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Chapter 8: Problem 67

There are many Lewis structures you could draw for sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (each \(\mathrm{H}\) is bonded to an \(\mathrm{O}\) ). (a) What Lewis structure(s) would you draw to satisfy the octet rule? (b) What Lewis structure(s) would you draw to minimize formal charge?

Short Answer

Expert verified
To satisfy the octet rule, the Lewis structure for H₂SO₄ would be: \( H-O-S-O \) | X X O O However, to minimize formal charges, the Lewis structure for H₂SO₄ would be: \( H-O-S(O_{3})(O_{4}) \) || | || O_{1} O_{2} In this structure, O₁ and O₂ have double bonds with sulfur, O₃ and O₄ have single bonds, and all atoms have formal charges of 0 or -1.

Step by step solution

01

Calculate valence electrons

H has 1, S has 6, and O has 6 valence electrons each. Since there are 2 H atoms, 1 S atom, and 4 O atoms, we will have a total of 2(1) + 6 + 4(6) = 2 + 6 + 24 = 32 valence electrons.
02

Arrange atoms and add bonds

We will place the S atom in the center, with the 4 O atoms surrounding it and the two H atoms bonded to two of the O atoms. This gives us the following initial structure: \( H-O-S-O \) | X X O O Each S-O bond and O-H bond counts as 2 electrons. This accounts for 2(4) + 2(2) = 12 of the 32 valence electrons.
03

Distribute remaining electrons

We have 32 (total) - 12 (used for bonds) = 20 electrons left. Each oxygen should have 6 electrons around it to fulfill the octet rule, so we will add 4 electrons to each of the two O atoms not bonded to H and 4 more electrons to each of the two O atoms bonded to H.
04

Lewis structure satisfying the octet rule

The resulting Lewis structure for H₂SO₄ that satisfies the octet rule is: \( H-O-S-O \) | X X O O b. Drawing Lewis structure(s) to minimize formal charge#
05

Calculate formal charges

We can calculate the formal charge for each atom in the Lewis structure for H₂SO₄ that we obtained in step (a). For H, O, and S, the formal charges are: \( H: 1 - 1 = 0 \) (Each hydrogen has 0 formal charge.) \( O_{1,2}: 6 - (2 + 6) = -2 \) (The oxygen atoms bonded to hydrogen have -2 formal charge each) \( O_{3,4}: 6 - (2 + 4) = 0 \) (The oxygen atoms not bonded to hydrogen have 0 formal charge.) \( S: 6 - (0 + 4) = +2 \) (Sulfur has a formal charge of +2.)
06

Adjust bonds to minimize formal charge

To minimize the formal charge on the oxygen atoms (O₁ and O₂) bonded to hydrogen, we can convert one lone pair from each of these oxygens into a bond pair with sulfur, creating a double bond (S=O). This will reduce the formal charge on sulfur to zero and on both oxygen atoms to -1.
07

Lewis structure minimizing formal charge

The resulting Lewis structure for H₂SO₄ that minimizes formal charge is: \( H-O-S(O_{3})(O_{4}) \) || | || O_{1} O_{2} Here, O₁ and O₂ have double bonds with sulfur, O₃ and O₄ have single bonds, and all atoms now have formal charges of 0 or -1.

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Most popular questions from this chapter

The hypochlorite ion, \(\mathrm{ClO}^{-},\) is the active ingredient in bleach. The perchlorate ion, \(\mathrm{ClO}_{4}^{-},\) is a main component of rocket propellants. Draw Lewis structures for both ions.

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