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VSEPR and Bond Hybridization- AP Chemistry Notes- New Syllabus 2024-2025

VSEPR and Bond Hybridization study Notes – AP Chemistry

VSEPR and Bond Hybridization study Notes – AP Chemistry – per latest AP Chemistry Syllabus.

LEARNING OBJECTIVE

    • Based on the relationship between Lewis diagrams, VSEPR theory, bond orders, and bond polarities:
      i. Explain structural properties of molecules.
      ii. Explain electron properties of molecules.

Key Concepts:

    • VSEPR Theory and Electron Pair Repulsions
    • Lewis Diagrams and VSEPR Predictions
      i. Molecular geometry
      ii. Bond angles
      iii. Relative bond energies
      iv. Relative bond lengths
      v. Dipole moment
      vi. Hybridization
    • Hybridization and Ideal Bond Angles
    • Orbital Overlap, Sigma and Pi Bonds

AP Chemistry-Concise Summary Notes- All Topics

VSEPR Theory and Electron Pair Repulsions

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the 3D arrangement of atoms in a molecule by minimizing repulsions between regions of electron density (bonding and lone pairs) around the central atom.

Key Properties:

  • Coulombic Repulsion: Electron pairs (bonding and nonbonding) repel each other, spreading out as far as possible.
  • Regions of Electron Density: Each single bond, multiple bond, or lone pair counts as one region of electron density.

Basic Electron Geometries:

    • 2 regions → linear (180°).
    • 3 regions → trigonal planar (120°).
    • 4 regions → tetrahedral (109.5°).
    • 5 regions → trigonal bipyramidal (90°, 120°).
    • 6 regions → octahedral (90°).

Lone Pair Effects: Lone pairs repel more strongly than bonding pairs, compressing bond angles (e.g., bent and trigonal pyramidal structures).

Key Idea: VSEPR theory explains molecular geometry by minimizing electron–electron repulsions, allowing accurate prediction of bond angles and shapes from a Lewis diagram.

Example :

Predict the geometry and bond angle of \(\mathrm{NH_3}\) using VSEPR theory.

▶️ Answer/Explanation

Step 1: Draw the Lewis structure: nitrogen has 5 valence electrons, bonded to 3 hydrogens, leaving 1 lone pair.

Step 2: Total regions of electron density = 4 (3 bonds + 1 lone pair).

Step 3: Electron geometry = tetrahedral. But one region is a lone pair, so molecular geometry = trigonal pyramidal.

Step 4: Ideal tetrahedral angle = 109.5°, but lone pair repulsion compresses H–N–H angles to ~107°.

Final Answer: \(\mathrm{NH_3}\) has a trigonal pyramidal shape with bond angles slightly less than 109.5° (~107°).

Lewis Diagrams and VSEPR Predictions

(i): Molecular Geometry

Molecular geometry is the three-dimensional arrangement of atoms in a molecule. It is determined by the number of regions of electron density (bonding + lone pairs) around the central atom, as predicted by VSEPR theory.

Regions of Electron Density: Each single bond, multiple bond, or lone pair counts as one region.

Basic Electron Domain Geometries:

Number of Electron RegionsElectron Domain GeometryIdeal Bond Angles
2Linear180°
3Trigonal Planar120°
4Tetrahedral109.5°
5Trigonal Bipyramidal90°, 120°
6Octahedral90°

Molecular Geometry (Adjusted for Lone Pairs):

Molecular GeometryBonding / Lone PairsIdeal Bond AnglesExamples
Linear2 bonds, 0 lone pairs180°\(\mathrm{CO_2}\)
Trigonal Planar3 bonds, 0 lone pairs120°\(\mathrm{BF_3}\)
Bent (Trigonal Planar)2 bonds, 1 lone pair< 120°\(\mathrm{SO_2}\)
Tetrahedral4 bonds, 0 lone pairs109.5°\(\mathrm{CH_4}\)
Trigonal Pyramidal3 bonds, 1 lone pair< 109.5° (~107°)\(\mathrm{NH_3}\)
Bent (Tetrahedral)2 bonds, 2 lone pairs< 109.5° (~104.5°)\(\mathrm{H_2O}\)
Trigonal Bipyramidal5 bonds, 0 lone pairs90°, 120°\(\mathrm{PCl_5}\)
Seesaw4 bonds, 1 lone pair< 90°, < 120°\(\mathrm{SF_4}\)
T-shaped3 bonds, 2 lone pairs< 90°\(\mathrm{ClF_3}\)
Linear (Trigonal Bipyramidal)2 bonds, 3 lone pairs180°\(\mathrm{XeF_2}\)
Octahedral6 bonds, 0 lone pairs90°\(\mathrm{SF_6}\)
Square Pyramidal5 bonds, 1 lone pair< 90°\(\mathrm{BrF_5}\)
Square Planar4 bonds, 2 lone pairs90°\(\mathrm{XeF_4}\)

Key Idea: Molecular geometry results from electron domain geometry modified by lone pairs. Lone pairs push bonding pairs closer together, altering the observed molecular shape.

Example :

Predict the molecular geometry of \(\mathrm{SF_4}\).

▶️ Answer/Explanation

Step 1: Lewis structure → Sulfur has 6 valence electrons, bonded to 4 fluorines → 4 bonds + 1 lone pair = 5 regions of electron density.

Step 2: Electron geometry for 5 regions = trigonal bipyramidal.

Step 3: One position is occupied by a lone pair → shape is seesaw.

Final Answer: \(\mathrm{SF_4}\) has a seesaw geometry.

(ii): Bond Angles

Bond angles are the angles formed between adjacent bonds around a central atom. They are determined by electron domain geometry and modified by lone pairs and multiple bonds due to differences in repulsion strength.

Ideal Bond Angles (no lone pairs):

Electron Domain GeometryIdeal Bond Angles
Linear180°
Trigonal Planar120°
Tetrahedral109.5°
Trigonal Bipyramidal90° (axial–equatorial), 120° (equatorial–equatorial)
Octahedral90°

Lone Pair Effects:

  • Lone pairs repel more strongly than bonding pairs.
  • This reduces bond angles between bonding atoms compared to ideal values.
  • Example: \(\mathrm{NH_3}\) ~107° (vs. 109.5°); \(\mathrm{H_2O}\) ~104.5° (vs. 109.5°).

Multiple Bond Effects:

  • Double and triple bonds have greater electron density → repel more strongly than single bonds.
  • This can slightly increase angles adjacent to the multiple bond and reduce others.

Key Idea: Bond angles depend on electron domain geometry but are modified by lone pairs and multiple bonds. Lone pairs compress bond angles, while multiple bonds expand angles adjacent to them.

Example :

Compare the bond angles in \(\mathrm{CH_4}\), \(\mathrm{NH_3}\), and \(\mathrm{H_2O}\).

▶️ Answer/Explanation

Step 1: \(\mathrm{CH_4}\): 4 bonding pairs, no lone pairs → tetrahedral, 109.5° bond angles.

Step 2: \(\mathrm{NH_3}\): 3 bonding pairs + 1 lone pair → trigonal pyramidal. Lone pair repels more strongly → bond angles ~107°.

Step 3: \(\mathrm{H_2O}\): 2 bonding pairs + 2 lone pairs → bent. Two lone pairs push bonding pairs closer → bond angles ~104.5°.

Final Answer: \(\mathrm{CH_4}\): 109.5° > \(\mathrm{NH_3}\): 107° > \(\mathrm{H_2O}\): 104.5°.

(iii): Relative Bond Energies Based on Bond Order

 Bond energy is the energy required to break a bond in a molecule. It reflects the strength of attraction between bonded atoms. Bond order — the number of shared electron pairs between two atoms — is the primary factor determining bond energy.

Bond Order and Energy:

  • Bond order = 1 (single bond) → lowest bond energy.
  • Bond order = 2 (double bond) → higher bond energy.
  • Bond order = 3 (triple bond) → highest bond energy.

Explanation: Higher bond order means more shared electrons, creating stronger attraction between nuclei and electrons, requiring more energy to break.

Bond Strength Trend: \(\mathrm{single < double < triple}\).

Other Factors: Bond energy also depends on electronegativity difference and atomic size, but bond order is the dominant factor.

Key Idea: Higher bond order increases bond energy because more electrons are shared between atoms, strengthening the attraction and making the bond harder to break.

Example :

Rank the bonds \(\mathrm{C–C}\), \(\mathrm{C=C}\), and \(\mathrm{C \equiv C}\) in order of increasing bond energy, and explain.

▶️ Answer/Explanation

Step 1: A single bond (\(\mathrm{C–C}\)) has bond order = 1, weakest bond.

Step 2: A double bond (\(\mathrm{C=C}\)) has bond order = 2, stronger than single bond.

Step 3: A triple bond (\(\mathrm{C \equiv C}\)) has bond order = 3, strongest bond.

Final Answer: Increasing bond energy: \(\mathrm{C–C < C=C < C \equiv C}\).

(iv): Relative Bond Lengths

Bond length is the average distance between the nuclei of two bonded atoms. It depends primarily on bond order (number of electron pairs shared) and the atomic radii of the bonded atoms.

Bond Order and Length:

  • Single bond → longest.
  • Double bond → shorter.
  • Triple bond → shortest.

Reason: Higher bond order increases electron density between nuclei, pulling atoms closer together.

Atomic Radius Effect:

  • Bonds between larger atoms are longer because nuclei are farther apart.
  • Example: \(\mathrm{C–H}\) bond is shorter than \(\mathrm{C–Cl}\) bond due to smaller H radius.

Bond Length and Strength Relationship:

  • Shorter bonds are stronger (higher bond energy).
  • Longer bonds are weaker (lower bond energy).

Key Idea: Bond length decreases as bond order increases and increases with larger atomic radii. Shorter bonds are generally stronger than longer bonds.

Example :

Compare the bond lengths in \(\mathrm{C–C}\), \(\mathrm{C=C}\), and \(\mathrm{C \equiv C}\), and explain why \(\mathrm{C–Cl}\) is longer than \(\mathrm{C–H}\).

▶️ Answer/Explanation

Step 1: Bond order effect: \(\mathrm{C–C}\) (single) is longest, \(\mathrm{C=C}\) (double) is shorter, \(\mathrm{C \equiv C}\) (triple) is shortest.

Step 2: Atomic size effect: Cl is much larger than H, so the \(\mathrm{C–Cl}\) bond length is greater than the \(\mathrm{C–H}\) bond length.

Step 3: Thus, bond length depends on both bond order (number of bonds) and the size of atoms involved.

Final Answer: \(\mathrm{C–C > C=C > C \equiv C}\) in length, and \(\mathrm{C–Cl}\) is longer than \(\mathrm{C–H}\) due to chlorine’s larger radius.

(v): Presence of a Dipole Moment

A dipole moment occurs when a molecule has an uneven distribution of electron density, resulting in partial positive and negative charges. Dipole moments arise from bond polarity (differences in electronegativity) combined with molecular geometry (vector addition of bond dipoles).

Bond Polarity:

  • Bond dipole forms when two atoms have different electronegativities.
  • The more electronegative atom gains partial negative charge (\(\delta^-\)), the less electronegative atom gains partial positive charge (\(\delta^+\)).

Molecular Geometry:

  • Dipole moments are vectors; molecular geometry determines whether they cancel or add.
  • Symmetrical molecules → bond dipoles cancel → nonpolar (e.g., \(\mathrm{CO_2}\)).
  • Asymmetrical molecules → bond dipoles reinforce → polar (e.g., \(\mathrm{H_2O}\)).

Dipole Strength:

  • Magnitude depends on bond polarity and distance between charges (bond length).
  • Measured in Debye units (D).

Key Idea: A molecule has a dipole moment if it contains polar bonds arranged asymmetrically. Bond polarity + geometry together determine molecular polarity.

Example :

Explain why \(\mathrm{CO_2}\) is nonpolar but \(\mathrm{H_2O}\) is polar, even though both contain polar bonds.

▶️ Answer/Explanation

Step 1: \(\mathrm{CO_2}\): Carbon–oxygen bonds are polar. Geometry is linear (180°), so the two bond dipoles are equal and opposite, canceling out.

Step 2: \(\mathrm{H_2O}\): Oxygen–hydrogen bonds are polar. Geometry is bent (~104.5°), so the bond dipoles do not cancel but add to form a net dipole.

Final Answer: \(\mathrm{CO_2}\) is nonpolar due to symmetry; \(\mathrm{H_2O}\) is polar due to asymmetry.

(vi): Hybridization of Valence Orbitals

Hybridization is the mixing of atomic orbitals (s, p, sometimes d) on the central atom to form new equivalent hybrid orbitals used for bonding. The number of electron domains (regions of electron density) around the central atom determines the type of hybridization and the resulting molecular geometry.

Electron Domains and Hybridization:

Electron RegionsHybridizationGeometryIdeal Bond Angles
2spLinear180°
3sp²Trigonal Planar120°
4sp³Tetrahedral109.5°
5sp³dTrigonal Bipyramidal90°, 120°
6sp³d²Octahedral90°

Orbital Overlap: Hybrid orbitals form σ bonds with other atoms; unhybridized p orbitals can form π bonds in multiple bonds.

Consistency with VSEPR: Hybridization provides the orbital-level explanation for the bond angles predicted by VSEPR.

Examples:

  • \(\mathrm{CH_4}\): sp³ hybridization, tetrahedral.
  • \(\mathrm{C_2H_4}\): sp² hybridization (each C), trigonal planar.
  • \(\mathrm{C_2H_2}\): sp hybridization (each C), linear.

Key Idea: Hybridization links electron domain geometry (VSEPR) with orbital theory, explaining observed molecular geometries and bond angles at the atomic orbital level.

Example :

Determine the hybridization of the central atom in \(\mathrm{CH_4}\), \(\mathrm{NH_3}\), and \(\mathrm{BeCl_2}\).

▶️ Answer/Explanation

Step 1: \(\mathrm{CH_4}\): Carbon has 4 regions (4 bonds) → sp³ → tetrahedral geometry (109.5°).

Step 2: \(\mathrm{NH_3}\): Nitrogen has 4 regions (3 bonds + 1 lone pair) → sp³ → trigonal pyramidal geometry (~107°).

Step 3: \(\mathrm{BeCl_2}\): Beryllium has 2 regions (2 bonds) → sp → linear geometry (180°).

Final Answer: – \(\mathrm{CH_4}\): sp³ (tetrahedral, 109.5°). – \(\mathrm{NH_3}\): sp³ (trigonal pyramidal, ~107°). – \(\mathrm{BeCl_2}\): sp (linear, 180°).

Hybridization and Ideal Bond Angles

 Hybridization describes the mixing of atomic orbitals on the central atom to form new hybrid orbitals for bonding. The type of hybridization determines the geometry of the molecule and the ideal bond angles.

sp Hybridization:

  • 2 regions of electron density → linear geometry.
  • Orbitals: one s + one p → two sp hybrids.
  • Bond angles: 180°.
  • Example: \(\mathrm{CO_2}\), \(\mathrm{BeCl_2}\).

sp² Hybridization:

  • 3 regions of electron density → trigonal planar geometry.
  • Orbitals: one s + two p → three sp² hybrids.
  • Bond angles: 120°.
  • Example: \(\mathrm{BF_3}\), \(\mathrm{C_2H_4}\) (each C).

sp³ Hybridization:

  • 4 regions of electron density → tetrahedral geometry.
  • Orbitals: one s + three p → four sp³ hybrids.
  • Bond angles: 109.5°.
  • Example: \(\mathrm{CH_4}\), \(\mathrm{NH_3}\), \(\mathrm{H_2O}\).

Adjustment of Bond Angles:

  • Lone pairs repel more strongly than bonding pairs → actual bond angles are slightly less than ideal.
  • Example: \(\mathrm{NH_3}\) bond angle = ~107° (less than 109.5°); \(\mathrm{H_2O}\) bond angle = ~104.5°.

Key Idea: Hybridization links Lewis structures and VSEPR shapes. The type of hybridization (sp, sp², sp³) directly predicts the ideal bond angles and geometry of molecules.

Example :

Determine the hybridization and bond angles for the central atom in \(\mathrm{CH_4}\), \(\mathrm{C_2H_2}\), and \(\mathrm{C_2H_4}\).

▶️ Answer/Explanation

Step 1: \(\mathrm{CH_4}\) – Carbon has 4 regions of electron density. – Hybridization = sp³. – Geometry = tetrahedral. – Bond angles = 109.5°.

Step 2: \(\mathrm{C_2H_2}\) – Each carbon is triple-bonded to the other and single-bonded to H (2 regions of electron density). – Hybridization = sp. – Geometry = linear. – Bond angle = 180°.

Step 3: \(\mathrm{C_2H_4}\) – Each carbon is double-bonded to the other and bonded to 2 H atoms (3 regions of electron density). – Hybridization = sp². – Geometry = trigonal planar. – Bond angles = 120°.

Final Answer: – \(\mathrm{CH_4}\): sp³, 109.5°. – \(\mathrm{C_2H_2}\): sp, 180°. – \(\mathrm{C_2H_4}\): sp², 120°.

Orbital Overlap, Sigma and Pi Bonds

Chemical bonds form through the overlap of atomic orbitals. A sigma (σ) bond results from head-on orbital overlap, while a pi (π) bond results from side-to-side overlap of unhybridized p orbitals. Multiple bonds contain one σ bond and one or more π bonds.

Sigma (σ) Bonds:

  • Formed by end-to-end (head-on) overlap of orbitals (s–s, s–p, or p–p hybrid overlaps).
  • Stronger overlap → greater bond strength.
  • Present in every covalent bond (single, double, triple).

Pi (π) Bonds:

  • Formed by side-to-side overlap of unhybridized p orbitals.
  • Weaker than σ bonds due to less effective overlap.
  • Found only in multiple bonds: double bonds = 1 σ + 1 π, triple bonds = 1 σ + 2 π.

Bond Energy and Bond Strength:

  • σ bonds have greater bond energy than π bonds.
  • Multiple bonds (double, triple) are stronger and shorter than single bonds because of the added π bonds.

Restricted Rotation:

  • σ bonds allow free rotation around the bond axis.
  • π bonds restrict rotation because breaking the side-to-side overlap would require energy.
  • This leads to the existence of geometric isomers (cis/trans isomers).

Key Idea: Sigma bonds provide the strongest bonding framework, while pi bonds add strength but also restrict rotation. This explains differences in bond length, energy, and molecular geometry, as well as the existence of geometric isomerism.

Example :

Explain the difference between the bonding in ethane (\(\mathrm{C_2H_6}\)), ethene (\(\mathrm{C_2H_4}\)), and ethyne (\(\mathrm{C_2H_2}\)).

▶️ Answer/Explanation

Step 1: Ethane (\(\mathrm{C_2H_6}\)) – Carbon–carbon bond is a single bond → 1 σ bond (sp³–sp³ overlap). – All C–H bonds are σ bonds. – Free rotation possible around the C–C bond.

Step 2: Ethene (\(\mathrm{C_2H_4}\)) – Carbon–carbon bond is a double bond → 1 σ + 1 π bond (sp²–sp² overlap + p–p side overlap). – C–H bonds are σ bonds. – Restricted rotation around the double bond → possible cis/trans isomers.

Step 3: Ethyne (\(\mathrm{C_2H_2}\)) – Carbon–carbon bond is a triple bond → 1 σ + 2 π bonds (sp–sp overlap + two p–p overlaps). – C–H bonds are σ bonds. – Bond is very short and strong, no free rotation.

Final Answer: – \(\mathrm{C_2H_6}\): single bond (σ only), free rotation. – \(\mathrm{C_2H_4}\): double bond (σ + π), restricted rotation. – \(\mathrm{C_2H_2}\): triple bond (σ + 2π), strongest and shortest, no rotation.

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