Bond Angle Deviations in VSEPR: Causes, Trends & Examples

Bond Angle Deviations in VSEPR

When we predict the shape of a molecule using VSEPR (Valence Shell Electron Pair Repulsion) theory, we begin with ideal geometries — perfect tetrahedra (109.5°), perfect trigonal planes (120°), perfect octahedra (90°). But nature rarely produces perfection. The actual bond angles in real molecules almost always deviate from these ideal values, sometimes by just a degree or two, and sometimes by twenty degrees or more.

Understanding why these deviations occur, in which direction they go, and how large they are is one of the most important skills in structural chemistry. The deviations are not random — they follow clear, predictable patterns rooted in the fundamental nature of electron pair repulsions.

This write-up provides a comprehensive treatment of bond angle deviations in VSEPR theory: the theory itself, the hierarchy of repulsions, the factors that cause deviations, and extensive worked examples.

1.1 Origins

The VSEPR model has its intellectual roots in the work of several chemists:

1.2 Core Philosophy

VSEPR theory is fundamentally a steric theory — it treats electron pairs (both bonding and lone pairs) as charge clouds that repel each other and arrange themselves to minimize repulsion (and thus minimize energy) around a central atom.

The theory makes no assumptions about orbital shapes, hybridization, or quantum mechanics. It is purely electrostatic and steric in nature, which is both its greatest strength (simplicity) and its greatest weakness (lack of quantitative rigor).

Postulate 1: Electron Pair Domains

The electron pairs in the valence shell of a central atom arrange themselves to be as far apart as possible. Each bonding pair (BP) and each lone pair (LP) constitutes an electron domain (also called a steric number unit).

Postulate 2: Repulsion Hierarchy

Not all electron pairs repel each other equally. The hierarchy of repulsions is:

This is the single most important principle for understanding bond angle deviations.

Postulate 3: Geometry Determination

The electron domain geometry (arrangement of all electron pairs) is determined by the steric number (SN):

Postulate 4: Molecular Geometry vs. Electron Domain Geometry

The molecular geometry (shape described by atomic positions only) may differ from the electron domain geometry because lone pairs are "invisible" in molecular shape but still exert repulsive forces.

Postulate 5: Multiple Bonds as Single Domains

A double bond or triple bond counts as one electron domain (though it occupies more space than a single bond).

The steric number (SN) is defined as:

This determines both the electron domain geometry and, combined with the number of lone pairs, the molecular geometry:

4.1 Linear (SN = 2): 180°

Two electron domains place themselves on opposite sides of the central atom to maximize separation. This is the only arrangement that achieves maximum distance for two points on a sphere.

4.2 Trigonal Planar (SN = 3): 120°

Three electron domains arrange in a plane with 120° between each pair. This is the solution to the Thomson problem for three points on a sphere — the vertices of an equilateral triangle inscribed in a great circle.

4.3 Tetrahedral (SN = 4): 109.5°

Four electron domains arrange at the vertices of a regular tetrahedron. The angle between any two vertices of a regular tetrahedron, measured from the center, is:

This can be derived from the dot product of position vectors. If the four vertices of a tetrahedron are at positions:

Then the angle between any two vectors from the origin is:

4.4 Trigonal Bipyramidal (SN = 5): 90° and 120°

Five electron domains arrange with three in the equatorial plane (120° apart) and two on the axial positions (90° to the equatorial plane, 180° to each other).

This geometry is not a regular polyhedron — the axial and equatorial positions are inequivalent:

Since 90° repulsions are much stronger than 120° repulsions, the equatorial positions are less crowded (fewer 90° interactions), making them preferred for lone pairs and larger substituents.

4.5 Octahedral (SN = 6): 90° and 180°

Six electron domains arrange at the vertices of a regular octahedron. Each domain has 4 neighbors at 90° and 1 neighbor at 180°. All positions are equivalent.

The ideal bond angles described above assume that all electron domains are identical. In reality, electron domains differ in their repulsive strength. The key insight of VSEPR is:

Lone pairs repel more strongly than bonding pairs. This single fact is responsible for the majority of bond angle deviations in chemistry.

1.1 Why Do Lone Pairs Repel More Strongly?

There are several complementary explanations:

a) Spatial extent: A lone pair is held by only one nucleus. It is not constrained by a second atom and therefore spreads out over a larger volume. A bonding pair is held between two nuclei and is more confined (elongated toward the bonded atom). The larger, more diffuse lone pair exerts greater repulsion on neighboring electron domains.

b) Electron density distribution: In a bonding pair, the electron density is concentrated in the internuclear region (between the two atoms). In a lone pair, the electron density is concentrated closer to the central atom. This means lone pair electron density is closer to other electron domains on the central atom, resulting in stronger repulsion.

c) Effective solid angle: Lone pairs subtend a larger effective solid angle at the central atom than bonding pairs. They "take up more room" in the valence shell.

d) Analogy: Think of a bonding pair as a balloon that is pinched at one end (tied to the bonded atom) — it is elongated and narrow. A lone pair is like a balloon that is free — it is rounder and takes up more space. A round balloon pushes its neighbors more than a narrow one.

1.2 Quantifying the Repulsion Hierarchy

While VSEPR is primarily qualitative, some semi-quantitative estimates of repulsion strengths have been proposed. Gillespie suggested that the relative repulsion energies follow approximately:

These numbers vary depending on the specific system, but the ordering is always the same. The LP–LP repulsion is roughly 1.5 to 2 times stronger than BP–BP repulsion.

The general rule for predicting bond angle deviations is:

Lone pairs compress the bond angles between bonding pairs.

More precisely:

The magnitude of the deviation increases with the number of lone pairs and their proximity to the bond angle in question.

1.1 Methane (CH₄) — Perfect Tetrahedron

        H
        |
   H — C — H        All bond angles = 109.5°
        |
        H

This is the reference point for all tetrahedral deviations.

1.2 Ammonia (NH₃) — Trigonal Pyramidal

        H
        |
   H — N — H        Bond angle H–N–H = 107.8°
       :
      (lone pair)

Why is the bond angle less than 109.5°?

The lone pair on nitrogen occupies more space than a bonding pair. It pushes the three N–H bonding pairs closer together:

The lone pair "squeezes" the bond angle from 109.5° down to 107.8°.

1.3 Water (H₂O) — Bent

   H — O — H        Bond angle H–O–H = 104.5°
      :  :
   (2 lone pairs)

Why is the bond angle even smaller than in NH₃?

Now there are two lone pairs, each exerting strong LP–BP repulsion. The two bonding pairs are squeezed even more:

The cumulative effect of two lone pairs compresses the bond angle to 104.5° — a deviation of −5.0° from ideal.

Trend across the series:

Each additional lone pair compresses the bond angle further. This is one of the most fundamental trends in structural chemistry.

1.4 Why Not a Larger Deviation?

One might ask: if lone pairs repel so strongly, why doesn't the bond angle in water compress to, say, 90°? The answer is that as the bond angle decreases, the BP–BP repulsion increases (bonding pairs are forced closer together). An equilibrium is reached where the compressive force of LP–BP repulsion is balanced by the repulsive force of BP–BP repulsion at the smaller angle.

2.1 Boron Trifluoride (BF₃) — Perfect Trigonal Planar

       F
       |
   F — B — F        All bond angles = 120°

2.2 Sulfur Dioxide (SO₂) — Bent

   O = S = O         Bond angle O–S–O = 119.5°
      :
   (1 lone pair)

Deviation: 119.5° (−0.5° from ideal)

The deviation is smaller than in NH₃ (which also has 1 lone pair but in a tetrahedral arrangement) because:

2.3 Ozone (O₃) — Bent

   O = O — O         Bond angle O–O–O = 116.8°
      :
   (1 lone pair on central O)

The deviation is larger than in SO₂ because the bonding pairs in ozone are effectively single bonds (bond order 1.5), which are narrower than the double bonds in SO₂ (bond order ~2). Narrower bonding pairs leave more room for the lone pair to compress the angle.

2.4 General Trend for SN = 3

The trigonal bipyramidal geometry is unique because it has two distinct types of positions: axial and equatorial. This creates a rich set of possibilities for lone pair placement and bond angle deviations.

3.1 Phosphorus Pentachloride (PCl₅) — Perfect Trigonal Bipyramidal

         Cl (axial)
          |
  Cl — P — Cl (equatorial)     Axial–equatorial: 90°
         / \                    Equatorial–equatorial: 120°
        Cl   Cl                Axial–axial: 180°

3.2 Sulfur Tetrafluoride (SF₄) — Seesaw

         F (axial)
          |
   F — S — F (equatorial)      Lone pair in equatorial position
          |
         F (axial)

Why equatorial?

If the lone pair were axial:

If the lone pair is equatorial:

Since 90° interactions are the most destabilizing, the equatorial position (with only 2 such interactions) is preferred over the axial position (with 3 such interactions).

Bond angle deviations in SF₄:

The equatorial bond angle is compressed dramatically (from 120° to 101.5°) because the lone pair in the equatorial plane pushes the equatorial bonding pairs together. The axial–axial angle also bends slightly away from the lone pair.

3.3 Chlorine Trifluoride (ClF₃) — T-shaped

         F (axial)
          |
   F — Cl — F (equatorial)
          :
        (2 lone pairs in equatorial positions)

Why both equatorial?

With 2 lone pairs, the options are:

The equatorial arrangement minimizes 90° interactions.

Bond angle deviations in ClF₃:

Wait, that's not right. Let me reconsider. In ClF₃, the T-shape means:

The axial–equatorial angle (the angle between an axial F and the equatorial F) is about 87.5° (slightly less than 90° due to lone pair repulsion).

The angle between the two axial F atoms is approximately 175° (slightly less than 180°, bent away from the lone pairs).

Actually, let me reconsider the geometry more carefully. In a T-shaped molecule:

The F(axial)–Cl–F(equatorial) angle is slightly less than 90° (~87.5°).

The F(axial)–Cl–F(axial) angle is slightly less than 180° (~175°).

The lone pairs in the equatorial plane compress the axial bonds slightly toward each other (reducing the 180° angle) and also compress the axial-equatorial angle slightly below 90°.

3.4 Xenon Difluoride (XeF₂) — Linear

   F — Xe — F
         : : :
      (3 lone pairs in equatorial positions)

Bond angle: F–Xe–F = 180° (exact)

This is a case where the bond angle remains at the ideal value despite the presence of lone pairs. The reason is that the three lone pairs are all in the equatorial plane and repel each other symmetrically. The axial bonding pairs are pushed toward 180° by the equatorial lone pairs.

3.5 Summary for SN = 5

4.1 Sulfur Hexafluoride (SF₆) — Perfect Octahedral

4.2 Bromine Pentafluoride (BrF₅) — Square Pyramidal

         F (apical)
          |
   F — Br — F
   |         |
   F ——————— F
      (1 lone pair, trans to apical F)

Bond angle deviations:

The lone pair pushes all bonding pairs closer together, compressing the 90° angles.

4.3 Xenon Tetrafluoride (XeF₄) — Square Planar

   F       F
    \     /
     Xe          (2 lone pairs above and below the plane)
    /     \
   F       F

Bond angle: F–Xe–F = 90° (exact)

This is a remarkable case: despite having two lone pairs, the bond angles remain at the ideal values. Why?

The two lone pairs are trans to each other (180° apart). They push the four bonding pairs into the equatorial plane, but since the lone pair repulsion is symmetric (equal from above and below), the in-plane angles are not distorted.

This is analogous to XeF₂, where the symmetric arrangement of lone pairs preserves the ideal bond angles.

4.4 Comparison: BrF₅ vs. XeF₄

It seems paradoxical that adding a lone pair (going from BrF₅ to XeF₄) restores the ideal angle. The explanation is that the second lone pair in XeF₄ is placed trans to the first, creating a symmetric arrangement that does not distort the equatorial plane. In BrF₅, the single lone pair creates an asymmetric push that compresses the bonding pairs.

Beyond the primary effect of lone pairs, several other factors cause bond angles to deviate from ideal values.

As discussed extensively above, lone pairs compress bond angles. The magnitude of compression depends on:

1.1 Number of Lone Pairs

More lone pairs → greater compression:

1.2 Proximity of Lone Pairs to the Bond Angle

Lone pairs that are adjacent to (i.e., sharing a vertex with) the bond angle in question have the greatest effect. Lone pairs that are farther away have less effect.

In trigonal bipyramidal geometry, an equatorial lone pair compresses the equatorial–equatorial angle much more than the axial–axial angle, because the equatorial lone pair is directly between the two equatorial bonding pairs.

1.3 Electronegativity of the Central Atom

A more electronegative central atom holds its lone pairs more tightly (closer to the nucleus), making them effectively "larger" and more repulsive. This leads to greater bond angle compression.

Comparison: Group 15 trihydrides (EH₃)

Wait — this seems backwards! Nitrogen is more electronegative, yet its bond angle deviation is *smaller*. What's going on?

The explanation is subtle and involves two competing effects:

Effect 1: Lone pair size. A more electronegative central atom holds bonding electrons closer to itself, making the bonding pairs shorter and more concentrated. This increases BP–BP repulsion at small angles, which resists compression. The lone pair is also held more tightly, but the net effect is that the bonding pairs resist being squeezed.

Effect 2: Bonding pair electronegativity. When the central atom is very electronegative (N), the bonding pairs are pulled toward the center, making them more "lone-pair-like" (larger, more diffuse near the central atom). This increases BP–BP repulsion, which resists compression.

When the central atom is less electronegative (P, As, Sb), the bonding pairs are pulled away from the center toward the hydrogen atoms. This makes the bonding pairs smaller and less repulsive near the central atom, allowing the lone pair to compress the bond angle more freely. The bonding pairs behave more like "thin" balloons that can be squeezed together easily.

The result: Less electronegative central atoms show larger deviations from the tetrahedral angle, approaching 90° (the angle expected for pure *p*-orbital bonding with no *s*-character).

1.4 The "Pure p" Limit

An important theoretical limit: if the bonding used only *p* orbitals (with zero *s*-character), the bond angle would be 90°. The lone pair would occupy the *s* orbital (which is spherically symmetric and has no directional preference).

As the central atom becomes heavier and less electronegative (P → As → Sb), the bond angles approach 90°, suggesting that the bonding is increasingly described by pure *p* orbitals with the lone pair in the *s* orbital. This connects VSEPR to hybridization theory:

Group 16 dihydrides (EH₂) show the same trend:

Again, the bond angles decrease down the group, approaching 90°.

2.1 The General Rule

More electronegative peripheral (bonded) atoms cause the bond angle to decrease.

When the atoms bonded to the central atom are very electronegative, they pull the bonding electron density away from the central atom. This makes the bonding pairs:

The result is a smaller bond angle.

2.2 Example: Effect of Halide Substitution on Water

F₂O has a smaller angle than H₂O because fluorine is extremely electronegative — it pulls the bonding pairs away from oxygen, making them thinner and allowing the lone pairs to compress the angle further.

Cl₂O has a larger angle than H₂O because chlorine is less electronegative than fluorine and also has lone pairs that create steric repulsion with the oxygen lone pairs. The bulky chlorine atoms with their lone pairs push the O–Cl bonds apart.

2.3 Example: Nitrogen Trihalides (NX₃)

NF₃ has the smallest bond angle because fluorine is the most electronegative halogen — it pulls the bonding pairs away from nitrogen most effectively, reducing BP–BP repulsion and allowing greater compression by the lone pair.

2.4 Comparison: NH₃ vs. NF₃

This is a beautiful illustration of the electronegativity effect: the same central atom (N), the same number of lone pairs (1), but a 5.6° difference in bond angle due solely to the electronegativity of the peripheral atoms.

3.1 The General Rule

Multiple bonds (double and triple bonds) occupy more space than single bonds and exert greater repulsion.

A double bond has a larger electron density cloud than a single bond (4 electrons vs. 2 electrons in the bonding region). A triple bond is even larger (6 electrons). These larger electron domains repel other domains more strongly.

3.2 Effect on Bond Angles

When a molecule has both single bonds and multiple bonds, the multiple bonds "push" the single bonds closer together:

The angle between two single bonds is compressed when one or more double/triple bonds are present on the same central atom.

Conversely:

The angle involving a double bond is expanded at the expense of the angle between single bonds.

3.3 Example: Formaldehyde (CH₂O)

        O
        ‖
   H — C — H

Actual bond angles:

The C=O double bond is a larger electron domain than the C–H single bonds. It pushes the two C–H bonds closer together, compressing the H–C–H angle and expanding the H–C=O angles.

3.4 Example: Phosgene (COCl₂)

        O
        ‖
   Cl — C — Cl

Actual bond angles:

The C=O double bond compresses the Cl–C–Cl angle even more than in formaldehyde because the Cl atoms are larger than H atoms and the compression is amplified.

3.5 Example: Carbon Dioxide vs. Water

In CO₂, the two double bonds are equivalent and repel each other equally → 180°. In water, the two lone pairs dominate and compress the bond angle.

3.6 General Table: Effect of Multiple Bonds

4.1 The General Rule

Larger (bulkier) substituents on the central atom increase bond angles because they occupy more physical space and repel each other more strongly.

This effect is distinct from the electronegativity effect and is primarily steric in nature.

4.2 Example: Substituted Methanes

In symmetric molecules, steric effects don't change the angle because all substituents are identical. The effects become apparent when substituents differ in size.

4.3 Example: Trimethylamine vs. Tri-tert-butylamine

The bulky tert-butyl groups force the C–N–C angle to open up to relieve steric strain.

4.4 Example: Water vs. Dimethyl Ether

The methyl groups are bulkier than hydrogen atoms and push the C–O–C angle wider.

5.1 The Effect

When lone pairs on adjacent atoms (not the central atom) are oriented near each other, they create additional repulsion that can distort bond angles. This is distinct from the central atom lone pair effect.

5.2 Example: Hydrogen Peroxide (H₂O₂)

   H — O — O — H
      :     :

H₂O₂ has a dihedral angle of approximately 111.5° (not planar). The molecule adopts a skewed conformation to minimize lone pair–lone pair repulsion between the two oxygen atoms.

If the molecule were planar (cis or trans), the lone pairs on adjacent oxygens would be closer together, increasing repulsion. The skewed geometry allows the lone pairs to be farther apart.

5.3 Example: Hydrazine (N₂H₄)

   H₂N — NH₂

Similar to H₂O₂, hydrazine adopts a gauche conformation (dihedral angle ~90°) rather than a planar (trans) conformation, to minimize lone pair–lone pair repulsion between the two nitrogen atoms.

6.1 Cyclopropane (C₃H₆)

In cyclopropane, the C–C–C bond angle is forced to be 60° — far from the ideal tetrahedral angle of 109.5°. This enormous deviation creates severe ring strain (also called angle strain or Baeyer strain):

6.2 Cyclobutane (C₄H₈)

6.3 Cyclopentane (C₅H₁₀)

6.4 Cyclohexane (C₆H₁₂)

6.5 Summary of Ring Strain

7.1 Anomalous Geometries

Some molecules have geometries that cannot be explained by simple VSEPR:

7.2 Stereochemically Inactive Lone Pairs

In some heavy element compounds, the lone pair does not exert the expected steric effect:

The inert pair effect in heavy *p*-block elements (Tl, Pb, Bi) causes the *ns²* lone pair to become more tightly held and less stereochemically active, leading to bond angles that are closer to the ideal values than VSEPR would predict.

Henry Bent formulated a powerful rule in 1961 that connects hybridization, electronegativity, and bond angles:

"Atomic s-character concentrates in orbitals directed toward electropositive substituents, and p-character concentrates in orbitals directed toward electronegative substituents."

Since *s* orbitals are lower in energy and more tightly held, an atom preferentially directs *s*-rich hybrid orbitals toward less electronegative (more electropositive) substituents.

Consequences of Bent's Rule:

a) Bond angles open up when substituents are electropositive:

If the central atom directs more *s*-character toward one bond (because the substituent is electropositive), that bond has more *s*-character, which corresponds to a larger bond angle. The remaining bonds have more *p*-character and smaller angles between them.

b) Bond angles close when substituents are electronegative:

If the substituent is electronegative, the central atom directs more *p*-character toward it, making that bond more *p*-like and the angle involving it smaller.

Example: Application of Bent's Rule

Why is the H–O–H angle in water 104.5° instead of 90°?

If oxygen used pure *p* orbitals for bonding, the angle would be 90°. The lone pairs would be in the *s* orbital and one *p* orbital. But hydrogen is slightly electropositive relative to oxygen, so oxygen directs some *s*-character toward the O–H bonds (by Bent's rule). This increases the bond angle above 90°.

The hybridization of oxygen in water is approximately sp⁴ (or about 20% *s*-character in each bonding orbital), corresponding to a bond angle of about 104.5°.

Why is the F–O–F angle in OF₂ (103.1°) smaller than H–O–H (104.5°)?

Fluorine is more electronegative than hydrogen. By Bent's rule, oxygen directs more *p*-character toward the O–F bonds, making them more *p*-like. This reduces the bond angle.

Russell Dragо proposed an alternative model that emphasizes the role of electrostatic interactions and orbital overlap in determining bond angles. His approach is particularly useful for transition metal complexes and main-group compounds where VSEPR fails.

Drago argued that the VSEPR model overemphasizes lone pair effects and that in many cases, bond angle variations are better explained by changes in orbital overlap and electronegativity rather than by lone pair repulsion alone.

Modern computational chemistry can calculate bond angles with high accuracy using methods such as:

These calculations confirm the general trends predicted by VSEPR but provide quantitative accuracy that VSEPR cannot.

Example: Computed vs. Experimental Bond Angles

The agreement between high-level calculations and experiment is typically within 1°.

*NO₂ has an odd electron, making it a special case.

Molecules or ions with the same number of electrons and the same steric number tend to have similar geometries and bond angle deviations:

Note: H₃O⁺ has a bond angle of 113°, which is larger than NH₃'s 107.8° despite both having one lone pair. This is because oxygen is more electronegative than nitrogen, pulling the bonding pairs closer and making them more compact, which reduces LP–BP repulsion relative to BP–BP repulsion, allowing the angle to open up slightly.

Wait, that seems contradictory to what I said earlier. Let me reconsider.

Actually, H₃O⁺ has a bond angle of about 113°. The reason it's larger than NH₃ (107.8°) is that H₃O⁺ has a formal positive charge on oxygen, which effectively reduces the electron density in the bonding pairs. With less electron density, the BP–BP repulsion is reduced, and the LP–BP repulsion (which is still strong) pushes the bonds apart, opening the angle.

Hmm, actually I need to be more careful. Let me look at this again.

H₃O⁺: O has 3 bonds and 1 lone pair. The bond angle is about 113°.

NH₃: N has 3 bonds and 1 lone pair. The bond angle is about 107.8°.

The positive charge on H₃O⁺ means there's less electron density overall. The bonding pairs are more tightly held toward the oxygen (since O is more electronegative and the positive charge increases effective nuclear charge). This makes the bonding pairs more compact and concentrated near the oxygen, increasing their effective size near the central atom. This would increase BP–BP repulsion, opening the angle.

Alternatively, the positive charge reduces the total number of electrons, making the lone pair less diffuse and reducing LP–LP and LP–BP repulsions relative to BP–BP repulsions.

In any case, the comparison is instructive.

Walsh diagrams (A.D. Walsh, 1953) provide a molecular orbital perspective on bond angles. They show how MO energies change as a function of bond angle.

For AH₂ molecules (like H₂O, H₂S):

For molecules with 8 valence electrons (H₂O), the Walsh diagram predicts a bent geometry with an angle around 104-105°. For molecules with fewer valence electrons (BeH₂, 4 valence electrons), the Walsh diagram predicts a linear geometry.

This MO-based approach provides a more rigorous foundation for understanding bond angles than VSEPR alone.

In trigonal bipyramidal molecules like PF₅, the axial and equatorial positions are not permanently fixed. Through a process called Berry pseudorotation, the molecule can interconvert axial and equatorial positions via a square pyramidal transition state:

Trigonal bipyramidal → Square pyramidal (TS) → Trigonal bipyramidal (with swapped axial/equatorial)

This process has a very low activation energy (~15 kJ/mol for PF₅) and occurs rapidly at room temperature. It explains why, for example, the ¹⁹F NMR spectrum of PF₅ shows only one signal at room temperature (all five fluorines are equivalent on the NMR timescale due to rapid pseudorotation) but two signals at low temperature (axial and equatorial fluorines become distinguishable when pseudorotation is frozen out).

For molecules with lone pairs (like SF₄), Berry pseudorotation also occurs but is modified by the preference of lone pairs for equatorial positions.

1.1 Transition Metal Complexes

VSEPR is designed for main-group compounds. Transition metal complexes with partially filled *d* orbitals often have geometries that VSEPR cannot predict:

1.2 Electron-Deficient Compounds

Molecules like B₂H₆ (diborane) have bridging hydrogen atoms with 3-center 2-electron bonds. VSEPR cannot handle these bonding arrangements.

1.3 Very Heavy Elements

As discussed above, relativistic effects in heavy elements (6th period and beyond) can make lone pairs stereochemically inactive, leading to geometries that VSEPR does not predict.

1.4 Weakly Bound Complexes

Van der Waals complexes and very weakly bound species may have geometries determined by dispersion forces rather than electron pair repulsion.

VSEPR and hybridization theory are complementary:

The connection is made through Bent's rule: lone pairs preferentially occupy orbitals with higher *s*-character (because *s* orbitals are lower in energy and closer to the nucleus). This leaves the bonding orbitals with more *p*-character, which corresponds to smaller bond angles.

For example, in water:

Step 1: Determine the steric number.

Step 2: Determine the electron domain geometry.

Step 3: Determine where the lone pair goes.

Step 4: Predict the molecular geometry.

Step 5: Predict bond angle deviations.

Experimental values: Eq–Eq: ~103°, Ax–Ax: ~172°

Additional factor: Se is less electronegative than S, so the bonding pairs are pulled slightly more toward F, reducing BP–BP repulsion and allowing slightly more compression than in SF₄.

Both molecules:

NH₃:

NF₃:

By Bent's rule: In NH₃, nitrogen directs more *s*-character toward H (electropositive), increasing the bond angle. In NF₃, nitrogen directs more *p*-character toward F (electronegative), decreasing the bond angle.

Both molecules:

H₂O:

H₂S:

Step 1: Count valence electrons.

Step 2: Determine bonding.

Step 3: Geometry.

Step 4: Bond angle.

In XeF₂:

The 3 equatorial lone pairs repel each other symmetrically (each pair at 120° from the others in the equatorial plane). They push the axial bonding pairs toward 180°. Since the lone pair arrangement is symmetric with respect to the axial axis, there is no asymmetric force to bend the F–Xe–F angle away from 180°.

This is a case where the lone pairs enforce the ideal angle rather than distorting it.

NO₂ is an odd-electron molecule (17 valence electrons):

This is a case where multiple bond repulsion dominates over lone pair compression.

Bond angle deviations in VSEPR theory are not random perturbations — they are systematic, predictable consequences of the fundamental principle that electron domains repel each other with different strengths depending on their nature. The hierarchy LP–LP > LP–BP > BP–BP, combined with the effects of electronegativity, bond order, steric bulk, and ring strain, provides a remarkably powerful framework for understanding molecular geometry.

The key takeaways are:

These principles, applied systematically, allow chemists to predict and rationalize the three-dimensional structures of molecules with confidence — a skill that underlies all of structural chemistry, from simple inorganic compounds to the complex architectures of biological macromolecules.

Key References: