Compound structure and properties

ELECTRONEGATIVITY AND THE BOND CONTINUUM

Electronegativity is an atom’s pull on the electrons it shares in a bond. It rises left to right across a period and falls down a group, and Unit 1 already told you why: across a period the effective nuclear charge grows with the valence electrons in the same shell; down a group the shared electrons sit a shell farther from the nucleus. Fluorine, top right, pulls hardest; cesium, bottom left, barely pulls at all.

Two atoms in a bond pull against each other, and the difference in their electronegativities says who wins:

The boundaries are conventions, not walls. The exam’s own words: “the difference between ionic and covalent bonding is not distinct but rather a continuum.”

Read the bar in the direction the exam does.

• Similar electronegativities share evenly: a nonpolar covalent bond. C–H counts here — carbon is slightly more electronegative than hydrogen, but the bond is effectively nonpolar, and the CED says so explicitly.
• Unequal electronegativities share unevenly: a polar covalent bond, with a partial negative charge (${\delta }^{-}$) on the more electronegative atom. In single bonds, a larger difference means a larger bond dipole.
• A large difference, and in practice a metal with a nonmetal, transfers electrons: ionic. The nonmetal takes them; the two ions attract.
• A metal alone: metallic, with valence electrons delocalized over the whole solid — not attached to any one atom.

The properties are the verdict; the difference is the guess. The CED is direct about this: “examination of the properties of a compound is the best way to characterize the type of bonding.” An ionic solid melts high, shatters when struck, and conducts electricity only when molten or dissolved (its ions must be free to move). A molecular substance melts low and does not conduct. A metal conducts as a solid and bends instead of shattering. When a prompt gives you a property table, argue from the table and use electronegativity as the supporting reason, not the other way around.

POTENTIAL ENERGY AGAINST DISTANCE: THE CURVE THAT IS THE BOND

Bring two atoms toward each other and plot the potential energy of the pair against the distance between their nuclei. Far apart, nothing happens: the energy is zero by convention. As they approach, each nucleus begins to attract the other atom’s electrons, and the energy falls — attraction lowers potential energy. Closer still, the two nuclei repel each other and the two electron clouds repel, and the energy climbs steeply. Between the two lies a minimum, and the minimum is the bond:

Two bonds on one graph. The deeper and farther-left the minimum, the stronger and shorter the bond. The steep left wall is nucleus–nucleus repulsion; the gentle right tail is attraction fading with distance.

Three readings, in order. The horizontal position of the minimum is the bond length — the distance at which attraction and repulsion balance. The vertical depth of the minimum, measured from zero, is the bond energy — what it costs to pull the atoms apart to infinity. And the curvature near the bottom says how stiff the bond is, which is a Unit 7 idea you will meet in vibrational spectra; for now, read length and energy.

What moves the minimum. Two things, and the CED names both. Bond order: a double bond is shorter and stronger than a single bond between the same atoms, a triple shorter and stronger still — more shared electrons, more attraction to both nuclei. Atomic size: larger atoms have their bonding electrons farther from the nuclei, so the bond is longer and, typically, weaker. Between those two levers, most bond-length and bond-energy comparisons are settled.

Two numbers worth knowing as anchors, not for memorizing but for sanity: the H–H bond, the simplest there is, is about 74 pm long with a bond energy near 436 kJ/mol; the N$\equiv$N triple bond is about 110 pm and near 945 kJ/mol, which is why nitrogen gas is so hard to react. A bond energy in the tens of kJ/mol is not a covalent bond; one in the thousands is not either.

COULOMB’S LAW FOR IONS: CHARGE FIRST, THEN DISTANCE

Unit 1 used Coulomb’s law inside one atom. Unit 2 uses it between two ions, and the form is the same:

Two levers, and they are not equal. Charge multiplies. A pair of $2+$ and $2-$ ions attracts four times as strongly as a pair of $1+$ and $1-$ ions at the same distance — $2\times 2$ against $1\times 1$. That single factor is why magnesium oxide melts near $2850^{\circ }$C and sodium chloride near $800^{\circ }$C. Distance divides. Smaller ions sit closer, so their charges act over a shorter $r$ and the attraction is stronger; that is why lithium fluoride melts higher than potassium iodide even though both are $1+∕1-$.

The order of operations when comparing two ionic compounds. Compare the charges first. Only when the charges tie do you compare the sizes — and then the sum of the two ionic radii is the distance to use, because the ions touch. Write the reason in that order, and say what the stronger attraction predicts: a higher melting point, a larger lattice energy, a harder crystal.

Where the ionic radii come from. A cation is smaller than its atom (electrons removed, the rest pulled in); an anion is larger (electrons added, repulsion expands the cloud); and down a group both grow, a shell at a time. You will usually be handed the radii; the habit is to add the two that touch, not to compare one ion’s size in isolation.

IONIC SOLIDS: A LATTICE, AND WHY IT SHATTERS

An ionic solid is not molecules. It is a three-dimensional array in which every cation is surrounded by anions and every anion by cations — the arrangement that puts opposite charges as close as possible and like charges as far apart as possible. The CED asks for exactly that statement and no more: you will not be asked to name a crystal structure, but you will be asked to draw or explain a particulate model that is consistent with Coulomb’s law.

Brittleness, explained by the model. Slide one layer of the lattice a single ion over and every attraction across the plane becomes a repulsion.

The properties the model predicts, and the reason for each. High melting point: every ion is held by several strong Coulombic attractions at once, and all of them must be overcome. Brittle: the figure above — a shear puts like charges opposite one another. Conducts only when molten or dissolved: conduction needs mobile charges; in the solid the ions are locked in place, so it is an insulator until the lattice is broken up. Say the property, then the particulate reason; that pairing is the whole of skill 4.C.

METALS AND ALLOYS: A SEA OF ELECTRONS, AND WHAT YOU DROP INTO IT

A metal’s valence electrons are not attached to individual atoms. The model the CED asks for is a lattice of positive metal ions in a sea of delocalized valence electrons: the electrons belong to the whole solid, and the attraction between the mobile sea and the fixed cations is the metallic bond.

That one picture predicts the metallic properties. Conducts as a solid: the electrons are already free to move. Malleable and ductile: slide one layer of cations over another and every cation is still sitting in the same electron sea — nothing changes sign, so nothing repels, so the metal bends instead of breaking. Compare that to the ionic lattice one card up: the same push that shatters salt merely dents copper.

Three lattices. In steel, carbon atoms are small enough to sit in the spaces between iron atoms; in brass, zinc atoms are close enough in size to copper to take copper’s places.

Alloys, and how to tell which kind. Compare the atomic radii. Atoms of very different size make an interstitial alloy: the small ones fill the interstices — the gaps — between the large ones, as carbon does in iron. Atoms of comparable size make a substitutional alloy: one replaces the other in the lattice, as zinc does for copper in brass. Either way the added atoms interrupt the smooth sliding of layers, so an alloy is typically harder and less malleable than the pure metal, while the electron sea stays and the alloy still conducts. That “harder because the layers can no longer slide freely” is the sentence the exam wants.

LEWIS DIAGRAMS: THE PROCEDURE, NOT THE GUESS

A Lewis diagram is an inventory of valence electrons arranged so that each atom gets what it needs — eight for most main-group atoms, two for hydrogen. Draw it by procedure every time, because the procedure is what you audit when something looks wrong:

Six steps, and step 6 is the one that is skipped. A diagram that looks right and has the wrong count is wrong.

Worked once, fast, on $C{O}_2$. Step 1: C brings 4, each O brings 6: $4+2\left(6\right)=16$. Step 2: carbon is the center; two C–O single bonds use 4, leaving 12. Step 3: three lone pairs on each oxygen use 12, leaving 0. Step 4: nothing left for carbon — and carbon has only 4 electrons around it. Step 5: convert one lone pair from each oxygen into a second bond: O$=$C$=$O, carbon now sees 8, each oxygen still sees 8 (two lone pairs and a double bond). Step 6: $2\left(4\right)+2\left(4\right)=16$ on the page. $\surd$

The exceptions the CED expects you to handle. Hydrogen wants two electrons, not eight. Boron and beryllium centers are often short of an octet and stable that way ($B{F}_3$ has six around boron, all formal charges zero — do not force a double bond that gives fluorine a positive charge). A center in Period 3 or below can hold more than eight — $S{F}_4$, $\mathrm{PC}{l}_5$, $\mathrm{Xe}{F}_4$ — and the CED’s instruction is to draw the diagram and predict the shape without any theory of how the extra pairs are held: hybridization involving d orbitals is not assessed. And a molecule with an odd number of valence electrons, such as NO with 11, cannot satisfy every octet; that is a limitation of the model, and the exam may ask you to say so.

RESONANCE AND FORMAL CHARGE: WHEN ONE DIAGRAM IS NOT ENOUGH

Formal charge is bookkeeping: the charge an atom would carry if every bond were shared exactly evenly.

where $V$ is the number of valence electrons of the free atom, $N$ the number of nonbonding electrons on it in the diagram, and $B$ the number of bonding electrons it shares. Compute it for every atom of every candidate diagram. The formal charges must sum to the overall charge of the species — that is a check, and it is free.

Case one — equivalent diagrams: resonance. Draw the nitrate ion, $N{O}_3^{-}$, and you must put the double bond on one oxygen. But the three oxygens are identical, so there are three equally valid diagrams, and no one of them is the molecule:

The real ion is the average: three identical N–O bonds, each of bond order $\frac{4}{3}$, all the same length — shorter than a single bond and longer than a double. That measured fact is the evidence the exam asks you to connect to resonance.

The double-headed arrow means “the true structure is the blend,” never “the molecule flips between these.” What resonance predicts is measurable: all three N–O bonds in nitrate are the same length (about 124 pm, between the single-bond and double-bond values), and each carries one-third of the negative charge.

Case two — nonequivalent diagrams: formal charge helps rank. When the same atomic skeleton has different electron arrangements that are not equivalent by symmetry, their contributions may differ. Prefer the diagram whose formal charges are closest to zero; if charges are unavoidable, prefer the one that puts the negative charge on the more electronegative atom. Adjacent like charges are a sign the diagram is poor. The cyanate ion, $\mathrm{OC}{N}^{-}$, is the standard test, and Problem 7 runs it.

Case three — the model’s limit. Some species have no diagram that satisfies every rule: an odd electron count (NO, $N{O}_2$) leaves one atom short of an octet whatever you do. The exam’s expectation is that you say what the model cannot do, not that you invent an electron.

VSEPR: COUNT DOMAINS, THEN COUNT ATOMS

Valence-shell electron-pair repulsion is a single idea: electron domains around a central atom spread out as far as they can, because electrons repel. Everything else is counting.

An electron domain is one region of electron density on the central atom: a single bond, a double bond, a triple bond, or a lone pair each count as one. Count them from the Lewis diagram — which is why the Lewis diagram comes first, always.

Domains give the electron-domain geometry; atoms give the molecular geometry. Two names, and the exam asks for the second. The lone pairs occupy positions in the domain arrangement but are invisible in the shape, which is named by the atoms alone:

Why the angle shrinks. A lone pair is held by one nucleus, not two, so it spreads out and pushes the bonding domains closer together. Ammonia’s H–N–H angle is about $107^{\circ }$, not $109.5^{\circ }$; water’s, with two lone pairs, about $104.5^{\circ }$. When a question asks for a bond angle, give the number and the direction of the deviation and its cause — that is the whole answer, and the CED singles out “confusing molecular geometry with bond angles” as a named loss.

Five and six domains. With five domains the positions are not all alike: three sit in a plane (equatorial, $120^{\circ }$ apart) and two stand above and below (axial, $90^{\circ }$ to the plane). Lone pairs take the roomier equatorial positions first, which is why $S{F}_4$ is a seesaw and $\mathrm{Cl}{F}_3$ is T-shaped. With six domains all positions are equivalent, and two lone pairs sit opposite each other, which is why $\mathrm{Xe}{F}_4$ is square planar and, being symmetric, nonpolar.

POLARITY: BOND DIPOLES ARE ARROWS, AND ARROWS ADD

A bond dipole exists whenever two bonded atoms differ in electronegativity: draw it as an arrow pointing toward the more electronegative atom. A molecular dipole is the vector sum of the bond dipoles over the shape. So polarity is a two-step question, and both steps are needed:

Same test three times: draw the arrows on the shape, then ask whether symmetry cancels them. $\mathrm{CC}{l}_4$ would be nonpolar; replace one Cl with H and the cancellation fails.

The rule that covers most cases. A molecule with polar bonds is nonpolar only when its shape is symmetric and every position around the center carries the same atom: linear $A{B}_2$, trigonal planar $A{B}_3$, tetrahedral $A{B}_4$, trigonal bipyramidal $A{B}_5$, octahedral $A{B}_6$, and square planar $A{B}_4$ with two opposite lone pairs. A lone pair on the center, or one different outer atom, almost always leaves a net dipole. Say both halves — “polar bonds, and a shape that does not cancel them” — because each half alone is a partial answer.

HYBRIDIZATION, SIGMA, AND PI: ONE VOCABULARY FOR THE SAME COUNT

The CED wants three things named and nothing derived. Hybridization is a label for the arrangement of electron domains on an atom, and the label follows the domain count you already made for VSEPR:

Five or six domains: name the shape and stop. The CED does not assess hybridization involving d orbitals, and it does not assess how hybrid orbitals are constructed — only the sp, $s{p}^2$, $s{p}^3$ labels and what they imply about angle.

Sigma and pi. Every bond has exactly one sigma ($\sigma$) bond, made by head-on overlap along the line between the nuclei. A double bond is one sigma plus one pi ($\pi$) bond, made by side-by-side overlap above and below that line; a triple bond is one sigma plus two pi. So count sigma bonds by counting bonds, and count pi bonds by counting the extra lines in the multiple bonds. Two consequences the exam uses:

• A sigma bond is stronger than a pi bond — the head-on overlap is larger. That is why a C$=$C double bond (about 614 kJ/mol) is strong, but less than twice a C–C single bond (about 347 kJ/mol).
• A pi bond locks rotation. The side-by-side overlap breaks if one end twists, so the two ends of a double bond cannot rotate freely, and a molecule with a double bond can have geometric isomers — cis and trans — that a single bond, rotating freely, cannot.

Why the double bond matters: the pi overlap is what a twist would destroy, so 1,2-dichloroethene comes in two distinct molecules. The single-bonded 1,2-dichloroethane does not.