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Chapter 9: Problem 65

In the sulphate ion, \(\mathrm{SO}_{4}^{2-}\), the sulphur atom is the central atom with the other 4 oxygen atoms attached to it. (a) Draw a Lewis structure for the sulphate ion. (b) What hybridization is exhibited by the \(\mathrm{S}\) atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

Short Answer

Expert verified
The Lewis structure of \(\mathrm{SO}_{4}^{2-}\) consists of a central Sulfur atom with two double bonds to two oxygen atoms and single bonds to the other two oxygen atoms. The Sulfur atom is sp³ hybridized, and there are three other equivalent resonance structures. The sulphate ion has a total of 4 pi electrons.

Step by step solution

01

Drawing a Lewis structure

First, we need to count the total number of valence electrons present in the sulphate ion, which is \(\mathrm{SO}_{4}^{2-}\). Sulfur has 6 valence electrons, and each oxygen atom has 6 valence electrons. Also, we need to consider the extra 2 electrons due to the negative charge on the ion. So, the total number of valence electrons is 6(S) + 6x4(O) + 2(charge) = 32 electrons. Now, let's draw a Lewis structure by arranging atoms with Sulfur at the center and Oxygen atoms surrounding it. Single bonds are drawn between Sulfur and each Oxygen atom initially. We also add three lone pairs of electrons to each Oxygen atom. Then we check for the octet rule on each atom.
02

Determine the hybridization of the Sulfur atom

To determine the hybridization, we need to know the electron geometry around the central atom, which in our case is Sulfur. We can figure out the electron geometry by counting the sigma bonds and lone pairs around the Sulfur atom.
03

Identify if there are equivalent resonance structures for the ion

Resonance structures occur when there is more than one valid Lewis structure for a molecule or ion. In the case of the sulphate ion, we need to identify these structures if they exist.
04

Count the number of electrons in the pi system of the ion

Pi systems consist of pi bonds, which are formed when p-orbitals overlap. We will count the number of electrons involved in these pi bonds to determine the total electrons in the pi system of the sulphate ion. Now the complete solution:
05

Drawing a Lewis structure

The Lewis structure of \(\mathrm{SO}_{4}^{2-}\) consists of a central Sulfur atom, connected with double bonds to two oxygen atoms, and single bonds to the other two oxygen atoms. Each singly-bonded oxygen atom has three lone pairs of electrons, whereas each doubly-bonded oxygen atom has two lone pairs of electrons. The Sulfur atom follows the expanded octet rule in this case.
06

Determine the hybridization of the Sulfur atom

As Sulfur is connected to four oxygen atoms, its electron geometry is tetrahedral. Therefore, the Sulfur atom is sp³ hybridized.
07

Identify if there are equivalent resonance structures for the ion

In the sulphate ion, resonance structures exist. We can draw three other equivalent resonance structures by changing the positions of double bonds among the oxygen atoms.
08

Count the number of electrons in the pi system of the ion

In the sulphate ion, there are two double bonds with one pi bond each. Each pi bond consists of 2 pi electrons, which means that there are a total of 2 x 2 = 4 pi electrons in the sulphate ion \(\mathrm{SO}_{4}^{2-}\).

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Most popular questions from this chapter

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 \mathrm{p}}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger O- \(O\) bond than \(\mathrm{O}_{2}\) itself.

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

For each statement, indicate whether it is true or false. (a) \(\ln\) order to make a covalent bond, the orbitals on each atom in the bond must overlap. (b) A \(p\) orbital on one atom cannot make a bond to an \(s\) orbital on another atom. \((\mathbf{c})\) Lone pairs of electrons on an atom in a molecule influence the shape of a molecule. (d) The 1 s orbital has a nodal plane. \((\mathbf{e})\) The \(2 p\) orbital has a nodal plane.

(a) An \(\mathrm{AB}_{6}\) molecule has no lone pairs of electrons on the \(\mathrm{A}\) atom. What is its molecular geometry? (b) An \(\mathrm{AB}_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four \(\mathrm{B}\) atoms). What is the electron-domain geometry around the A atom? (c) For the \(\mathrm{AB}_{4}\) molecule in part (b), predict the molecular geometry.
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