Corrected exercises: hybridization and the quantum model of bonding (202-SN1-RE)

Idea INHERITED from Lewis: a covalent bond is the sharing of two valence electrons. NEW idea: these two electrons occupy a region of space obtained by the overlap of two atomic valence orbitals, a region called a molecular orbital. Only UNPAIRED valence electrons can take part.

Hydrogen has the configuration $1s^{1}$: a single unpaired electron in a spherical $1s$ orbital. The two atoms bring their $1s$ orbitals together until they overlap, and the two electrons, with opposite spins, occupy the shared region centred on the internuclear axis. This is a $\sigma$ bond, written $\sigma_{1s-1s}$. The potential energy curve below shows why an equilibrium distance exists: to the left of the minimum the nuclei repel each other, to the right the overlap weakens, and the minimum is the compromise between the two, close enough for the orbitals to overlap, far enough for the repulsion to stay bearable.

For $HCl$: chlorine has one unpaired electron in a $3p$ orbital, since $3p^{5}$ means two filled orbitals and one half-filled orbital. The overlap is between the $1s$ orbital of hydrogen and this $3p$ orbital of chlorine, along the axis of the lobe: a $\sigma_{1s-3p}$ bond. For $F_{2}$: each fluorine has one unpaired electron in a $2p$ orbital, and the two $2p$ orbitals overlap end to end: a $\sigma_{2p-2p}$ bond. In all three cases the overlap is HEAD-ON, so the bond is $\sigma$.

Nitrogen is $\left[He\right]2s^{2}2p^{3}$: by Hund's rule the three $2p$ electrons each occupy a different orbital, so there are THREE unpaired electrons. The model predicts three bonds, which is exactly what $NH_{3}$ has: the prediction is correct. Carbon is $\left[He\right]2s^{2}2p^{2}$: the $2s$ orbital is full and two $2p$ orbitals hold one electron each, that is, only TWO unpaired electrons. The model therefore predicts two bonds, that is, $CH_{2}$, whereas carbon is universally TETRAVALENT and forms $CH_{4}$. As it stands, the model contradicts the whole of organic chemistry: this is what makes hybridization necessary.

Carbon: $1s^{2}2s^{2}2p^{2}$. The $2s$ orbital holds a pair, and by Hund's rule the two $2p$ electrons go into two different $p$ orbitals with parallel spins. That leaves one EMPTY $2p$ orbital. The count gives TWO unpaired electrons.

One of the two $2s$ electrons is moved into the vacant $2p$ orbital. The excited state is written $1s^{2}2s^{1}2p^{3}$: the four valence orbitals, one $2s$ and three $2p$, now hold one electron each. There are therefore FOUR unpaired electrons, which makes tetravalence possible.

$sp^{3}$ hybridization is the mathematical combination of the $2s$ orbital and the THREE $2p$ orbitals. It produces FOUR identical hybrid orbitals, of equal energy, intermediate between those of $s$ and $p$, and pointing toward the four corners of a tetrahedron. Each holds one unpaired electron and can form a $\sigma$ bond. The name $sp^{3}$ gives the recipe: one part $s$, three parts $p$.

Because the promotion allows FOUR bonds to form instead of two. Each covalent bond releases energy; two extra bonds therefore release far more than the promotion cost. The system pays an advance to collect a higher return. In numbers: the promotion energy is of the order of a few hundred kJ/mol, whereas two $C-H$ bonds return nearly 830 kJ/mol.

The four hybrid orbitals point to the corners of a regular TETRAHEDRON, so $CH_{4}$ is tetrahedral with angles of $109.5^{\circ}$. This is exactly what VSEPR theory predicted by placing four bonding pairs as far apart as possible. The two theories therefore give the same geometry, but for different reasons: VSEPR invokes the REPULSION between electron pairs, hybridization describes the SHAPE of the orbitals that carry those pairs. The second is more fundamental, the first quicker to apply.

$sp^{2}$ hybridization combines the $2s$ orbital with only TWO $2p$ orbitals, producing three coplanar hybrid orbitals at $120^{\circ}$ and leaving one PURE $2p$ orbital perpendicular to that plane. In $C_{2}H_{4}$, each carbon forms three $\sigma$ bonds with its hybrids: two to hydrogens and one to the other carbon. The two remaining pure $2p$ orbitals, parallel to each other, overlap SIDEWAYS and form a $\pi$ bond. The molecule therefore has five $\sigma$ bonds and one $\pi$ bond, and it is PLANAR, with angles of $120^{\circ}$.

$sp$ hybridization combines the $2s$ with a SINGLE $2p$, giving two hybrids pointing in opposite directions at $180^{\circ}$ and leaving TWO pure $2p$ orbitals perpendicular to each other. In $C_{2}H_{2}$, each carbon forms two $\sigma$ bonds, one to hydrogen and one to the other carbon. The two pairs of pure $p$ orbitals form TWO $\pi$ bonds in two perpendicular planes. Tally: three $\sigma$ and two $\pi$ bonds, and the molecule is LINEAR.

Two single bonds would be worth $2\times 347=694$ kJ/mol, whereas the double bond is worth only 614: it is therefore WEAKER than two single bonds. Three single bonds would be worth $3\times 347=1041$, whereas the triple bond is worth only 839. Breaking it down: the $\sigma$ bond is worth 347, the first $\pi$ adds $614-347=267$, the second adds $839-614=225$. A $\pi$ bond is therefore clearly WEAKER than a $\sigma$ bond, which makes sense: the sideways overlap of two p orbitals is less effective than head-on overlap, since the lobes meet along their sides rather than end to end. That is also why $\pi$ bonds are the ones attacked first in addition reactions.

A $\sigma$ bond has rotational symmetry about the internuclear axis: turning a group about that axis changes nothing in the overlap, hence nothing in the energy. Rotation is free, and costs only a few kJ/mol of steric strain. A $\pi$ bond, on the other hand, requires the two $p$ orbitals to stay PARALLEL. Turning by $90^{\circ}$ makes them perpendicular, the overlap drops to zero and the $\pi$ bond is broken: the 267 kJ/mol calculated in c would have to be supplied. Rotation is therefore blocked at ordinary temperature.

The existence of GEOMETRIC isomerism, called $E/Z$ or cis-trans. Since the two carbons of a double bond cannot rotate relative to each other, two different arrangements of the substituents correspond to two DISTINCT compounds, which can be separated and have different physical and biological properties. This is what distinguishes a cis fatty acid from a trans fatty acid, and it is the mechanism of vision: absorbing a photon supplies exactly the energy needed to flip retinal from the 11-cis form to the all-trans form.

Count the number of DIRECTIONS around the atom: each $\sigma$ bond counts as one direction, each lone pair as well, and a multiple bond counts as only ONE direction, since its $\pi$ bonds lie in the remaining pure p orbitals. Four directions give $sp^{3}$, three give $sp^{2}$, two give $sp$. This is exactly the same count as the number of electron domains in VSEPR theory, which is no coincidence: the two models describe the same reality.

The oxygen of $H_{2}O$ forms two $\sigma$ bonds and carries two lone pairs, that is, FOUR directions: it is $sp^{3}$. The basic geometry is therefore tetrahedral, which would predict $109.5^{\circ}$. The measured angle is smaller, $104.5^{\circ}$, because a LONE pair, held by no second nucleus, takes up more room than a bonding pair and repels more strongly. The two lone pairs therefore squeeze the angle between the two bonds. The same effect gives $107^{\circ}$ in $NH_{3}$, which has only one lone pair: the squeeze there is half as large.

In $H_{2}C=O$, carbon forms three $\sigma$ bonds, two to the hydrogens and one to oxygen, and no lone pair: three directions, so $sp^{2}$. Oxygen forms one $\sigma$ bond and carries two lone pairs: also three directions, so it is $sp^{2}$ as well. The remaining pure $p$ orbitals, one on each atom, form the $\pi$ bond of the double bond. Tally: three $\sigma$ and one $\pi$ bond, and the molecule is PLANAR with angles close to $120^{\circ}$.

In $H_{2}C=NH$, carbon has three directions, so $sp^{2}$; nitrogen has two $\sigma$ bonds, to carbon and to hydrogen, plus one lone pair, that is, three directions, so $sp^{2}$ too. In $HCN$, carbon has two directions, one $\sigma$ bond to $H$ and one to $N$: it is $sp$. Nitrogen has one $\sigma$ bond and one lone pair, that is, two directions: it is $sp$ as well. The molecule is linear and has two $\sigma$ bonds, one $H-C$ and the other in the triple bond $C\equiv N$, and two $\pi$ bonds, both in the triple bond.

The nitrogen of $HCN$ carries ONE lone pair. Being $sp$, it has two hybrid orbitals: one is used for the $\sigma$ bond with carbon, the other holds precisely this lone pair, pointing away from the molecule along its axis. Its two $p$ orbitals are fully taken up by the two $\pi$ bonds. This lone pair, available and well exposed along the axis, is what makes the cyanide ion an excellent nucleophile and a deadly poison, since it binds to the iron of cytochrome c oxidase.

$PCl_{5}$: phosphorus carries five bonds and no lone pair, that is, 5 domains, so $sp^{3}d$ hybridization and a trigonal bipyramidal geometry. The mixing involves one $3s$ orbital, three $3p$ orbitals and one $3d$ orbital, that is, five atomic orbitals giving exactly five hybrid orbitals: the number of orbitals is always conserved in hybridization. $SF_{6}$: six bonds, no lone pair, 6 domains, $sp^{3}d^{2}$ hybridization and an octahedral geometry; one $3s$, three $3p$ and two $3d$ give six hybrids.

$SF_{4}$ has 4 bonds and 1 lone pair, $ClF_{3}$ has 3 and 2, $XeF_{2}$ has 2 and 3: in all three cases the total is 5 domains, so the hybridization is $sp^{3}d$ for all three. Their geometries nonetheless differ completely, seesaw, T-shaped and linear, because geometry describes only the position of the ATOMS whereas hybridization describes the arrangement of ALL the domains, lone pairs included. This is the distinction never to lose sight of: hybridization is read from the number of domains, geometry from the number of bonds.

In $SF_{4}$, the five $sp^{3}d$ orbitals divide into four that form the $\sigma$ bonds with the fluorines and one that holds the lone pair. That last one is one of the three equatorial orbitals, because that is where the lone pair suffers the fewest repulsions at 90°. In $XeF_{4}$, the six $sp^{3}d^{2}$ orbitals divide into four $\sigma$ bonds and two lone pairs, the latter occupying two opposite positions of the octahedron, hence at 180° from each other.

Nitrogen is in the second period: its valence shell $n=2$ contains only the $2s$ and $2p$ subshells, there is no $2d$ orbital, and hybridization therefore cannot go beyond $sp^{3}$, that is, four domains. Phosphorus, in the third period, has $3d$ orbitals available and has no such limit; its larger size also lets it hold five atoms around it. As for the modern criticism, it is well founded: the $3d$ orbitals of sulfur or phosphorus are far too high in energy to mix effectively with the $3s$ and $3p$, and the correct description calls on three-centre, four-electron bonds that are largely ionic. That does not condemn the model, provided you know what you are asking of it. $sp^{3}d$ hybridization is not a description of the electronic reality, it is a counting rule that correctly matches a number of domains to a geometry, and on that ground it is never wrong. This is exactly the status of the VSEPR model, which predicts shapes very well without being a quantum theory. A model is judged by its domain of validity, not by its literal fidelity.

The bond order is $\dfrac{n_{\text{bonding}}-n_{\text{antibonding}}}{2}$, where we count the ELECTRONS occupying the bonding and antibonding orbitals. We divide by two because it takes a pair to make a bond. An order of 1, 2 or 3 corresponds to a single, double or triple bond; a fractional order is possible and signals an intermediate bond; a bond order of ZERO means there is no bond, hence that the molecule does not exist.

Each hydrogen brings its single $1s$ electron. The two $1s$ orbitals combine into a bonding $\sigma_{1s}$ and an antibonding $\sigma^{*}_{1s}$. The two electrons go into the bonding orbital, the lower in energy. Order $=\dfrac{2-0}{2}=1$: a single bond, a stable molecule, as experiment confirms.

For $He_{2}$, each helium brings TWO electrons: two go into the bonding orbital, but the next two have to occupy the antibonding orbital. Order $=\dfrac{2-2}{2}=0$: no bond, the molecule DOES NOT EXIST, which explains why helium is a monatomic gas. For $H_{2}^{+}$, only one electron remains, in the bonding orbital: order $=\dfrac{1-0}{2}=0.5$. This species EXISTS, it is observed in the gas phase, but with a bond about half as strong as that of $H_{2}$ and a longer internuclear distance. The model therefore correctly predicts a half bond, a notion that Lewis cannot represent with lines.

Order $=\dfrac{8-4}{2}=2$: a double bond, in agreement with the measured length and energy. But the diagram puts the last two electrons SEPARATELY in the two degenerate $\pi^{*}$ orbitals, with parallel spins, by Hund's rule. Oxygen gas therefore has TWO UNPAIRED ELECTRONS: it is a PARAMAGNETIC species, attracted by a magnetic field.

Experiment settles the matter without ambiguity: liquid oxygen poured between the poles of an electromagnet stays suspended there, the classic demonstration of paramagnetism. Yet the $O=O$ Lewis structure shows eight electrons, all paired, and therefore predicts a diamagnetic species: it CONTRADICTS experiment. Conclusion about the two models: Lewis and valence bond theory remain excellent for quickly predicting a geometry and counting bonds, which is why they are taught first; but only molecular orbital theory accounts for magnetism, fractional orders and electronic spectra. Neither is wrong, they simply do not have the same domain of validity, and the second is the more fundamental.

$N_{2}$: $BO=\dfrac{8-2}{2}=3$. $O_{2}$: $BO=\dfrac{8-4}{2}=2$. $F_{2}$: $BO=\dfrac{8-6}{2}=1$. The correlation is perfect and goes both ways as expected: as the bond order falls from 3 to 1, the length increases from 110 to 142 pm and the dissociation energy collapses from 945 to 159 kJ/mol. The ratio is spectacular: the triple bond of dinitrogen is six times as strong as the single bond of difluorine, which explains at once the chemical inertness of atmospheric dinitrogen, the energy cost of the Haber process, and conversely the extreme reactivity of difluorine.

$N_{2}$: all electrons are paired, the molecule is diamagnetic. $O_{2}$: the last two electrons occupy the two degenerate $\pi^{*}$ orbitals separately, with parallel spins in accordance with Hund's rule, so oxygen gas is PARAMAGNETIC. $F_{2}$: both $\pi^{*}$ orbitals are full, the molecule is diamagnetic. Only molecular orbital theory predicts the paramagnetism of oxygen gas; the Lewis structure $O=O$ shows all electrons paired and so wrongly predicts a diamagnetic compound. This is the most famous failure of the Lewis model, and it can be seen with the naked eye: liquid oxygen stays hanging between the poles of a magnet.

$O_{2}^{+}$ has 11 valence electrons, that is, 8 bonding and 3 antibonding: $BO=\dfrac{8-3}{2}=2.5$. $O_{2}$: $BO=2$. $O_{2}^{-}$, 13 electrons: $BO=\dfrac{8-5}{2}=1.5$. $O_{2}^{2-}$, 14 electrons: $BO=\dfrac{8-6}{2}=1$. The lengths follow exactly the reverse order, 112 < 121 < 128 < 149 pm: each added electron populates an antibonding orbital, weakening and lengthening the bond. The reasoning is counterintuitive for anyone thinking in Lewis terms, where adding an electron does not “break” anything; it becomes obvious as soon as bonding and antibonding orbitals are distinguished. The ion $O_{2}^{2-}$, the peroxide ion, does have a single bond, which is consistent with the fragility of hydrogen peroxide.

The inversion comes from the mixing of the $2s$ orbital with the $2p_{z}$ orbital of the same atom, which is effective only if their energies are close. On the left of the period, from boron to nitrogen, the $2s$-$2p$ gap is small, the mixing is strong and it pushes the $\sigma_{2p}$ upward, above the $\pi_{2p}$. Toward oxygen and fluorine, the increasing nuclear charge lowers the $2s$ strongly, the gap widens, the mixing becomes negligible and the normal order is restored. The observation that settles the matter is the magnetism of diboron: $B_{2}$ has 6 valence electrons, and the question is where the last two go. If the $\sigma_{2p}$ were the lowest, they would pair up in it and the molecule would be diamagnetic; if it is the two degenerate $\pi_{2p}$ orbitals that are lowest, Hund's rule puts them there separately and the molecule is paramagnetic. Now $B_{2}$ is experimentally paramagnetic: it is indeed the inverted order that applies on the left of the period. A single magnetic susceptibility measurement thus decides between two diagrams.

An $sp^{3}$ hybrid mixes 1 $s$ orbital and 3 $p$ orbitals, that is, 4 orbitals in all: the share of $s$ is $\dfrac{1}{4}=25\%$. An $sp^{2}$ mixes 1 $s$ and 2 $p$: $\dfrac{1}{3}=33\%$. An $sp$ mixes 1 $s$ and 1 $p$: $\dfrac{1}{2}=50\%$. Since the $s$ orbital is on average closer to the nucleus than the $p$ orbital, a hybrid rich in $s$ is shorter and more contracted. Its overlap with the $1s$ orbital of hydrogen therefore occurs at a smaller distance, and the $C-H$ bond shortens steadily: 109 pm on an $sp^{3}$ carbon, 108 pm on an $sp^{2}$, 106 pm on an $sp$. The same contraction makes these bonds slightly stronger. In the figure, the angle between the bonds opens with the $s$ character: the $s$ orbital is spherical, and the more it weighs in the hybrid, the wider the bonds spread.

An acid is stronger the more stable its conjugate base. Removing the proton from a $C-H$ bond leaves a carbanion whose lone pair occupies the hybrid orbital that carried the bond: an $sp^{3}$ for ethane, an $sp^{2}$ for ethylene, an $sp$ for acetylene. The more $s$ character this orbital contains, the more tightly the lone pair is held near the nucleus, hence the more the negative charge is stabilized. The acetylide ion, with its 50% $s$ character, is by far the most stable of the three, which shows up as a gap of 25 $pK_{a}$ units from ethane, a factor of $10^{25}$ in the acidity constant. It is worth taking the measure of this: a mere change of hybridization, without altering a single atom or introducing any electron-withdrawing group, shifts the acidity by twenty-five orders of magnitude.

The rule is that a base deprotonates an acid effectively only if the $pK_{a}$ of its conjugate acid is HIGHER than that of the acid to be deprotonated. The conjugate acid of hydroxide is water, with a $pK_{a}$ of 16, far below the 25 of acetylene: the equilibrium lies on the wrong side by about nine orders of magnitude, and sodium hydroxide is not suitable. The conjugate acid of sodium amide is ammonia, with a $pK_{a}$ of 38, well above 25: the reaction lies toward the acetylide by about thirteen orders of magnitude, so it is quantitative. This is why acetylides are prepared with sodium amide in liquid ammonia, and never in aqueous solution.

The reasoning is the same as in b), but reversed. A lone pair held tightly by the nucleus is stable, hence not very available to capture a proton: basicity DECREASES as $s$ character increases. We therefore predict methylamine ($sp^{3}$, 25% $s$) more basic than pyridine ($sp^{2}$, 33%), itself more basic than acetonitrile ($sp$, 50%). The $pK_{a}$ values of the conjugate acids confirm this: 10.6, then 5.2, then about $-10$, a fall of more than twenty units. What $s$ character lets you predict at a stroke is therefore remarkable: the length of a bond, its strength, the acidity of a $C-H$ bond, the basicity of a lone pair and even the apparent electronegativity of carbon, higher for an $sp$ carbon than for an $sp^{3}$. A single parameter, the proportion of $s$ orbital in the hybrid, governs this whole series of seemingly unrelated properties.