2.3 Chemical Bonding Models: Lewis Structures, VSEPR Theory, and Molecular Polarity

Key Takeaways

  • Lattice energy quantifies ionic stability; the Born-Haber cycle uses Hess's law to determine experimental lattice energies, while the Kapustinskii equation provides a semi-empirical calculation based on ionic radii and formal charges without crystal symmetry parameters.

  • Formal charge (FC=V−N−B/2FC = V - N - B/2) guides optimal Lewis structures; hypervalent species of Period 3 and below reflect polar covalent multicenter bonding (such as 3-center 4-electron systems) rather than substantial dd-orbital participation.

  • VSEPR theory predicts molecular geometries from steric numbers (2 to 7) based on electron pair repulsion hierarchies: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair.

  • Trigonal bipyramidal arrangements (SN=5SN = 5) place lone pairs preferentially in equatorial positions to minimize destabilizing 90∘90^\circ interactions, producing seesaw, T-shaped, and linear geometries.

  • Molecular polarity is the vector sum of individual bond dipoles; high molecular symmetry (D∞h,Td,D3h,Oh,D4hD_{\infty h}, T_d, D_{3h}, O_h, D_{4h}) cancels dipole vectors to yield zero net dipole moment even in molecules with highly polar bonds.

Last updated: October 2026

Chemical Bonding Models: Lewis Structures, VSEPR Theory, and Molecular Polarity

Chemical bonding lies at the heart of inorganic structure and reactivity. The PRC Chemist Licensure Examination tests candidates on both thermodynamic models of bonding (Born-Haber cycles, lattice energies, and the Kapustinskii equation) and structural models (Lewis structures, formal charges, resonance, VSEPR geometry predictions, and molecular symmetry).


1. Ionic Bonding Energetics and Lattice Energy

An ionic bond consists of non-directional electrostatic attractions between oppositely charged spherical ions arranged in an extended crystalline lattice. The thermodynamic stability of an ionic solid is quantified by its lattice energy (U0U_0).

Definitions and Conventions

  • Lattice Enthalpy of Formation (ΔHlatt∘\Delta H^\circ_{latt}): Enthalpy change when 1 mole of an ionic crystalline solid is formed from its constituent gaseous ions (exothermic, ΔHlatt∘<0\Delta H^\circ_{latt} < 0): Mz+(g)+Xz−(g)→MX(s)ΔHlatt∘<0M^{z+}(g) + X^{z-}(g) \rightarrow MX(s) \quad \Delta H^\circ_{latt} < 0
  • Lattice Dissociation Energy (U0U_0): Energy required to separate 1 mole of solid into infinitely separated gaseous ions (endothermic, U0=−ΔHlatt∘>0U_0 = -\Delta H^\circ_{latt} > 0).

The Born-Haber Thermochemical Cycle

Because lattice energy cannot be measured directly by experiment, it is calculated from accessible thermodynamic quantities using a Born-Haber cycle based on Hess's Law of constant heat summation:

ΔHf∘=ΔHsub∘(M)+IE1(M)+12ΔHdiss∘(X2)+ΔHEA∘(X)+ΔHlatt∘\Delta H^\circ_f = \Delta H^\circ_{sub}(M) + IE_1(M) + \frac{1}{2} \Delta H^\circ_{diss}(X_2) + \Delta H^\circ_{EA}(X) + \Delta H^\circ_{latt}

Worked Thermodynamic Calculation: Sodium Chloride (NaClNaCl) Given experimental standard enthalpies at 298 K298\text{ K}:

  1. Standard enthalpy of formation of NaCl(s)NaCl(s): ΔHf∘=−411.2 kJ/mol\Delta H^\circ_f = -411.2\text{ kJ/mol}
  2. Enthalpy of sublimation of sodium: Na(s)→Na(g)ΔHsub∘=+107.3 kJ/molNa(s) \rightarrow Na(g) \quad \Delta H^\circ_{sub} = +107.3\text{ kJ/mol}
  3. First ionization energy of sodium: Na(g)→Na+(g)+e−IE1=+495.8 kJ/molNa(g) \rightarrow Na^+(g) + e^- \quad IE_1 = +495.8\text{ kJ/mol}
  4. Bond dissociation enthalpy of chlorine: 12Cl2(g)→Cl(g)12D(Cl2)=12(+242.6)=+121.3 kJ/mol\frac{1}{2} Cl_2(g) \rightarrow Cl(g) \quad \frac{1}{2} D(Cl_2) = \frac{1}{2}(+242.6) = +121.3\text{ kJ/mol}
  5. Electron affinity of chlorine: Cl(g)+e−→Cl−(g)ΔHEA∘=−348.6 kJ/molCl(g) + e^- \rightarrow Cl^-(g) \quad \Delta H^\circ_{EA} = -348.6\text{ kJ/mol}

Applying Hess's Law: −411.2=107.3+495.8+121.3−348.6+ΔHlatt∘-411.2 = 107.3 + 495.8 + 121.3 - 348.6 + \Delta H^\circ_{latt} −411.2=+375.8+ΔHlatt∘-411.2 = +375.8 + \Delta H^\circ_{latt} ΔHlatt∘=−411.2−375.8=−787.0 kJ/mol\Delta H^\circ_{latt} = -411.2 - 375.8 = -787.0\text{ kJ/mol} Thus, the lattice dissociation energy of NaClNaCl is U0=+787.0 kJ/molU_0 = +787.0\text{ kJ/mol}.

Theoretical Models: Born-Landé and Kapustinskii Equations

  1. The Born-Landé Equation: Accounts for attractive Coulombic forces and short-range Born repulsive forces: U0=NAM∣z+z−∣e24πε0r0(1−1n)U_0 = \frac{N_A M |z_+ z_-| e^2}{4\pi \varepsilon_0 r_0} \left(1 - \frac{1}{n}\right) where NAN_A is Avogadro's number, MM is the geometric Madelung constant (1.74761.7476 for NaClNaCl rock-salt structure; 1.76271.7627 for CsClCsCl; 1.63811.6381 for zinc blende; 1.6411.641 for wurtzite), z+z_+ and z−z_- are ionic charge numbers, r0r_0 is equilibrium internuclear separation, and nn is the Born exponent (55 to 1212).

  2. The Kapustinskii Equation: Enables calculation of lattice energy without knowing the crystal structure or Madelung constant by approximating MM as a function of the number of ions: U0=1.202×105⋅ν⋅∣z+z−∣r++r−(1−34.5r++r−)[kJ/mol]U_0 = \frac{1.202 \times 10^5 \cdot \nu \cdot |z_+ z_-|}{r_+ + r_-} \left(1 - \frac{34.5}{r_+ + r_-}\right) \quad \text{[kJ/mol]} where ν\nu is the number of ions per formula unit (e.g., ν=2\nu = 2 for NaClNaCl, ν=3\nu = 3 for CaCl2CaCl_2) and ionic radii r+,r−r_+, r_- are entered in picometers (pm\text{pm}).

Worked Example: Comparing NaClNaCl vs MgOMgO (rMg2+=72 pm,rO2−=140 pmr_{Mg^{2+}} = 72\text{ pm}, r_{O^{2-}} = 140\text{ pm}, sum =212 pm= 212\text{ pm}): U0(MgO)=1.202×105⋅2⋅(2×2)212(1−34.5212)=961600212(1−0.1627)=4535.8×0.8373≈3798 kJ/molU_0(MgO) = \frac{1.202 \times 10^5 \cdot 2 \cdot (2 \times 2)}{212} \left(1 - \frac{34.5}{212}\right) = \frac{961600}{212} (1 - 0.1627) = 4535.8 \times 0.8373 \approx 3798\text{ kJ/mol} Because lattice energy scales with ∣z+z−∣|z_+ z_-|, the quadrupled charge product in MgOMgO (∣(+2)(−2)∣=4|(+2)(-2)| = 4 versus ∣(+1)(−1)∣=1|(+1)(-1)| = 1 in NaClNaCl) increases lattice energy nearly five-fold, producing refractory behavior (melting point of MgO=2852∘CMgO = 2852^\circ\text{C} vs NaCl=801∘CNaCl = 801^\circ\text{C}).


2. Covalent Bonding, Formal Charges, and Modern Hypervalence

Formal Charge Calculation and Resonance

The formal charge (FCFC) represents the hypothetical electric charge assigned to an atom in a molecule, assuming shared bonding electrons are split equally between bonded atoms:

FC=V−N−B2FC = V - N - \frac{B}{2}

where VV is valence electrons in the isolated neutral atom, NN is number of nonbonding valence electrons (lone pair electrons), and BB is total number of shared bonding electrons (so B/2B/2 equals number of covalent bonds).

Rules for Selecting Optimal Resonance Contributors:

  1. Structures with formal charges closest to zero are most stable.
  2. Negative formal charges must reside on the most electronegative atoms.
  3. Structures with adjacent like charges (e.g., +1+1 adjacent to +1+1) are severely destabilized.

Case Study: The Thiocyanate Ion ([SCN]−[SCN]^-): Total valence electrons =6(S)+4(C)+5(N)+1(charge)=16= 6(S) + 4(C) + 5(N) + 1(\text{charge}) = 16.

  • Form I: [:S¨−C≡N:]−  ⟹  FC(S)=6−6−1=−1;  FC(C)=4−0−4=0;  FC(N)=5−2−3=0[:\ddot{S}-C\equiv N:]^- \implies FC(S) = 6 - 6 - 1 = -1; \; FC(C) = 4 - 0 - 4 = 0; \; FC(N) = 5 - 2 - 3 = 0
  • Form II: [:S¨=C=N¨:]−  ⟹  FC(S)=6−4−2=0;  FC(C)=4−0−4=0;  FC(N)=5−4−2=−1[:\ddot{S}=C=\ddot{N}:]^- \implies FC(S) = 6 - 4 - 2 = 0; \; FC(C) = 4 - 0 - 4 = 0; \; FC(N) = 5 - 4 - 2 = -1
  • Form III: [:S≡C−N¨:2−]−  ⟹  FC(S)=6−2−3=+1;  FC(C)=4−0−4=0;  FC(N)=5−6−1=−2[:S\equiv C-\ddot{N}:^{2-}]^- \implies FC(S) = 6 - 2 - 3 = +1; \; FC(C) = 4 - 0 - 4 = 0; \; FC(N) = 5 - 6 - 1 = -2 Evaluation: Form II is the major resonance contributor because Nitrogen is significantly more electronegative (χ=3.04\chi = 3.04) than Sulfur (χ=2.58\chi = 2.58). Placing the negative charge on Nitrogen (Form II) is energetically superior to Sulfur (Form I). Form III is negligible due to charge separation (+1+1 on S, −2-2 on N).

Modern Interpretation of Hypervalence (3c-4e Bonding)

Historically, Lewis octet expansion in Period 3+ elements (PCl5,SF6,XeF2PCl_5, SF_6, XeF_2) was rationalized via Pauling's sp3dsp^3d and sp3d2sp^3d^2 hybridization invoking outer dd-orbitals. However, modern quantum chemical calculations show that 3d3d orbitals in phosphorus and sulfur are far too high in energy and radially diffuse to engage in covalent bonding.

Instead, hypervalent bonding is explained by 3-center 4-electron (3c-4e) bonding (Pimentel-Rundle model):

  • In linear XeF2XeF_2 (AX2E3AX_2E_3), three equatorial lone pairs occupy sp2sp^2-like hybrid orbitals in the equatorial plane.
  • The axial F−Xe−FF-Xe-F linkage is formed by the linear overlap of one unhybridized 5pz5p_z orbital on XeXe with a 2pz2p_z orbital on each FF atom.
  • This linear combination generates three molecular orbitals: one bonding (σ\sigma), one nonbonding (σnb\sigma_{nb}), and one antibonding (σ∗\sigma^*): ψb=12ϕF1+12ϕXe+12ϕF2\psi_b = \frac{1}{2} \phi_{F1} + \frac{1}{\sqrt{2}} \phi_{Xe} + \frac{1}{2} \phi_{F2} ψnb=12(ϕF1−ϕF2)[density localized entirely on fluorine atoms]\psi_{nb} = \frac{1}{\sqrt{2}} (\phi_{F1} - \phi_{F2}) \quad \text{[density localized entirely on fluorine atoms]}
  • Four electrons occupy ψb\psi_b and ψnb\psi_{nb}. Because the nonbonding orbital has nodes at Xenon, the net bonding electron pair is delocalized over two bonds, giving each Xe−FXe-F bond a bond order of 0.50.5 and placing significant partial negative charge on the electronegative Fluorine atoms without requiring dd-orbital participation!

3. Valence Shell Electron Pair Repulsion (VSEPR) Theory

VSEPR theory posits that valence electron pairs surrounding a central atom repel one another electrostatically and adopt a three-dimensional spatial orientation that maximizes their mutual angular separation.

Electron Pair Repulsion Hierarchy

Lone Pair - Lone Pair (lp-lp)>Lone Pair - Bonding Pair (lp-bp)>Bonding Pair - Bonding Pair (bp-bp)\text{Lone Pair - Lone Pair (lp-lp)} > \text{Lone Pair - Bonding Pair (lp-bp)} > \text{Bonding Pair - Bonding Pair (bp-bp)} Physical Rationale: A lone pair is electrostatically attracted to only one atomic nucleus; its electron cloud is broad, diffuse, and concentrated close to the central atom. In contrast, a bonding pair is stretched between two positive nuclei, confining its electron density to an internuclear cylinder. Consequently, lone pairs exert greater angular repulsive pressure, compressing adjacent bond angles:

  • CH4CH_4 (AX4AX_4, no lone pairs): ideal tetrahedral angle =109.5∘= 109.5^\circ
  • NH3NH_3 (AX3EAX_3E, 1 lone pair): compressed to 107.3∘107.3^\circ
  • H2OH_2O (AX2E2AX_2E_2, 2 lone pairs): compressed to 104.5∘104.5^\circ

Steric Number (SN) and Complete Geometry Classification

The Steric Number is defined as: SN=(number of bonded atoms)+(number of lone pairs on central atom)SN = (\text{number of bonded atoms}) + (\text{number of lone pairs on central atom})

SNSNElectron GeometryLone PairsVSEPR FormulaMolecular GeometryIdeal AnglesExamples
2Linear0AX2AX_2Linear180∘180^\circBeCl2,CO2,HCN,[N3]−BeCl_2, CO_2, HCN, [N_3]^-
3Trigonal Planar0AX3AX_3Trigonal Planar120∘120^\circBF3,SO3,[NO3]−,[CO3]2−BF_3, SO_3, [NO_3]^-, [CO_3]^{2-}
1AX2EAX_2EBent<120∘  (≈119∘)<120^\circ \; (\approx 119^\circ)SO2,[NO2]−,O3SO_2, [NO_2]^-, O_3
4Tetrahedral0AX4AX_4Tetrahedral109.5∘109.5^\circCH4,CCl4,[NH4]+,[SO4]2−CH_4, CCl_4, [NH_4]^+, [SO_4]^{2-}
1AX3EAX_3ETrigonal Pyramidal<109.5∘  (≈107∘)<109.5^\circ \; (\approx 107^\circ)NH3,PCl3,[H3O]+NH_3, PCl_3, [H_3O]^+
2AX2E2AX_2E_2Bent<109.5∘  (≈104.5∘)<109.5^\circ \; (\approx 104.5^\circ)H2O,OF2,SCl2,[NH2]−H_2O, OF_2, SCl_2, [NH_2]^-
5Trigonal Bipyramidal0AX5AX_5Trigonal Bipyramidal120∘ eq,90∘ ax120^\circ \text{ eq}, 90^\circ \text{ ax}PCl5,PF5,AsF5PCl_5, PF_5, AsF_5
1AX4EAX_4ESeesaw<120∘,<90∘,<180∘<120^\circ, <90^\circ, <180^\circSF4,SeF4,TeCl4SF_4, SeF_4, TeCl_4
2AX3E2AX_3E_2T-shaped<90∘  (≈87.5∘)<90^\circ \; (\approx 87.5^\circ)ClF3,BrF3,[ICl3]ClF_3, BrF_3, [ICl_3]
3AX2E3AX_2E_3Linear180∘180^\circXeF2,[I3]−,[ICl2]−XeF_2, [I_3]^-, [ICl_2]^-
6Octahedral0AX6AX_6Octahedral90∘,180∘90^\circ, 180^\circSF6,[PF6]−,[SiF6]2−SF_6, [PF_6]^-, [SiF_6]^{2-}
1AX5EAX_5ESquare Pyramidal<90∘  (≈85∘)<90^\circ \; (\approx 85^\circ)BrF5,IF5,[SbF5]2−BrF_5, IF_5, [SbF_5]^{2-}
2AX4E2AX_4E_2Square Planar90∘,180∘90^\circ, 180^\circXeF4,[ICl4]−XeF_4, [ICl_4]^-
7Pentagonal Bipyramidal0AX7AX_7Pentagonal Bipyramidal72∘ eq,90∘ ax72^\circ \text{ eq}, 90^\circ \text{ ax}IF7IF_7
2AX5E2AX_5E_2Pentagonal Planar72∘72^\circ[XeF5]−[XeF_5]^-

The Steric Number 5 Equatorial Preference Rule

In a trigonal bipyramid, positions are non-equivalent:

  • An axial position has three 90∘90^\circ neighbors (high repulsive interaction).
  • An equatorial position has only two 90∘90^\circ neighbors and two 120∘120^\circ neighbors.

Because electron repulsion decreases precipitously with angle (repulsion at 90∘≫90^\circ \gg repulsion at 120∘120^\circ), lone pairs ALWAYS occupy equatorial positions in SN=5SN=5!

  • 1 lone pair (AX4EAX_4E, e.g., SF4SF_4): lone pair is equatorial   ⟹  \implies Seesaw.
  • 2 lone pairs (AX3E2AX_3E_2, e.g., ClF3ClF_3): both lone pairs are equatorial   ⟹  \implies T-shaped.
  • 3 lone pairs (AX2E3AX_2E_3, e.g., XeF2XeF_2): all three lone pairs reside in the equatorial plane, cancelling repulsions   ⟹  \implies Linear (180∘180^\circ).

4. Molecular Polarity and Symmetry

A chemical bond between atoms of differing electronegativity possesses a bond dipole moment: μ=q⋅r\mu = q \cdot r measured in Debye (1 D=3.336×10−30 C⋅m1\text{ D} = 3.336 \times 10^{-30}\text{ C}\cdot\text{m}). The molecular dipole moment μ⃗net\vec{\mu}_{net} is the vector sum of all individual bond dipoles and lone-pair dipole contributions: μ⃗net=∑iμ⃗i\vec{\mu}_{net} = \sum_{i} \vec{\mu}_i

A molecule has a net dipole moment (is polar, μ>0\mu > 0) only if its 3D geometry lacks symmetry elements that invert or cancel dipole vectors:

  • Nonpolar Molecules (μ=0\mu = 0): High symmetry cancels all bond dipoles:
    • Linear AX2AX_2 (CO2,BeCl2,CS2CO_2, BeCl_2, CS_2)
    • Trigonal planar AX3AX_3 (BF3,SO3BF_3, SO_3)
    • Tetrahedral AX4AX_4 (CH4,CCl4CH_4, CCl_4)
    • Trigonal bipyramidal AX5AX_5 (PCl5PCl_5)
    • Linear AX2E3AX_2E_3 (XeF2XeF_2)
    • Octahedral AX6AX_6 (SF6SF_6)
    • Square planar AX4E2AX_4E_2 (XeF4XeF_4)
  • Polar Molecules (μ>0\mu > 0): Asymmetric distributions of dipoles and lone pairs:
    • Bent AX2EAX_2E (SO2,μ=1.63 DSO_2, \mu = 1.63\text{ D})
    • Trigonal pyramidal AX3EAX_3E (NH3,μ=1.47 DNH_3, \mu = 1.47\text{ D}; PCl3PCl_3 is also polar)
    • Bent AX2E2AX_2E_2 (H2O,μ=1.85 DH_2O, \mu = 1.85\text{ D}; OF2,μ=0.30 DOF_2, \mu = 0.30\text{ D})
    • Seesaw AX4EAX_4E (SF4,μ=0.63 DSF_4, \mu = 0.63\text{ D})
    • T-shaped AX3E2AX_3E_2 (ClF3,μ=0.56 DClF_3, \mu = 0.56\text{ D})
    • Square pyramidal AX5EAX_5E (BrF5,μ=1.51 DBrF_5, \mu = 1.51\text{ D})
Loading diagram...
VSEPR Geometries Derived from Steric Number 5 (Trigonal Bipyramidal)
Test Your Knowledge

What is the molecular geometry and approximate bond angle of Chlorine Trifluoride (ClF3), and what theoretical principle determines this shape?

A

Trigonal pyramidal with bond angles of 107 degrees, because three bonding pairs experience equal repulsive forces from a single axial lone pair.

B

Seesaw with bond angles compressed to 102 degrees, because one equatorial lone pair repels four axial bonding pairs.

C

T-shaped, about 87.5 degrees, because both lone pairs take equatorial positions of a trigonal bipyramid.

D

Trigonal planar with 120-degree bond angles, because the three fluorine atoms occupy equatorial positions in a planar sp2 framework.

Test Your Knowledge

Which of the following ionic solids is predicted by the Kapustinskii and Born-Landé equations to have the largest magnitude of lattice energy?

A

Barium oxide (BaO)

B

Magnesium oxide (MgO)

C

Sodium chloride (NaCl)

D

Calcium chloride (CaCl2)

Test Your Knowledge

When evaluating resonance contributors for the sulfate ion (SO42-), what distinguishes the expanded octet structure containing two S=O double bonds from the octet-compliant structure with four S-O single bonds?

A

The octet-compliant structure with four single bonds gives sulfur a formal charge of 0, making it thermodynamically superior to expanded octet structures.

B

The sulfate ion can only be represented with four equivalent double bonds, giving sulfur a formal charge of -2.

C

The two-double-bond structure gives sulfur a formal charge of 0, while the all-single-bond structure gives +2.

D

Placing all negative formal charges on the central sulfur atom stabilizes the resonance hybrid because sulfur is less electronegative than oxygen.

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