Question

Determine the oxidation number for the indicated element in each of the following substances: (a) $\mathrm{S}$ in ${\mathrm{SO}}_3,$ (b) Ti in ${\mathrm{TiCl}}_4$, (c) $P$ in AgPF ${}_6$, (d) $\mathrm{N}$ in ${\mathrm{HNO}}_3$ (e) $\mathrm{S}$ in ${\mathrm{H}}_2{\mathrm{SO}}_3,(\mathbf{f})\mathrm{O}$ in ${\mathrm{OF}}_2$

Short answer

The oxidation numbers for the indicated elements in the given substances are: (a) S in SO3: +6, (b) Ti in TiCl4: +4, (c) P in AgPF6: +5, (d) N in HNO3: +5, (e) S in H2SO3: +4, (f) O in OF2: +2.

Step-by-step solution

(a) S in SO3

: We have the molecule SO3. Oxygen has an oxidation number of -2. There are three oxygen atoms, and we can express the sum of all oxidation numbers as $(\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{S})+(3\cdot (-2))=0$. We solve for the oxidation number of S: $x+(-6)=0$ $x=6$ The oxidation number of S in SO3 is +6.

(b) Ti in TiCl4

: In this case, we have the molecule TiCl4. Chlorine has an oxidation number of -1, and there are four chlorine atoms. The sum of oxidation numbers can be expressed as $(\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{Ti})+(4\cdot (-1))=0$. We solve for the oxidation number of Ti: $x+(-4)=0$ $x=4$ The oxidation number of Ti in TiCl4 is +4.

(c) P in AgPF6

: We have the molecule AgPF6. The oxidation number of Ag is +1 (since it's a monatomic ion), and F has an oxidation of -1. There are six fluorine atoms, and we solve for the oxidation number of P: $+1+(\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{P})+(6\cdot (-1))=0$ Now, solve for the oxidation number of P: $1+x-6=0$ $x=5$ The oxidation number of P in AgPF6 is +5.

(d) N in HNO3

: In the molecule HNO3, hydrogen has an oxidation number of +1, oxygen has an oxidation number of -2, and there are three oxygen atoms. We solve for the oxidation number of N: $(+1)+(\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{N})+(3\cdot (-2))=0$ $1+x-6=0$ $x=5$ The oxidation number of N in HNO3 is +5.

(e) S in H2SO3

: For the molecule H2SO3, hydrogen has an oxidation number of +1, oxygen has an oxidation number of -2, and there are three oxygen atoms. We solve for the oxidation number of S: $(2\cdot (+1))+(\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{S})+(3\cdot (-2))=0$ $2+x-6=0$ $x=4$ The oxidation number of S in H2SO3 is +4.

(f) O in OF2

: In the molecule OF2, fluorine has an oxidation number of -1, and there are two fluorine atoms. Since OF2 is an exception to the rule that oxygen always has an oxidation state of -2, oxygen will have an oxidation number different from -2. We solve for the oxidation number of O: $(\mathrm{oxidation}\mathrm{number}\mathrm{of}\mathrm{O})+(2\cdot (-1))=0$ $x-2=0$ $x=2$ The oxidation number of O in OF2 is +2.

Key concepts

Oxidation State

Oxidation state, often called oxidation number, is an essential concept in chemistry. It helps us understand how atoms interact in a compound. Think of it as a bookkeeping system that allows chemists to keep track of electron transfers between atoms. This is useful when examining redox reactions, where oxidation states change. In any chemical compound, the oxidation state of an element represents the theoretical charge an atom might have if all bonds were ionic. It provides insight into how electrons are distributed in a molecule. An increase in oxidation state indicates a loss of electrons (oxidation), while a decrease suggests a gain of electrons (reduction). Recognizing oxidation states helps in the study of chemical behavior and mechanisms. Predicting how compounds will react during chemical processes becomes possible when understanding the shifts in oxidation states.

Chemical Compounds

Chemical compounds are formed when two or more elements bond together in a fixed ratio. These compounds can be ionic or covalent, influencing the way we assign oxidation states. Each atom in a compound can have its own oxidation number. The sum of these oxidation numbers for all atoms in a neutral compound is zero; for a polyatomic ion, it equals the ion's charge.When assessing a compound like ${\mathrm{SO}}_3$, the sulfur atom and oxygen atoms interact to form a stable molecule. Oxygen, typically known to have an oxidation number of -2, pairs with sulfur to form this compound. By understanding the known oxidation numbers of elements like oxygen, we can deduce the oxidation state of sulfur to ensure the sum equates to zero. This approach allows chemists to determine the energetic stability and reactivity of compounds through their oxidation states.Grasping these principles is crucial for navigating and predicting the outcomes of chemical reactions.

Element Oxidation

Element oxidation involves the process by which an element increases its oxidation state, usually through the loss of electrons. This concept is part of a broader study known as redox (reduction-oxidation) reactions. In these reactions, one element's gain of electrons (reduction) complements another's loss of electrons (oxidization). Each element's tendency to oxidize or reduce is unique, influenced by its position in the periodic table. Elements in the same group often share similar characteristics. For instance, transitioning metals can have multiple oxidation states due to their ability to lose different numbers of electrons. Understanding element oxidation is key in fields like electrochemistry, where batteries and fuel cells rely on redox reactions for energy. It also plays a vital role in processes such as corrosion, metabolic pathways in biology, and industrial manufacturing of chemicals.

Chemistry Education

Chemistry education emphasizes concepts like oxidation state to cultivate a comprehensive understanding of chemical reactions. By learning to determine and interpret oxidation states, students enhance their problem-solving skills and chemical literacy. Educational strategies often include:

• Hands-on experiments where students calculate oxidation numbers in real compounds.
• Integration of technology, such as simulations, that allow visualization of electron transfers.
• Problem-based learning activities focusing on redox reactions and their implications in environmental chemistry.

These methods not only solidify students' grasp of fundamental concepts but also engage them in critical thinking and application of the knowledge to real-world scenarios. By mastering these foundational ideas, students prepare for advanced topics in chemistry and related fields, fostering a generation of informed and capable thinkers.