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Chapter 22: Problem 47

Write the Lewis structure for each of the following species, and indicate the structure of each: (a) \(\mathrm{SeO}_{3}{ }^{2-} ;\) (b) \(\mathrm{S}_{2} \mathrm{Cl}_{2} ;(\mathrm{c})\) chlorosulfonic acid, \(\mathrm{HSO}_{3} \mathrm{Cl}\) (chlorine is bonded to sulfur).

Short Answer

Expert verified
The Lewis structures for the given species are as follows: (a) \(\mathrm{SeO}_3^{2-}\): O \ Se = O / O (b) \(\mathrm{S}_2\mathrm{Cl}_2\): S - Cl | S - Cl (c) \(\mathrm{HSO}_3\mathrm{Cl}\) (Chlorine is bonded to Sulfur): O \ S - Cl / \ H - O

Step by step solution

01

Count valence electrons

To determine the total number of valence electrons, add the valence electrons of each atom and the additional electrons from the negative charge. - Selenium (Se): 6 valence electrons (it's in group 16 of the periodic table) - Oxygen (O): 6 valence electrons each - The 2- charge adds 2 more electrons Total number of valence electrons = 6(Se) + 3*6(O) + 2(charge) = 6 + 18 + 2 = 26
02

Place the central atom and arrange the others around it

Put the least electronegative atom in the center (Se in this case) and arrange the oxygens around it. Se | O
03

Connect the atoms with single bonds

Connect each oxygen to the central selenium atom using a single bond. O \ Se / O
04

Distribute the remaining electrons

Distribute the remaining electrons to complete the octet for each atom. We have used 6 electrons thus far for the single bonds, so there are 20 left to distribute. The Lewis structure becomes: O \ Se = O / O Each oxygen atom has now an octet configuration, and there are no more electrons to distribute. (b) \(\mathrm{S}_2\mathrm{Cl}_2\)
05

Count valence electrons

Determine the total number of valence electrons, adding the valence electrons of each atom. - Sulfur (S): 6 valence electrons each - Chlorine (Cl): 7 valence electrons each Total number of valence electrons = 2*6(S) + 2*7(Cl) = 12 + 14 = 26
06

Arrange and connect the atoms

Since this molecule contains two of each atom, we should connect one S with one Cl to create S-Cl bonds: S - Cl | S - Cl
07

Distribute the remaining electrons

Distribute the remaining electrons to complete the octet for each atom. We have used 4 electrons for the single bonds, so there are 22 left. The final Lewis structure is: S - Cl | S - Cl Each atom has now an octet configuration. (c) \(\mathrm{HSO}_3\mathrm{Cl}\) (Chlorine is bonded to Sulfur)
08

Count valence electrons

Determine the total number of valence electrons, adding the valence electrons of each atom. - Hydrogen (H): 1 valence electron - Sulfur (S): 6 valence electrons - Oxygen (O): 6 valence electrons each - Chlorine (Cl): 7 valence electrons Total number of valence electrons = 1(H) + 6(S) + 3*6(O) + 7(Cl) = 1 + 6 + 18 + 7 = 32
09

Arrange and connect the atoms

Put the least electronegative atom in the center, which is S in this case. Then arrange the other atoms around it and connect them with single bonds: H \ S - O / \ O Cl
10

Distribute the remaining electrons

Distribute the remaining electrons to complete the octet for each atom. We have used 8 electrons for the single bonds, so there are 24 left. The final Lewis structure is: O \ S - Cl / \ H - O Each atom has now an octet configuration.

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Most popular questions from this chapter

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the hightemperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\)

(a) Draw the Lewis structures for at least four species that have the general formula $$ [: X=Y:]^{n} $$ where \(\mathrm{X}\) and \(\mathrm{Y}\) may be the same or different, and \(n\) may have a value from +1 to -2 . (b) Which of the compounds is likely to be the strongest Bronsted base? Explain. [Sections \(22.1,22.7,\) and 22.9\(]\)

Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.

Write a balanced equation for each of the following reactions: (a) Burning magnesium metal in a carbon dioxide atmosphere reduces the \(\mathrm{CO}_{2}\) to carbon. (b) In photosynthesis, solar energy is used to produce glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and \(\mathrm{O}_{2}\) from carbon dioxide and water. (c) When carbonate salts dissolve in water, they produce basic solutions.

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the (a) \(\mathrm{NH}_{4}^{+},(\mathbf{b}) \mathrm{NO}_{2}^{-}\), nitrogen: (c) \(\mathrm{N}_{2} \mathrm{O},(\mathrm{d}) \mathrm{NO}_{2}\)
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