Chapter 3. Valence Shell Electron Pair Repulsion Theory (VSEPR)
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1. VSEPR
1. VSEPR (Valence Shell Electron Pair Repulsion Model)
⑴ Overview: Lewis Structures - Valence Shell Electron Pair Repulsion Theory (VSEPR) - Valence Bond Theory - Molecular Orbital Theory
① Model for predicting the three-dimensional structure of molecules using the principle of electron-electron repulsion.
② Three-dimensional model
⑵ Rules
① 1st. Arrange shared and unshared electron pairs as far apart as possible to minimize electron-pair repulsion.
② 2nd. Treat double and triple bonds as a single unit.
③ 3rd. Consider unshared electron pairs equivalent to atoms; a single unpaired electron is equivalent to a single unshared electron pair.
④ 4th. Unshared electron pairs exert greater repulsion than shared electron pairs, resulting in smaller bond angles on the opposite side of the unshared pair.
○ CH4: C-H bond angle is 109.5°
○ NH3: N-H bond angle is 107°
○ H2O: O-H bond angle is 105°
⑤ 5th. By considering atomic size, electronegativity, and more, detailed molecular structures can be predicted.
○ VSEPR Rule: Increasing bond polarity → Decreasing atomic distance → Increasing repulsion → Increasing bond angle
○ Example: N is smaller than P, leading to larger electron repulsion and greater bond angles in NH3 compared to PH3
⑶ Steric Number
① Definition: Number of atoms bonded to the central atom + Number of unshared electron pairs on the central atom
② Steric Number 2: sp
○ Linear (Example: BeF2, CO2, C2H2): Bond angle of 180°
Figure 1. Structure of BeF2
③ Steric Number 3: sp2
○ Trigonal planar (Example: BF3, C2H4): Bond angle of 120°
Figure 2. Structure of BF3
○ Bent shape: When there is one pair of unshared electrons
④ Steric Number 4: sp3
○ Tetrahedral (Example: CH4): Tetrahedral angle, so bond angle is 109.5°
○ Trigonal pyramidal: When there is one pair of unshared electrons (Example: NH3)
○ Bent shape: When there are two pairs of unshared electrons (Example: H2O)
⑤ Steric Number 5: sp3d. Unshared electron pairs cause significant repulsion, so they are placed in the horizontal plane to minimize repulsion.
○ Trigonal bipyramidal
○ See-saw shape: When there is one pair of unshared electrons, they lie in the horizontal plane
○ T-shaped: When there are two pairs of unshared electrons, they lie in the horizontal plane
○ Linear shape: When there are three pairs of unshared electrons, they lie in the horizontal plane (Example: ICl2-, XeF2)
Figure 3. Structure of XeF2
⑥ Steric Number 6: sp3d2. Unshared electron pairs cause significant repulsion, so they are placed in the vertical direction to minimize repulsion.
○ Octahedral
○ Square pyramid: When there is one pair of unshared electrons, they lie in the vertical direction (Example: SF5-, IF5)
○ Planar square: When there are two pairs of unshared electrons, they lie in the vertical direction (Example: XeF4, SF42-)
Figure 4. Structure of XeF4
⑷ Cautionary Notes
① In cases where the steric number is 5, the vertically oriented bonding electron pairs experience greater repulsion than the horizontally oriented bonding electron pairs, resulting in longer bond lengths.
② In cases where the steric number is 6, the concepts of vertical and horizontal orientation do not apply, so the discussion in ① does not hold.
③ Occurrence of violations of the octet rule
○ Reason for satisfying the octet rule: Violating the octet rule in the second and third period electron shells leads to increased electron repulsion.
○ Larger atoms allocate more space to outer electrons, allowing them to be stable without forming octets.
○ Second-period atoms lack d orbitals, preventing extension of the octet rule.
○ Third-period atoms have d orbitals, allowing extension of the octet rule: PCl5 (trigonal bipyramidal), SF4 (seesaw), ClF3 (T-shaped), I3- (linear)
Figure 5. Structure of PCl5
Figure 6. Structure of SF4
Figure 7. Structure of ClF3
ClF2- also has a similar structure.
Figure 8. Structure of I3-
○ Even if the octet rule is not satisfied, the number of valence electrons must still be satisfied.
2. Dipole Moment
⑴ Overview
① Dipole: A state where equal and opposite charges are placed at two points separated by a certain distance.
② Dipole moment (Double Dipole Moment)
○ Represents the polarity of a chemical bond with a bond moment.
○ Indicates the extent of electron pair distortion.
○ For example, two atoms with different electronegativities forming a bond will have a dipole moment.
③ Formula: Dipole Moment (Unit: D) = μ = q × d
○ q: Charge of the atom
○ d: Distance between charges, i.e., distance between δ+ and δ-
○ Dipole moment is also influenced by the difference in electronegativity and the distance between nuclei.
④ Dipole moments are represented by arrows ⤉
○ In physics, the direction of the dipole moment is from the (-) pole to the (+) pole.
Figure 9. Dipole Moment in Physics (indicated by p)
○ In chemistry, the direction of the dipole moment is from the (+) pole to the (-) pole, reflecting the arrangement of atoms with nuclei at the center.
⑤ Unit: D (debye)
○ 1 debye = 10-18 esu·cm
○ 1 e- = 4.80 × 10-10 esu
⑵ Major Bond Dipole Moments
Table. 1. Major Bond Dipole Moments
⑶ Molecular Dipole Moment
① Definition: The sum of all bond dipole moments in a molecule.
② Considerations when calculating
○ All polar bonds and the direction of their polarity
○ Count the number of axes centered around individual atoms to determine the molecular geometry.
○ Evaluate whether dipole moments reinforce or cancel each other out in individual spatial regions.
③ Trends
○ Nonpolar molecules: μ = 0
○ Charge is distributed evenly throughout the molecule and reacts to external electric fields.
○ Examples: CH4 (methane), C2H6 (ethane), C6H6 (benzene), CO2, BF3, CCl4, CnH2n+2 (alkane), 1,4-dichlorobenzene
○ Polar molecules: μ ≠ 0
○ Charge is skewed towards one side, resulting in δ+ and δ- regions within the molecule.
○ Alignment of the molecule occurs in response to external electric fields.
○ Polar molecules readily dissolve in other polar molecules.
○ Example 1: H2O (water): μ = 1.85 D
○ Example 2: CH3OH (methanol): μ = 1.70 D
○ Example 3: NH3 (ammonia): μ = 1.47 D
○ Example 4: imidazole: μ = 5.6 D
○ Example 5: oxazole: μ = 1.4 D
○ Example 6: thiazole: μ = 1.6 D
○ Example 7: C2F2H2 is nonpolar in the trans configuration and polar in the cis configuration.
○ Effective resonance reduces dipole moment since formal charges are distributed.
○ Examples: chlorobenzene < chlorohexane
○ Dipole moment increases due to electron delocalization caused by aromaticity and other factors.
○ Generally, higher dipole moments correspond to higher boiling points.
○ Exception: carbon tetrachloride (μ = 0, bp = 77°C) > chloroform (μ = 1.0 D, bp = 62°C)
○ Reason: Carbon tetrachloride’s van der Waals forces are stronger.
④ Caution 1: Consideration of Hyperconjugation
○ Carbon-carbon bonds have zero electronegativity difference, but hyperconjugation allows sp3 carbon to donate electrons to sp2 carbon.
○ sp3 carbon cannot donate electrons to other sp3 carbon.
○ μ for R-X: R-Cl > R-F > R-Br > R-I
○ μ for HX: HF > HCl > HBr > HI
○ Boiling point: HF (19.5°C) > HI (-35.36°C) > HBr (-66°C) > HCl (-85.05°C)
○ Reason for HI > HBr > HCl: Polarizability
○ Reason for higher boiling point of HF: Hydrogen bonding
⑤ Caution 2: Dipole Moments in Aromatic Heterocyclic Compounds
Figure 10. Dipole Moments in Heterocyclic Compounds
○ Special attention should be paid to pyrrole (top left): Pyrrole’s nitrogen acts as an electron donor.
⑷ Python code for calculating dipole moments
Input: 2019-01-02 20:31
Revised: 2022-02-02 21:21