![]() These four 2 p atomic orbitals can be combined to give four molecular orbitals: two \(\pi\) (bonding) orbitals and two \(\pi\)* (antibonding) orbitals.Ĭ With 4 electrons available, only the \(\pi\) orbitals are filled. We have two unhybridized 2 p atomic orbitals left on carbon and two on nitrogen, each occupied by a single electron. Of 10 valence electrons (5 from N, 4 from C, and 1 from H), 4 are used for \(\sigma\) bonding:ī We are now left with 2 electrons on N (5 valence electrons minus 1 bonding electron minus 2 electrons in the lone pair) and 2 electrons on C (4 valence electrons minus 2 bonding electrons). This leaves one sp hybrid on each atom to either bond to hydrogen (C) or hold a lone pair of electrons (N). Because the nitrogen atom can also be described as sp hybridized, we can use one sp hybrid on each atom to form a C–N \(\sigma\) bond. Describe the bonding in HCN.Ī Because HCN is a linear molecule, it is likely that the bonding can be described in terms of sp hybridization at carbon. Fill the orbitals with the remaining electrons in order of increasing energy.Use any remaining unhybridized p orbitals to form \(\pi\) and \(\pi\)* orbitals. Determine the number of remaining valence electrons.Use the hybrid orbitals to form the \(\sigma\)-bonded framework of the molecule and determine the number of valence electrons that are used for \(\sigma\) bonding. From the geometry given, predict the hybridization in HCN.Given: chemical compound and molecular geometryĪsked for: bonding description using hybrid atomic orbitals and molecular orbitals ![]() The two electrons remaining are enough to fill only the bonding \(\pi\) orbital.ĭescribe the bonding in HCN using a combination of hybrid atomic orbitals and molecular orbitals. As in the diatomic molecules discussed previously, the singly occupied 2 p z orbitals in ethylene can overlap to form a bonding/antibonding pair of \(\pi\) molecular orbitals. Figure 4.11.1: Molecular Orbital Energy-Level Diagram for \(\pi\) Bonding in Ethylene. Because each 2 p z orbital has a single electron, there are only two \(\pi\) electrons, enough to fill only the bonding (\(\pi\)) level, leaving the \(\pi\)* orbital empty. The \(\pi\)* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis. With the formation of a \(\pi\) bonding orbital, electron density increases in the plane between the carbon nuclei. According to Molecular Orbital Theory, these two orbitals can be combined to form a \(\pi\) bonding orbital and a \(\pi\)* antibonding orbital, which produces the energy-level diagram shown in Figure 4.11.1. 2 p z) on neighbouring carbon atoms (part (b) in Figure 4.6.1). In Valence Bond Theory we described the \(\pi\) bonding in ethylene (C 2H 4) as the result of overlapping parallel 2p atomic orbitals (e.g.
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