The question of an element’s “charge” in a compound is understood through the concept of the oxidation state. This value is a bookkeeping tool used by chemists that represents the hypothetical charge an atom would have if electrons were completely transferred to the more electronegative atom in each bond. Unlike an ion’s actual charge, the oxidation state is a formal value that helps track the gain or loss of electrons in chemical reactions, especially in molecules where atoms share electrons rather than fully transferring them. Sulfur is unique because its position on the periodic table allows it to exhibit a wide variety of oxidation states, depending entirely on the atoms it bonds with.
Sulfur’s Ground State and Most Common Ionic Charge
Sulfur in its elemental form, typically found as a ring-shaped molecule containing eight atoms (\(S_8\)), has an oxidation state of zero. Because the sulfur atoms in this structure are identical, they share electrons equally, resulting in no charge difference.
The most common negative charge sulfur adopts when forming an ionic compound is \(-2\), resulting in the sulfide ion (\(S^{2-}\)). Sulfur naturally has six valence electrons, and gaining two electrons achieves a stable, full outer shell, mimicking the electron configuration of argon. This \(-2\) state is typically observed when sulfur bonds with metals or elements less electronegative than itself, such as in iron sulfide (FeS) or hydrogen sulfide (\(H_2S\)).
The Range of Possible Oxidation States for Sulfur
Sulfur’s unique chemistry allows it to exhibit a full spectrum of oxidation states ranging from a minimum of \(-2\) to a maximum of \(+6\). This extensive range is possible because sulfur atoms possess valence electrons in the \(3s\) and \(3p\) orbitals, and have accessible empty \(3d\) orbitals. The availability of these empty orbitals means sulfur can expand its electron sharing capacity beyond the typical octet rule when bonding with highly electron-attracting elements.
Positive oxidation states are observed when sulfur reacts with more electronegative elements, such as oxygen or the halogens, which pull shared electrons away from the sulfur atom, resulting in a positive oxidation state. The intermediate \(+4\) state is a common occurrence, found in compounds like sulfur dioxide (\(SO_2\)) and the sulfite ion (\(SO_3^{2-}\)).
The highest possible state, \(+6\), occurs when sulfur shares all six of its valence electrons with even more electronegative partners, like in the sulfate ion (\(SO_4^{2-}\)) or in sulfuric acid (\(H_2SO_4\)). Other, less common intermediate states, such as \(+1\) and \(+2\), can also be observed in certain complex structures, like the polysulfides or in compounds involving bonds between two or more sulfur atoms.
Methodology for Calculating Sulfur’s Charge
Determining the charge of sulfur in a compound requires applying a consistent set of rules for assigning oxidation states to the atoms involved. The foundational principle is that the sum of all oxidation states in a neutral compound must equal zero, or equal the overall charge if the compound is a polyatomic ion.
To calculate sulfur’s charge, one must first assign known, fixed oxidation states to the other elements in the compound. Oxygen is almost always assigned an oxidation state of \(-2\) in compounds. Hydrogen is typically assigned \(+1\) when bonded to a non-metal, and the alkali metals (Group 1) and alkaline earth metals (Group 2) are always \(+1\) and \(+2\), respectively.
Consider the calculation for sulfur’s oxidation state in the sulfate ion (\(SO_4^{2-}\)), which has an overall charge of \(-2\). The four oxygen atoms each have a state of \(-2\), contributing a total of \(-8\) to the ion’s charge. If \(x\) represents the unknown oxidation state of sulfur, the equation is \(x + 4(-2) = -2\). Solving for \(x\) reveals that the oxidation state of sulfur in the sulfate ion is \(+6\).
For a neutral compound like sulfur dioxide (\(SO_2\)), the sum of the oxidation states must be zero. The two oxygen atoms contribute a total charge of \(-4\). Therefore, if \(x\) is the oxidation state of sulfur, the equation is \(x + 2(-2) = 0\). This calculation demonstrates that sulfur in sulfur dioxide has an oxidation state of \(+4\).