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Chapter 10: Problem 27

The following polyatomic anions involve covalent bonds between O atoms and the central nonmetal atom. Propose an acceptable Lewis structure for each. (a) \(\mathrm{SO}_{3}^{2-} ;\) (b) \(\mathrm{NO}_{2}^{-} ;\) (c) \(\mathrm{CO}_{3}^{2-} ;\) (d) \(\mathrm{HO}_{2}^{-}\).

Short Answer

Expert verified
The Lewis structures for \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{NO}_{2}^{-}\), \(\mathrm{CO}_{3}^{2-}\), and \(\mathrm{HO}_{2}^{-}\) can be drawn by total the valence electrons, adding any extra for the charge of the ion, and then arranging them to fill octets as much as possible.

Step by step solution

01

\(SO_{3}^{2-}\) Lewis structure

Sulfur (S) has 6 valence electrons and each oxygen (O) also has 6. Since there are three O atoms, the total number of valence electrons is \(6+3*6 = 24\). Adding two more for the charge of the ion, we have 26 electrons available to fill the octets of the atoms. Place the S in the center with a single bond to each of the O atoms, using up 6 electrons, and then distribute the remaining electrons (20 of them) as lone pairs on the O atoms. There should be 8 electrons (two lone pairs and the bond) around each O, and 8 around the S (which can expand its octet as needed).
02

\(NO_{2}^{-}\) Lewis structure

Nitrogen (N) has 5 valence electrons and each oxygen (O) has 6, for a total of \(5+2*6 = 17\), and adding one more for the charge of the ion gives 18 electrons. With N in the center, single bond each O and leave 10 electrons. Distribute these as lone pairs on the oxygens, and put any remaining (two) on the nitrogen. Note that one of the O has a full octet, but the N and the other O have only 7 electrons, so this species is a radical.
03

\(CO_{3}^{2-}\) Lewis structure

Carbon (C) has 4 valence electrons and each oxygen (O) has 6, so the total is \(4+3*6 = 22\), and adding two for the charge of the ion gives 24 electrons to use. Place the carbon in the center, single bond it to all three of the oxygens, and distribute the remaining electrons (18 in total) as lone pairs on the oxygens. There should be 8 electrons around each O, and the C is short of two electrons. To solve this issue, create a double bond with one O atom, giving a full octet to each atom and making each oxygen equivalent to the others.
04

\(HO_{2}^{-}\) Lewis structure

Hydrogen (H) has 1 valence electron, and each oxygen (O) has 6, for a total of \(1+2*6 = 13\), and adding one for the charge of the ion gives 14 electrons to use. Place one O in center, single bond to both H and the other O, and then distribute the remaining electrons (10 in total). Each O should have 8 electrons around it (a full octet), with any remaining electron (after filling the octets of the O atoms) going on the hydrogen.

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Most popular questions from this chapter

Which of the following molecules would you expect to be polar: (a) \(\mathrm{HCN} ;\) (b) \(\mathrm{SO}_{3} ;\) (c) \(\mathrm{CS}_{2} ;\) (d) OCS; (e) \(\operatorname{SOCl}_{2} ;\) (f) \(\operatorname{SiF}_{4} ;\) (g) \(\operatorname{POF}_{3}\) ? Give reasons for your conclusions.

Briefly describe each of the following ideas: (a) formal charge; (b) resonance; (c) expanded valence shell; (d) bond energy.

Of the following species, the one with a triple covalent bond is (a) \(\mathrm{NO}_{3}^{-} ;\) (b) \(\mathrm{CN}^{-} ;\) (c) \(\mathrm{CO}_{2} ;\) (d) \(\mathrm{AlCl}_{3}\).

Without referring to tables in the text, indicate which of the following bonds you would expect to have the greatest bond length, and give your reasons. (a) \(\mathrm{O}_{2}\); (b) \(\mathrm{N}_{2} ;\) (c) \(\mathrm{Br}_{2} ;\) (d) \(\mathrm{BrCl}\).

Comment on the similarities and differences in the molecular structure of the following four-atom species: \(\mathrm{NO}_{3}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{SO}_{3}^{2-},\) and \(\mathrm{ClO}_{3}^{-}\).
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