Question

There are many Lewis structures you could draw for sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (each \(\mathrm{H}\) is bonded to an 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

(a) To satisfy the octet rule, draw a Lewis structure with S as the central atom, single bonds between S and each of the 4 O atoms, and single bonds between 2 O atoms and their respective H atoms, with the remaining 24 electrons distributed around the O atoms. In this structure, all atoms obey the octet rule. (b) To minimize formal charges, create double bonds between S and 2 of the O atoms while keeping the single bonds between the other 2 O atoms and their respective H atoms. In this structure, all formal charges are 0, satisfying both the octet rule and minimizing formal charges.

Step-by-step solution

Calculate Total Valence Electrons

First, we need to find the total number of valence electrons in H2SO4. The valence electrons for each element are: H = 1, S = 6, and O = 6. In sulfuric acid, there are 2 H atoms, 1 S atom, and 4 O atoms. Therefore, the total valence electrons are: Total Valence Electrons = (2 x 1) + (1 x 6) + (4 x 6) = 2 + 6 + 24 = 32

Place Central Atom and Identify Bonding Atoms

Sulfur (S) is the central atom as it has the lowest electronegativity in the compound. The outer atoms are the oxygen (O) and hydrogen (H) atoms. Place the S atom in the center and connect it with 4 O atoms using single bonds. Since H atoms can only bond once, connect each of the 2 remaining O atoms with 1 H atom each.

Distribute Remaining Electrons

To satisfy the octet rule, distribute the remaining electrons to the outer atoms (O and H). Currently, there are 4 single bonds, 1 between S and each O atom. A single bond accounts for 2 electrons, so 8 electrons are already used in the structure. There are 32 - 8 = 24 electrons left to distribute. Place the remaining 24 electrons around the outer O atoms (6 electrons around each O atom). Since H atoms can only have two electrons around them, there is no need to distribute the electrons to H. (a) At this point, all atoms satisfy the octet rule: S: 6 valence electrons + 4 single bonds = 6 + (4 x 2) = 14 electrons O: 6 valence electrons + 1 single bond + 6 non-bonding electrons = 6 + 2 + 6 = 14 electrons H: 1 valence electron + 1 single bond = 1 + 2 = 2 electrons

Calculate Formal Charges

Next, we calculate the formal charge on each atom: Formal Charge = Valence Electrons - (Non-bonding Electrons + 0.5 x Bonding Electrons) For the S atom: Formal Charge = 6 - (0 + 0.5 x 8) = 6 - 4 = +2 - For the O atoms bonded to the H atoms (we have two of them): Formal Charge = 6 - (6 + 0.5 x 2) = 6 - 7 = -1 - For the other O atoms (we have two of them): Formal Charge = 6 - (6 + 0.5 x 2) = 6 - 7 = -1 (b) To minimize the formal charge, create double bonds between S and 2 of the O atoms: S = O \\ O-S-O // O Now, the formal charges become 0 for all atoms: For the S atom: Formal Charge = 6 - (0 + 0.5 x 12) = 6 - 6 = 0 For all O atoms: Formal Charge = 6 - (4 + 0.5 x 4) = 6 - 6 = 0 Thus, the Lewis structure that satisfies the octet rule and minimizes the formal charge is the one with double bonds between S and two O atoms and single bonds between S and the remaining O atoms, each of which is bonded to an H atom.

Key concepts

Chemical Bonding

Understanding chemical bonding is fundamental to grasping the structure of sulfuric acid, \( \mathrm{H}_{2}\mathrm{SO}_{4} \). Chemical bonds form when atoms share or transfer electrons to achieve a stable electron configuration, usually the octet rule for many elements. In the Lewis structure of \( \mathrm{H}_{2}\mathrm{SO}_{4} \), sulfur bonds with oxygen and hydrogen atoms to fulfill this stable configuration.

At the core of sulfuric acid's structure is the element sulfur (S), which acts as a central atom. It is surrounded by oxygen (O) atoms, two of which are attached to hydrogen (H) atoms. We know that electron sharing occurs through the formation of covalent bonds. Initially, each S-O bond is a single bond; however, to minimize formal charge, some of these bonds become double bonds. This is a key characteristic that affects the shape and reactivity of the molecule. Double bonds are stronger and shorter than single bonds and play a critical role in the overall stability of the molecule.

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom that can participate in the formation of chemical bonds. In the exercise, the number of valence electrons is used to construct the Lewis structure of sulfuric acid. Hydrogen has 1 valence electron, sulfur has 6, and oxygen has 6. The summation of these electrons gives us the total to work with—32 valence electrons for \( \mathrm{H}_{2}\mathrm{SO}_{4} \).

The arrangement of these electrons is the first step in drawing our Lewis structure. Sulfuric acid is unique because while hydrogen generally forms single bonds, sulfur and oxygen can form multiple bond types. In structuring the molecule, it's essential to place these valence electrons not only to form bonds but also to fill the octets of the oxygen atoms, adhering to the octet rule. The remaining electrons are assigned as lone pairs or part of double bonds to minimize the formal charge and satisfy the octet requirement for each atom.

Formal Charge

Formal charge is a bookkeeping tool used in chemistry to determine the charge distribution within a molecule. The lower the formal charge, the more stable the molecule. To compute the formal charge, the formula is:\[\text{Formal Charge} = \text{Valence Electrons} - (\text{Non-bonding Electrons} + 0.5 \times \text{Bonding Electrons})\]

In the example of \( \mathrm{H}_{2}\mathrm{SO}_{4} \), we've seen that initially, the sulfur atom had a formal charge of +2, and the oxygen atoms had a formal charge of -1. To achieve a more stable structure with lower formal charges, we rearrange electrons to form double bonds. This reduces the formal charge on sulfur to 0, while also bringing the oxygen atoms' formal charges to 0. A Lewis structure that minimizes formal charge tends to be the most representative of the molecule's actual electronic structure, and this consideration is crucial when predicting the behavior and reactivity of molecules.