What is the oxidation state of \(\mathrm{Cr}\) in each compound? (a) \(\mathrm{CrO}\) (b) \(\mathrm{CrO}_{3}\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Short Answer

Expert verified
The oxidation states of Cr are: (a) +2 in CrO, (b) +6 in CrO3, (c) +3 in Cr2O3.

Step by step solution

01

Understanding oxidation states and rules

The oxidation state of an atom is a measure of the degree of oxidation of an atom, expressed as an integer. Oxygen usually has an oxidation state of -2 in compounds, except in peroxides, where it is -1, and in OF2, where it is +2. The sum of oxidation states for all atoms in a neutral molecule must be zero. For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.
02

Determine the oxidation state of Cr in CrO

In the compound CrO, oxygen has an oxidation state of -2. There is one Cr atom and one oxygen atom. To balance the oxidation states and make the sum zero (as the compound is neutral), Cr must have an oxidation state of +2. This is calculated by setting up the equation: Cr + (-2) = 0, which solves to Cr = +2.
03

Determine the oxidation state of Cr in CrO3

In CrO3, there are three oxygen atoms, each having an oxidation state of -2, making a total of -6. To balance this and make the compound neutral, the oxidation state of Cr must be +6. The equation would be Cr + 3(-2) = 0, which solves to Cr = +6.
04

Determine the oxidation state of Cr in Cr2O3

In Cr2O3, there are two Cr atoms and three O atoms. The total oxidation number for oxygen is 3(-2) = -6. If the total oxidation number must be zero, then the total for the two Cr atoms must be +6. Therefore, each Cr atom has an oxidation state of +3. The equation would be 2(Cr) + 3(-2) = 0, which solves to 2(Cr) = +6 and Cr = +3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Oxidation
Chemical oxidation refers to the process where a substance loses electrons and increases its oxidation state. This process is fundamental in the field of inorganic chemistry where oxidation states help in understanding the behavior of elements in compounds. For instance, when evaluating the compound CrO, the oxidation state of chromium (Cr) is determined by considering the common oxidation state of oxygen which is typically -2. Since the compound is neutral, Cr must balance this with a +2, showing that Cr has undergone oxidation if it has increased from a lower oxidation state.

To really grasp this idea, it's essential to visualize electrons as items being given away. Just as with personal finances, if you give something away, you lose value—in this case, electrons. The chromium atom 'pays with electrons' to oxygen, which is eager to 'accept the payment' as it is electron-greedy or what chemists call electronegative.
Inorganic Chemistry
Inorganic chemistry is a branch of chemistry that deals with inorganic compounds, which typically do not contain a carbon-hydrogen (C-H) bond, in contrast to organic compounds. Understanding the concept of oxidation states, as applied to inorganic compounds like those containing chromium, is a cornerstone in this field.

In exercises determining the oxidation state of elements such as chromium in CrO, CrO3, and Cr2O3, inorganic chemistry principles are applied. These principles take into account rules about common oxidation states of different elements, as well as how these states balance within a molecule to adhere to charge neutrality or the specific charge of polyatomic ions. Thus, inorganic chemistry not only involves understanding the individual behaviors of elements but also how they interact within a compound to maintain a stable electronic arrangement.
Stoichiometry
Stoichiometry is a section of chemistry that quantifies the relationships between the amounts of reactants and products in a chemical reaction. In the context of oxidation states, stoichiometry can be used to calculate the correct ratio of atoms in a compound and ensure the overall charge is balanced. For example, looking at Cr2O3, stoichiometry is crucial to determine that there are two chromium atoms for every three oxygen atoms.

The stoichiometric calculation can be straightforward but requires attention to detail. To solve for the oxidation state of chromium in Cr2O3 accurately, one must understand that the oxidation states must sum to zero for a neutral molecule and use the known oxidation state of oxygen to solve for the unknown chromium state. This meticulous balancing act is at the heart of stoichiometry and can become second nature to students through practice. It is not unlike a 'chemical equation accountancy' where atoms and their oxidation states must be tallied precisely to ensure balance.
Redox Reactions
Redox reactions, or oxidation-reduction reactions, are processes where the oxidation states of atoms are changed due to the transfer of electrons. One substance is oxidized (loses electrons) while another is reduced (gains electrons). These reactions are fundamental to numerous processes in chemistry and biology, such as respiration and combustion.

When one is asked to determine the oxidation state of chromium in various compounds, they are indirectly dealing with the principles that govern redox reactions. The relative change in oxidation states between reactants and products helps chemists understand the flow of electrons in a reaction. For the compounds CrO, CrO3, and Cr2O3, identifying the oxidation state of chromium provides insight into potential redox behavior, which is critical for predicting the reactivity and properties of these compounds. To ensure students develop a strong foundation in this area, it's important to not just memorize rules but also to understand the movement of electrons which symbolically represent the essence of these reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which metal has the least tendency to be oxidized? (a) \(\mathrm{Ag}\) (b) \(\mathrm{Na}\) (c) \(\mathrm{Ni}\) (d) \(\mathrm{Pb}\)

Consider the unbalanced redox reaction. $$ \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Zn}^{2+}(a q) $$ Balance the equation in acidic solution and determine how much of a \(0.500 \mathrm{M} \mathrm{KMnO}_{4}\) solution is required to completely dissolve \(2.85 \mathrm{~g}\) of \(\mathrm{Zn}\).

The electrolytic cell represented in Figure \(16.17\) can be used to plate silver onto other metal surfaces. The plating reaction is: \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\). Notice from the reaction that \(1 \mathrm{~mol} \mathrm{e}^{-}\)plates out \(1 \mathrm{~mol} \mathrm{Ag}(\mathrm{s})\). Use this stoichiometric relationship to determine how much time is required with an electrical current \(0.100\) amp to plate out \(1.0 \mathrm{~g}\) Ag. The amp is a unit of electrical current equivalent to \(1 \mathrm{C} / \mathrm{s}\). (Hint: Recall that the charge of an electron is 1.60 \(\times 10^{-19} \mathrm{C}\) )

Assign an oxidation state to each element in each reaction and use the change in oxidation state to determine which element is being oxidized and which element is being reduced. (a) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g)\)

Balance each redox reaction using the half-reaction method. (a) \(\mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\operatorname{Sn}(s)\) (b) \(\mathrm{Mg}(\mathrm{s})+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cr}(s)\) (c) \(\mathrm{Al}(\mathrm{s})+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Ag}(s)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free