Problem 6
Question
Determine the oxidation number of the boldface element in the following formulas for ions a. \({N H}_{4}+\) b. \(A s O_{4} 3-\) c. \({C r O}_{4}^{2-}\)
Step-by-Step Solution
Verified Answer
The oxidation numbers of the boldface elements in the given ions are: a. Nitrogen (N) in \({NH_4}^+\) is -3, b. Arsenic (As) in \(A s O_{4}^{3-}\) is +5, and c. Chromium (Cr) in \({C r O}_{4}^{2-}\) is +6.
1Step 1: Identify the oxidation number of Hydrogen
Since hydrogen is not the boldface element, we should first determine its oxidation number. In most compounds, hydrogen has an oxidation number of +1.
2Step 2: Apply rule 3 to find the oxidation number of Nitrogen
The sum of oxidation numbers in the \({NH_4}^+\) ion is equal to the charge on the ion, which is +1. We can set up an equation to find the oxidation number of nitrogen (N), denoted as x:
\(x + 4\times(+1) = +1\)
3Step 3: Solve for the oxidation number of Nitrogen
Now, we will solve the equation for x:
\(x + 4 = 1\)
\(x = -3\)
So, the oxidation number of Nitrogen (N) in \({NH_4}^+\) is -3.
b. \(A s O_{4}^{3-}\)
4Step 1: Identify the oxidation number of Oxygen
Since oxygen is not the boldface element, we should first determine its oxidation number. In most compounds, oxygen has an oxidation number of -2.
5Step 2: Apply rule 3 to find the oxidation number of Arsenic
The sum of oxidation numbers in the \(A s O_{4}^{3-}\) ion is equal to the charge on the ion, which is -3. We can set up an equation to find the oxidation number of arsenic (As), denoted as x:
\(x + 4\times(-2) = -3\)
6Step 3: Solve for the oxidation number of Arsenic
Now, we will solve the equation for x:
\(x - 8 = -3\)
\(x = +5\)
So, the oxidation number of Arsenic (As) in \(A s O_4^{3-}\) is +5.
c. \({C r O}_{4}^{2-}\)
7Step 1: Identify the oxidation number of Oxygen
In most compounds, oxygen has an oxidation number of -2.
8Step 2: Apply rule 3 to find the oxidation number of Chromium
The sum of oxidation numbers in the \({C r O}_{4}^{2-}\) ion is equal to the charge on the ion, which is -2. We can set up an equation to find the oxidation number of chromium (Cr), denoted as x:
\(x + 4\times(-2) = -2\)
9Step 3: Solve for the oxidation number of Chromium
Now, we will solve the equation for x:
\(x - 8 = -2\)
\(x = +6\)
So, the oxidation number of Chromium (Cr) in \({C r O}_{4}^{2-}\) is +6.
Key Concepts
Understanding Chemical CompoundsGrasping Ion ChargeOxidation State Determination
Understanding Chemical Compounds
A chemical compound consists of two or more different elements that are chemically bonded together. These elements are present in fixed proportions and display unique properties as compounds that are different from their constituent elements.
The elements can bond in various ways, which can be ionic or covalent.
Ionically bonded compounds arise when there is a transfer of electrons between atoms, usually between metals and non-metals, resulting in the formation of ions.
Covalent bonds involve the sharing of electrons to achieve a full outer electron shell, resulting in more stable molecules.
The elements can bond in various ways, which can be ionic or covalent.
Ionically bonded compounds arise when there is a transfer of electrons between atoms, usually between metals and non-metals, resulting in the formation of ions.
- Ionic bonds are strong due to the attraction between positively and negatively charged ions, forming a crystalline lattice structure.
Covalent bonds involve the sharing of electrons to achieve a full outer electron shell, resulting in more stable molecules.
- This bonding method allows for a wide variety of complex molecules.
Grasping Ion Charge
The concept of ion charge is foundational in chemistry, and it refers to the electric charge that an ion has, due to the loss or gain of electrons.
An ion can either be a cation (positively charged) or an anion (negatively charged).
The sum of charges in a chemical compound is always neutral, meaning that the total positive charges are equal to the total negative charges.
When writing formulas for compounds, the ions are combined in such a way to reflect a balance of charges, enhancing the stability of the compound.
An ion can either be a cation (positively charged) or an anion (negatively charged).
- Cations form when an atom loses electrons, resulting in more protons than electrons.
- Anions form when an atom gains electrons, resulting in more electrons than protons.
The sum of charges in a chemical compound is always neutral, meaning that the total positive charges are equal to the total negative charges.
When writing formulas for compounds, the ions are combined in such a way to reflect a balance of charges, enhancing the stability of the compound.
- For instance, in \(NH_4^+\), the total positive charge is +1, which is characteristic for ammonium ions.
Oxidation State Determination
Determining the oxidation state of an element within a compound is an essential skill in chemistry because it helps in identifying electron transfer in reactions.
The oxidation state of an element can be thought of as an indicator of the degree of oxidation (loss of electrons) of an element in a compound.
The oxidation state of an element can be thought of as an indicator of the degree of oxidation (loss of electrons) of an element in a compound.
- It is represented as a number that can be positive, negative, or zero, showing how many electrons an atom has gained, lost, or shared when forming a compound.
- The sum of the oxidation states for all atoms in a molecule or ion equals the total charge of the molecule or ion.
- Elements in their elemental form have an oxidation state of zero.
- For monatomic ions, the oxidation state is equal to the charge of the ion.
- Oxygen generally has an oxidation state of -2, and hydrogen is usually +1 in compounds.
Other exercises in this chapter
Problem 4
Challenge Identify the oxidizing agent and the reducing agent in each reaction. a. \(M g+I_{2} \rightarrow M g l_{2}\) b. \(H_{2} S+C l_{2} \rightarrow S+2 H C
View solution Problem 5
Determine the oxidation number of the boldface element in the following formulas for compounds a. Nac?o\(_{4}\) b. AIPO_ \(_{4}\) c. \({HNO}_{2}\)
View solution Problem 7
Determine the oxidation number of nitrogen in each of these molecules or ions. a. \({NH}_{3}\) b. \({KCN}\) c. \({N}_{2} {H}_{4}\)
View solution Problem 8
Challenge Determine the net change of oxidation number of each of the elements in these redox equations. a. \(C+O_{2} \rightarrow C O_{2}\) b. \(C l_{2}+Z n l_{
View solution