Give the oxidation state of sulfur in each substance: (a) \(\mathrm{S}_{8}\) (b) \(\mathrm{SF}_{4}\) (c) \(\mathrm{SF}_{6}\) (d) \(\mathrm{H}_{2} \mathrm{~S}\) (e) \(\mathrm{FeS}_{2}\) (f) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (g) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
a) 0, b) +4, c) +6, d) -2, e) -1, f) +6, g) +2 (average, usually 0 and +4)

Step by step solution

01

Determine the Oxidation State in \(\text{S}_8\)

Elemental sulfur in the form of \(\text{S}_8\) has an oxidation state of 0. This is because each sulfur atom is bonded to other sulfur atoms in the same element form with no charge.
02

Find the Oxidation State in \(\text{SF}_4\)

Fluorine typically has an oxidation state of \(-1\). In \(\text{SF}_4\), let the oxidation state of sulfur be \(x\):\[ x + 4(-1) = 0 \] Solve for \(x\):\[ x - 4 = 0 \]\[ x = +4 \]
03

Calculate the Oxidation State in \(\text{SF}_6\)

Similarly, in \(\text{SF}_6\), let \(x\) be the oxidation state of sulfur. Fluorine is \(-1\) and there are six fluorines:\[ x + 6(-1) = 0 \] Solve for \(x\):\[ x - 6 = 0 \]\[ x = +6 \]
04

Determine the Oxidation State in \(\text{H}_2\text{S}\)

Hydrogen typically has an oxidation state of \(+1\). In \(\text{H}_2\text{S}\), let the oxidation state of sulfur be \(x\):\[ 2(+1) + x = 0 \] Solve for \(x\):\[ 2 + x = 0 \]\[ x = -2 \]
05

Find the Oxidation State in \(\text{FeS}_2\)

In \(\text{FeS}_2\), iron typically has an oxidation state of \text{+2}\. Let the oxidation state of each sulfur atom be \(x\):\[ \text{+2} + 2x = 0 \] Solve for \(x\):\[ 2x = -2 \]\[ x = -1 \] Thus, each sulfur atom in \(\text{FeS}_2\) has an oxidation state of \-1\.
06

Calculate the Oxidation State in \(\text{H}_2\text{SO}_4\)

Hydrogen has an oxidation state of \(+1\), and oxygen has \(-2\). In \(\text{H}_2\text{SO}_4\), let the oxidation state of sulfur be \(x\):\[ 2(+1) + x + 4(-2) = 0 \] Solve for \(x\):\[ 2 + x - 8 = 0 \]\[ x - 6 = 0 \]\[ x = +6 \]
07

Determine the Oxidation State in \(\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}\)

The water molecules do not affect the oxidation state calculation. Sodium has \(+1\), and oxygen has \(-2\). Let’s denote the average oxidation state of each sulfur atom as \(x\):\[ 2(+1) + 2x + 3(-2) = 0 \] Solve for \(x\):\[ 2 + 2x - 6 = 0 \]\[ 2x = 4 \]\[ x = +2 \] Thus, the average oxidation state of sulfur is \(+2\). However, in \(\text{Na}_2\text{S}_2\text{O}_3\), sulfur commonly has \(0\) and \(+4\) oxidation states.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation State Rules
To determine the oxidation state of an element in a compound, you must follow some basic rules. These rules help simplify the process of identifying how many electrons an atom has gained, lost, or shared. For sulfur compounds, here are the key rules:
1. **Pure elements** have an oxidation state of 0. Hence, in elemental sulfur \(\mathrm{S}_8\), each sulfur atom has an oxidation state of 0.
2. **Fluorine** always has an oxidation state of -1 in its compounds. This helps to determine the oxidation state of sulfur in compounds like \(\mathrm{SF}_4\) and \(\mathrm{SF}_6\).
3. **Hydrogen** usually has an oxidation state of +1, and **oxygen** typically has an oxidation state of -2. These rules assist in calculating the oxidation states in compounds like \(\mathrm{H}_2\mathrm{SO}_4\).
4. The sum of oxidation states in a **neutral molecule** must be 0. In an **ion**, the sum of oxidation states must equal the ion's charge.
Understanding these rules makes it easy to break down any complex sulfur compound and correctly assign oxidation states.
Redox Chemistry
Redox (reduction-oxidation) reactions involve the transfer of electrons between two substances. Sulfur, being an element that can have several different oxidation states, plays a significant role in redox chemistry.
In a redox reaction:
  • The element that **loses electrons** is said to be **oxidized** and its oxidation state increases.
  • The element that **gains electrons** is said to be **reduced** and its oxidation state decreases.
This movement of electrons is what drives many chemical reactions, including those involving sulfur. For example, converting \(\mathrm{FeS}_2\) (Iron(II) sulfide) to sulfuric acid \(\mathrm{H}_2\mathrm{SO}_4\) involves sulfur atoms changing their oxidation states from -1 in \(\mathrm{FeS}_2\) to +6 in \({\mathrm{SO}_4}^{2-}\). These changes reflect sulfur's flexibility in forming different compounds.
Sulfur Compounds
Sulfur compounds are widely varied due to their ability to take on multiple oxidation states. Here are some common sulfur compounds and their properties:
1. **Elemental Sulfur (\(\mathrm{S}_8\))**: Often found as a yellow solid, it consists of molecules each containing 8 sulfur atoms. Its oxidation state is 0 as it is in its elemental form.
2. **Sulfur Tetrafluoride (\(\mathrm{SF}_4\))**: A toxic gas where sulfur has an oxidation state of +4.
3. **Sulfur Hexafluoride (\(\mathrm{SF}_6\))**: A gaseous compound where sulfur is in the +6 oxidation state. It is often used as an insulating gas in electrical equipment.
4. **Hydrogen Sulfide (\(\mathrm{H}_2\mathrm{S}\))**: A colorless gas with a characteristic foul odor of rotten eggs. In this, sulfur has an oxidation state of -2.
5. **Iron(II) Disulfide (\(\mathrm{FeS}_2\))**: Also known as pyrite, this crystalline solid has sulfur in a -1 oxidation state.
6. **Sulfuric Acid (\(\mathrm{H}_2\mathrm{SO}_4\))**: A strong acid with sulfur at a +6 oxidation state, widely used in industry.
7. **Sodium Thiosulfate (\(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \cdot 5\mathrm{H}_2\mathrm{O}\))**: A compound in which sulfur commonly exhibits oxidation states of 0 and +4, often utilized in photographic processing.

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