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Trihalomethanes (THMs) are a group of chemical compounds that consist of a central carbon atom bonded to three halogen atoms (chlorine, bromine, iodine, or fluorine) and one hydrogen atom. They are usually formed as byproducts in the process of disinfecting drinking water with chemicals such as chlorine. THMs are considered environmental pollutants and can have adverse health effects, so their presence in drinking water is monitored and regulated.

Let's choose two example THMs for which we will draw the Lewis structures. We can choose Trichloromethane (CHCl3) and Bromodichloromethane (CHBrCl2).

To draw the Lewis structure for Trichloromethane (CHCl3), follow these steps: 1. Count the total number of valence electrons: Carbon (C) has 4, Hydrogen (H) has 1, and each Chlorine (Cl) has 7. The total number of valence electrons is 4 + 1 + 3 * 7 = 26. 2. Place the least electronegative atom (C) in the center, and arrange the other atoms around it (H and Cl). 3. Add single bonds between the central atom and the surrounding atoms, using 2 valence electrons for each bond: 26 - 4 * 2 = 18 valence electrons remain. 4. Distribute the remaining valence electrons as lone pairs around the surrounding atoms (H and Cl), starting with the most electronegative ones (each Cl should complete an octet, while H only needs a duet): 18 - 3 * 2 * 2 = 6 valence electrons remain after completing the octets of the Cl atoms. 5. Distribute the leftover valence electrons as lone pairs on the central atom (C): 6 - 2 = 4 valence electrons remain on the C atom, completing its octet.

To draw the Lewis structure for Bromodichloromethane (CHBrCl2), follow these steps: 1. Count the total number of valence electrons: Carbon (C) has 4, Hydrogen (H) has 1, Bromine (Br) has 7, and each Chlorine (Cl) has 7. The total number of valence electrons is 4 + 1 + 7 + 2 * 7 = 26. 2. Place the least electronegative atom (C) in the center, and arrange the other atoms around it (H, Br, and Cl). 3. Add single bonds between the central atom and the surrounding atoms, using 2 valence electrons for each bond: 26 - 4 * 2 = 18 valence electrons remain. 4. Distribute the remaining valence electrons as lone pairs around the surrounding atoms (H, Br, and Cl), starting with the most electronegative ones (each Cl and Br should complete an octet, while H only needs a duet): 18 - 3 * 2 * 2 = 6 valence electrons remain after completing the octets of the Cl and Br atoms. 5. Distribute the leftover valence electrons as lone pairs on the central atom (C): 6 - 2 = 4 valence electrons remain on the C atom, completing its octet.