Chemical Bonding and Molecular Structure

Both ${\text{NH}}_3$ and ${\text{NF}}_3$ are trigonal pyramidal with a lone pair on nitrogen. Fluorine is far more electronegative than hydrogen, so ${\text{NF}}_3$ should have the larger dipole moment. Does it?

The opposite. ${\text{NH}}_3$ measures $1.47$ D; ${\text{NF}}_3$ manages only $0.24$ D.

The lone pair carries a dipole moment of its own, pointing away from the nitrogen. In ammonia the three $\text{N}-\text{H}$ bond moments also point towards nitrogen, so bond moments and lone-pair moment reinforce. In ${\text{NF}}_3$ the bond moments point away from nitrogen towards fluorine, directly opposing the lone pair, and the two nearly cancel.

Advanced bonding questions are almost always of this kind. Two molecules that look alike behave differently, and the reason is a vector, an orbital mixing, or a lone pair that a first-pass analysis ignored.

1. Dipole moments as vector sums

$\vec{\mu }=\sum {\vec{\mu }}_{bond}+{\vec{\mu }}_{lonepair},1\text{D}=3.336\times 10^{-30}\text{C m}$

Symmetry does most of the work. Any molecule whose bond vectors sum to zero is non-polar however polar its individual bonds are: ${\text{CO}}_2$, ${\text{BF}}_3$, ${\text{CCl}}_4$, ${\text{PCl}}_5$, ${\text{SF}}_6$ are all zero. Adding a lone pair breaks that symmetry, which is why ${\text{SO}}_2$ is polar and ${\text{CO}}_2$ is not.

For two identical bonds at angle $\theta$,

${\mu }_{net}=2{\mu }_{bond}\cos \frac{\theta }{2}$

Illustration 1

${\text{H}}_2\text{O}$ has a bond angle of $104.5^{\circ }$ and a dipole moment of $1.85$ D. Estimate the $\text{O}-\text{H}$ bond moment, ignoring the lone pairs.

${\mu }_{net}=2{\mu }_{bond}\cos \left(52.25^{\circ }\right)$

$1.85=2{\mu }_{bond}\left(0.6122\right)\Rightarrow {\mu }_{bond}=1.51$ D

The true bond moment is smaller than this, because the two lone pairs contribute a moment in the same direction as the resultant. Ignoring them always overestimates the bond contribution.

Illustration 2

Explain why ${\text{CO}}_2$ is non-polar while ${\text{SO}}_2$ has a dipole moment of $1.63$ D, despite both being triatomic.

${\text{CO}}_2$ is linear, so the two $\text{C}=\text{O}$ moments are equal and opposite and cancel exactly.

${\text{SO}}_2$ has a lone pair on sulphur, making it bent at about $119^{\circ }$. The two bond moments no longer cancel, and the lone-pair moment adds to the resultant.

The presence or absence of a lone pair is the whole difference. Counting electron domains before predicting polarity is not optional.

2. Fajans' rules: how covalent is an ionic bond?

No bond is purely ionic. A cation polarises the anion's electron cloud, pulling density into the internuclear region, and the degree of that distortion is covalent character. Four factors increase it:

The fourth is the one Advanced tests. ${\text{Cu}}^{+}$ and ${\text{Na}}^{+}$ have nearly the same radius, yet $\text{CuCl}$ is far more covalent than $\text{NaCl}$, because the $3d^{10}$ shell shields the nuclear charge poorly.

Illustration 3

Arrange $\text{LiCl}$, $\text{NaCl}$, $\text{LiI}$ and $\text{NaI}$ in order of increasing covalent character, and predict which has the lowest melting point.

Smaller cation and larger anion both raise covalent character.

$\text{NaCl}<\text{LiCl}<\text{NaI}<\text{LiI}$

$\text{LiI}$ has the smallest cation with the largest anion and is the most covalent, so it has the lowest melting point and the greatest solubility in organic solvents.

Covalent character shows up as low melting point, volatility and solubility in non-polar solvents — three observable consequences of one structural idea.

3. VSEPR, and where Bent's rule takes over

The domain count gives the geometry; lone pairs then compress the bond angles, because a lone pair occupies more angular space than a bonding pair:

$\text{lone-lone}>\text{lone-bond}>\text{bond-bond}\text{repulsion}$

So methane is $109.5^{\circ }$, ammonia $107^{\circ }$ and water $104.5^{\circ }$ — one and then two lone pairs squeezing the same tetrahedral arrangement.

Bent's rule goes further and explains cases VSEPR cannot:

More electronegative substituents prefer the hybrid orbital with less $s$ character.

Since $s$ character is concentrated near the nucleus and electronegative atoms pull electron density away anyway, the atom saves its valuable $s$ character for lone pairs and for bonds to less electronegative partners. This predicts that ${\text{OF}}_2$ ($103.3^{\circ }$) has a smaller angle than ${\text{H}}_2\text{O}$ ($104.5^{\circ }$), and that in ${\text{PF}}_3{\text{Cl}}_2$ the fluorines occupy the axial positions, which have more $p$ character.

Illustration 4

Predict the shapes of ${\text{XeF}}_2$, ${\text{XeF}}_4$, ${\text{ClF}}_3$ and ${\text{SF}}_4$, giving the electron and molecular geometries.

${\text{XeF}}_2$: $5$ domains ($2$ bonds, $3$ lone pairs), trigonal bipyramidal electron geometry, linear shape with lone pairs equatorial.

${\text{XeF}}_4$: $6$ domains ($4$ bonds, $2$ lone pairs), octahedral electron geometry, square planar.

${\text{ClF}}_3$: $5$ domains ($3$ bonds, $2$ lone pairs), T-shaped.

${\text{SF}}_4$: $5$ domains ($4$ bonds, $1$ lone pair), see-saw.

Lone pairs always take equatorial positions in a trigonal bipyramid, because an equatorial site has only two neighbours at $90^{\circ }$ against three for an axial site.

Illustration 5

Both ${\text{PF}}_3$ ($97.8^{\circ }$) and ${\text{PH}}_3$ ($93.5^{\circ }$) have bond angles far below tetrahedral. Explain why, and say what Bent's rule alone would predict for the comparison between them.

Both angles are far below $109.5^{\circ }$ because phosphorus uses nearly pure $p$ orbitals for bonding, keeping its $s$ character in the lone pair — a general feature of period 3 and below.

Bent's rule alone would predict the fluoride to have the narrower angle, since fluorine is more electronegative and should demand orbitals of higher $p$ character. The observed angle is in fact slightly wider, because repulsion between the three bulky fluorine atoms opposes that effect and wins.

Bent's rule gives the direction of an effect, not always its magnitude. Where two effects oppose, the observed value settles which won.

4. Molecular orbital theory, and the s-p mixing inversion

Bond order and magnetism come straight from filling a molecular orbital diagram:

$\text{bond order}=\frac{N_b-N_a}{2}$

The critical detail is that the ordering of the ${\sigma }_{2p_z}$ and ${\pi }_{2p}$ levels changes at oxygen.

For ${\text{B}}_2$, ${\text{C}}_2$ and ${\text{N}}_2$ the $2s$ and $2p_z$ orbitals are close enough in energy to mix, which pushes ${\sigma }_{2p_z}$ above the degenerate ${\pi }_{2p}$ pair. From ${\text{O}}_2$ onward the $2s$-$2p$ gap has widened, mixing is negligible, and the normal order resumes.

The payoff is immediate. ${\text{B}}_2$ has two electrons in the degenerate $\pi$ pair and is therefore paramagnetic; ${\text{C}}_2$ fills that pair and is diamagnetic with a bond order of $2$ made of two $\pi$ bonds and no $\sigma$. Neither result is obtainable from valence bond theory.

Illustration 6

Compute the bond orders and predict the magnetism of ${\text{O}}_2$, ${\text{O}}_2^{+}$, ${\text{O}}_2^{-}$ and ${\text{O}}_2^{2-}$.

${\text{O}}_2$ has $12$ valence electrons: $N_b=8$, $N_a=4$, bond order $2$, with two unpaired electrons in ${\pi }^{\ast }$ — paramagnetic.

${\text{O}}_2^{+}$: one ${\pi }^{\ast }$ electron removed, bond order $2.5$, one unpaired, paramagnetic.

${\text{O}}_2^{-}$: one added, bond order $1.5$, one unpaired, paramagnetic.

${\text{O}}_2^{2-}$: two added, bond order $1$, diamagnetic.

Bond length runs the other way: ${\text{O}}_2^{+}<{\text{O}}_2<{\text{O}}_2^{-}<{\text{O}}_2^{2-}$, since higher bond order always means shorter and stronger.

Illustration 7

Explain why ${\text{N}}_2$ has a first ionisation energy of $15.6$ eV while ${\text{O}}_2$ has only $12.1$ eV, even though a nitrogen atom is harder to ionise than an oxygen atom.

In ${\text{N}}_2$ the highest occupied orbital is the bonding ${\sigma }_{2p}$, pushed up by $s$-$p$ mixing but still a bonding orbital.

In ${\text{O}}_2$ the highest occupied orbital is an antibonding ${\pi }^{\ast }$, which lies well above it.

Removing an electron from an antibonding orbital is easier, and it even strengthens the bond, which is why ${\text{O}}_2^{+}$ is shorter than ${\text{O}}_2$.

The molecular property inverts the atomic one. This is one of the cleanest demonstrations that molecular orbitals, not atomic ones, are what a molecule actually has.

5. Heteronuclear molecules and isoelectronic reasoning

For a heteronuclear pair, the more electronegative atom's orbitals lie lower, so bonding orbitals resemble it more and antibonding orbitals resemble the other. Bond order counting is otherwise unchanged.

Isoelectronic species share bond orders. $\text{CO}$, ${\text{NO}}^{+}$, ${\text{CN}}^{-}$ and ${\text{N}}_2$ all have $14$ electrons and a bond order of $3$. $\text{NO}$ has $15$ and a bond order of $2.5$ with one unpaired electron — which is why it dimerises at low temperature and why removing that electron to give ${\text{NO}}^{+}$ shortens the bond.

Illustration 8

$\text{NO}$ has a bond length of $115$ pm and ${\text{NO}}^{+}$ of $106$ pm. Explain.

$\text{NO}$ has $15$ valence electrons, the last of which occupies a ${\pi }^{\ast }$ orbital, giving bond order $2.5$.

Removing it gives ${\text{NO}}^{+}$ with $14$ electrons and bond order $3$, isoelectronic with ${\text{N}}_2$.

Losing an antibonding electron strengthens and shortens the bond by $9$ pm.

This is the standard test of whether a student is counting antibonding electrons. Ionisation normally weakens a bond; here it does the reverse.

6. Back bonding

When an atom with a lone pair sits next to an atom with an empty orbital, the pair can be donated sideways into that vacancy, adding partial multiple-bond character.

Three standard cases:

${\text{BF}}_3$. Each fluorine donates a $2p$ lone pair into boron's empty $2p$, giving partial double-bond character. The $\text{B}-\text{F}$ bond is shorter than a single bond, and ${\text{BF}}_3$ is a weaker Lewis acid than ${\text{BCl}}_3$, because the back bonding it enjoys must be given up to accept a pair.

Trisilylamine, ${\left({\text{SiH}}_3\right)}_3\text{N}$, is planar. Nitrogen donates its lone pair into silicon's empty $d$ orbitals, so it rehybridises to $s{p}^2$. Trimethylamine, with no such acceptor, stays pyramidal.

Siloxanes. The $\text{Si}-\text{O}-\text{Si}$ angle is about $144^{\circ }$, far wider than the $111^{\circ }$ of $\text{C}-\text{O}-\text{C}$, for the same reason.

Illustration 9

Arrange ${\text{BF}}_3$, ${\text{BCl}}_3$ and ${\text{BBr}}_3$ in order of increasing Lewis acidity, and explain.

Back bonding requires good orbital overlap, which needs similar orbital sizes. Boron's $2p$ overlaps well with fluorine's $2p$, poorly with chlorine's $3p$ and worse with bromine's $4p$.

${\text{BF}}_3$ therefore has the most back bonding and is the most reluctant to accept a lone pair.

${\text{BF}}_3<{\text{BCl}}_3<{\text{BBr}}_3$

This inverts the electronegativity prediction, which would make ${\text{BF}}_3$ the strongest acid. The overlap argument beats the inductive one, and knowing which wins is the point of the question.

7. Hydrogen bonding: within or between

An $\text{H}$ atom attached to $\text{N}$, $\text{O}$ or $\text{F}$ carries enough positive charge to be attracted to a lone pair on another such atom. Whether the bond forms within a molecule or between molecules changes the physical properties completely.

Intramolecular hydrogen bonding uses up the donor and acceptor within one molecule, so it lowers boiling point and solubility in water. Intermolecular bonding links molecules together and raises both.

Hence $o$-nitrophenol boils at $214^{\circ }$C and is steam-volatile, while $p$-nitrophenol boils at $279^{\circ }$C. The same logic explains why $\text{HF}$ boils at $19.5^{\circ }$C against $\text{HCl}$'s $-85^{\circ }$C, and why ice is less dense than water.

Illustration 10

Explain why $o$-nitrophenol can be separated from $p$-nitrophenol by steam distillation.

The ortho isomer forms a six-membered internal hydrogen bond between the hydroxyl hydrogen and a nitro oxygen, so it cannot hydrogen bond to its neighbours or to water.

It is therefore more volatile and less water-soluble, and distils over with the steam.

The para isomer cannot reach its own nitro group, so it hydrogen bonds intermolecularly, has a much higher boiling point, and remains behind.

Geometry, not the functional groups, decides the outcome. Both isomers contain identical groups in identical numbers.

8. Resonance, formal charge and lattice energy

Resonance structures are contributors to one real structure, not species in equilibrium. The most important contributors have: complete octets, minimum formal charge separation, and negative formal charge on the more electronegative atom.

$\text{formal charge}=V-L-\frac{B}{2}$

where $V$ is the valence electron count, $L$ the lone-pair electrons and $B$ the bonding electrons.

For ionic solids the lattice enthalpy dominates stability:

$U\propto \frac{z_{+}{z}_{-}}{r_{+}+r_{-}}$

and a Born-Haber cycle relates it to measurable quantities: sublimation, ionisation, dissociation, electron gain and formation enthalpies.

Illustration 11

Compute formal charges on each atom in the two resonance forms of ${\text{SO}}_2$ with expanded octets and with a dative bond, and say which contributes more.

With one double and one dative single bond: sulphur is $6-2-3=+1$; the doubly bonded oxygen is $6-4-2=0$; the dative oxygen is $6-6-1=-1$.

With two double bonds and an expanded octet: all three atoms have formal charge zero.

The expanded-octet form contributes more, because it has no charge separation.

Sulphur is a third-period element with accessible $d$ orbitals, so octet expansion is permitted. The same argument would be invalid for ozone, where the charge-separated form is the only option.

Illustration 12

Explain why $\text{MgO}$ ($2852^{\circ }$C) melts far higher than $\text{NaF}$ ($993^{\circ }$C), despite similar ionic radii.

Lattice enthalpy is proportional to the product of the ionic charges.

$\text{MgO}$ pairs $2+$ with $2-$, giving a product of $4$.

$\text{NaF}$ pairs $1+$ with $1-$, a product of $1$.

The fourfold increase in the electrostatic term dominates the small difference in interionic distance.

Charge matters far more than size in lattice energy, because it enters as a product while radius enters only as a sum in the denominator.

Summary

• Dipole moment is a vector sum including the lone pair: ${\text{NH}}_3$ is $1.47$ D but ${\text{NF}}_3$ only $0.24$ D, because in ${\text{NF}}_3$ the two contributions oppose.
• ${\mu }_{net}=2{\mu }_{bond}\cos \left(\theta /2\right)$ for two identical bonds; symmetry makes ${\text{CO}}_2$, ${\text{BF}}_3$, ${\text{CCl}}_4$ and ${\text{SF}}_6$ non-polar.
• Fajans: small cation, large anion, high charge and a pseudo noble gas ($d^{10}$) cation all raise covalent character.
• $\text{CuCl}$ is far more covalent than $\text{NaCl}$ at almost identical cation size, because $3d^{10}$ shields poorly.
• VSEPR: lone-lone $>$ lone-bond $>$ bond-bond repulsion; lone pairs always go equatorial in a trigonal bipyramid.
• Bent's rule: electronegative substituents take the orbitals with less $s$ character, which is why ${\text{OF}}_2$ has a narrower angle than ${\text{H}}_2\text{O}$.
• Bond order $=\frac{N_b-N_a}{2}$; $s$-$p$ mixing puts ${\sigma }_{2p}$ above ${\pi }_{2p}$ up to ${\text{N}}_2$, and below it from ${\text{O}}_2$ on.
• ${\text{O}}_2$ is paramagnetic with bond order $2$; the series ${\text{O}}_2^{+}>{\text{O}}_2>{\text{O}}_2^{-}>{\text{O}}_2^{2-}$ runs $2.5,2,1.5,1$.
• ${\text{N}}_2$ ionises harder than ${\text{O}}_2$ because its top electron is bonding while oxygen's is antibonding.
• Isoelectronic species share bond order: ${\text{N}}_2$, $\text{CO}$, ${\text{NO}}^{+}$ and ${\text{CN}}^{-}$ all have $14$ electrons and bond order $3$.
• Back bonding shortens $\text{B}-\text{F}$, makes ${\text{BF}}_3$ the weakest boron Lewis acid, and flattens trisilylamine to planar.
• Intramolecular hydrogen bonding lowers boiling point and water solubility; intermolecular raises both — hence steam distillation of $o$-nitrophenol.
• Best resonance contributors have full octets and least charge separation; lattice enthalpy $\propto \frac{z_{+}{z}_{-}}{r_{+}+r_{-}}$, with charge dominating.