Hybridization and molecular geometry

Types of chemical bonds: σ and π

The spatial structure of a molecule largely depends on how the electron clouds of atoms overlap during the formation of chemical bonds.

• $\_sigma$-bond (sigma bond) occurs when orbitals overlap along the axis connecting the atomic nuclei. It can be formed by the interaction of two $s$-orbitals ($H_2$), an $s$- and a $p$-orbital ($HCl$), or two $p$-orbitals ($Cl_2$).
• $\_pi$-bond (pi bond) is formed due to the lateral overlap of unhybridized $p$-orbitals ($p_y$ or $p_z$) above and below the bond plane. This occurs when the space on the axis between the nuclei is already occupied by a $\_sigma$-bond.

The number of $\_pi$-bonds determines bond multiplicity: a double bond consists of one $\_sigma$- and one $\_pi$-bond, while a triple bond consists of one $\_sigma$- and two $\_pi$-bonds. The presence of $\_pi$-bonds significantly increases molecule stability (e.g., in nitrogen $N_2$ with a triple bond, the energy reaches 946 kJ/mol).

Main types of hybridization

Depending on the number of participating orbitals, there are three main types of hybridization typical, for example, for the carbon atom:

1. $sp$-hybridization. One $s$- and one $p$-orbital mix. Two identical hybrid orbitals are formed, extending in opposite directions. The angle between them is 180°, which determines the linear geometry (examples include acetylene $C_2H_2$ and carbon dioxide $CO_2$). The remaining two unhybridized $p$-orbitals can form two $\_pi$-bonds.
2. $sp^2$-hybridization. Involves one $s$- and two $p$-orbitals. Three hybrid orbitals arise, lying in the same plane at an angle of 120° to each other. The molecular structure is planar (e.g., ethylene $C_2H_4$). The remaining perpendicular $p$-orbital forms one $\_pi$-bond.
3. $sp^3$-hybridization. Includes one $s$- and all three $p$-orbitals of the same level. The result is four hybrid orbitals directed toward the vertices of a tetrahedron at equal angles of 109.5°. The molecule has a tetrahedral structure (examples include methane $CH_4$ or a diamond crystal).

Special cases: bonds without hybridization and d-elements

Not all molecules are formed with hybridization. Some atoms (e.g., sulfur and phosphorus) in the ground state have unpaired $p$-electrons on the outer shell located in mutually perpendicular $p$-orbitals.

When hydrogen compounds with phosphorus (phosphine $PH_3$) and sulfur (hydrogen sulfide $H_2S$) are formed, chemical bonds are created by pure, unhybridized $p$-orbitals. Since the initial $p$-orbitals are at an angle of 90° to each other, the bond angles in these molecules are close to right angles (around 93°).

For $d$-block metals, $d$-orbitals are also valence orbitals. If two $d$-, one $s$-, and three $p$-orbitals participate in hybridization (the $d^2sp^3$-hybridization type), 6 hybrid orbitals are formed, directed toward the vertices of an octahedron. This is typical for octahedral coordination complexes.

Class and exam questions

What else to study on this topic

• Electronic structure of phosphorus and sulfur atoms in the ground state
• Diagram of bond formation in the acetylene molecule (C2H2)
• The effect of multiple bonds on molecular energy (using nitrogen, oxygen, and fluorine as examples)
• Orbital overlap during the formation of d-block metal complexes

Other topics in "Chapter 5. Structure of Matter"

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