The collection

Every essay — page 18

Page 18 of 36, continuing through the fields in the same order.

Orbitals Where the atoms go Bonding models What symmetry decides Beyond the octet What a spectrum settles When the molecule does not stop What the shape is for What is taught wrongly Series Named objects Orbitals Refutations Search

Where the atoms go

VSEPR as a repulsion minimisation rather than a table, the angles that fall out of it, and the case where five sites are not all alike.

A 6-ring at 111°: the twist-boat. A closed ring of 6 equal bonds meeting at 111°, drawn from the coordinates the closure conditions produce. Its torsions are 16.9°, -63.6°, 44.2° and repeat; its puckering amplitude is 0.508 bond lengths at a phase of 344°, with q₃ exactly zero, so it lies on the equator. Turning the ring changes none of those numbers, which is what makes them a description of the shape rather than of the view.

The ring that cannot hold still

Cyclohexane's chair is rigid and its boat is not, and that is a statement about the rank of a matrix rather than about strain. Hold every bond length and every bond angle fixed and count what is left: the chair has nothing, and the boat sits on a continuous loop of shapes with the same bonds and the same angles.

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The bond sum, the measurement, and what is left over. For each pyramid: the vector sum of three bond moments estimated from the electronegativity difference and the measured geometry, the measured dipole, and the difference between them. Positive is towards the lone pair. Ammonia's bonds point that way and nitrogen trifluoride's point the other, which is why the trifluoride's far more polar bonds give it a dipole six times smaller. The leftover is 0.58 D for ammonia and at least 1.48 D for the trifluoride, so it is not one lone pair's property.

The lone pair is not the missing term

Ammonia's dipole is 1.47 debye and nitrogen trifluoride's is 0.235, although the N–F bonds are far more polar than the N–H ones. The bond sums explain the reversal exactly and point in opposite directions — and the lone pair that is supposed to make up the difference has to be worth 0.58 debye in one molecule and at least 1.48 in the other.

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Which site a long bond takes, and it is not the one the rule says. The energy of the axial placement minus the energy of the equatorial one, against the odd bond's length. Above the line the equatorial placement wins and below it the axial one does. The textbook sentence is that a bulkier group goes equatorial where there is more room; given a longer bond and nothing else, the repulsion model puts it AXIAL, at every length, by a margin that grows. A site pushed further out interacts with everything less, so what is left to decide is the arrangement of the four sites left behind — and three equatorial with one axial is the better set of four.

The sites are not the same size

Every arrangement in this collection puts its sites on one sphere, which is an assumption about bond lengths made silently. Give the repulsion model a bond length and it predicts that the long bond goes axial — the opposite of the rule the model is always cited for.

7 figures $The twist the ring forces on the bond. For each ring size, the largest torsion about the double bond that the ring will close on — 180° being a flat trans arrangement and 90° being a π bond broken outright. The six-ring will not close at any torsion tested. The eight-ring reaches 139.3°, against 136° measured in trans-cyclooctene by diffraction: a model with bond lengths and bond angles in it and no energy anywhere agrees with the crystal to a few degrees.$

The double bond a ring cannot hold

A trans double bond needs a ring of nine carbons to sit flat, and the eight-ring will hold one twisted by 40.7 degrees — against 136 degrees measured in trans-cyclooctene. The same eight-ring threshold, applied by counting ring sizes, sorts nine bridgehead alkenes correctly with no bridgehead anywhere in the argument.

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Face to face repels, edge to face attracts. Two benzene molecules 5 Å apart, one turned against the other, with only the quadrupole interaction between them. Stacked, the two negative faces meet and the energy is +8.8 kJ/mol; perpendicular, one molecule's positive rim meets the other's negative face and it is -4.6. The sign changes on the way, so the preference is not that one arrangement is weaker — it is that they are opposite. This is a molecule with no dipole moment at all.

Zero dipole is not no interaction

Benzene's dipole moment is exactly zero at every origin, and its quadrupole moment is large enough to decide a crystal structure. Two benzenes face to face repel by nine kilojoules a mole; edge to face they attract, and the sign changes on the way between.

7 figures $water, drawn with its nuclei the size they are. The molecule with a disc round each nucleus whose radius is the computed root-mean-square displacement of that nucleus in the vibrational ground state. The hydrogens' discs are a substantial fraction of the bond length, and this is at no temperature at all.$

The atoms are not at the points

Every structure in this collection is a set of points, and every bond angle it argues about is a property of that set. Computed from the fitted force fields it already has, water's hydrogens are 0.094 Å from where they are drawn and its bond angle has a spread of 8.9 degrees — larger than the difference between 104.5 and the tetrahedral value that half the essays here are about. This is at no temperature at all.

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The dipole moment, and how many bands there are. For each of five molecules: the dipole moment of the point-charge model, the number of modes whose dipole derivative does not vanish, and the largest derivative. The molecules with no dipole at all have the most active bands, which is the whole of the argument.

A dipole is not what an infrared spectrum sees

Carbon dioxide has no dipole moment at all and three of its four modes are infrared active. Methane has none and six of nine; boron trifluoride none and five of six. Water, which has the largest dipole of the five, has three modes and three bands — and its dipole predicted neither number.

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Flat rings: the angles, the torsions and what is measured. For each ring from three to eight, held flat: its interior angle, how far that is from tetrahedral, the angle strain that follows, the torsional strain of having every bond eclipsed, and the measured strain energy. The five-ring is the row the essay is about — its angles are almost ideal and it is strained.

The strain that is not in the angles

Cyclopentane's flat bond angles are 108°, a degree and a half from tetrahedral, and its angle strain computed from a standard bending constant is 0.4 kJ mol⁻¹. Its measured strain is twenty-six. The missing sixty kilojoules are torsional — one ethane barrier for every bond in the ring, which no account built on bond angles mentions.

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The atoms that meet across a ring

Twelve closed conformers of a ten-membered ring at one bond angle, so every one has identical angle strain by construction. Two of them differ by 0.026 kilojoules a mole in torsional energy and by 0.80 ångström in how close two atoms on opposite sides of the ring come — 2.331 against 3.131, where two carbons are in contact at about 3.4. An account built from angles and torsions calls those two structures the same.

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What boron trifluoride's bands are strong in, and what they move. Every infrared-active mode of boron trifluoride, with its band strength and the root-mean-square displacement of its atoms in the zero point, each scaled to its own largest. The two do not order the modes the same way — the rank correlation between them is 0.2 — and the strongest band belongs to the mode at 719 cm⁻¹, in which 89.93 per cent of the motion is the lightest atom's. Every mode moves the same weighted amount of mass, exactly, so that is not what separates them either.

The mode that moves least radiates most

Boron trifluoride's strongest infrared band is the one in which the fluorines barely move: ninety per cent of the motion belongs to the boron, which is a fifth of the molecule's mass. The mode that moves the most mass is nine and a half times weaker. Across five molecules the rank correlation between band strength and how far the atoms actually go runs from +1 to −0.66, and every normal mode carries exactly the same weighted motion by construction.

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Three bent bonds, at a hundred and one degrees to each other. A carbon–carbon triple bond in its localised description: three equivalent bent bonds, spaced by 101.54 degrees, each tilted 63.43 degrees off the axis and each carrying 0.17 of an s orbital — an sp⁵ hybrid. Its charge sits 0.32 ångström off the axis, where every canonical orbital's sits on it. The mixing that makes the three equivalent is a rotation in the three-dimensional occupied space, so the density is untouched.

Three bent bonds, and the same hybrid

A triple bond localises into three bent bonds at 101.537 degrees to one another, each an sp⁵ hybrid with exactly one sixth s character. A double bond's two bent components are sp⁵ hybrids at 101.537 degrees. The two are the same hybrid, built out of different frameworks and in different numbers, and the reason is that a third of a half is a half of a third.

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An effect that cannot distort a molecule, deciding how far it distorts. The energy along one distortion coordinate, three times. With only the first-order term the minimum is at 0.6; with only the second-order term there is no minimum away from zero at all, because the gap of 1.5 is above the critical 1 for a closed shell on its own. With both, the molecule distorts to 1.06 — well past the first-order answer — and gains 0.09 more than the two separate stabilisations add up to.

Two distortions in one coordinate

A second-order Jahn–Teller effect that cannot distort a molecule by itself — its gap is half again above the critical value — nearly doubles the distortion when a first-order effect is already acting. The molecule goes to 1.075 instead of 0.600 and gains twice the energy, and it does it while the gap the second-order term divides by is opening rather than closing.

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