Hybridization Knowledge-Based Study

Study notes

1. The Concept of Orbital Mixing

   In valence bond theory, simple overlap of pure atomic orbitals (such as s and p orbitals) cannot adequately explain the observed geometries and bond equivalencies of polyatomic molecules. For example, in methane ($CH_4$), carbon has the ground-state valence electron configuration $2s^2 2p^2$. This configuration predicts only two covalent bonds or bonds of unequal lengths and energies formed by differing orbitals. Experimentally, however, all four $C-H$ bonds in methane are identical in bond length, strength, and spatial orientation, pointing toward the corners of a regular tetrahedron with bond angles of $109.5^\circ$.

   To account for this, Linus Pauling introduced the concept of hybridization: the mathematical mixing of nonequivalent atomic orbitals of comparable energy on a central atom to generate a new set of equivalent orbitals known as hybrid orbitals.

   • Conservation of Orbitals: The total number of hybrid orbitals formed always equals the number of pure atomic orbitals combined.
   • Directional Character: Hybrid orbitals possess greater directional concentration than pure atomic orbitals, leading to more effective orbital overlap and stronger covalent bonds.
   • Energy Balance: Hybridization requires an initial input of energy to promote electrons and mix orbitals, which is energetically offset by the formation of stronger, more stable bonds.

   Hybridization Schemes

   The type of hybrid orbitals formed depends on the number and type of atomic orbitals involved in mixing.

   1. $sp^3$ Hybridization

   Formed by mixing one s orbital and three p orbitals, yielding four equivalent $sp^3$ hybrid orbitals arranged in a tetrahedral geometry directed toward the vertices of a tetrahedron with ideal bond angles of $109.5^\circ$.

   • Example: Methane ($CH_4$), where carbon forms four $\sigma$ bonds with hydrogen.
   • Character: 25% s-character and 75% p-character.

   2. $sp^2$ Hybridization

   Formed by combining one s orbital and two p orbitals, producing three equivalent $sp^2$ hybrid orbitals oriented in a trigonal planar arrangement with $120^\circ$ bond angles. One pure p orbital remains unhybridized, oriented perpendicular to the hybrid orbital plane.

   • Example: Boron trifluoride ($BF_3$) and ethene ($C_2H_4$).
   • Character: 33.3% s-character and 66.7% p-character.

   3. $sp$ Hybridization

   Produced by mixing one s orbital and one p orbital, generating two equivalent $sp$ hybrid orbitals oriented in a linear arrangement with a $180^\circ$ bond angle. Two mutually perpendicular pure p orbitals remain unhybridized.

   • Example: Beryllium chloride ($BeCl_2$) and ethyne ($C_2H_2$).
   • Character: 50% s-character and 50% p-character.

   Using Steric Number to Identify Hybrid State

   The hybridization of a central atom can be systematically deduced from its Lewis structure using the steric number (SN), also referred to as the number of electron domains:

   $$\text{Steric Number} = (\text{Number of bonded atoms}) + (\text{Number of lone pairs on the central atom})$$

   Because hybrid orbitals accommodate both $\sigma$ bonds and lone pairs, the steric number corresponds directly to the required number of hybrid orbitals:

   • Steric Number = 2: Requires two hybrid orbitals $ightarrow$ $sp$ hybridization (Linear electron-domain geometry, $180^\circ$).
   • Steric Number = 3: Requires three hybrid orbitals $ightarrow$ $sp^2$ hybridization (Trigonal planar electron-domain geometry, $120^\circ$).
   • Steric Number = 4: Requires four hybrid orbitals $ightarrow$ $sp^3$ hybridization (Tetrahedral electron-domain geometry, $109.5^\circ$).

   Effect of Lone Pairs on Molecular Geometry

   While the electron-domain geometry is defined by the hybridization state, lone pairs alter the resulting molecular geometry and compress adjacent bond angles due to greater electron-electron repulsion:

   • Water ($H_2O$): Oxygen has 2 bonding pairs and 2 lone pairs ($\text{SN} = 4$). The hybridization is $sp^3$, but the molecular geometry is bent with a bond angle of $104.5^\circ$.
   • Ammonia ($NH_3$): Nitrogen has 3 bonding pairs and 1 lone pair ($\text{SN} = 4$). The hybridization is $sp^3$, but the molecular geometry is trigonal pyramidal with a bond angle of $107.3^\circ$.

   Role of Unhybridized Orbitals in Multiple Bonding

   Covalent bonds formed between atoms are classified according to the geometry of orbital overlap:

   • Sigma ($\sigma$) Bond: Formed by direct end-to-end (head-on) overlap of orbitals along the internuclear axis. Hybrid orbitals or pure s orbitals form $\sigma$ bonds. Every single covalent bond is a $\sigma$ bond.
   • Pi ($\pi$) Bond: Formed by sideways (lateral) overlap of parallel, unhybridized p orbitals above and below the internuclear axis. Electron density is concentrated above and below the plane of the nuclei.

   Multiple Bond Composition

   In molecules with double or triple bonds, only one bond can be formed by head-on overlap along the internuclear axis. Any additional bond must be a $\pi$ bond:

   • Single bond: $1\sigma$ bond (typically involving hybrid orbitals).
   • Double bond: $1\sigma$ bond + $1\pi$ bond (formed by unhybridized parallel p orbitals).
   • Triple bond: $1\sigma$ bond + $2\pi$ bonds (formed by two perpendicular pairs of unhybridized p orbitals).

   For example, in ethene ($H_2C=CH_2$), each carbon atom is $sp^2$ hybridized. Three $sp^2$ orbitals form three $\sigma$ bonds (two with hydrogen atoms and one between the two carbon atoms). The remaining unhybridized $2p$ orbital on each carbon atom overlaps side-by-side to produce the carbon-carbon $\pi$ bond.