(d) Count valence electrons: $\mathrm{N_{2}}$ $10$, $\mathrm{CN^{-}}$ $4+5+1=10$, $\mathrm{NO^{+}}$ $5+6-1=10$ — all isoelectronic with $\mathrm{N_{2}}$: bond order $3$. $\mathrm{O_{2}^{2-}}$ ($14$) and $\mathrm{F_{2}}$ ($14$) have bond order $1$.
Polarity of Bonds, Dipole Moment and Fajans' Rules
A bond between two identical atoms ($\mathrm{H_{2}}$, $\mathrm{Cl_{2}}$) shares its electron pair equally: it is a non-polar covalent bond. Between different atoms the more electronegative atom pulls the pair towards itself: in $\mathrm{H{-}Cl}$ the chlorine end carries a small negative charge $\delta-$ and the hydrogen end an equal positive charge $\delta+$. Such a polar covalent bond has two separated charges — an electric dipole.
📘 Definition 4.7 — Dipole moment
The dipole moment $\mu$ of a bond (or molecule) is the product of the magnitude of the separated charge and the distance between the centres of positive and negative charge:
$$\mu=Q\times r$$
It is usually expressed in debye: $1\ \mathrm{D}=3.33564\times10^{-30}\ \mathrm{C\,m}$. Dipole moment is a vector. Chemists draw it as an arrow with a cross at the tail, pointing from the positive end to the negative end: $\overset{\delta+}{\mathrm{H}}\!\!-\!\!\overset{\delta-}{\mathrm{Cl}}$ with $\mathrm{+\!\!\!\rightarrow}$ from H to Cl. (Physics uses the opposite direction, from $-$ to $+$; in this course we use the chemists' arrow.) The dipole moment of a polyatomic molecule is the vector sum of its bond dipoles, so it depends on the shape — Section 4.2.
Just as most covalent bonds are partly ionic, most ionic bonds are partly covalent. A small, highly charged cation pulls the electron cloud of a neighbouring anion towards itself; the cloud is polarised, and some electron density is now shared between the two ions. The ability of a cation to do this is its polarising power; the ease with which an anion's cloud is distorted is its polarisability.
📐 Rule 4.4 — Fajans' rules (covalent character of ionic bonds)
An ionic bond has more covalent character when:
- the cation is small — its charge is concentrated and polarises the anion strongly ($\mathrm{LiCl}$ is more covalent than $\mathrm{KCl}$);
- the anion is large — its outer electrons are far from its own nucleus and easily pulled away ($\mathrm{LiI}$ more covalent than $\mathrm{LiF}$);
- the charges are large on either ion ($\mathrm{AlCl_{3}}\gt\mathrm{MgCl_{2}}\gt\mathrm{NaCl}$ in covalent character);
- the cation has a pseudo-noble-gas configuration, $(n-1)d^{10}ns^{0}$ (e.g. $\mathrm{Cu^{+}}$, $\mathrm{Ag^{+}}$, $\mathrm{Zn^{2+}}$), rather than a noble-gas configuration: $d$ electrons shield the nuclear charge poorly, so such a cation polarises more strongly than a noble-gas cation of the same size and charge ($\mathrm{CuCl}$ is more covalent than $\mathrm{NaCl}$, although $\mathrm{Cu^{+}}$, $96\ \mathrm{pm}$, and $\mathrm{Na^{+}}$, $95\ \mathrm{pm}$, are almost the same size).
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — percentage ionic character from a dipole moment
Not required by the CBSE syllabus. Dipole moment is listed by JEE Main, NEET and JEE Advanced; this estimate of ionic character is a common application of it, although no syllabus names it.
If $\mathrm{H{-}Cl}$ were 100% ionic, a full charge $e=1.602\times10^{-19}\ \mathrm{C}$ would sit at each end of the $127\ \mathrm{pm}$ bond, giving $\mu_{\mathrm{ionic}}=e\,d=1.602\times10^{-19}\times127\times10^{-12}=2.035\times10^{-29}\ \mathrm{C\,m}=6.10\ \mathrm{D}$. The measured value is $1.03\ \mathrm{D}$, so $$\%\ \text{ionic character}=\frac{\mu_{\mathrm{observed}}}{\mu_{\mathrm{ionic}}}\times100=\frac{1.03}{6.10}\times100\approx17\%.$$ The method is popular in JEE problems; it treats the charges as points, so the result is only an estimate.✏️ Example 6 — Dipole moments and Fajans' rules
(a) In $\mathrm{HCl}$ (bond length $127\ \mathrm{pm}$) the partial charges are about $\pm0.17\,e$. Calculate the dipole moment in $\mathrm{C\,m}$ and in debye ($e=1.602\times10^{-19}\ \mathrm{C}$, $1\ \mathrm{D}=3.33564\times10^{-30}\ \mathrm{C\,m}$). (b) Arrange in order of increasing covalent character: (i) $\mathrm{NaCl}$, $\mathrm{AlCl_{3}}$, $\mathrm{MgCl_{2}}$; (ii) $\mathrm{LiBr}$, $\mathrm{LiF}$, $\mathrm{LiI}$, $\mathrm{LiCl}$. (c) Anhydrous $\mathrm{AlCl_{3}}$ sublimes at about $180\ ^{\circ}\mathrm{C}$, while $\mathrm{AlF_{3}}$ stays solid up to about $1290\ ^{\circ}\mathrm{C}$. Explain.
💡 Show solution▶
(a) $\mu=Q\,r=0.17\times1.602\times10^{-19}\times127\times10^{-12}=3.46\times10^{-30}\ \mathrm{C\,m}$. In debye: $3.46\times10^{-30}/3.33564\times10^{-30}=1.04\ \mathrm{D}$ (close to the measured $1.03\ \mathrm{D}$).
(b)(i) Same anion; cation charge rises and size falls from $\mathrm{Na^{+}}$ ($95\ \mathrm{pm}$) to $\mathrm{Mg^{2+}}$ ($65$) to $\mathrm{Al^{3+}}$ ($50$): $\mathrm{NaCl}\lt\mathrm{MgCl_{2}}\lt\mathrm{AlCl_{3}}$.
(b)(ii) Same cation; anion size rises $\mathrm{F^{-}}\ 136\lt\mathrm{Cl^{-}}\ 181\lt\mathrm{Br^{-}}\ 195\lt\mathrm{I^{-}}\ 216\ \mathrm{pm}$, and with it polarisability: $\mathrm{LiF}\lt\mathrm{LiCl}\lt\mathrm{LiBr}\lt\mathrm{LiI}$.
(c) $\mathrm{Al^{3+}}$ is small and triply charged — a strong polariser. The large $\mathrm{Cl^{-}}$ ion is easily polarised, so $\mathrm{AlCl_{3}}$ is largely covalent (molecular, volatile). The small $\mathrm{F^{-}}$ ion holds its electrons tightly and is hardly polarised, so $\mathrm{AlF_{3}}$ stays essentially ionic, with a high melting point.
Exercises
I
Lewis structures and formal charge.
- 1 Write the Lewis symbols of S, Mg, Si and Br, and state the number of valence electrons of each.
- 2 Find the total number of valence electrons to be shown in the Lewis structures of $\mathrm{NO_{3}^{-}}$, $\mathrm{SO_{4}^{2-}}$, $\mathrm{NH_{4}^{+}}$ and $\mathrm{PO_{4}^{3-}}$.
- 3 Draw the Lewis structure of hydrogen peroxide, $\mathrm{H_{2}O_{2}}$.
- 4 Draw Lewis structures of ethene, $\mathrm{C_{2}H_{4}}$, and ethyne, $\mathrm{C_{2}H_{2}}$.
- 5 Draw a Lewis structure of nitric acid, $\mathrm{HNO_{3}}$ (H on an oxygen; N bonded to three O), and find the formal charge on every atom.
- 6 Find the formal charges in the nitrite ion $[\mathrm{O{=}N{-}O}]^{-}$ (one lone pair on N).
- 7 Find the formal charges on C and N in the cyanide ion, $[\mathrm{:C{\equiv}N:}]^{-}$.
- 8 For $\mathrm{SO_{2}}$ compare $\mathrm{O{=}S{-}O}$ (octet on S) with $\mathrm{O{=}S{=}O}$ (expanded octet). Which has the smaller formal charges? Both $\mathrm{S{-}O}$ bonds are actually $143\ \mathrm{pm}$ — what does that tell you?
💡 Show solutions▶
1 S: 6 dots (two pairs, two single); Mg: 2; Si: 4 (all single); Br: 7 (three pairs, one single).
2 $\mathrm{NO_{3}^{-}}$: $5+18+1=24$; $\mathrm{SO_{4}^{2-}}$: $6+24+2=32$; $\mathrm{NH_{4}^{+}}$: $5+4-1=8$; $\mathrm{PO_{4}^{3-}}$: $5+24+3=32$.
3 $2+12=14$ electrons: $\mathrm{H{-}O{-}O{-}H}$ (3 bonds, 6 electrons) with two lone pairs on each O (8 electrons). Each O has an octet.
4 $\mathrm{C_{2}H_{4}}$ ($12$ electrons): $\mathrm{H_{2}C{=}CH_{2}}$, no lone pairs. $\mathrm{C_{2}H_{2}}$ ($10$): $\mathrm{H{-}C{\equiv}C{-}H}$, no lone pairs. Each C has an octet.
5 $1+5+18=24$ electrons: $\mathrm{H{-}O{-}N({=}O){-}O}$; O(H) has 2 lone pairs, $=\mathrm{O}$ 2, the single-bonded terminal O 3. FC: H $0$; O(H) $6-4-2=0$; N $5-0-4=+1$; $=\mathrm{O}$ $6-4-2=0$; $-\mathrm{O}$ $6-6-1=-1$. Sum $0$ ✓.
6 $18$ electrons. $=\mathrm{O}$: $6-4-2=0$; N: $5-2-3=0$; $-\mathrm{O}$: $6-6-1=-1$. Sum $-1$ ✓.
7 C: $4-2-3=-1$; N: $5-2-3=0$. The charge sits on carbon.
8 $\mathrm{O{=}S{-}O}$: S $6-2-3=+1$, the single-bonded O $-1$, $=\mathrm{O}$ $0$. $\mathrm{O{=}S{=}O}$: all $0$ — smaller formal charges. Equal bond lengths show that neither drawing alone is right: the molecule is a resonance hybrid ($\mathrm{O{=}S{-}O}\leftrightarrow\mathrm{O{-}S{=}O}$, with the expanded structure also contributing).
II
Exceptions to the octet rule and the ionic bond.
- 1 Give one example each of a molecule with an incomplete octet, an odd-electron molecule and a molecule with an expanded octet.
- 2 How many electrons surround the central atom in $\mathrm{BCl_{3}}$, $\mathrm{XeF_{4}}$ (two lone pairs on Xe) and $\mathrm{ClF_{3}}$ (two lone pairs on Cl)?
- 3 Phosphorus forms $\mathrm{PCl_{3}}$ and $\mathrm{PCl_{5}}$, but nitrogen forms only $\mathrm{NCl_{3}}$. Why?
- 4 Show by electronic configurations how $\mathrm{Ca}$ and $\mathrm{O}$ reach noble-gas configurations in $\mathrm{CaO}$.
- 5 State three factors that favour the formation of an ionic bond.
- 6 $\Delta_{i}H_{1}(\mathrm{K})=419$, $\Delta_{eg}H(\mathrm{Cl})=-349$ and the lattice enthalpy of $\mathrm{KCl}$ is $717\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta H$ for $\mathrm{K(g)+Cl(g)\rightarrow K^{+}(g)+Cl^{-}(g)}$ and for $\mathrm{K(g)+Cl(g)\rightarrow KCl(s)}$.
- 7 Why do ionic solids have high melting points and conduct electricity when molten but not when solid?
- 8 Applied. A furnace lining must survive $2000\ ^{\circ}\mathrm{C}$. Which would you test, $\mathrm{NaCl}$ or $\mathrm{MgO}$? Justify with lattice enthalpy.
💡 Show solutions▶
1 E.g. $\mathrm{BF_{3}}$ (6 around B); $\mathrm{NO}$ (11 valence electrons); $\mathrm{SF_{6}}$ (12 around S).
2 $\mathrm{BCl_{3}}$: $3\times2=6$. $\mathrm{XeF_{4}}$: $4\times2+2\times2=12$. $\mathrm{ClF_{3}}$: $3\times2+2\times2=10$.
3 N ($n=2$) has only four valence orbitals ($2s$, three $2p$) and cannot hold more than 8 electrons. P ($n=3$) has vacant $3d$ orbitals and can expand its octet to 10.
4 $\mathrm{Ca}\ [\mathrm{Ar}]4s^{2}\rightarrow\mathrm{Ca^{2+}}\ [\mathrm{Ar}]+2e^{-}$; $\mathrm{O}\ [\mathrm{He}]2s^{2}2p^{4}+2e^{-}\rightarrow\mathrm{O^{2-}}\ [\mathrm{He}]2s^{2}2p^{6}=[\mathrm{Ne}]$.
5 Low ionisation enthalpy of the metal; very negative electron gain enthalpy of the non-metal; high lattice enthalpy of the solid.
6 $419+(-349)=+70\ \mathrm{kJ\,mol^{-1}}$; then $70-717=-647\ \mathrm{kJ\,mol^{-1}}$.
7 Strong electrostatic attractions act in all directions through the whole lattice, so much energy is needed to free the ions. In the solid the ions are fixed in place; in the melt they can move and carry charge.
8 $\mathrm{MgO}$: its ions are doubly charged and small, so its lattice enthalpy (about $3800\ \mathrm{kJ\,mol^{-1}}$) is nearly five times that of NaCl; it melts at about $2850\ ^{\circ}\mathrm{C}$, while NaCl melts at $801\ ^{\circ}\mathrm{C}$.
III
Bond parameters and resonance. Use the table in this section where needed.
- 1 Estimate the $\mathrm{H{-}Br}$ bond length from the covalent radii $r_{\mathrm{H}}=37$ and $r_{\mathrm{Br}}=114\ \mathrm{pm}$. The measured value is $141\ \mathrm{pm}$; why is it shorter?
- 2 Arrange $\mathrm{C{-}C}$, $\mathrm{C{\equiv}C}$ and $\mathrm{C{=}C}$ in order of increasing bond length and of increasing bond enthalpy.
- 3 Which needs more energy to dissociate, $\mathrm{N_{2}}$ or $\mathrm{O_{2}}$? Explain with bond order.
- 4 Breaking all four $\mathrm{C{-}H}$ bonds in one mole of gaseous methane needs $1652\ \mathrm{kJ}$. Find the mean $\mathrm{C{-}H}$ bond enthalpy.
- 5 Draw the canonical structures of the nitrate ion and find the $\mathrm{N{-}O}$ bond order.
- 6 Write three canonical structures for $\mathrm{CO_{2}}$ with formal charges. Which contributes most?
- 7 Why are both oxygen–oxygen bonds in ozone the same length ($128\ \mathrm{pm}$)?
- 8 Give the Lewis bond orders of $\mathrm{O_{2}}$ and $\mathrm{O_{2}^{2-}}$ and predict which has the longer bond.
💡 Show solutions▶
1 $37+114=151\ \mathrm{pm}$. The bond is polar (Br $2.8$, H $2.1$); its partial ionic character draws the atoms closer than the sum of covalent radii.
2 Length: $\mathrm{C{\equiv}C}\ (120)\lt\mathrm{C{=}C}\ (134)\lt\mathrm{C{-}C}\ (154\ \mathrm{pm})$. Enthalpy: $\mathrm{C{-}C}\ (348)\lt\mathrm{C{=}C}\ (614)\lt\mathrm{C{\equiv}C}\ (839\ \mathrm{kJ\,mol^{-1}})$.
3 $\mathrm{N_{2}}$: bond order $3$ ($946\ \mathrm{kJ\,mol^{-1}}$) against $2$ for $\mathrm{O_{2}}$ ($498$). Higher bond order, stronger bond.
4 $1652/4=413\ \mathrm{kJ\,mol^{-1}}$.
5 $24$ electrons: one $\mathrm{N{=}O}$ and two $\mathrm{N{-}O^{-}}$; the double bond can be on any of the three O atoms — three equivalent structures. Bond order $4/3\approx1.33$ (N carries $+1$, two O carry $-1$ in each structure).
6 $\mathrm{O{=}C{=}O}$ (all FC $0$); $\mathrm{{}^{+}O{\equiv}C{-}O^{-}}$ and $\mathrm{{}^{-}O{-}C{\equiv}O^{+}}$ (FC $+1$, $0$, $-1$). The first, with zero formal charges, contributes most; the bond length ($115\ \mathrm{pm}$) is a little shorter than a normal $\mathrm{C{=}O}$ ($121\ \mathrm{pm}$) because the other two also contribute.
7 Ozone is a resonance hybrid of $\mathrm{O{=}O{-}O}$ and $\mathrm{O{-}O{=}O}$; each bond has order $1.5$ and a length between the single ($148$) and double ($121\ \mathrm{pm}$) values.
8 $\mathrm{O_{2}}$ $2$ ($12$ valence electrons, $\mathrm{O{=}O}$); $\mathrm{O_{2}^{2-}}$ $1$ ($14$, isoelectronic with $\mathrm{F_{2}}$). The peroxide ion has the longer bond (about $149$ against $121\ \mathrm{pm}$).
IV
Bond polarity, dipole moment and Fajans' rules ($1\ \mathrm{D}=3.33564\times10^{-30}\ \mathrm{C\,m}$, $e=1.602\times10^{-19}\ \mathrm{C}$; Pauling electronegativities: H $2.1$, C $2.5$, N $3.0$, O $3.5$, F $4.0$, Si $1.8$, Cl $3.0$).
- 1 Which of these bonds are polar: $\mathrm{H{-}H}$, $\mathrm{H{-}F}$, $\mathrm{C{-}Cl}$, $\mathrm{Cl{-}Cl}$, $\mathrm{N{-}H}$?
- 2 Mark $\delta+$ and $\delta-$ on the bonds $\mathrm{O{-}H}$, $\mathrm{C{-}F}$ and $\mathrm{Si{-}H}$.
- 3 Express the dipole moment of water, $1.85\ \mathrm{D}$, in $\mathrm{C\,m}$.
- 4 A diatomic molecule has $\mu=1.20\ \mathrm{D}$ and bond length $150\ \mathrm{pm}$. Find the partial charges as a fraction of $e$.
- 5 Arrange $\mathrm{KCl}$, $\mathrm{LiCl}$, $\mathrm{NaCl}$, $\mathrm{BeCl_{2}}$ in order of increasing covalent character.
- 6 Why is $\mathrm{CuCl}$ more covalent than $\mathrm{NaCl}$?
- 7 $\mathrm{SnCl_{4}}$ is a liquid that fumes in moist air, while $\mathrm{SnCl_{2}}$ is a solid. Which is more covalent, and why?
- 8 Applied. Lithium-battery electrolytes need lithium salts that dissolve in organic solvents. Why is $\mathrm{LiI}$ far more soluble than $\mathrm{LiF}$ in such solvents?
💡 Show solutions▶
1 Polar: $\mathrm{H{-}F}$, $\mathrm{C{-}Cl}$, $\mathrm{N{-}H}$ (different electronegativities). Non-polar: $\mathrm{H{-}H}$, $\mathrm{Cl{-}Cl}$.
2 $\overset{\delta-}{\mathrm{O}}{-}\overset{\delta+}{\mathrm{H}}$; $\overset{\delta+}{\mathrm{C}}{-}\overset{\delta-}{\mathrm{F}}$; $\overset{\delta+}{\mathrm{Si}}{-}\overset{\delta-}{\mathrm{H}}$ — silicon ($1.8$) is less electronegative than hydrogen ($2.1$).
3 $1.85\times3.33564\times10^{-30}=6.17\times10^{-30}\ \mathrm{C\,m}$.
4 $Q=\mu/r=\dfrac{1.20\times3.33564\times10^{-30}}{150\times10^{-12}}=2.67\times10^{-20}\ \mathrm{C}$; $Q/e=2.67\times10^{-20}/1.602\times10^{-19}=0.167$, i.e. about $\pm0.17\,e$.
5 $\mathrm{KCl}\lt\mathrm{NaCl}\lt\mathrm{LiCl}\lt\mathrm{BeCl_{2}}$: cation size falls ($133$, $95$, $60$, $31\ \mathrm{pm}$) and Be also carries $2+$.
6 $\mathrm{Cu^{+}}$ is $[\mathrm{Ar}]3d^{10}$ (pseudo-noble-gas). The $d$ electrons shield its nuclear charge poorly, so it polarises $\mathrm{Cl^{-}}$ much more than $\mathrm{Na^{+}}$ ($[\mathrm{Ne}]$) of almost the same size.
7 $\mathrm{SnCl_{4}}$: $\mathrm{Sn^{4+}}$ has a higher charge (and is smaller) than $\mathrm{Sn^{2+}}$, so it polarises the chloride ions far more (Fajans' rule 3). High covalent character gives a molecular liquid.
8 $\mathrm{I^{-}}$ ($216\ \mathrm{pm}$) is large and easily polarised by the small $\mathrm{Li^{+}}$, so LiI has much more covalent character; covalent compounds dissolve better in organic solvents. $\mathrm{F^{-}}$ is small and hard to polarise; LiF is ionic with a high lattice enthalpy.
🎯 Exam corner
E1. NEET In the Lewis structure of ozone with one $\mathrm{O{=}O}$ and one $\mathrm{O{-}O}$ bond, the formal charge on the central oxygen atom is (A) $-1$ (B) $0$ (C) $+1$ (D) $+2$
💡 Show solution▶
Idea The central O has one lone pair ($L=2$) and three bonds ($B=6$): $\text{FC}=6-2-3=+1$.
Answer (C).
E2. JEE Main Which pair of species is isoelectronic and has the same bond order? (A) $\mathrm{CO}$, $\mathrm{NO^{+}}$ (B) $\mathrm{O_{2}}$, $\mathrm{N_{2}}$ (C) $\mathrm{CN^{-}}$, $\mathrm{O_{2}^{2-}}$ (D) $\mathrm{F_{2}}$, $\mathrm{N_{2}}$
💡 Show solution▶
Count Valence electrons: $\mathrm{CO}$ $10$, $\mathrm{NO^{+}}$ $10$, $\mathrm{O_{2}}$ $12$, $\mathrm{N_{2}}$ $10$, $\mathrm{CN^{-}}$ $10$, $\mathrm{O_{2}^{2-}}$ $14$, $\mathrm{F_{2}}$ $14$. Only (A) is an isoelectronic pair; both are triple-bonded (order $3$).
Answer (A).
E3. Board · 3 marks Explain, with one example each, three limitations of the octet rule.
💡 Show solution▶
1 mark Incomplete octet: the central atom has fewer than 8 electrons, e.g. B in $\mathrm{BCl_{3}}$ has 6.
1 mark Odd-electron molecules: an odd total of valence electrons, e.g. $\mathrm{NO}$ (11), so not every atom can have an octet.
1 mark Expanded octet: period-3 and heavier elements use $d$ orbitals to hold more than 8, e.g. S in $\mathrm{SF_{6}}$ has 12. (Also acceptable: noble-gas compounds such as $\mathrm{XeF_{2}}$; no information about shape or energy.)
E4. JEE Main Numerical answer. The dipole moment of $\mathrm{HBr}$ is $0.79\ \mathrm{D}$ and its bond length is $141\ \mathrm{pm}$. Using $e=1.602\times10^{-19}\ \mathrm{C}$ and $1\ \mathrm{D}=3.33564\times10^{-30}\ \mathrm{C\,m}$, find the percentage ionic character of the bond to the nearest integer.
💡 Show solution▶
100% ionic $\mu_{\mathrm{ionic}}=e\,d=1.602\times10^{-19}\times141\times10^{-12}=2.259\times10^{-29}\ \mathrm{C\,m}=6.77\ \mathrm{D}$.
Ratio $\dfrac{0.79}{6.77}\times100=11.7\%$.
Answer 12.
E5. JEE Advanced One or more options correct. (A) All three $\mathrm{N{-}O}$ bonds in $\mathrm{NO_{3}^{-}}$ have the same length. (B) The $\mathrm{C{-}O}$ bond order in $\mathrm{CO_{3}^{2-}}$ is $1.5$. (C) The formal charge on nitrogen in $\mathrm{NH_{4}^{+}}$ is $+1$. (D) Boron in $\mathrm{BF_{3}}$ obeys the octet rule.
💡 Show solution▶
(A) True: the ion is a resonance hybrid of three equivalent structures.
(B) False: $4$ bonds over $3$ positions gives $1.33$.
(C) True: $5-0-\tfrac12(8)=+1$.
(D) False: B has only 6 electrons.
Answer (A), (C).
E6. Board · Case study The chlorides of the period-3 elements change character across the period: $\mathrm{NaCl}$ melts at $801\ ^{\circ}\mathrm{C}$, $\mathrm{MgCl_{2}}$ at $714\ ^{\circ}\mathrm{C}$, $\mathrm{AlCl_{3}}$ sublimes at about $180\ ^{\circ}\mathrm{C}$, and $\mathrm{SiCl_{4}}$ is a liquid boiling at $58\ ^{\circ}\mathrm{C}$. (i) Use Fajans' rules to explain the trend from NaCl to $\mathrm{AlCl_{3}}$. (2 marks) (ii) Why does anhydrous $\mathrm{AlCl_{3}}$ dissolve in organic solvents? (1 mark) (iii) Which would be more covalent, $\mathrm{AlCl_{3}}$ or $\mathrm{AlI_{3}}$? (1 mark)
💡 Show solution▶
(i) From $\mathrm{Na^{+}}$ to $\mathrm{Mg^{2+}}$ to $\mathrm{Al^{3+}}$ the cation charge rises ($1$, $2$, $3$) and its size falls ($95$, $65$, $50\ \mathrm{pm}$), so its polarising power rises sharply. The chloride ion is distorted more and more, covalent character increases, and the compound changes from an ionic lattice to molecular units with weak forces between them — so the melting point falls.
(ii) It is largely covalent; covalent substances dissolve in non-polar or weakly polar organic solvents.
(iii) $\mathrm{AlI_{3}}$: $\mathrm{I^{-}}$ is larger and more polarisable than $\mathrm{Cl^{-}}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Ordinary oxygen and ozone are both made only of oxygen atoms. Why is it the thin ozone layer, and not the far more abundant $\mathrm{O_{2}}$, that shields life on Earth from harmful UV-B radiation — and how did chlorofluorocarbons (CFCs) damage that shield?
Why it matters. UV-B causes skin cancer and cataracts and damages crops and plankton. After the Antarctic "ozone hole" was discovered in 1985, the Montreal Protocol (1987) phased out CFCs worldwide; it is often called the most successful environmental treaty ever, and the ozone layer is expected to recover around the middle of this century.
What to do.
- Draw the Lewis structures of $\mathrm{O_{2}}$ and of the two canonical structures of $\mathrm{O_{3}}$. What are the bond orders? Which molecule should have the weaker oxygen–oxygen bond?
- One photon breaks one bond. Using $\mathrm{O{=}O}$ $498\ \mathrm{kJ\,mol^{-1}}$, find the longest wavelength that can split an $\mathrm{O_{2}}$ molecule ($\lambda=N_{A}hc/E$; $h=6.626\times10^{-34}\ \mathrm{J\,s}$, $c=3.0\times10^{8}\ \mathrm{m\,s^{-1}}$, $N_{A}=6.022\times10^{23}\ \mathrm{mol^{-1}}$). UV-B lies between $280$ and $315\ \mathrm{nm}$: can $\mathrm{O_{2}}$ absorb it by breaking its bond?
- Splitting $\mathrm{O_{3}}\rightarrow\mathrm{O_{2}}+\mathrm{O}$ needs only about $105\ \mathrm{kJ\,mol^{-1}}$. Find the threshold wavelength. Explain why a weak bond on its own is not enough — the molecule must also absorb the light (ozone absorbs very strongly between $200$ and $310\ \mathrm{nm}$).
- A typical $\mathrm{C{-}Cl}$ bond enthalpy is $330\ \mathrm{kJ\,mol^{-1}}$. Find the threshold wavelength and explain, from a reliable source, how chlorine atoms released from CFCs high in the stratosphere destroy ozone in a chain.
Tools: Definitions 4.5 and 4.6, the photon energy $E=hc/\lambda$ from Section 2.2, and a source such as a national meteorological or environmental agency's ozone pages.
Hand in: Your Lewis structures, the three threshold wavelengths, and a half-page explanation of the ozone shield and its repair.
Going further. Replacements for CFCs (HFCs) do not destroy ozone but are powerful greenhouse gases; the Kigali Amendment (2016) phases them down. Find out which property of the $\mathrm{C{-}F}$ bond makes HFCs so long-lived in the atmosphere.
Instructor note: $N_{A}hc=0.1197\ \mathrm{J\,m\,mol^{-1}}$. $\mathrm{O_{2}}$: $0.1197/498\,000=2.40\times10^{-7}\ \mathrm{m}=240\ \mathrm{nm}$ — only UV-C splits $\mathrm{O_{2}}$, so UV-B passes through it. $\mathrm{O_{3}}$: $0.1197/105\,000=1.14\times10^{-6}\ \mathrm{m}\approx1140\ \mathrm{nm}$ in principle, but absorption (the Hartley band) makes UV the effective range. C–Cl: $0.1197/330\,000=3.63\times10^{-7}\ \mathrm{m}=363\ \mathrm{nm}$; in practice CFCs absorb only below about $220\ \mathrm{nm}$, which reaches them only high in the stratosphere. Bond orders: $\mathrm{O_{2}}$ $2$, $\mathrm{O_{3}}$ $1.5$. C–F is one of the strongest single bonds (about $485\ \mathrm{kJ\,mol^{-1}}$).
🧪 Self-check quiz
0/5 correct
Q1. The formal charge on nitrogen in the nitrate ion, $\mathrm{NO_{3}^{-}}$, is
Q2. Which molecule has an expanded octet on its central atom?
Q3. The average C–O bond order in the carbonate ion is
Q4. Which ionic solid has the largest lattice enthalpy?
Q5. According to Fajans' rules, which compound has the greatest covalent character?