Question

Assign oxidation numbers to all the atoms in each of the following. a. \(\operatorname{Sr} \mathrm{Cr}_{2} \mathrm{O}_{7}\) b. \(\mathrm{CuCl}_{2}\) c. \(O_{2}\) \(\mathbf{d .} \mathbf{H}_{2} \mathbf{O}_{2}\) e. \(\mathrm{MgCO}_{3}\) f. Ag g. \(\mathrm{PbSO}_{3}\) h. \(\mathrm{PbO}_{2}\) i. \(\quad \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) \(\mathbf{j} . \mathrm{CO}_{2}\) \(\mathbf{k} .\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) I. \(\quad \mathrm{Cr}_{2} \mathrm{O}_{3}\)

Short answer

a. Sr = +2, Cr = +6, O = -2 b. Cu = +2, Cl = -1 c. O = 0 d. H = +1, O = -1 e. Mg = +2, C = +4, O = -2 f. Ag = 0 g. Pb = +2, S = +6, O = -2 h. Pb = +4, O = -2 i. Na = +1, C = +3, O = -2 j. C = +4, O = -2 k. N = -3, H = +1, Ce = +3, S = +6, O = -2 l. Cr = +3, O = -2

Step-by-step solution

a. Assigning oxidation numbers for \(\operatorname{SrCr}_{2}\mathrm{O}_{7}\)

The compound \(\operatorname{SrCr}_{2}\mathrm{O}_{7}\) is ionic containing Sr, Cr, and O atoms. To find the oxidation numbers, we start with oxygen. In most cases, the oxidation number of oxygen is -2. Since there are seven oxygen atoms, their total contribution is -14. Sr is an alkaline earth metal and forms a +2 charge as it loses two electrons. So, the oxidation number of Sr is +2. To find the oxidation number for Cr, we just need to balance the overall charge of the compound. \(2n_{Cr} + 2 - 14 = 0\) Solve for \(n_{Cr}\): \(n_{Cr} = \frac{14 - 2}{2}\) Which gives: \(n_{Cr} = 6\) So, the oxidation numbers are: Sr = +2, Cr = +6, O = -2.

b. Assigning oxidation numbers for \(\mathrm{CuCl}_{2}\)

The compound \(\mathrm{CuCl}_{2}\) is ionic and contains Cu and Cl atoms. Copper (Cu) is a transition metal and has a variable charge. In this case, Cu has an oxidation number of +2 because there are two chloride ions, each with an oxidation number of -1. The sum of the oxidation numbers of this compound must equal zero. So, the oxidation numbers are: Cu = +2, Cl = -1.

c. Assigning oxidation numbers for \(O_{2}\)

\(O_2\) is a diatomic molecule consisting of only oxygen atoms. Since there are no other elements involved, the oxidation number of each oxygen atom in its free state is zero. So, the oxidation number is: O = 0.

d. Assigning oxidation numbers for \(\mathrm{H}_{2}\mathrm{O}_{2}\)

In \(\mathrm{H}_{2}\mathrm{O}_{2}\), hydrogen usually has an oxidation number of +1, and there are two hydrogen atoms that contribute a total of +2. To balance the overall charge of the compound, the oxidation number of each oxygen atom must be -1. So, the oxidation numbers are: H = +1, O = -1.

e. Assigning oxidation numbers for \(\mathrm{MgCO}_{3}\)

In \(\mathrm{MgCO}_{3}\), magnesium (Mg) has an oxidation number of +2, and there are three oxygen atoms, each with an oxidation number of -2. Carbon forms the ion CO₃²⁻ and since the total charge on CO₃²⁻ is -2, the oxidation number of carbon (C) must be +4. So, the oxidation numbers are: Mg = +2, C = +4, O = -2.

f. Assigning oxidation number for Ag

Ag is a monatomic element in its free state. The oxidation number is zero. So, the oxidation number is: Ag = 0.

g. Assigning oxidation numbers for \(\mathrm{PbSO}_{3}\)

In \(\mathrm{PbSO}_{3}\), Pb has an oxidation number of +2, since the compound \(\mathrm{SO}_{3}\) has an overall charge of -2 and there is one Pb. Within the compound \(\mathrm{SO}_{3}\), there are three O atoms, each with an oxidation number of -2, so the oxidation state of sulfur (S) must be +6 to balance the total charge. So, the oxidation numbers are: Pb = +2, S = +6, O = -2.

h. Assigning oxidation numbers for \(\mathrm{PbO}_{2}\)

In \(\mathrm{PbO}_{2}\), Pb must have an oxidation number of +4 to balance the total charge contributed by the two oxygen atoms, which is -4 (2 × -2). So, the oxidation numbers are: Pb = +4, O = -2.

i. Assigning oxidation numbers for \(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)

In \(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\), sodium (Na) has an oxidation number of +1. There are two sodium atoms that contribute a total of +2 charge. Each of the four oxygen atoms has an oxidation number of -2. To balance the total charge of the compound, carbon (C) must have an oxidation state of +3 for both C atoms. So, the oxidation numbers are: Na = +1, C = +3, O = -2.

j. Assigning oxidation numbers for \(\mathrm{CO}_{2}\)

In \(\mathrm{CO}_{2}\), each oxygen atom has an oxidation number of -2. To balance the total charge, the oxidation number of carbon (C) must be +4. So, the oxidation numbers are: C = +4, O = -2.

k. Assigning oxidation numbers for \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\)

In \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\), ammonium (NH₄⁺), cerium (Ce), and sulfate (SO₄²⁻) are the three separate ions. In the ammonium ion (NH₄⁺), hydrogen has an oxidation number of +1 and nitrogen (N) has an oxidation number of -3. In the sulfate ion (SO₄²⁻), oxygen has an oxidation number of -2, sulfur has an oxidation number of +6. Ce has an oxidation number of +3 to balance the other charges. So, the oxidation numbers are: N = -3, H = +1, Ce = +3, S = +6, O = -2.

l. Assigning oxidation numbers for \(\mathrm{Cr}_{2}\mathrm{O}_{3}\)

In \(\mathrm{Cr}_{2}\mathrm{O}_{3}\), each oxygen atom has an oxidation number of -2. The three oxygen atoms contribute a total of -6 charge. To balance the total charge of the compound, each chromium (Cr) atom must have an oxidation number of +3. So, the oxidation numbers are: Cr = +3, O = -2.

Key concepts

Chemical Compounds

Chemical compounds are substances that form when two or more elements are chemically bonded together. They play a vital role in various fields such as chemistry, biology, physics, and even everyday life.
Compounds are classified based on the type of elements they are made of and the bonds that hold these elements together. Understanding the formation and characteristics of chemical compounds helps us comprehend everything from how medicines work to the processes in ecosystems. In this exercise, every mentioned formula represents a chemical compound, each unique in its structure and properties.
For example:

• \(\operatorname{SrCr}_{2}\mathrm{O}_{7}\) contains strontium, chromium, and oxygen, forming a compound with both ionic and covalent characteristics.
• \(\mathrm{CuCl}_{2}\) is a simpler compound composed of copper and chlorine.
• \(\mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) includes sodium, carbon, and oxygen, showcasing the diversity in elements forming compounds.

These compounds illustrate how elements can combine in unique ways to form substances with new properties, emphasizing the importance of chemical compounds in science.

Ionic Compounds

Ionic compounds are formed when atoms transfer electrons between each other, leading to the creation of positively and negatively charged ions. These ions are held together by strong electrostatic forces known as ionic bonds.
Typically, ionic compounds consist of a metal and a non-metal. The metal loses electrons and becomes a positive ion, while the non-metal gains those electrons, becoming a negative ion.For example, in \(\mathrm{CuCl}_{2}\), copper (Cu) is the metal, losing electrons to form \(\mathrm{Cu}^{2+}\) ions, while chlorine (Cl) gains those electrons, forming \(\mathrm{Cl}^{-}\) ions. This results in an ionic bond between the different ions.Key characteristics of ionic compounds include:

• They often have high melting and boiling points because of their strong bonds.
• They are generally soluble in water.
• In solid form, they do not conduct electricity, but they become conductive when dissolved in water.

Ionic compounds like \(\mathrm{SrCr}_{2}\mathrm{O}_{7}\) and \(\mathrm{PbSO}_{3}\) are critical in understanding both chemical reactions and material properties. Understanding these compounds helps us appreciate their role in natural processes and technological applications.

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between chemical species, leading to changes in their oxidation states. This concept is pivotal in explaining numerous chemical reactions, ranging from simple laboratory reactions to complex biological processes.
In a redox reaction, one part of the reaction involves oxidation, where a substance loses electrons, and the other part involves reduction, where a substance gains electrons. Tracking oxidation numbers allows us to identify which elements are oxidized and which are reduced in a reaction.For instance, in the compound \(\mathrm{H}_{2}\mathrm{O}_{2}\), we can observe changes in oxidation numbers when it undergoes decomposition:* Initially, hydrogen's oxidation number is +1, and oxygen's is -1.* Upon decomposition into water (\(\mathrm{H}_{2}\mathrm{O}\)) and oxygen gas (\(\mathrm{O}_{2}\)), we notice redistribution of electrons.Redox reactions play a key role in:

• Energy generation (such as in batteries and fuels).
• Photosynthesis and cellular respiration.
• Industrial processes like metal extraction and synthesis of chemicals.

Recognizing how redox reactions operate helps explain crucial biological and synthetic processes, highlighting the intricate dance of electrons that sustains both life and technology.