Chemical Bonds & Molecular Orbitals
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Questions and Answers

Which statement accurately describes the relationship between the number of atomic orbitals and molecular orbitals in bond formation?

  • The number of molecular orbitals formed is always less than the number of contributing atomic orbitals due to energy loss during bond formation.
  • The number of molecular orbitals formed is independent of the number of contributing atomic orbitals and depends only on the specific atoms involved.
  • The number of molecular orbitals formed equals the number of contributing atomic orbitals, adhering to the principle of orbital conservation. (correct)
  • The number of molecular orbitals formed is always greater than the number of contributing atomic orbitals due to the creation of new hybrid orbitals.

In the context of molecular orbital theory, what characterizes a 'bonding' molecular orbital?

  • No electron density between nuclei, resulting in no change in stability.
  • Increased electron density between nuclei, leading to stabilization. (correct)
  • Decreased electron density between nuclei, leading to stabilization.
  • Increased electron density outside nuclei, leading to destabilization.

What is a key requirement for pi ($\pi$) bonding to occur between two atoms?

  • The involvement of s-orbitals to create a strong sigma ($\sigma$) bond as a foundation.
  • The presence of $sp^3$ hybridized orbitals allowing for maximum overlap.
  • The alignment of p-orbitals on adjacent atoms parallel to each other, permitting sideways overlap. (correct)
  • The free rotation of atoms involved to maximize end-on-end orbital overlap.

Why is the concept of antibonding orbitals important in molecular orbital theory, even though they are often unoccupied?

<p>They can become occupied if the molecule absorbs energy or gains extra electrons, influencing stability. (B)</p> Signup and view all the answers

Consider a molecule where the central atom has three surrounding atoms and one lone pair. According to the shortcut method, what is the hybridization of the central atom?

<p>$sp^3$ (B)</p> Signup and view all the answers

How does the s-character of a hybrid orbital influence the bond strength and length?

<p>Higher s-character results in stronger and shorter bonds because s-orbitals are closer to the nucleus. (D)</p> Signup and view all the answers

Which of the carbon-carbon single bonds listed is the strongest?

<p>$sp$-$sp^3$ C-C bond (D)</p> Signup and view all the answers

What is the molecular geometry of $H_2O$?

<p>Bent (B)</p> Signup and view all the answers

How does the presence of lone pairs on a central atom affect the molecular geometry compared to the orbital geometry?

<p>Lone pairs cause the molecular geometry to be different from the orbital geometry. (B)</p> Signup and view all the answers

In certain molecules, an atom with a lone pair that is adjacent to a pi ($\pi$) bond undergoes rehybridization. What is the primary reason for this rehybridization?

<p>To place the lone pair in a p-orbital, allowing for better overlap with the p orbitals of the pi bond. (C)</p> Signup and view all the answers

Flashcards

Sigma (σ) Molecular Orbital

Constructive end-on-end overlap between atomic orbitals, resulting in a stabilizing effect due to shared electron density.

Antibonding (σ*) Molecular Orbital

Destructive end-on-end overlap between atomic orbitals, leading to instability due to minimal electron density between nuclei.

Pi (π) Bonding

Occurs when p orbitals overlap in a side-on fashion. Requires neighboring atoms with electrons in unhybridized p-orbitals.

Hybridization

A process where valence orbitals mix to form hybrid orbitals, influencing molecular geometry and bonding properties.

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sp³ Hybridization

Results from one s orbital hybridizing with three p orbitals, forming four hybrid orbitals with tetrahedral geometry (109° angles).

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sp² Hybridization

Results from one s orbital hybridizing with two p orbitals, creating three hybrid orbitals with trigonal planar geometry (120° angles).

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sp Hybridization

Results from one s orbital hybridizing with one p orbital forming two hybrid orbitals with linear geometry (180° angles).

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Molecular Geometry

The spatial arrangement of atoms in a molecule.

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Rehybridization Exception

Atoms with a lone pair adjacent to a pi bond undergo rehybridization to allow the lone pair to occupy a p orbital.

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S-Character and Bond Strength

Bonds with more s-character are stronger and shorter due to s orbitals being closer to the nucleus.

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Study Notes

  • Chemical bonds involve shared electrons, which is favorable due to Coulomb's Law.
  • In atoms, electrons are in atomic orbitals (1s, 2s, 2p, etc.).
  • Bonds form through atomic orbital overlap, creating molecular orbitals; the number of orbitals is conserved.
  • Two atomic orbitals overlapping results in two molecular orbitals.
  • Constructive overlap between atomic orbitals leads to a bonding σ (sigma) molecular orbital.
  • Electron density is between nuclei, stabilizing the molecule.
  • Destructive overlap between atomic orbitals results in an antibonding σ* (sigma star) molecular orbital.
  • There is no electron density between nuclei when destructive overlap occurs, destabilizing the bond.
  • Unstable configurations arise when positively charged nuclei are close without "glue" electrons.

Molecular Orbital Diagram for H₂

  • The molecular orbital diagram for H₂ depicts bonding and antibonding orbitals.
  • Antibonding orbitals have high energy, while bonding orbitals have low energy.
  • ΔE represents the energy difference between atomic and molecular orbitals.
  • σ* is the lowest unoccupied molecular orbital (LUMO).
  • σ is the highest occupied molecular orbital (HOMO).

Pi (π) Bonding

  • Pi bonding occurs through "side-on" overlap of p orbitals.
  • Occurs when neighboring atoms contain electrons in unhybridized p-orbitals, seen in alkenes and alkynes.
  • Electron density is shared between neighboring p orbitals, forming a π bond.
  • P orbitals must be aligned in the same plane for pi bonding.
  • Atoms involved in pi bonding cannot freely rotate due to orbital overlap.
  • Pi bonding isn't possible when p orbitals are at 90° to each other.
  • Two pi bonds can exist on the same atom through side-on overlap of two sets of p orbitals at 90° to each other with alkynes as an example.

Bonding and Antibonding Orbitals

  • The number of molecular orbitals must equal the number of atomic orbitals.
  • Combining two atomic orbitals always yields one bonding and one antibonding molecular orbital.

Antibonding Orbitals

  • In H₂, a third electron would occupy the antibonding orbital.
  • Antibonding orbitals fill when electrons gain energy or with extra electrons.

Hybridization

  • Second-row elements (B, C, N, O) possess four valence orbitals being one 2s and three 2p orbitals (px, py, pz).
  • Atoms arrange in tetrahedral (~109°) or trigonal (~120°) formations, so bonding can't involve pure p-orbitals.
  • Valence orbitals mix to form hybrid orbitals of both s and p character.
  • Hybrid orbitals arrange to minimize electron repulsion.

Types of Hybridization

  • sp³: 1 s orbital and 3 p orbitals lead to a tetrahedral geometry (109°).
  • sp²: 1 s orbital and 2 p orbitals lead to a trigonal planar geometry (120°).
  • sp: 1 s orbital and 1 p orbital lead to a linear geometry (180°).

sp³ Hybridization

  • In CH₄, the four C-H bonds have identical lengths (1.50 Å) and bond angles (109.5°).
  • The 2s orbital and three 2p orbitals hybridize into four identical sp³ orbitals, each with 25% s-character and 75% p-character.
  • Four sp³ orbitals result in a "tetrahedral" orbital geometry.

sp² Hybridization

  • Combining one s orbital with two p-orbitals creates three hybrid orbitals with 33% s-character and 66% p-character, and one unhybridized p-orbital.
  • Each sp² orbital is 120° apart ("trigonal planar" geometry).
  • It's the geometry for central atoms in trivalent boron compounds (e.g., BH₃), carbocations, and atoms with single pi bonds (e.g., alkenes).

sp Hybridization

  • One s orbital hybridized with one p-orbital results in two hybrid orbitals 180° apart and two "leftover" unhybridized p orbitals.
  • Each sp orbital pair is 180° apart ("linear" geometry).
  • Found in the central atom of beryllium compounds (e.g., BeCl₂) and atoms with triple bonds.

Shortcut for Hybridization

  • Add the number of surrounding atoms (A) and lone pairs (LP).
  • If A+LP = 4, the hybridization is sp³.
  • If A+LP = 3, the hybridization is sp².
  • If A+LP = 2, the hybridization is sp.

Exception to Hybridization

  • If an atom with a lone pair is adjacent to a pi bond, rehybridization occurs to put the lone pair in a p orbital, so it's sp² not sp³.

Orbital vs. Molecular Geometry

  • In CH₄, the geometry of orbitals around the central atom ("orbital geometry") and the geometry of hydrogens ("molecular geometry") are identical.
  • Molecules like NH₃ and H₂O have four electron pair domains arranged tetrahedrally around the central atom.
  • Lone pairs are not bonded to an atom.
  • While orbital geometry is tetrahedral, molecular geometry is trigonal pyramidal in NH₃ and bent in H₂O.
  • Lone pairs compress bond angles.

Examples of Hybridization

Example Hybridization Orbital Geometry Molecular Geometry
CH₄ sp³ tetrahedral tetrahedral
:NH₃ sp³ tetrahedral trigonal pyramidal
H₂O: sp³ tetrahedral bent
:NH₂⁻ sp³ tetrahedral bent
CH₃⁺ sp³ tetrahedral trigonal pyramidal
BH₃ sp² trigonal planar trigonal planar
CH₃⁺ sp² trigonal planar trigonal planar
BeCl₂ sp linear linear
H₂C=CH₂ sp² trigonal planar trigonal planar
C=O sp² trigonal planar trigonal planar
C≡N sp trigonal planar trigonal planar
Benzene Ring sp² trigonal planar trigonal planar

Hybridization and Bond Strength

  • Higher s-character in a bond results in a stronger and shorter bond.
  • sp-hybridized C-H bonds are stronger compared to sp²-hybridized, and sp² stronger than sp³-hybridized C-H bonds.
  • With C-C bonds, a sp³-sp bond is stronger than a sp³-sp² bond, while sp³-sp² is stronger than a sp³-sp³ C-C bond.

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Learn about chemical bonds formed through shared electrons and atomic orbital overlap, which is energetically favorable because of Coulomb's Law. Explore molecular orbitals of Hydrogen (H₂) resulting from constructive and destructive interference. Understand the difference between bonding and antibonding orbitals.

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