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Chemical Bonding and Structure - Hybridization (HL only)

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Hybridization is the conceptual process of mixing atomic orbitals (such as ss and pp orbitals) to form new hybrid orbitals that are degenerate (equal in energy) and suitable for the pairing of electrons to form chemical bonds.

sp3sp^3 Hybridization: Formed by mixing one 2s2s and three 2p2p orbitals. This results in four sp3sp^3 hybrid orbitals arranged in a tetrahedral geometry with bond angles of approximately 109.5109.5^\circ (e.g., in CH4CH_4 or NH3NH_3).

sp2sp^2 Hybridization: Formed by mixing one 2s2s and two 2p2p orbitals. This results in three sp2sp^2 hybrid orbitals arranged in a trigonal planar geometry with bond angles of 120120^\circ. One unhybridized pp orbital remains perpendicular to the plane (e.g., in C2H4C_2H_4 or BF3BF_3).

spsp Hybridization: Formed by mixing one 2s2s and one 2p2p orbital. This results in two spsp hybrid orbitals arranged in a linear geometry with bond angles of 180180^\circ. Two unhybridized pp orbitals remain (e.g., in C2H2C_2H_2 or CO2CO_2).

Sigma (σ\sigma) bonds: These are formed by the head-on overlap of atomic or hybrid orbitals along the internuclear axis. All single covalent bonds are σ\sigma bonds.

Pi (π\pi) bonds: These are formed by the sideways overlap of unhybridized pp orbitals. A double bond consists of one σ\sigma and one π\pi bond; a triple bond consists of one σ\sigma and two π\pi bonds.

The hybridization of an atom can be predicted by the number of electron domains: 22 domains sp\rightarrow sp, 33 domains sp2\rightarrow sp^2, 44 domains sp3\rightarrow sp^3.

📐Formulae

Number of electron domains=(number of σ bonds)+(number of lone pairs)\text{Number of electron domains} = (\text{number of } \sigma \text{ bonds}) + (\text{number of lone pairs})

Formal Charge (FC)=V(N+12B)\text{Formal Charge (FC)} = V - (N + \frac{1}{2}B)

Bond Order=Total number of bonding pairsTotal number of resonance structures\text{Bond Order} = \frac{\text{Total number of bonding pairs}}{\text{Total number of resonance structures}}

💡Examples

Problem 1:

Identify the hybridization of the carbon atoms in ethene (C2H4C_2H_4) and describe the bonding involved.

Solution:

Each carbon atom in C2H4C_2H_4 is bonded to two hydrogen atoms and one carbon atom (via a double bond). This gives each carbon 33 electron domains. Therefore, the hybridization is sp2sp^2.

Explanation:

In sp2sp^2 hybridization, each carbon forms three σ\sigma bonds using its sp2sp^2 hybrid orbitals. The remaining unhybridized 2p2p orbital on each carbon atom overlaps sideways to form one π\pi bond. Thus, the C=CC=C double bond consists of one σ\sigma bond and one π\pi bond.

Problem 2:

Determine the hybridization of the central nitrogen atom in the nitrate ion (NO3NO_3^-).

Solution:

In NO3NO_3^-, the central nitrogen atom has 33 resonance structures, but in each, it is surrounded by 33 electron domains (bonding regions to three oxygen atoms, with no lone pairs on NN). Thus, the hybridization is sp2sp^2.

Explanation:

Since there are 33 electron domains around the nitrogen, it requires three hybrid orbitals. Mixing one ss and two pp orbitals gives three sp2sp^2 orbitals, resulting in a trigonal planar electron domain geometry.

Problem 3:

Calculate the formal charge of Carbon in the CO2CO_2 molecule (O=C=OO=C=O).

Solution:

FC=4(0+12(8))=44=0FC = 4 - (0 + \frac{1}{2}(8)) = 4 - 4 = 0

Explanation:

Carbon (V=4V=4) has zero lone pair electrons (N=0N=0) and is sharing 8 electrons in four bonds (B=8B=8, two double bonds). A formal charge of 00 indicates a stable distribution of electrons.

Hybridization (HL only) - Revision Notes & Key Formulas | IB Grade 12 Chemistry