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Chapter 7: Problem 21

Draw resonance structures for (a) \(\mathrm{Cl}-\mathrm{NO}_{2}\) (b) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{N}-\mathrm{N}\) (c) \(\mathrm{SO}_{3}\)

Short Answer

Expert verified
Question: Determine the resonance structures for each of the following molecules: Cl-NO2, H2C-N-N, and SO3. Answer: For Cl-NO2, there are two resonance structures: 1. Cl-N=O-O 2. Cl-N-O=O For H2C-N-N, there are no resonance structures. For SO3, there are three resonance structures: 1. O=S-O-O 2. O-O=S-O 3. O-O-O=S

Step by step solution

01

(1. Drawing the Lewis structure of Cl-NO2)

To draw the Lewis structure for \(\mathrm{Cl}-\mathrm{NO}_{2}\), first, count the total number of valence electrons present in the molecule. Chlorine has 7 valence electrons, nitrogen has 5 valence electrons, and oxygen has 6 valence electrons. So, the total number of valence electrons is 7 + 5 + 2(6) = 24. Next, determine the central atom, which, in this case, is nitrogen. Start by placing the nitrogen atom in the center and connect it to chlorine and the two oxygen atoms. Now, place the remaining valence electrons in pairs as needed to satisfy the octet rule for each atom.
02

(2. Drawing the resonance structures for Cl-NO2)

Now we can determine if any resonance structures can be formed. The chlorine atom has three lone pairs, and nitrogen has one lone pair. Oxygen atoms each have two lone pairs. There is a possibility to form two resonance structures by moving one of the lone pairs of electrons from the chlorine atom to form a double bond with nitrogen, while breaking one of the double bonds of nitrogen with an oxygen atom. In this way, the nitrogen atom maintains its octet. These two resonance structures can be shown as below: Resonance structure 1: Cl-N=O-O Resonance structure 2: Cl-N-O=O
03

(3. Drawing the Lewis structure of H2C-N-N)

To draw the Lewis structure for \(\mathrm{H}_{2} \mathrm{C}-\mathrm{N}-\mathrm{N}\), count the total number of valence electrons. Carbon has 4 valence electrons, nitrogen has 5 valence electrons, and hydrogen has 1 valence electron. The total number of valence electrons is 2(1) + 4 + 2(5) = 16. Place the carbon atom in the center, connect it to the nitrogen atom, and connect the nitrogen atom to the other nitrogen atom. Attach the hydrogen atoms to the carbon. Fill the remaining valence electrons in pairs as needed to satisfy the octet rule.
04

(4. Drawing the resonance structures for H2C-N-N)

Check for possible resonance structures. Here, there is no need to reorganize any lone pairs of electrons, and no resonance structures can be formed for \(\mathrm{H}_{2} \mathrm{C}-\mathrm{N}-\mathrm{N}\).
05

(5. Drawing the Lewis structure of SO3)

To draw the Lewis structure for \(\mathrm{SO}_{3}\), first count the total number of valence electrons present in the molecule. Sulfur has 6 valence electrons, and oxygen has 6 valence electrons. So, the total number of valence electrons is 6 + 3(6) = 24. Place the sulfur atom in the center and connect it to three oxygen atoms. Fill the remaining electrons as pairs to satisfy the octet rule.
06

(6. Drawing the resonance structures for SO3)

Now, check for possible resonance structures. In this case, there are three resonance structures, which can be formed by moving one double bond between sulfur and one of the oxygen atoms while making the other two sulfur-oxygen bonds single. Resonance structure 1: O=S-O-O Resonance structure 2: O-O=S-O Resonance structure 3: O-O-O=S These are the resonance structures for each of the given molecules.

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Most popular questions from this chapter

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