Covalent Bonding: Orbitals - Hybridization

Essential Question: How do atoms hybridize their atomic orbitals to form stable molecular structures with specific geometries?

Key Vocabulary

• Hybridization: The process of mixing atomic orbitals to form new hybrid orbitals that are suitable for bonding and electron pair repulsion minimization.
• $sp$ Hybridization: The mixing of one $s$ and one $p$ atomic orbital to form two $sp$ hybrid orbitals, resulting in a linear electron geometry with a $180^\circ$ angle.
• $sp^2$ Hybridization: The mixing of one $s$ and two $p$ atomic orbitals to form three $sp^2$ hybrid orbitals, resulting in a trigonal planar electron geometry with $120^\circ$ angles.
• $sp^3$ Hybridization: The mixing of one $s$ and three $p$ atomic orbitals to form four $sp^3$ hybrid orbitals, resulting in a tetrahedral electron geometry with $109.5^\circ$ angles.
• $dsp^3$ Hybridization: The mixing of one $d$, one $s$, and three $p$ atomic orbitals to form five $dsp^3$ hybrid orbitals, leading to a trigonal bipyramidal electron geometry.
• $d^2sp^3$ Hybridization: The mixing of two $d$, one $s$, and three $p$ atomic orbitals to form six $d^2sp^3$ hybrid orbitals, resulting in an octahedral electron geometry.
• Sigma ($\sigma$) bond: A type of covalent bond formed by the direct, head-on overlap of atomic orbitals, where electron density is concentrated along the internuclear axis.
• Pi ($\pi$) bond: A type of covalent bond formed by the lateral (sideways) overlap of two unhybridized $p$ orbitals, resulting in electron density above and below the internuclear axis.
• Effective Electron Pairs: The total number of lone pairs and bonding groups (single, double, or triple bonds count as one group) around a central atom, used to determine electron geometry.
• VSEPR Model (Valence Shell Electron Pair Repulsion): A model used to predict the geometry of individual molecules by minimizing the repulsion between electron pairs around the central atom.

What is $sp$ Hybridization?

• Definition: This type of hybridization results from the mixing of one $s$ atomic orbital and one $p$ atomic orbital.
• Resulting Orbitals: Forms a set of two equivalent $sp$ hybrid orbitals. Each $sp$ hybrid orbital has 50% $s$ character and 50% $p$ character.
• Orientation and Geometry: These two $sp$ orbitals are oriented at an angle of $180^\circ$ to each other, giving a linear arrangement to minimize electron repulsion. This linear electron geometry is characteristic of $sp$ hybridization.
• Requirement: An atom with two effective pairs of electrons (e.g., two bonding groups and zero lone pairs, or one bonding group and one lone pair, though commonly two bonding groups) will always require $sp$ hybridization to achieve a linear arrangement.
• Orbital Energy-Level Diagram: For carbon, the $2s$ orbital and one of the three $2p$ orbitals hybridize to form two $sp$ orbitals. The other two $2p$ orbitals remain unhybridized. In an energy diagram (Figure 9.16), the $2s$ and one $2p$ orbital energies average out to the new $sp$ orbital energy, which is lower than the $2p$ and higher than the $2s$ original energies. The two unhybridized $2p$ orbitals retain their original energy levels, perpendicular to the $sp$ orbitals.

How does $sp$ Hybridization apply to Carbon Dioxide ($CO_2$)?

• Lewis Structure: $O=C=O$ (A central carbon atom double-bonded to two oxygen atoms, with two lone pairs on each oxygen).
• Carbon Atom Hybridization:
  • The carbon atom has two effective electron pairs, treating each double bond as one effective repulsive unit. This is because there are two regions of high electron density around the carbon (one for each C=O bond).
  • This requires a linear arrangement ($180^\circ$) for minimum repulsion, which is achieved by $sp$ hybridization.
  • The two $sp$ hybrid orbitals on carbon are used to form ${\sigma}$ (sigma) bonds with the oxygen atoms. Imagine these $sp$ orbitals extending directly from the carbon towards each oxygen, forming head-on overlaps.
  • Two $2p$ orbitals on carbon remain unhybridized. These unhybridized $p$ orbitals are oriented perpendicular to each other and perpendicular to the linear $sp$ orbitals (Figure 9.17 description: picture a carbon with two linear $sp$ orbitals and two $p$ orbitals intersecting at the carbon, one along the y-axis and one along the z-axis, both perpendicular to the x-axis where the $sp$ orbitals lie).
  • These two unhybridized $2p$ orbitals are used to form ${\pi}$ (pi) bonds with the oxygen atoms.
• Oxygen Atom Hybridization:
  • Each oxygen atom has three effective electron pairs (one double bond + two lone pairs). This dictates a trigonal planar electron geometry around each oxygen.
  • This arrangement is achieved by $sp^2$ hybridization for each oxygen atom. This means one $s$ and two $p$ orbitals on oxygen hybridize to form three $sp^2$ orbitals.
  • One $p$ orbital on each oxygen remains unhybridized (Figure 9.18 description: visualize an oxygen with three $sp^2$ orbitals in a plane, and one unhybridized $p$ orbital perpendicular to that plane). This unhybridized $p$ orbital is specifically oriented to overlap with one of the carbon's unhybridized $p$ orbitals to form a ${\pi}$ bond.
• Overall Bonding Description:
  • The $sp$ orbitals on carbon form ${\sigma}$ bonds with one of the $sp^2$ orbitals on each of the two oxygen atoms. This forms the backbone of the $O-C-O$ linear structure.
  • The remaining two $sp^2$ orbitals on each oxygen atom hold the two lone pairs.
  • The two ${\pi}$ bonds between the carbon and each oxygen atom are formed by the overlap of parallel unhybridized $2p$ orbitals (one from carbon and one from each oxygen). For example, the carbon's unhybridized $p<em>y$ orbital overlaps with oxygen's unhybridized $p</em>y$ orbital to form one ${\pi}$ bond, and the carbon's unhybridized $p<em>z$ orbital overlaps with the other oxygen's unhybridized $p</em>z$ orbital to form the second ${\pi}$ bond (Figure 9.19 description: shows the linear ${\sigma}$ frame with two sets of $p$ orbital overlaps — one vertical and one horizontal — above and below the internuclear axis).
  • Each C=O double bond thus consists of one ${\sigma}$ bond and one ${\pi}$ bond.
• Theoretical Models: More rigorous models suggest that each oxygen atom uses two $p$ orbitals simultaneously to form the ${\pi}$ bonds, contributing to unusually strong C=O bonds by increasing electron density shared between the atoms, leading to a resonance-like stabilization.

How does $sp$ Hybridization apply to Acetylene ($C<em>2H</em>2$) (Ethyne)?

• Lewis Structure: $H-C \equiv C-H$ (A linear molecule with a triple bond between carbons and single bonds to hydrogen).
• Carbon Atom Hybridization:
  • Each carbon atom has two effective electron pairs: one triple bond to the other carbon and one single bond to a hydrogen atom. Each multiple bond (triple bond) counts as one repulsive unit, making a total of two units.
  • This requires a linear arrangement ($180^\circ$), thus each carbon atom is $sp$ hybridized.
  • Each carbon uses one of its two $sp$ orbitals to form a ${\sigma}$ bond with a hydrogen atom. This is a head-on overlap between a carbon $sp$ orbital and a hydrogen $1s$ orbital (Figure 9.17 analogy for carbon's $sp$ orbitals).
  • The other $sp$ orbital on each carbon overlaps head-on with the similar $sp$ orbital on the adjacent carbon atom to form the C-C ${\sigma}$ bond.
  • Two unhybridized $p$ orbitals remain on each carbon atom. These $p$ orbitals are mutually perpendicular and also perpendicular to the C-C bond axis. (Think of them as $p<em>y$ and $p</em>z$ orbitals if the C-C bond is along the x-axis).
  • These two unhybridized $p$ orbitals on each carbon overlap laterally (sideways) with the corresponding unhybridized $p$ orbitals on the adjacent carbon to form the two ${\pi}$ bonds. One pair of $p$ orbitals (e.g., $p<em>y$) forms one ${\pi}$ bond, and the other pair (e.g., $p</em>z$) forms the second ${\pi}$ bond.
• Overall Bonding: The carbon-carbon triple bond consists of one ${\sigma}$ bond (from $sp-sp$ overlap) and two ${\pi}$ bonds (from $p-p$ overlaps). The C-H bonds are ${\sigma}$ bonds (from $sp-s$ overlap).

How does $sp$ Hybridization apply to Nitrogen ($N_2$)?

• Lewis Structure: $:N \equiv N:$ (A nitrogen molecule with a triple bond between two nitrogen atoms and a lone pair on each nitrogen).
• Nitrogen Atom Hybridization:
  • Each nitrogen atom is surrounded by two effective electron pairs: one triple bond and one lone pair. This means two regions of electron density.
  • This dictates a linear arrangement and therefore $sp$ hybridization for each nitrogen atom.
  • Each nitrogen has two $sp$ hybrid orbitals and two unhybridized $p$ orbitals (Figure 9.20a description: analogous to carbon, but with nitrogen's valence electrons. One $s$ and one $p$ form two $sp$ orbitals, leaving two $p$ orbitals unhybridized).
  • One $sp$ orbital on each nitrogen atom is used to form the N-N ${\sigma}$ bond through head-on overlap with an $sp$ orbital from the other nitrogen (Figure 9.20b).
  • The other $sp$ orbital on each nitrogen atom holds a lone pair (Figure 9.20b description: the linear N-N framework, with each N having an $sp$ orbital pointing away from the bond, housing a lone pair).
  • The two unhybridized $p$ orbitals on each nitrogen atom (e.g., $p<em>y$ and $p</em>z$) overlap in parallel (laterally) to form the two ${\pi}$ bonds (Figure 9.20c description: illustrates the sideways overlaps of the $p<em>y$ and $p</em>z$ orbitals, creating electron density above/below and in front/behind the internuclear axis).
• Overall Bonding: The N-N triple bond consists of one ${\sigma}$ bond and two ${\pi}$ bonds. Each nitrogen atom also possesses a lone pair housed in one of its $sp$ hybrid orbitals (Figure 9.20d description: a complete picture showing the ${\sigma}$ bond, two ${\pi}$ bonds, and the lone pairs in the $sp$ orbitals, resulting in a linear molecular shape).

What is $dsp^3$ Hybridization?

• Definition: This hybridization occurs when one $d$ orbital, one $s$ orbital, and three $p$ orbitals are combined.
• Resulting Orbitals: Forms a set of five equivalent $dsp^3$ hybrid orbitals. These orbitals are capable of forming five bonds from a central atom.
• Orientation and Geometry: These five $dsp^3$ orbitals adopt a trigonal bipyramidal arrangement (Figure 9.21 description: visualize a central atom with three orbitals arranged in a plane at $120^\circ$ (equatorial positions) and two orbitals arranged perpendicular to that plane, pointing up and down along an axis (axial positions)).
• Requirement: A central atom with five effective pairs of electrons (either five bonding pairs or a combination of bonding pairs and lone pairs) will require $dsp^3$ hybridization to achieve a trigonal bipyramidal arrangement, exceeding the octet rule.
• Theoretical Consideration: While this model is convenient for predicting geometry, more rigorous theoretical models suggest that it mistakenly invokes the significant use of $d$ orbitals in the actual bonding of fifth-row (or higher) molecules that exceed the octet rule. These more advanced models propose a more complex bonding mechanism. However, the simple $dsp^3$ model remains widely used due to its utility in predicting molecular geometries and its pedagogical convenience.

How does $dsp^3$ Hybridization apply to Phosphorus Pentachloride ($PCl_5$)?

• Lewis Structure: Central phosphorus atom bonded to five chlorine atoms (no lone pairs on phosphorus).
• Phosphorus Atom Hybridization:
  • The phosphorus atom is surrounded by five electron pairs (all bonding pairs to chlorine atoms).
  • This requires a trigonal bipyramidal arrangement for minimum repulsion, necessitating $dsp^3$ hybridization of the phosphorus atom.
  • The five $dsp^3$ orbitals on phosphorus are used to share electrons (form ${\sigma}$ bonds) with the five chlorine atoms. Three of these bonds will be equatorial (in a plane) and two will be axial (above and below the plane) (Figure 9.22 description: The central P atom is shown with five $dsp^3$ orbitals forming the trigonal bipyramidal shape, pointing towards the five Cl atoms).
• Chlorine Atom Hybridization:
  • Each chlorine atom is surrounded by four electron pairs (one bond to phosphorus + three lone pairs).
  • This requires a tetrahedral arrangement for the electron pairs around each chlorine, implying $sp^3$ hybridization for each chlorine atom. Each chlorine uses one of its $sp^3$ orbitals for bonding.
• Overall Bonding: The five P-Cl ${\sigma}$ bonds are formed by sharing electrons between a $dsp^3$ orbital on the phosphorus atom and an $sp^3$ orbital on each chlorine atom.

How does $dsp^3$ Hybridization apply to the Triiodide Ion ($I_3^-$)?

• Lewis Structure: A central iodine atom with two bonded iodine atoms and three lone pairs. The structure is linear, but the electron geometry is trigonal bipyramidal. ($[:I-I-I:]^-$ with three lone pairs on central I).
• Central Iodine Atom Hybridization:
  • The central iodine atom has five pairs of electrons: two bonding pairs (to the outer iodine atoms) and three lone pairs.
  • To minimize repulsions, these five electron pairs adopt a trigonal bipyramidal electron arrangement, calling for $dsp^3$ hybridization for the central iodine.
  • In a trigonal bipyramidal arrangement, lone pairs typically occupy equatorial positions to minimize $90^\circ$ repulsion angles. Thus, three of these $dsp^3$ hybrid orbitals hold the three lone pairs in the equatorial plane.
  • The remaining two $dsp^3$ hybrid orbitals on the central iodine atom extend along the axial positions and overlap with $sp^3$ orbitals of the outer iodine atoms to form two ${\sigma}$ bonds, resulting in a linear molecular geometry.
• Outer Iodine Atoms Hybridization:
  • The outer iodine atoms each have four pairs of electrons (one bond to central iodine + three lone pairs).
  • This requires a tetrahedral arrangement around each outer iodine, calling for $sp^3$ hybridization.

What is $d^2sp^3$ Hybridization?

• Definition: This hybridization involves combining two $d$ orbitals, one $s$ orbital, and three $p$ orbitals.
• Resulting Orbitals: Forms a set of six equivalent $d^2sp^3$ hybrid orbitals, oriented symmetrically around the central atom.
• Orientation and Geometry: These six $d^2sp^3$ orbitals are arranged octahedrally (Figure 9.23 description: picture a central atom with six orbitals pointing towards the corners of an octahedron, with $90^\circ$ angles between adjacent orbitals).
• Requirement: A central atom with six effective pairs of electrons (either six bonding pairs or a combination of bonding pairs and lone pairs) will always require $d^2sp^3$ hybridization to achieve an octahedral electron arrangement, exceeding the octet rule.

How does $d^2sp^3$ Hybridization apply to Sulfur Hexafluoride ($SF_6$)?

• Lewis Structure: Central sulfur atom bonded to six fluorine atoms (no lone pairs on sulfur).
• Sulfur Atom Hybridization:
  • The sulfur atom is surrounded by six electron pairs, all involved in bonding to fluorine atoms.
  • This requires an octahedral arrangement and thus $d^2sp^3$ hybridization for the sulfur atom.
  • Each of the six $d^2sp^3$ orbitals on the sulfur atom forms a ${\sigma}$ bond to a fluorine atom através de overlap com um orbital do flúor.
• Fluorine Atom Hybridization:
  • Each fluorine atom has four electron pairs (one bond to sulfur + three lone pairs).
  • These are assumed to be $sp^3$ hybridized; each fluorine uses one $sp^3$ orbital to form a ${\sigma}$ bond with sulfur.

How does $d^2sp^3$ Hybridization apply to Xenon Tetrafluoride ($XeF_4$)?

• Lewis Structure: Central xenon atom bonded to four fluorine atoms and possessing two lone pairs. The molecular geometry is square planar.
• Xenon Atom Hybridization:
  • The xenon atom has six pairs of electrons: four bonding pairs to fluorine and two lone pairs.
  • To minimize repulsions, these six electron pairs are arranged octahedrally in space (electron geometry).
  • This requires $d^2sp^3$ hybridization for the xenon atom.
  • Xenon uses its six $d^2sp^3$ hybrid atomic orbitals to bond to the four fluorine atoms (forming four ${\sigma}$ bonds) and to hold the two lone pairs.
  • In an octahedral electron geometry with two lone pairs, the lone pairs will occupy positions opposite each other (e.g., axial positions) to minimize repulsions with the bonding pairs and with each other, leading to a square planar molecular geometry for the four fluorine atoms.

What is the Problem-Solving Strategy for the Localized Electron Model?

• Overall Approach: The localized electron model describes a molecule by starting with its proposed Lewis structure, allowing electrons to arrange themselves for maximum stability. Subsequently, it assumes that the central atoms adjust their atomic orbitals through hybridization to achieve the minimum energy structure that accommodates these electron pairs.
• Three Distinct Steps:
  1. Draw the Lewis structure(s): This provides the initial arrangement of valence electrons, identifying all bonding pairs and lone pairs around each atom, especially the central atom(s). This helps count the total number of effective electron groups.
  2. Determine the arrangement of electron pairs using the VSEPR model: Based on the number of effective electron pairs determined from the Lewis structure, apply the VSEPR (Valence Shell Electron Pair Repulsion) model to establish the electron geometry around the central atom(s). This geometry (e.g., linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral) minimizes electron-pair repulsions. (Figure 9.15 description: A chart showing electron group arrangements corresponding to VSEPR geometries: 2 groups=linear, 3 groups=trigonal planar, 4 groups=tetrahedral, 5 groups=trigonal bipyramidal, 6 groups=octahedral).
  3. Specify the hybrid orbitals needed to accommodate the electron pairs: Based on the VSEPR geometry determined in the previous step, determine the appropriate type of