Question

Write a Lewis structure for each of the following polyatomic ions. Show all bonding valence electron pairs as lines and all nonbonding valence electron pairs as dots. For those ions that exhibit resonance, draw the various possible resonance forms. a. chlorate ion, \(\mathrm{ClO}_{3}^{-}\) b. peroxide ion, \(\mathrm{O}_{2}^{2-}\) c. acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Short answer

The Lewis structures for the given polyatomic ions are as follows: a. Chlorate ion, \(\mathrm{ClO}_{3}^{-}\) has three resonance structures: Cl - O - O - O .. .. .. b. Peroxide ion, \(\mathrm{O}_{2}^{2-}\) has one structure: .. O - O .. .. c. Acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) has two resonance structures: .. .. O - O \ \ C - C - O <-> O - C - C / O / H .. H -/ \ H

Step-by-step solution

Determine the total number of valence electrons

We will first calculate the total number of valence electrons for the entire ion. Chlorine (Cl) has 7 valence electrons, oxygen (O) has 6 valence electrons, and there are 3 oxygen atoms. Since the ion has a charge of -1, we should add one more electron. The total number of valence electrons is: \( 7 (from\:Cl) + 3 \times 6 (from\:O) + 1 (from\:charge) = 26 \)

Determine the central atom

Chlorine will be the central atom, as it is least electronegative.

Attach the oxygen atoms to the central atom

Connect the chlorine atom with the 3 oxygen atoms using single bonds. Each single bond consists of 2 electrons: O \ Cl - O / O

Distribute the remaining valence electrons

We have used 6 electrons for the three single bonds, so we still have 20 electrons left: \( 26 - 6 = 20 \) We will now distribute the remaining 20 electrons to fulfill the octet rule for all atoms (8 electrons around each atom): O .. \ Cl - O .. / O ..

Check for resonance structures

We can see that the chlorate ion can form resonance structures by shifting the double bonds with the oxygen atoms: Cl - O - O - O <-> O - Cl - O - O <-> O - O - Cl - O So, the three resonance structures for chlorate ion are: Cl - O - O - O .. .. .. #b. Peroxide Ion, \(\mathrm{O}_{2}^{2-}\)#

Determine the total number of valence electrons

Oxygen has 6 valence electrons, and there are 2 oxygen atoms. Since the ion has a charge of -2, we should add 2 more electrons. The total number of valence electrons is: \( 2 \times 6 (from\:O) + 2 (from\:charge) = 14 \)

Attach the oxygen atoms to each other

Connect the two oxygen atoms using a single bond, using 2 electrons: O - O

Distribute the remaining valence electrons

We have used 2 electrons for the single bond, so we still have 12 electrons left: \( 14 - 2 = 12 \) Distribute the remaining 12 electrons to fulfill the octet rule for the oxygen atoms (8 electrons around each atom): .. O - O .. .. There is no need for resonance structures in the peroxide ion. #c. Acetate Ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)#

Determine the total number of valence electrons

Carbon has 4 valence electrons, hydrogen has 1 valence electron, and oxygen has 6 valence electrons. Since the ion has a charge of -1, we should add one more electron. The total number of valence electrons is: \( 2 \times 4 (from\:C) + 3 \times 1 (from\:H) + 2 \times 6 (from\:O) + 1 (from\:charge) = 18 \)

Determine the skeletal structure

Build a skeletal structure with 2 carbons and 2 oxygens attached to the central carbon. Then, attach the hydrogen atoms to the carbons: O \ C - C - O / | H H \ H

Distribute the remaining valence electrons

We have used 12 electrons for bonding (5 single bonds and 1 double bond). We still have 6 electrons left: \( 18 - 12 = 6 \) Distribute the remaining 6 electrons to fulfill the octet rule for the oxygen atoms: .. O \\ \ C - C - O / | H H \ H

Check for resonance structures

We can see that the acetate ion can form resonance structures by shifting the double bond with the oxygen atoms: .. O \\ <-> \\ \ O - C - C - O - C - C - O / / O H - C H .. \ \ H H Hence, the two resonance structures for the acetate ion are: .. .. O - O \ \ C - C - O <-> O - C - C / O / H .. H -/ \ H