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Chapter 18: Problem 55

(a) What are trihalomethanes (THMs)? (b) Draw the Lewis structures of two example THMs.

Short Answer

Expert verified
(a) Trihalomethanes (THMs) are chemical compounds comprising a central carbon atom bonded to three halogen atoms (chlorine, bromine, iodine, or fluorine) and one hydrogen atom. They are often formed as byproducts in the process of disinfecting drinking water and are considered environmental pollutants with potential adverse health effects. (b) Two example THMs are Trichloromethane (CHCl3) and Bromodichloromethane (CHBrCl2). The Lewis structure for Trichloromethane (CHCl3) is: H | Cl - C - Cl | Cl The Lewis structure for Bromodichloromethane (CHBrCl2) is: H | Br - C - Cl | Cl

Step by step solution

01

(a) Definition of Trihalomethanes (THMs)

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.
02

(b) Choosing Two Example THMs

Let's choose two example THMs for which we will draw the Lewis structures. We can choose Trichloromethane (CHCl3) and Bromodichloromethane (CHBrCl2).
03

(b1) Drawing the Lewis Structure for Trichloromethane (CHCl3)

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.
04

(b2) Drawing the Lewis Structure for Bromodichloromethane (CHBrCl2)

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.

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Most popular questions from this chapter

Why is the photodissociation of \(\mathrm{N}_{2}\) in the atmosphere relatively unimportant compared with the photodissociation of \(\mathrm{O}_{2}\) ? Human Activities and Earth's Atmosphere (Section 18.2)

The hydroxyl radical, \(\mathrm{OH}\), is formed at low altitudes via the reaction of excited oxygen atoms with water: $$ \mathrm{O}^{*}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{OH}(g) $$ (a) Write the Lewis structure for the hydroxyl radical. (Hint: It has one unpaired electron.) Once produced, the hydroxyl radical is very reactive. Explain why each of the following series of reactions affects the pollution in the troposphere: (b) \(\mathrm{OH}+\mathrm{NO}_{2} \longrightarrow \mathrm{HNO}_{3}\) (c) \(\mathrm{OH}+\mathrm{CO}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{OOH}\) \(\mathrm{OOH}+\mathrm{NO} \longrightarrow \mathrm{OH}+\mathrm{NO}_{2}\) (d) \(\mathrm{OH}+\mathrm{CH}_{4} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3}\) \(\mathrm{CH}_{3}+\mathrm{O}_{2} \longrightarrow \mathrm{OOCH}_{3}\) \(\mathrm{OOCH}_{3}+\mathrm{NO} \longrightarrow \mathrm{OCH}_{3}+\mathrm{NO}_{2}\) (e) The concentration of hydroxyl radicals in the troposphere is approximately \(2 \times 10^{6}\) radicals per \(\mathrm{cm}^{3}\). This estimate is based on a method called long path absorption spectroscopy (LPAS), similar in principle to the Beer's law measurement discussed in the Closer Look essay on p. 564 , except that the path length in the LPAS measurement is \(20 \mathrm{~km}\). Why must the path length be so large?

(a) How are the boundaries between the regions of the atmosphere determined? (b) Explain why the stratosphere, which is more than 20 miles thick, has a smaller total mass than the troposphere, which is less than 10 miles thick.

The enthalpy of evaporation of water is \(40.67 \mathrm{~kJ} / \mathrm{mol}\). Sunlight striking Earth's surface supplies \(168 \mathrm{~W}\) per square meter \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .\) (a) Assuming that evaporation of water is only due to energy input from the Sun, calculate how many grams of water could be evaporated from a 1.00 square meter patch of ocean over a 12 -hour day. (b) The specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). If the initial temperature of a 1.00 square meter patch of ocean is \(26^{\circ} \mathrm{C},\) what is its final temperature after being in sunlight for 12 hours, assuming no phase changes and assuming that sunlight penetrates uniformly to depth of \(10.0 \mathrm{~cm}\) ?

An impurity in water has an extinction coefficient of \(3.45 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(280 \mathrm{nm}\), its absorption maximum Closer Look, p. 564). Below 50 ppb, the impurity is not a problem for human health. Given that most spectrometers cannot detect absorbances less than 0.0001 with good reliability, is measuring the absorbance of water at \(280 \mathrm{nm}\) a good way to detect concentrations of the impurity above the 50 -ppb threshold?
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