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Compounds & Chemical Bonding

Two Classes of Bonds

Covalent bonding - Occurs in molecules - Involves the sharing of electrons
Ionic Bonding - Occurs in ionic solids - Involves the transfer of electrons from one atom to another - Simpler concept and will be discussed first

Ionic Bonds

Defined as attractive forces between oppositely charged particles.
Typically occurs between metals and nonmetals due to the following reasons: - Metals: - Have low ionization energies - Easily lose electrons to reach a stable state - Non-metals: - Possess very exothermic electron affinities
Formation of Ionic Lattice: - Stabilizes ions, as seen in the example of sodium (Na+) and chloride (Cl-).

Ionic Compounds

Characterized by the formation from a metal and a nonmetal.
Ionic Bond: Attraction between the positive and negative ions within the compound.
Questions Addressed about Ionic Compounds: - Why does electron transfer occur? - Why does sodium form Na+ and not Na2+ or Na–? - Why does chlorine form Cl– and not Cl2– or Cl+?

Ionic Crystals

Structure: Exist in a three-dimensional array of cations and anions known as a lattice structure.
Ionic Chemical Formulas: Always presented in empirical formula form, indicating the smallest whole number ratio of cations to anions.

Lattice Energy

Defined as the change in potential energy (PE) of a system when one mole of solid salt is created from its gaseous ions.
Represents the energy released during the formation of the ionic lattice.

Calculation of Lattice Energy

Single step: - Reaction: $ext{Na(s)} + rac{1}{2} ext{Cl}_2(g) ightarrow ext{NaCl}(s)$ - Standard Enthalpy Change: $ext{DHf}^ ext{°} = -411.1 ext{ kJ/mol}$
Stepwise Path: - Each step involves the following transformations: - $ext{Na(s)} ightarrow ext{Na(g)}$ - $ext{DHf}^ ext{°}( ext{Na, g}) = 107.8 ext{ kJ/mol}$ - $rac{1}{2} ext{Cl}_2(g) ightarrow ext{Cl(g)}$ - $ext{DHf}^ ext{°}( ext{Cl, g}) = 121.3 ext{ kJ/mol}$ - $ext{Na(g)} ightarrow ext{Na}^+(g) + e^-$ - Ionization Energy: $ext{IE}( ext{Na}) = 495.4 ext{ kJ/mol}$ - $ext{Cl(g)} + e^- ightarrow ext{Cl}^-(g)$ - Electron Affinity: $ext{EA}( ext{Cl}) = -348.8 ext{ kJ/mol}$ - Final ion formation: - $ext{Na}^+(g) + ext{Cl}^-(g) ightarrow ext{NaCl}(s)$ - Lattice Energy: $ext{DHlattice} = -787 ext{ kJ/mol}$

Stability and Exothermic Nature of Ionic Compounds

Lattice energy is always exothermic, meaning that energy is released during the formation of ionic lattices.
The magnitude of lattice energy becomes more exothermic (more negative value) as ions with opposite charges are brought closer together, leading to a more stable ionic solid arrangement.

Question on Lattice Energy

Problem: Assuming the interionic separation in the lattice is nearly identical, which species would have the greatest lattice energy? - Choices: - A. sodium chloride (NaCl) - B. calcium chloride (CaCl2) - C. calcium nitride (Ca3N2) - D. sodium oxide (Na2O) - E. calcium oxide (CaO)

Formation of Cations and Anions

Metals: - Located on the left-hand side of the Periodic Table. - Have small positive ionization energies, requiring little energy to remove an electron. - Have small or positive electron affinity, making it less favorable to attract an electron, making it energetically cheaper to form a cation.
Non-metals: - Located on the right-hand side of the Periodic Table. - Have large positive ionization energies; thus, it is difficult to remove an electron. - Have large negative electron affinities, allowing for easy addition of electrons, resulting in energy release. - Making it energetically cheaper to form an anion.

Review of Electron Configurations

Follow Aufbau ordering, which reflects the energy level of electrons in an atom.
Electronic configuration is based on “filling” an atom with electrons, adhering to the increasing values of n and, within any given energy level, by increasing values of ℓ.
This framework also assists in explaining how ions form.

Electron Configurations of Ions

Sodium (Na): - Neutral Atom: $1s^2 2s^2 2p^6 3s^1$ = [Ne] 3s1 - Ion Na+: $1s^2 2s^2 2p^6$ = [Ne] - Ionization Energies: - $ext{IE}_1 = 496 ext{ kJ/mol}$ - Relatively small; not too difficult to remove the first electron. - $ext{IE}_2 = 4563 ext{ kJ/mol}$ - Significantly larger; difficult to remove the second electron, indicating why Na2+ doesn’t form.
Calcium (Ca): - Neutral Atom: $[Ar] 4s^2$ - Ion Ca2+: $[Ar]$ - Ionization Energies: - $ext{IE}_1 = 590 ext{ kJ/mol}$ - $ext{IE}_2 = 1140 ext{ kJ/mol}$ - Relatively manageable, while: - $ext{IE}_3 = 4940 ext{ kJ/mol}$ - Too significant to remove the third electron, explaining why Ca3+ does not form.

Stability of Ionic Configurations

The noble gas core serves as a stabilizing factor limiting the number of electrons that metals can lose.
Ions typically end up with noble gas configurations, which applies to both cations and anions: - Chloride (Cl): - Neutral: $1s^2 2s^2 2p^6 3s^2 3p^5$ = [Ne] 3s2 3p5 - Ion Cl–: $1s^2 2s^2 2p^6 3s^2 3p^6$ = [Ar]

The Octet Rule

Atoms tend to gain or lose electrons until they achieve a stable outer shell of eight electrons (octet).
Well applicable to: - Group 1A and 2A metals - Aluminum (Al)
Not applicable to: - Hydrogen (H) and Helium (He) (limited to two electrons in n = 1 shell) - Transition metals and post transition metals.

Transition Metals

First electrons lost from the outermost s orbital.
Electron loss order: - Highest n level is prioritized before using d orbitals, as illustrated with Iron (Fe): - Neutral: $[Ar] 3d^6 4s^2$ - Ion Fe2+: $[Ar] 3d^6$ (loses 4s electrons first) - Ion Fe3+: $[Ar] 3d^5$ (also loses d electrons).

Predicting Cation Configurations

For Bismuth (Bi): - Electron configuration: $[Xe] 6s^2 4f^{14} 5d^{10} 6p^3$ . - Possible expected ions: Bi3+ and Bi5+.
For Iron (Fe): - Electron configuration: $[Ar] 4s^2 3d^6$ . - Possible expected ions: Fe2+ and Fe3+.

Predicting Anion Configurations

Non-metals gain electrons to attain noble gas configuration: - Oxygen: $[He] 2s^2 2p^4$ transforms to: $[He] 2s^2 2p^6$ by gaining 2 e– → O2–. - Nitrogen: $[He] 2s^2 2p^3$ transforms to: $[He] 2s^2 2p^6$ by gaining 3 e– → N3–.

Lewis Symbols

An electron bookkeeping method to track valence electrons.
Representation: Write the chemical symbol surrounded by dots corresponding to the number of valence electrons.
Group Numbers and Valence Electrons: - Group 1A: 1, 2A: 2, 3A: 3, 4A: 4, 5A: 5, 6A: 6, 7A: 7, 8A (Noble): 8.

Lewis Symbols and Bonding

Can utilize Lewis symbols to illustrate electron transfer in ionic bonding.

Covalent Compounds

Form individual separate molecules, characterized by shared electrons.
These molecules generally do not conduct electricity and usually have low melting points.

Covalent Bonds

Defined by the sharing of pairs of electrons between two atoms.
For example, in a reaction between two Hydrogen atoms: - Atoms approach each other, leading to electron sharing for bond formation.

Covalent Bond Mechanics

Attraction of valence electrons from one atom by the nucleus of the other atom leads to the sharing of electrons.
Electron Density Shift: - As the distance between nuclei decreases, the probability of locating either electron closer to either nucleus increases, inducing a closer position of nuclei.
At close distances, nuclei begin repelling due to their positive charge, and the bond stabilizes at the point of net attractive force.

Characterization of Covalent Bonds

Bond Length: The distance between the two atomic nuclei is represented by: $r_A + r_B$ .
Bond Energy: Refers to the energy released when the bond forms (decrease in PE), or the energy required to break the bond.

Driving Forces in Covalent Bonds

Question: What is the primary driving force behind the formation of covalent bonds? - A. Energy released from electron attraction.
- B. Energy gained from electron attraction. - C. Energy released by the attraction between nuclei and electrons. - D. Energy gained from the repulsion between nuclei.

Lewis Structures

These diagrams depict molecules using Lewis symbols and show the electronic structure of covalent bonds.
Dots represent electrons, while pairs shared signify covalent bonds, ensuring each atom achieves a conventional octet (or duet for hydrogen).

Octet Rule in Covalent Bonding

Atoms engage in covalent bonding to share electrons, allowing them to fulfill an outer shell of eight electrons, which ensures stability.

Examples of Diatomic Gases and Lewis Structures

Hydrogen (H2): - $H \bullet + H \bullet ightarrow H:H$ represents shared electron pairs as a line between H atoms.
Fluorine (F2): - Each Fluorine is stabilized with its octet by forming a single bond.
Other diatomic elements follow similar bonding rules and are usually gases due to weak intermolecular forces.

Multiple Bonds in Covalent Compounds

Single Bonds: Produced by sharing a single pair of electrons (e.g., N2).
Double Bonds: Produced by sharing two pairs of electrons (e.g., O2).
Triple Bonds: Produced by sharing three pairs of electrons (e.g., N2).

Electronegativity and Bond Polarity

Equal sharing occurs in bonds between two identical atoms; unequal sharing occurs in bonds between different elements.
One atom often attracts electrons more strongly, resulting in partial charges across the bond due to differences in electronegativities.

Polar Covalent Bonds

Defined as bonds with partial positive and negative charges at opposite ends, forming dipoles within molecules.

Electronegativity (EN)

Defined as the relative attraction of an atom for shared electrons in a bond.
Comparison of Electronegativity Values: - Example values include: - F: 4.1 - O: 3.5 - Cl: 3.0 - N: 3.0

Electronegativity Trends

EN increases from left to right across a period due to the increased effective nuclear charge (Zeff).
EN decreases from top to bottom within a group due to the increase in principal quantum number (n).

Ionic and Covalent Bonding

Represent two extremes of bonding types, with actual bonding often found between both extremes based on the elements involved.

Reactivity of Elements Based on Electronegativity

Correlates to elemental reactivity based on their electronegativity values.
Metals tend to have low EN, which makes them easier to oxidize, while those with high EN are harder oxidizers.

Drawing Lewis Structures

Lewis structures must account for octets, although exceptions exist (e.g., BeCl2, BCl3).
Steps to draw: 1. Determine bonding arrangement (skeletal structure). 2. Count total valence electrons for all atoms. 3. Place two electrons between bonded atoms recalling to fulfill octets for terminal atoms and potentially forming double or triple bonds if needed.

Examples of Lewis Structures

SiF4: - Valence electrons calculated as: $1 imes 4 + 4 imes 7 = 32$ e– total. - Structure drawn with Si as central atom and F atoms fulfilling their octets.
H2CO3: - Carbon center; electrons adjusted through bond formation to satisfy the octet condition.

Expanded Octets and Electron Deficient Structures

Atoms in Period 3 and below can hold up to 18 electrons, using d orbitals for bonding, while certain elements, like boron (B), may form electron-deficient structures.

Formal Charge (FC) Calculation

FC is a bookkeeping method defined as: - FC = ext{# valence e–} - [ ext{# unshared e–} + rac{1}{2}( ext{# bonding e–})]
Indicates the apparent charge and serves to determine the stability of various structures, favoring those with smaller or zero charges.

Resonance Structures

Scheme for explaining equivalent Lewis structures where none represent the actual configuration. - The actual structure is an average of all potential Lewis structures created through electron relocation.