Question

Obtain the oxidation number for the element noted in each of the following. (a) \(\mathbf{N}\) in \(\mathrm{N}_{2}\) (b) \(\mathrm{Cr}\) in \(\mathrm{CrO}_{4}^{2-}\) (c) \(\mathrm{Zn}\) in \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\) (d) As in \(\mathrm{H}_{2} \mathrm{AsO}_{3}^{-}\)

Short answer

(a) 0, (b) +6, (c) +2, (d) +3

Step-by-step solution

Recognize the Oxidation Number Rule for Elements

For an element in its pure form, like the nitrogen molecule \(\mathrm{N}_2\), the oxidation number is 0. This is because the molecule is in its diatomic form, and no other elements are involved.

Analyze the Compound Structure for \( \mathrm{CrO}_{4}^{2-} \)

In the polyatomic ion \( \mathrm{CrO}_{4}^{2-} \), oxygen typically has an oxidation number of -2. There are 4 oxygen atoms, contributing a total of \( 4 \times -2 = -8 \). The ion has a net charge of -2, so the oxidation number of Cr can be found by solving: \( x + (-8) = -2 \).

Solve for Chromium's Oxidation Number in \( \mathrm{CrO}_{4}^{2-} \)

Using the equation \( x + (-8) = -2 \), solve for \( x \). Add 8 to both sides: \( x = -2 + 8 \). This simplifies to \( x = +6 \). Thus, the oxidation number of Cr in \( \mathrm{CrO}_{4}^{2-} \) is +6.

Determine the Oxidation Number for \( \mathrm{Zn} \) in \( \mathrm{Zn} (\mathrm{OH})_{4}^{2-} \)

The hydroxide ion (OH) typically has a charge of -1, and there are 4 such ions, contributing \( 4 \times -1 = -4 \). The ion has an overall charge of -2. To find the oxidation number of Zn, solve \( x + (-4) = -2 \).

Solve for \( \mathrm{Zn} \)'s Oxidation Number

Using the equation \( x + (-4) = -2 \), solve for \( x \). Add 4 to both sides: \( x = -2 + 4 \). This simplifies to \( x = +2 \). Thus, the oxidation number of Zn in \( \mathrm{Zn}(\mathrm{OH})_{4}^{2-} \) is +2.

Calculate the Oxidation Number for As in \( \mathrm{H}_{2} \mathrm{AsO}_{3}^{-} \)

Oxygen typically has an oxidation number of -2. With 3 oxygen atoms, this totals \( 3 \times -2 = -6 \). Hydrogen has an oxidation number of +1, so 2 hydrogen atoms contribute \( 2 \times +1 = +2 \). The ion has a charge of -1, and we'll use \( x \) to denote the oxidation number of As. Set up the equation: \( x + 2 + (-6) = -1 \).

Solve for Arsenic's Oxidation Number

Using the equation \( x + 2 + (-6) = -1 \), simplify to solve for \( x \): \( x - 4 = -1 \). Add 4 to both sides, and \( x = -1 + 4 \), resulting in \( x = +3 \). Therefore, the oxidation number of As in \( \mathrm{H}_{2} \mathrm{AsO}_{3}^{-} \) is +3.

Key concepts

Element Oxidation State

The concept of an element's oxidation state is crucial in understanding its chemical behavior and reactivity. An element's oxidation state is essentially the "charge" it possesses when combined in a compound. This value provides insight into the distribution of electrons among the atoms in a molecule.

In pure elements, like diatomic nitrogen (2), the oxidation state is always zero. This is because the atoms share electrons equally in these forms. However, when elements form compounds, their oxidation states can vary depending on the accompanying elements and the molecule's overall charge. For instance, chromium in the polyatomic ion \( \text{CrO}_4^{2-} \) has an oxidation state of +6, as calculated by balancing the charges of the entire ion and considering the typical oxidation state of oxygen (-2).

Understanding oxidation states is vital for predicting how an element might react with others, aiding in balancing chemical equations, and explaining phenomena in terms of electron transfer.

Molecular Compounds

Molecular compounds are composed of nonmetal elements bonded together via covalent bonds, where electrons are shared between atoms. These covalent bonds are key differences from ionic compounds, where electrons are transferred from one element to another.

Oxidation numbers in molecular compounds provide a way of imagining the "distribution" of electrons between different atoms within the compound. For example, in 2, each nitrogen atom shares electrons equally, meaning the oxidation state remains zero for both atoms.

However, differing electronegativities in compounds cause unequal electron sharing. In such molecules, like the hydroxide ion ( ( ext{OH})_), hydrogen and oxygen share electrons, but because oxygen is more electronegative, it pulls electrons toward itself, resulting in (-1) oxidation state for oxygen.

• In highly electronegative environments, an element will possess a higher oxidation state.
• Conversely, a less electronegative element in a molecular compound will have a lower oxidation state.

Ionic Compounds

Ionic compounds are formed through the transfer of electrons from one atom to another, resulting in positively and negatively charged ions. These compounds typically consist of metal and nonmetal elements. This electron transfer leads metals to generally have positive oxidation states, while nonmetals have negative oxidation states.

For example, in the compound \( \text{Zn}( ext{OH})_4^{2-} \), zinc acts as a metal, forming a positive oxidation state of +2 to balance the four hydroxide ions each with an oxidation state of -1.

Ionic compounds are characterized by strong electrostatic interactions between ions:\- They form crystalline structures.
- The formulas of ionic compounds reflect the simplest ratio of ions that achieves electrical neutrality.
Understanding the oxidation numbers assists in deciphering the formula of ionic compounds and predicting their reactivity.

Chemical Formulas

Chemical formulas represent the composition of molecules and compounds, indicating the types and numbers of each type of atom present. They are essential for conveying the precise arrangement of atoms within compounds, which determines their reactivity and properties.

Oxidation numbers are intertwined with chemical formulas as they provide insight into how electrons are shared or transferred among the atoms in a compound. For instance, the formula \( \text{H}_2 ext{AsO}_3^- \) tells us there are two hydrogen atoms, one arsenic atom, and three oxygen atoms.

Here are some pointers to clarity around chemical formulas:

• Chemical formulas show the stoichiometry of molecules, indicating the relative numbers of each type of atom.
• They reflect both the molecular structure and the oxidation states of elements involved.

When interpreting a chemical formula, understanding the oxidation state of each element helps in predicting the possible chemical behavior and reactions of the compound.