WEBVTT 00:00:03.330 --> 00:00:07.610 Butane, C4H10, has the structural formula shown. 00:00:08.180 --> 00:00:11.280 What is the hybridization of the carbon atom in the butane? 00:00:11.900 --> 00:00:19.870 (A) sp3, (B) sp, (C) sp2, or (D) sp3d. 00:00:20.980 --> 00:00:27.570 Orbital hybridization is the concept of mixing atomic orbitals into new and different types of hybrid orbitals. 00:00:28.350 --> 00:00:33.460 Carbon atoms have the He 2s2 2p2 electron configuration. 00:00:34.540 --> 00:00:46.020 And it would be reasonable to assume that the 2s and 2p atomic orbitals have different interactions with covalently bonded hydrogen atoms in molecules like butane, considering their different energies and shapes. 00:00:46.370 --> 00:00:49.880 But experimental data suggests this is not the case at all. 00:00:50.950 --> 00:00:58.480 The 2s and 2p orbitals tend to mathematically combine to form four hybrid sp3 bonding orbitals. 00:00:59.480 --> 00:01:07.410 They all make equivalent bonds with the 1s orbital electrons of adjacent hydrogen atoms and spread out equally in three-dimensional space. 00:01:08.060 --> 00:01:10.250 Let’s see how this hybridization will occur. 00:01:11.060 --> 00:01:13.650 This is carbon’s valence electron configuration. 00:01:14.040 --> 00:01:18.260 Carbon has valence electrons in the 2s and 2p atomic orbitals. 00:01:19.300 --> 00:01:23.690 The first step in orbital hybridization is the promotion of an electron. 00:01:24.160 --> 00:01:30.880 For carbon, a 2s electron is promoted to an empty 2p orbital, which puts carbon in an excited state. 00:01:32.000 --> 00:01:36.740 Now, carbon has four unpaired electrons in the 2s and 2p orbitals. 00:01:38.220 --> 00:01:45.460 These four unpaired electrons in different orbitals are mixed together to create four identical orbitals of the same energy level. 00:01:45.980 --> 00:01:50.410 These four identical orbitals are neither s orbitals nor p orbitals. 00:01:50.820 --> 00:01:53.020 They are sp3 hybrid orbitals. 00:01:54.270 --> 00:01:58.150 Therefore, option (A), sp3, is the correct answer.