Untitled Flashcards Set

Sigma (σ) and Pi (π) Bonds

• Sigma Bonds:

  • Formed by the direct, head-on overlap of atomic orbitals, including combinations such as s-s, s-p, and p-p.

  • This overlap occurs along the axis connecting two atomic nuclei.

  • Every single bond between atoms consists of one sigma bond which facilitates rotational freedom.

• Pi Bonds:

  • Result from the sideways overlap of p orbitals.

  • These bonds are present in double and triple bonds; specifically, a double bond consists of one sigma bond and one pi bond, while a triple bond consists of one sigma bond and two pi bonds.

  • Pi bonds restrict the rotation of the bonded atoms, making the geometric arrangement crucial to the properties of the molecule.

• Key Point:

  • Sigma bonds support free rotation due to their direct alignment, whereas pi bonds create fixed orientation due to their nature of overlap.

• IB-Style Problem:

  • Question: How many sigma and pi bonds are present in ethyne (C₂H₂)?

  • Answer: Each C-H bond is characterized as a sigma bond. The C≡C bond contains one sigma bond and two pi bonds, leading to a total of 3 sigma bonds and 2 pi bonds.

Covalent Bonding Definition and Properties

• Definition: Covalent bonding occurs when two atoms share one or more pairs of electrons, leading to a stable electron configuration typically resembling the electron configuration of noble gases.

• Properties of Covalent Compounds:

  • Generally present low melting and boiling points, though network covalent structures like diamond exhibit much higher thermal stability.

  • Typically non-conductive in solid and liquid states except for substances like graphite, which conduct electricity in its layered form.

  • The solubility of covalent compounds varies; polar molecules tend to dissolve in polar solvents such as water, while nonpolar molecules prefer nonpolar solvents.

• IB-Style Problem:

  • Question: Why does CO₂ have a low boiling point compared to NaCl?

  • Answer: CO₂ exhibits weak intermolecular forces known as London dispersion forces, while NaCl is ionic and consists of strong electrostatic forces that hold ions together, leading to a higher boiling point.

Hybridization and VSEPR Geometry

• Hybridization:

  • sp: Results in a linear arrangement with 180° bond angles (found in molecules like BeCl₂).

  • sp²: Results in a trigonal planar configuration with 120° bond angles (seen in molecules like BF₃).

  • sp³: Produces a tetrahedral shape with bond angles of 109.5° (as in CH₄ - methane).

• IB-Style Problem:

  • Question: Determine the hybridization of the central atom and predict the geometry of PCl₅.

  • Answer: PCl₅ exhibits five electron domains, leading to sp³d hybridization. Its geometry is classified as trigonal bipyramidal due to the arrangement of electron pairs.

Octet Rule and Lewis Structures

• The octet rule explains that atoms gain, lose, or share electrons to achieve a complete set of eight electrons in their outer shell, promoting stability.

• Steps to Draw Lewis Structures:

  1. Count the total number of valence electrons available for bonding.

  2. Arrange the atoms and connect them using single bonds initially.

  3. Complete the octets of the outer atoms using lone pairs before completing the central atom's octet.

  4. If the octet is not complete, consider forming multiple bonds by shifting lone pairs.

• IB-Style Problem:

  • Question: Draw the Lewis structure for the nitrate ion (NO₃⁻).

  • Solution: The total number of electrons is 24. The structure comprises one nitrogen atom with a double bond to one oxygen atom, and two oxygen atoms connected via single bonds while exhibiting resonance.

Light Wavelength and Bond Dissociation

• Stronger bonds necessitate higher energy (shorter wavelength light) for breaking.

• O₂ exhibits a double bond and requires ultraviolet light (~242 nm) for dissociation.

• O₃, although having weaker bonds than O₂, dissociates with longer wavelengths (~330 nm) due to its different energy requirements.

• IB-Style Problem:

  • Question: Why does the ozone layer absorb UV light effectively?

  • Answer: Ozone is capable of dissociating at specific UV wavelengths, offering protection to the Earth from harmful radiation.

Exceptions to the Octet Rule

• Certain molecules are electron-deficient, such as Boron in BF₃, which has only 6 valence electrons around Boron, or Beryllium in BeCl₂, exhibiting only 4.

• Atoms in Period 3 or beyond can have expanded octets, allowing structures like PCl₅ and SF₆ to stabilize with more than 8 electrons in their valence shells.

• IB-Style Problem:

  • Question: Why can sulfur in SF₆ exceed the octet rule?

  • Answer: Sulfur can utilize d orbitals in bonding, which permits it to maintain more than eight electrons within its valence shell.

Formal Charge and Expanded Octets

• Formal Charge (FC) can aid in determining the most stable resonance structure for a molecule and is calculated using the formula: [ FC = Valence : electrons - (Non-bonding : electrons + \frac{Bonding : electrons}{2}) ]

• IB-Style Problem:

  • Question: Determine the formal charge of each atom in the carbonate ion (CO₃²⁻).

  • Answer: In the carbonate ion, the central carbon atom has a formal charge of 0, while the two oxygen atoms possess formal charges of -1 each.

Coordinate Covalent Bonds

• Definition: A unique type of covalent bond where one atom donates both electrons in the bond formation (e.g., in NH₄⁺ and H₃O⁺ ions).

• IB-Style Problem:

  • Question: Describe the bonding in the ammonium ion (NH₄⁺).

  • Answer: In NH₄⁺, nitrogen donates a lone pair to bond with a proton (H⁺), forming a coordinate covalent bond that stabilizes the ion.

Resonance Structures

• Resonance refers to situations in which a molecule can be represented by two or more different valid Lewis structures, contributing to the overall bonding character of the molecule.

• IB-Style Problem:

  • Question: Draw all resonance structures for benzene (C₆H₆).

  • Answer: Benzene can be depicted through resonance structures that feature alternating double bonds and single bonds, showcasing delocalized π electrons within the ring structure.

Polar Bonds and Polar Molecules

• Bond polarity arises from differences in electronegativity between atoms, causing uneven electron distribution.

• Molecular polarity, on the other hand, is determined by the symmetry of the molecule; dipole moments may cancel out leading to nonpolar molecules such as CO₂.

• IB-Style Problem:

  • Question: Explain why H₂O is polar but CCl₄ is nonpolar.

  • Answer: H₂O has a bent shape which does not allow dipoles to cancel; CCl₄, being tetrahedral and symmetrical, allows dipoles to counteract each other resulting in a nonpolar character.

Intermolecular Forces (IMFs)

• Types of IMFs:

  • London Forces: These are the weakest intermolecular forces present in nonpolar molecules like CH₄.

  • Dipole-Dipole Forces: Occur in polar molecules (e.g., HCl).

  • Hydrogen Bonding: The strongest type of IMF, present in molecules featuring N-H, O-H, or F-H bonds.

• IB-Style Problem:

  • Question: Rank CH₃OH, CH₄, and H₂O by boiling point.

  • Answer: The boiling points rank as follows: H₂O > CH₃OH > CH₄, since H₂O’s hydrogen bonds are particularly strong compared with the weaker London forces found in CH₄.

Boiling Point Trends

• The boiling point of a substance is influenced by molecular size, the strength of IMFs, and the structural arrangement of the molecules involved.

• IB-Style Problem:

  • Question: Explain why ethanol (C₂H₅OH) has a higher boiling point than propane (C₃H₈).

  • Answer: Ethanol can form hydrogen bonds due to its hydroxyl (-OH) group, while propane is limited to London dispersion forces, which are significantly weaker.

Metallic Bonding

• Metallic bonds are characterized by a sea of delocalized electrons that provide excellent conductivity and malleability, making metals versatile in practical applications.

• Alloys: They are typically harder than pure metals due to the presence of different atom sizes, which prevents the layers of atoms from slipping over one another easily.

• IB-Style Problem:

  • Question: Why is brass harder than pure copper?

  • Answer: The inclusion of zinc atoms disrupts the regular copper lattice, resulting in greater strength in brass compared to pure copper.

Allotropes of Carbon

• Diamond: Each carbon atom forms four sigma bonds through sp³ hybridization, resulting in a very strong, three-dimensional structure.

• Graphite: Consists of layers where each carbon atom forms three sigma bonds (sp² hybridization) and possesses delocalized π electrons, enabling electrical conductivity between layers.

• IB-Style Problem:

  • Question: Why does graphite conduct electricity but diamond does not?

  • Answer: Graphite’s structure allows for delocalized electrons to move freely between layers, while diamond’s rigid structure lacks free-moving electrons.

Bonding Triangle and Electronegativity

• The bonding triangle aids in predicting bond types by assessing differences in electronegativity between bonded atoms.

• IB-Style Problem:

  • Question: Classify the bond in NaCl and HF.

  • Answer: NaCl is classified as ionic due to the significant electronegativity difference, while HF is identified as polar covalent due to intermediate electronegativity between the bonded atoms.

Writing and Balancing Equations

• IB-Style Problem:

  • Question: Balance the combustion reaction of propane (C₃H₈).

  • Solution: The balanced chemical equation for the combustion of propane is as follows:[ C_3H_8 + 5O_2 → 3CO_2 + 4H_2O ] This reaction illustrates the stoichiometric ratios necessary for full combustion under typical conditions.