CHEMICAL BONDING Drawing Lewis structures for simple organic compounds Draw the Lewis structure for iodoethane (C2H5I): Be certain you include any lone pairs.
The Correct Answer and Explanation is:
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Explanation
To draw the Lewis structure for iodoethane (C₂H₅I), we follow a systematic process to ensure all atoms have a stable electron configuration.
First, we determine the total number of valence electrons. Carbon (C) is in Group 14 and has 4 valence electrons. Hydrogen (H) is in Group 1 and has 1 valence electron. Iodine (I), a halogen in Group 17, has 7 valence electrons. For C₂H₅I, the total count is (2 × 4) + (5 × 1) + (1 × 7) = 8 + 5 + 7 = 20 valence electrons.
Next, we arrange the atoms to form the molecular skeleton. In simple organic compounds, carbon atoms typically form a chain, so we connect the two carbons with a single bond (C-C). Hydrogen and halogen atoms are almost always terminal, meaning they are on the outside of the structure and form only one bond. We attach the five hydrogens and the one iodine to the carbon backbone, resulting in a structure of CH₃CH₂I.
Then, we draw single bonds to connect all the atoms in this skeleton. This structure has one C-C bond, five C-H bonds, and one C-I bond, for a total of seven single bonds. Since each bond consists of two electrons, we have used 7 × 2 = 14 electrons.
Finally, we distribute the remaining electrons as lone pairs to satisfy the octet rule for all atoms, except for hydrogen which follows the duet rule. We have 20 total valence electrons and have used 14, leaving 20 – 14 = 6 electrons. The hydrogen atoms are stable with their single bonds. Each carbon atom has four single bonds, fulfilling its octet with eight electrons. The iodine atom has one single bond (two electrons) and needs six more to complete its octet. We place the remaining six electrons on the iodine atom as three lone pairs. This uses all 20 valence electrons and satisfies the octet rule for every atom, resulting in the correct Lewis structure.
