Comprehensive Notes on Orbital Hybridization and Molecular Geometries and Bonding Types

Electronic configurations such as $1s^{2}$ , $2s^{2}$ , and $2p^{6}$ are specific to isolated atoms like carbon or oxygen.
Atoms rarely exist in isolation; they typically join to form molecules.
To form bonds, an atom requires a "free spot" or space for an up-arrow electron to receive a down-arrow electron from another atom.
Standard ground-state configurations often limit the number of possible bonds. For example, a configuration with only one unpaired electron theoretically allows only one bond.
Atoms aim to maximize the number of bonds they can make to reach a more stable state.
Electrons cannot simply be "thrown" from a lower energy $s$ orbital to a higher energy $p$ orbital if they would not naturally occupy that space.
The solution is orbital hybridization: the rearrangement and combination of $s$ and $p$ orbitals into hybrid orbitals (specifically named "hybrid orbitals").

The name of a hybrid orbital exactly describes which atomic orbitals were combined.
The number of hybrid orbitals created is always equal to the number of atomic orbitals added together.
There is a limit to the number of standard orbitals available for hybridization in the lower shells: one $s$ orbital and three $p$ orbitals.
When counting electrons and bonds, simply count the "things" (atoms or lone pairs) surrounding the central atom to determine the hybridization type.
Hybridization results in a release of energy because it facilitates pairing, resulting in a lower overall energy level for the atom within a molecule.
Hybrid orbitals exist at an energy level between the original $s$ (lower energy) and $p$ (higher energy) orbitals.
Unused orbitals do not disappear; for example, if one $s$ and two $p$ orbitals are hybridized, the third $p$ orbital remains unhybridized and available.

$sp$ Hybridization:
  - Composition: One $s$ orbital + one $p$ orbital.
  - Quantity: Results in two $sp$ hybrid orbitals.
  - Geometry: Linear.
  - Separation: $180^{\circ}$ .
  - Example: Beryllium ( $Be$ ). In its natural state, it has two paired electrons in the $s$ orbital and empty $p$ orbitals. By hybridizing into two $sp$ orbitals, it can separate those two electrons into two individual bonds.
$sp^{2}$ Hybridization:
  - Composition: One $s$ orbital + two $p$ orbitals.
  - Quantity: Results in three $sp^{2}$ hybrid orbitals.
  - Geometry: Trigonal planar.
  - Unhybridized Orbitals: One empty $p$ orbital remains, typically sitting above and below the plane of the molecule.
  - Energy: All three $sp^{2}$ orbitals are equal in energy level.
$sp^{3}$ Hybridization:
  - Composition: One $s$ orbital + three $p$ orbitals.
  - Quantity: Results in four $sp^{3}$ hybrid orbitals.
  - Geometry: Tetrahedral.
  - Separation: Approximately $109.5^{\circ}$ .
  - Application: This is why Carbon ( $C$ ) consistently forms four bonds.
$sp^{3}d$ Hybridization:
  - Composition: One $s$ + three $p$ + one $d$ orbital.
  - Quantity: Results in five orbitals.
  - Geometry: Trigonal bipyramidal.
  - Requirement: Only possible for elements in Period 3 ( $n=3$ ) or higher, which have access to $d$ orbitals.
$sp^{3}d^{2}$ Hybridization:
  - Composition: One $s$ + three $p$ + two $d$ orbitals.
  - Quantity: Results in six orbitals.
  - Geometry: Octahedral.

Nitrogen ( $N$ ):
  - Configuration: $2s^{2}2p^{3}$ .
  - Hybridization: $sp^{3}$ .
  - Result: Even though it has four $sp^{3}$ orbitals, one is already double-filled (a lone pair), leaving only three spots for bonding. This represents a "weighted average" energy drop from the $p$ level to the hybrid level.
Oxygen ( $O$ ):
  - Configuration: $2s^{2}2p^{4}$ .
  - Hybridization: $sp^{3}$ .
  - Result: Contains two lone pairs and two spots available for bonding.
Period 2 Constraints: Atoms like Nitrogen and Oxygen are limited to four total orbitals ( $s + 3p$ ) because they lack the "backyard" or "parking lot" (empty $d$ orbitals) found in Period 3.
Period 3 and Violating the Octet Rule: Elements like Chlorine ( $Cl$ ) or Bromine ( $Br$ ) in $n=3$ can violate the octet rule by using their empty $d$ orbitals to create five or six hybrid orbitals.

Bonding is not simply the accumulation of single bonds. A double bond is not two single bonds.
Sigma ( $\sigma$ ) Bond:
  - Result of a head-on collision or overlap of orbitals.
  - All single bonds are sigma bonds.
  - They occur between two $s$ orbitals, an $s$ and a $p$ orbital, or two hybridized orbitals.
Pi ( $\pi$ ) Bond:
  - Result of side-by-side overlap of unhybridized, empty $p$ orbitals.
  - To form a pi bond, the atom must have available unhybridized $p$ orbitals (occurring in $sp$ and $sp^{2}$ hybridization, but not $sp^{3}$ ).
  - Visualization: Often described as a "hamburger bun" where two lobes (top and bottom) represent a single pi bond capable of holding two electrons.
Double Bonds: Consist of exactly one sigma bond and one pi bond.
Triple Bonds: Consist of one sigma bond and two pi bonds.

Bond Strength:
- Sigma bonds are generally stronger than pi bonds.
- A double bond is stronger than a single bond but not twice as strong (e.g., if a sigma bond has a strength of 1.0, a pi bond might contribute 0.8).
Restriction of Rotation:
- Single bonds (sigma bonds) allow for free rotation of the bonded atoms.
- Double bonds restrict rotation because the pi bond overlap (the "hamburger bun") holds the atoms in a fixed orientation. Atoms cannot rotate around the double bond without breaking the pi bond overlap.

Structure Overview: Central Carbon ( $C$ ) attached to three Hydrogens ( $H$ ) and one Oxygen ( $O$ ). The Oxygen is also attached to one Hydrogen.
Carbon Hybridization:
  - Surrounded by four "things" (four bonds).
  - Geometry: Tetrahedral.
  - Hybridization: $sp^{3}$ .
Oxygen Hybridization:
  - Surrounded by four "things": two bonds (to $C$ and $H$ ) and two lone pairs.
  - Geometry: Tetrahedral (electron group geometry).
  - Hybridization: $sp^{3}$ .
Bonding details:
  - Sigma bonds exist between Carbon ( $sp^{3}$ ) and Hydrogen ( $s$ ).
  - A sigma bond exists between Carbon ( $sp^{3}$ ) and Oxygen ( $sp^{3}$ ).
  - The Carbon-Oxygen-Hydrogen bond angle is slightly less than $109.5^{\circ}$ due to lone pair repulsion.