Problem 60 - Chemistry and Chemical Reactivity

Question

Sulfur forms a range of compounds with fluorine. Draw Lewis electron dot structures for \(\mathrm{S}_{2} \mathrm{F}_{2}\) (connectivity is FSSF), \(\mathrm{SF}_{2}, \mathrm{SF}_{4}, \mathrm{SF}_{6},\) and \(\mathrm{S}_{2} \mathrm{F}_{10} .\) What is the formal oxidation number of sulfur in each of these compounds?

Step-by-Step Solution

Answer

S in S2F2 = +1, SF2 = +2, SF4 = +4, SF6 = +6, S2F10 = +5.

Step 1: Draw the Lewis structure for S2F2

Start by identifying the total number of valence electrons in \(\mathrm{S}_{2} \mathrm{F}_{2}\). Sulfur (S) has 6 valence electrons, and fluorine (F) has 7. Thus, \(2 \times 6 + 2 \times 7 = 26\) valence electrons total. Arrange the atoms according to the given connectivity FSSF. Use two electrons (one pair) to connect each atom, then place the remaining electrons to satisfy the octet rule, starting with the outer atoms. Each fluorine atom forms a single bond with sulfur, and the two sulfurs are also directly bonded. Distribute remaining electrons to complete octets, forming lone pairs where necessary.

Step 2: Determine Oxidation State for S2F2

In \(\mathrm{S}_{2} \mathrm{F}_{2}\), assign oxidation states. Fluorine has an oxidation state of -1. Using the connectivity and knowing that total charge is 0, we can write the equation: \(2S + 2(-1) = 0\)(since there are two fluorines). This gives \(2S = +2\), so each S is +1.

Step 3: Draw the Lewis structure for SF2

Consider \(\mathrm{SF}_{2}\). Calculate the valence electrons: Sulfur has 6, and two fluorines contribute 7 each, giving \(6 + 2 \times 7 = 20\). Form a single bond between sulfur and each fluorine. Distribute remaining electrons to complete octets, placing lone pairs where necessary. Sulfur should have a lone pair after completing fluorine's octets.

Step 4: Determine Oxidation State for SF2

For \(\mathrm{SF}_{2}\), fluorine is -1, and there are two fluorines: \(S + 2(-1) = 0\). Thus, \(S = +2\).

Step 5: Draw the Lewis structure for SF4

For \(\mathrm{SF}_{4}\), calculate valence electrons. Sulfur has 6, fluorine has 7; \(6 + 4 \times 7 = 34\). Form single bonds between sulfur and each fluorine. Remaining electrons form lone pairs, and since sulfur can expand its octet, place a lone pair on sulfur to form a see-saw shape.

Step 6: Determine Oxidation State for SF4

In \(\mathrm{SF}_{4}\), each fluorine is -1, contributing \(4 imes (-1)\). Hence, \(S + 4(-1) = 0\), leading to \(S = +4\).

Step 7: Draw the Lewis structure for SF6

In \(\mathrm{SF}_{6}\), sulfur has 6 valence electrons and each fluorine has 7, totaling \(6 + 6 \times 7 = 48\). Form single bonds between sulfur and each fluorine, using all valence electrons.

Step 8: Determine Oxidation State for SF6

In \(\mathrm{SF}_{6}\), each fluorine is -1, so \(S + 6(-1) = 0\), meaning \(S = +6\).

Step 9: Draw the Lewis structure for S2F10

Consider \(\mathrm{S}_{2} \mathrm{F}_{10}\). Calculate the valence electrons: \(12 + 10 \times 7 = 82\). Connect two sulfur atoms, and each sulfur connects to five fluorines with single bonds, using the electrons to satisfy octets.

Step 10: Determine Oxidation State for S2F10

For \(\mathrm{S}_{2} \mathrm{F}_{10}\), since fluorine is -1, the equation is \(2S + 10(-1) = 0\). Solving gives \(2S = +10\) or \(S = +5\).

Key Concepts

Lewis Structures

Lewis structures are visual representations of molecules that display the bonding between atoms and the lone pairs of electrons that may exist. They help us understand the arrangement and number of valence electrons around different atoms in a molecule. To draw a Lewis structure, we need to know the total number of valence electrons available. For example, in the compound \(\mathrm{S}_{2} \mathrm{F}_{2}\) with the connectivity FSSF, sulfur (S) has six valence electrons and fluorine (F) has seven. Together, this contributes a sum of 26 valence electrons. When constructing the structure, pair the electrons to form bonds and arrange the remaining electrons to satisfy the octet rule for peripheral atoms like fluorine, seeking the most stable configuration for the structure.

Oxidation States

An oxidation state, or oxidation number, reflects the distribution of electrons among the bonded atoms in a chemical compound. It helps to predict the transfer of electrons in a chemical reaction. Generally, the oxidation state is represented by integers, where negative numbers indicate a gain of electrons and positive numbers indicate a loss.
The oxidation state of sulfur in \(\mathrm{S}_{2} \mathrm{F}_{2}\) can be computed by knowing that the oxidation state of fluorine is always -1. With two sulfur atoms and two fluorines, the equation for the total oxidation state is \(2\mathrm{S} + 2(-1) = 0\), resolving to sulfur having an oxidation state of +1 per atom. Understanding oxidation states is crucial as it affects the chemical behavior and reactivity of the compound.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a significant role in chemical bonding. These electrons determine how atoms interact with one another and form compounds. For instance, sulfur, a group 16 element, typically has six valence electrons, while fluorine has seven as a group 17 element.
When forming the Lewis structures for sulfur fluoride compounds like \(\mathrm{SF}_{2}\) or \(\mathrm{SF}_{4}\), it's essential to keep track of the valence electrons. The total number of valence electrons dictates how atoms share electrons to form bonds, ensuring each atom reaches a stable electronic configuration, usually resembling the nearest noble gas structure.

Octet Rule

The octet rule is a chemical guideline that states atoms tend to bond in such a manner that they each have eight electrons in their valence shell, achieving a stable electronic configuration similar to noble gases. This rule is predominantly applicable to the main-group elements, like sulfur and fluorine.
While drawing a Lewis structure, such as for \(\mathrm{SF}_{4}\) or \(\mathrm{SF}_{6}\), we aim to distribute electrons such that sulfur and fluorine attain a complete set of eight electrons around them. However, sulfur can exceed the octet rule, allowing it to hold more than eight electrons, thanks to its ability to utilize the 3d orbitals. This is evident in \(\mathrm{SF}_{6}\), where sulfur forms six bonds, containing a total of 12 electrons around it. This exception to the octet rule highlights sulfur's versatility in forming diverse sulfur fluoride compounds.