Question

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

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

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.

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.

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.

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:

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.

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.

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.

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 \(\mathrm{SO}_{4}^{2-}\).

Key concepts

Sulphate Ion

The sulphate ion, \(\mathrm{SO}_{4}^{2-}\), is a polyatomic ion made up of one sulfur atom surrounded by four oxygen atoms. The sulfur atom is the central component in this ion, functioning as the anchor to which other atoms are connected.
The ion overall holds a \(2-\) charge, which derives from having more electrons than protons, resulting in ionic stability.

• Sulfur has 6 valence electrons naturally, as does each oxygen atom.
• The negative charge contributes an additional 2 electrons, making the total 32 valence electrons.

When drawing the Lewis structure, these electrons are strategically placed around the sulfur and oxygen atoms to satisfy the octet rule for each, except sulfur which can use its d-orbitals to accommodate more electrons than typical. This results in a stable configuration for this ion.

Hybridization

Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals that are degenerate in energy and suitable for pairing electrons to form chemical bonds.
In the case of the sulphate ion, determining the hybridization involves examining the environment around the sulfur atom.

• There are four oxygen atoms connected to the sulfur atom.
• This formation indicates a tetrahedral geometry.

Because the sulfur atom forms four sigma bonds with the oxygen atoms, it utilizes one s orbital and three p orbitals to hybridize, resulting in sp³ hybridization. This type of hybridization helps explain the shape and bonding of the sulphate ion.

Resonance Structures

Resonance structures are different ways of arranging the electron pairs in molecules that can't be represented by a single Lewis structure. For the sulphate ion, there are multiple equivalent resonance structures possible. These arise from the different placements of double bonds between sulfur and oxygen.

• The resonance phenomenon occurs because electrons are delocalized, meaning they can be spread across different areas.
• In sulphate, this is represented by rotating the position of double bonds among the four oxygen atoms.

This electron delocalization contributes to the stability of the ion. It showcases how no single structure can accurately depict the binding within the ion, as the actual structure is a hybrid of all resonance possibilities.

Pi Bonds

Pi bonds are a type of covalent bond that form from the sideways overlap of p orbitals. These bonds exist above and below the axis of the bonded atoms -- unlike sigma bonds, which form directly between the atoms.

• In the Lewis structure of \(SO_4^{2-}\), there are two double bonds between sulfur and oxygen, which include pi bonds.
• Each double bond in the ion consists of one sigma bond and one pi bond.

Calculating the electrons in pi systems involves multiplying the pi bonds by the two electrons each contains, totaling four pi electrons within the sulphate ion. These contribute to the ion's overall structure and its chemical reactivity.