Module 4 - 6

Chemical Properties Determination
- Chemical properties are primarily determined by valence electrons (e–), which are the electrons in the outermost shell of an atom.
- When using the periodic table, do not include filled d and f orbitals.
Inert Gases Reactivity
- The lack of reactivity in inert gases is due to their filled outer s and p shells.
Lewis Convention
- The chemical symbol represents the nucleus and inner electrons.
- Outermost electrons are depicted as dots, where a line signifies an electron pair.
Limitations of Lewis Diagrams
- Lewis diagrams are useful primarily for atoms that involve s and p electrons, and are not ideal for d electrons.
Examples of Lewis Symbols: Na, C, F, and F2.

Octet Rule and Electronic Configuration
- The first octet period consists of 6 elements.
- Lewis diagrams for all atoms in the eight main groups of the periodic table are now known.
- To write a Lewis diagram, either the electronic configuration or position in the periodic table is required.
- Atoms tend to achieve a complete outer shell (octet status) by:
  a) losing or gaining electrons (ionic bonding) e.g., Na → Na+, Cl → Cl–
  b) sharing electrons (covalent bonding).

Steps for Drawing: (Refer to lab manual Expt 6, textbook p354)
1. Determine the total number of valence electrons, denoted as “V.”
- Do not include d electrons if d orbitals are full.
1. Draw a skeleton structure using single bonds.
2. Subtract the number of electrons used from “V.”
3. Distribute the remaining electrons, ensuring each atom achieves octet status (8 electrons), starting with the most electronegative atoms.
4. If not enough electrons to satisfy the octet for all atoms, use double or triple bonds until every (non-H) atom has 8 electrons.
5. If too many electrons remain, add extras as lone pairs to the central atom, while ensuring to follow the octet rule for those atoms in the second period (expanded octet allowed for atoms with n ≥ 3).

Example 1: NF3
- Valence Electrons Calculation:
  $V = (2 imes 4) + (4 imes 1) = 12$
- For N (5) + (3 x 7) for F = 26.
- Remaining electrons: $20$
Example 2: C2H4
- Total electrons used = 10, with 2 left to place, suggesting shared bonds.

Definition of Formal Charge:
- The formal charge of an atom within a molecule is a calculated value based on the arrangement of electrons.
- For N2O, represented as N-N-O, formal charges can be illustrated.
How to Calculate:
- Assign 1 electron from each shared electron pair (one per bond) and add total number of unshared lone pairs of electrons.
Charge Determination:
- If an atom has fewer outer electrons than it would as a neutral atom, it has a positive charge.
- Conversely, if it possesses more outer electrons than in its neutral state, it has a negative charge.
Summation of Charges:
- The sum of all atomic formal charges within a neutral molecule equals zero. For ions, it equals the charge of the ion.

Example: (CH3)3NO
- Valence calculation: (5 + 4*1) - 1 = 8 electrons used.
- Positive charge on nitrogen with only 4 core electrons rather than 5.

Definition:
- Multiple Lewis structures may represent the same molecule; for example, ozone (O3) demonstrates this concept.
- The true bonding state is not accurately represented by a single Lewis diagram but is instead a hybrid of multiple structures.
Implications:
- Resonance structures highlight delocalized bonding within species.

Example 1: CH3CO2–
- Shows three resonance structures, where no single structure represents the ion accurately.
Example 2: NO3–
- Similar behavior as CH3CO2– demonstrating equivalent N-O bonds.

Considerations:
- For the molecule NO:
  $V = 5 + 6 = 11$
- Notice the odd number of valence electrons, indicating a radical.
Notable Example: NO2
- $V = 5 + (2 imes 6) = 17$ indicates it is also a radical, employing electron sharing to suit electronegativity.

Example: BF3
- Valence calculation: $V = 3 + (3 imes 7) = 24$ .
- Boron only has 6 electrons, indicating it is a Lewis Acid.

Example: XeF4
- Public note: Xenon, despite being a noble gas, can form bonds.
- Total valence calculation: $V = 8 + (4 imes 7) = 36$ .
- Place remaining electrons and lone pairs, noting that Xe does not adhere to the octet rule, while Fluorine must.

Steps for Determining Geometry:
1. Draw a Lewis diagram, using one resonance form to depict structure correctly.
2. Count the number of bonds and lone pairs for the atom in question. For example, in SO2, sulfur has 1 single bond (B), 1 double bond (B), and 1 lone pair (E).
3. Determine stereochemistry based on distinct electron groups.

Geometry Framework:
- 2 groups: Linear (180°)
- 3 groups: Trigonal Planar (~120°)
- 4 groups: Tetrahedral (~109.5°)
- 5 groups: Trigonal Bipyramidal (90°, 120°)
- 6 groups: Octahedral (90°)

Criteria for Polarity:
- Diatomic molecules are non-polar if the two atoms are identical (e.g., H2, Br2).
- They are polar if they consist of different atoms (e.g., HCl, CO).
- Triatomic molecules must be examined for symmetry and bond polarity.

Assessing Geometry and Bond Polarities:
- Example: CO3^2– is non-polar, but SO3^2– is polar due to its geometry.

Summary of Frameworks and Angles:
- AB2: Linear (180°) is sp hybridized.
- AB3: Trigonal planar (~120°) is sp2 hybridized.
- AB4: Tetrahedral (~109.5°) is sp3 hybridized.
- AB5: Trigonal bipyramidal (~90°, ~120°) is sp3d hybridized.
- AB6: Octahedral (90°) is sp3d2 hybridized.

Example: CH4
- Tetrahedral structure requires four equivalent bonding orbitals, achieved via sp3 hybrid orbitals (25% s, 75% p).

Ethylene (C2H4)
- It has a planar structure characterized by sp2 hybridization where three sp2 hybrid orbitals (33% s, 67% p) facilitate the formation of sigma bonds and one unhybridized p orbital forms pi bonds.

Hybridization Summary:
- Carbon can undergo sp3, sp2, and sp hybridization depending on the bonding scenarios (single, double, or triple).
- Structural arrangement based on the number of sigma and pi bonds formed in each hybridization case demonstrates varied channels of molecular geometry.

Hybridization State Inquiry:
- Analyzing the hybridization state of bromine in BrCl3 shows a resembled structure, exemplifying the application in complex molecules.

This module concluded with comprehensive insights into molecular structure, bonding, resonance, and hybridization to better understand chemical behaviors and molecule configurations.