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. \({C r O}_{4}^{2-}\)

Short 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.

Step-by-step solution

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.

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: \(x + 4\times(+1) = +1\)

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. \(A s O_{4}^{3-}\)

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.

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: \(x + 4\times(-2) = -3\)

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. \({C r O}_{4}^{2-}\)

Identify the oxidation number of Oxygen

In most compounds, oxygen has an oxidation number of -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: \(x + 4\times(-2) = -2\)

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 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.

• Ionic bonds are strong due to the attraction between positively and negatively charged ions, forming a crystalline lattice structure.

On the other hand, covalent compounds form when atoms share pairs of electrons, typically between non-metals.
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.

Understanding the types of bonding in a compound is crucial for determining its oxidation states and reactivity, making it easier to predict and understand its chemical behavior.

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).

• 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 charge of an ion is crucial in chemical reactions as it determines how ions interact and bond with each other.
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.

Understanding the concept of ion charge helps in predicting how molecules will form, behave, and react with other substances.

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.

• 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.

To determine the oxidation state of a particular element in a compound, one uses several rules, including:

• 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.

These rules allow us to set up equations and solve for unknown oxidation numbers, just as was demonstrated in the original step-by-step solution for finding the oxidation numbers of nitrogen, arsenic, and chromium considering their surrounding elements.