Geometry of hybridization
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The document explains the concept of hybridization in carbon atoms, detailing sp3, sp2, and sp hybridization to describe the bonding in methane, ethene, and ethyne respectively. It illustrates how carbon can form equivalent orbitals by mixing its 2s and p orbitals to achieve stable bonds with hydrogen or other carbon atoms. The hybridization process results in the formation of sigma and pi bonds, facilitating the molecular structure and bond angles characteristic of these compounds.
Geometry of hybridization
- 1. 1 Geometry of Hybridization sp3 Four sp3 orbitals Hybridization describes the bonding atoms from an atom's point of view. For a tetrahedral coordinated carbon (e.g. methane CH4), the carbon should have 4 orbitals with the correct symmetry to bond to the 4 hydrogen atoms. Carbon's ground state configuration is 1s2 2s2 2p2 or more easily read: C ↑↓ ↑↓ ↑ ↑ 1s 2s 2p 2p 2p The carbon atom can use its two singly occupied p-type orbitals, to form two covalent bonds with two hydrogen atoms, yielding the singlet methylene CH2, the simplest carbene. The carbon atom can also bond to four hydrogen atoms by an excitation (or promotion) of an electron from the doubly occupied 2s orbital to the empty 2p orbital, producing four singly occupied orbitals. C* ↑↓ ↑ ↑ ↑ ↑ 1s 2s 2p 2p 2p The energy released by the formation of two additional bonds more than compensates for the excitation energy required, energetically favoring the formation of four C-H bonds. Quantum mechanically, the lowest energy is obtained if the four bonds are equivalent, which requires that they are formed from equivalent orbitals on the carbon. A set of four equivalent orbitals can be obtained that are linear combinations of the valence-shell (core orbitals are almost never involved in bonding) s and p wave functions, which are the four sp3 hybrids.
- 2. 2 C* ↑↓ ↑ ↑ ↑ ↑ 1s sp3 sp3 sp3 sp3 In CH4, four sp3 hybrid orbitals are overlapped by hydrogen 1s orbitals, yielding four σ (sigma) bonds (that is, four single covalent bonds) of equal length and strength. sp2 Three sp2 orbitals Ethene structure Other carbon compounds and other molecules may be explained in a similar way. For example, ethene (C2H4) has a double bond between the carbons. For this molecule, carbon sp2 hybridizes, because one π (pi) bond is required for the double bond between the carbons and only three σ bonds are formed per carbon atom. In sp2 hybridization the 2s orbital is mixed with only two of the three available 2p orbitals,
- 3. 3 C* ↑↓ ↑ ↑ ↑ ↑ 1s sp2 sp2 sp2 2p forming a total of three sp2 orbitals with one remaining p orbital. In ethylene (ethene) the two carbon atoms form a σ bond by overlapping one sp2 orbital from each carbon atom. The π bond between the carbon atoms perpendicular to the molecular plane is formed by 2p–2p overlap all with 120° bond angles. The hydrogen–carbon bonds are all of equal strength and length, in agreement with experimental data. sp Two sp orbitals The chemical bonding in compounds such as alkynes with triple bonds is explained by sp hybridization. In this model, the 2s orbital is mixed with only one of the three p orbitals, C* ↑↓ ↑ ↑ ↑ ↑ 1s sp sp 2p 2p resulting in two sp orbitals and two remaining p orbitals. The chemical bonding in acetylene (ethyne) (C2H2) consists of sp–sp overlap between the two carbon atoms forming a σ bond and two additional π bonds formed by p–p overlap. Each carbon also bonds at 180° angles.