Hybridization, Orbital Theory, Intermolecular Forces & Thermodynamics – Comprehensive Study Notes

Electron-Dot (Lewis) Structures & VSEPR Electron Geometry

• Always begin by drawing a correct Lewis structure; this sets the stage for every later decision (hybridization, geometry, orbital diagrams, polarity, etc.).
• Example set used in lecture:
  • \ce{CCl4}
    • Central C attached to four Cl atoms (may draw any orientation).
    • Electron groups around C = 4 (all bonding, 0 lone pairs).
    • Electron geometry = tetrahedral.
  • \ce{C2H4} (implied by the trigonal-planar discussion)
    • Each C is attached to three electron groups (2 single bonds + 1 double bond).
    • Electron geometry at each C = trigonal planar.
  • \ce{BrF3}
    • Five electron groups around Br (3 bonding pairs, 2 lone pairs).
    • Electron geometry = trigonal bipyramidal (molecular shape = T-shaped).

Choosing a Hybridization Scheme

• Rule: Hybridization is selected directly from the number of electron groups on the central atom (VSEPR).
  1. 2 groups → $sp$ (linear).
  2. 3 groups → $sp^2$ (trigonal planar).
  3. 4 groups → $sp^3$ (tetrahedral).
  4. 5 groups → $sp^3d$ (trigonal bipyramidal).
  5. 6 groups → $sp^3d^2$ (octahedral).
• Lecture focus:
  • $sp^3$ chosen for \ce{CCl4} because 4 electron groups.
  • $sp^2$ chosen for each C in \ce{C2H4} because 3 groups.
  • $sp^3d$ chosen for \ce{BrF3} because 5 groups.

Valence-Bond Orbital Diagrams (Carbon Examples)

1. $sp^3$ on Carbon (e.g., \ce{CCl4})

• Construct four equivalent $sp^3$ boxes on the same energy level.
• Carbon has 4 valence electrons → distribute one per box (Hund’s rule):
  $\boxed{\uparrow} \; \boxed{\uparrow} \; \boxed{\uparrow} \; \boxed{\uparrow}$
• Interpretation:
  • 4 singly occupied hybrids ⇒ 4 $\sigma$ bonds.
  • 0 $\pi$ bonds (no unhybridized p orbitals remain).

2. $sp^2$ on Carbon (e.g., \ce{C2H4})

• Three $sp^2$ hybrids on one level plus one unhybridized $p$ orbital at slightly higher energy.
• Electron fill (4 e⁻):
  Hybrids: $\boxed{\uparrow} \; \boxed{\uparrow} \; \boxed{\uparrow}$
  Unhybridized $p$: $\boxed{\uparrow}$
• Interpretation:
  • 3 $\sigma$ bonds (from hybrids).
  • 1 $\pi$ bond (from sideways overlap of two unhybridized $p$ on the two carbons) → produces the C=C double bond.
• Energy comment raised in class: hybrids are at lower energy than the remaining $p$; electron distribution obeys Aufbau + Hund, so one electron remains in the higher unhybridized $p$, generating the $\pi$ bond.

3. $sp^3d$ on Bromine (e.g., \ce{BrF3})

• Five $sp^3d$ boxes (same energy).
• Bromine valence electrons = 7.
  Fill: $\boxed{\uparrow\downarrow} \; \boxed{\uparrow\downarrow} \; \boxed{\uparrow} \; \boxed{\uparrow} \; \boxed{\uparrow}$
• Interpretation:
  • Lone pairs are recognized as paired electrons within hybrid boxes (first two boxes above) → 2 lone pairs.
  • Singly occupied hybrids = 3 $\sigma$ bonds (to F atoms).
  • 0 $\pi$ bonds (all d & p are hybridized; no leftover $p$).

Quick Sigma/Pi Bond Checklist

• Single bond ⇒ 1 $\sigma$.
• Double bond ⇒ 1 $\sigma$ + 1 $\pi$.
• Triple bond ⇒ 1 $\sigma$ + 2 $\pi$.

Molecular Orbital (MO) Theory Review

Bond-Order Formula

$\text{Bond order} = \frac{N<em>{\text{bonding}} - N</em>{\text{antibonding}}}{2}$

• $N_{\text{bonding}}$ = total electrons in bonding MOs.
• $N_{\text{antibonding}}$ = total electrons in antibonding MOs (marked with *).
• Interpretation:
  • \text{B.O.} > 0 → molecule/ion predicted to exist.
  • Larger $\text{B.O.}$ → stronger, shorter bond.

Example 1: 3-Electron Diatomic (e.g., \ce{He2^{+}} or \ce{Li2^{+}})

• Only $\sigma<em>{1s}$ and $\sigma</em>{1s}^{*}$ considered.
• Electron fill (3 e⁻): $\sigma<em>{1s}^2\, \sigma</em>{1s}^{*1}$.
  $\text{B.O.} = \frac{2 - 1}{2} = 0.5$ ⇒ should exist, weak bond.

Example 2: Fluorine Species (\ce{F2}, \ce{F2^{-}}, \ce{F2^{+}})

• 14 e⁻, 15 e⁻, 13 e⁻ respectively.
• Using the long p-block MO diagram ( $\sigma<em>{2s}$, $\sigma</em>{2s}^{*}$, $\sigma<em>{2p</em>z}$, $\pi<em>{2p</em>x}$, $\pi<em>{2p</em>y}$, etc.).
  1. \ce{F2}: $N<em>b = 8$, $N</em>{ab} = 6$ ⇒ $\text{B.O.}=1$.
  2. \ce{F2^{-}}: $N<em>b = 8$, $N</em>{ab} = 7$ ⇒ $\text{B.O.}=0.5$.
  3. \ce{F2^{+}}: $N<em>b = 8$, $N</em>{ab} = 5$ ⇒ $\text{B.O.}=1.5$ (strongest bond of the trio).
• Conclusion from lecture: \ce{F2^{+}} has the strongest bond because it has the highest bond order.

Magnetism from MO Diagrams

• Criterion:
  • At least one unpaired electron in any MO → paramagnetic (magnetic).
  • All electrons paired → diamagnetic (non-magnetic).
• Examples worked:
  • \ce{O2} (12 e⁻ in p MOs) shows two unpaired electrons in $\pi<em>{2p</em>x,y}^{*}$ ⇒ paramagnetic.
  • \ce{Ne2} (16 e⁻) has all electrons paired ⇒ diamagnetic.
  • \ce{F2} discussed; filling yields all electrons paired ⇒ diamagnetic.

Intramolecular vs. Intermolecular Forces

• Intramolecular = actual bonds inside the molecule (ionic or covalent).
• Intermolecular = attractions between separate molecules/ions.

Electronegativity & Bond Classification

• Difference $\Delta EN$ guideline:
  • $0 \le \Delta EN \le 0.4$ ⇒ non-polar covalent.
  • $0.5 \le \Delta EN \le 1.9$ ⇒ polar covalent.
  • $\Delta EN \ge 2.0$ ⇒ ionic (usually metal + non-metal).

Mapping Bond Type to Intermolecular Force (IMF)

• Ionic compounds → ion-ion attraction.
• Polar covalent molecules → dipole-dipole attraction.
  • Special case: hydrogen bonding when H is bound to \,\ce{F}\,, \ce{O}, or \ce{N} (lecturer also said “Cl” but standard rule is FON; follow instructor for exam).
• Non-polar covalent molecules → London dispersion forces (induced dipole-induced dipole).

Worked IMF Examples

• Exam hint: you must specify “polar” vs “non-polar” covalent; simply writing “covalent” is insufficient for full credit because IMF depends on polarity.

Stoichiometry & Unit Cancellation Strategy

1. Write the given quantity with units.
2. Multiply by conversion factors so that every unwanted unit cancels.
3. Final numerator carries the desired unit (e.g., $\text{kJ}$).

• Example (verbal): convert $6.52\,\text{g}$ of a compound → mol → kJ using molar mass then $\frac{\text{kJ}}{\text{mol}}$.

Boiling Point & Vapor Pressure Fundamentals

• Statement: “The temperature at which a liquid’s vapor pressure equals the external pressure is the boiling point.” → TRUE.
• Normal boiling point = temperature where $P_{\text{vap}} = 1\,\text{atm}$.

Clausius–Clapeyron Application (Propane Example)

• Data given:
  • Normal boiling point: $T<em>1 = -42\,^{\circ}\text{C} = 231\,\text{K}$ (where $P</em>1 = 1\,\text{atm}$).
  • $\Delta H<em>{\text{vap}} = 19.04\,\text{kJ mol}^{-1}$. • Desired temperature: $T</em>2 = 25\,^{\circ}\text{C} = 298\,\text{K}$.
• Two-point Clausius–Clapeyron equation:
  $\ln\left(\frac{P<em>2}{P</em>1}\right) = -\frac{\Delta H<em>{\text{vap}}}{R}\left(\frac{1}{T</em>2} - \frac{1}{T_1}\right)$
  with $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1} = 0.008314\,\text{kJ mol}^{-1}\text{K}^{-1}$.
• Plugging numbers (outline):
  1. Convert $\Delta H_{\text{vap}}$ to kJ consistent with R.
  2. Solve for $\ln(P<em>2)$, exponentiate to obtain $P</em>2$ in atm.
• Students should practice carrying units through to ensure cancellation.

Additional Instructor Comments & Exam Tips

• Hybridization labels can include the principal quantum number (e.g., $2sp^3$ for carbon) but are not strictly required unless specified by textbook/instructor.
• Maximum of three $p$ orbitals per atom; you cannot “create” extra orbitals—unused ones remain unhybridized.
• In MO problems, always select the correct MO diagram from your textbook handout; diagrams differ for \ce{B2}–\ce{N2} vs \ce{O2}–\ce{F2}.
• When judging magnetism: look for any unpaired electron in the filled MO diagram.
• For IMF identification, first determine the bond polarity then immediately map to the correct force; beware of the hydrogen-bonding special case.
• Practice writing full dimensional-analysis (factor-label) setups; graders look for correct unit cancellation as well as the final number.