Question

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3} \quad\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) d. As i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) e. \(\mathrm{HAsO}_{2}\)

Short answer

a. U: +6, O: -2 b. As: +3, O: -2 c. Na: +1, Bi: +5, O: -2 d. As: 0 e. H: +1, As: +3, O: -2 f. Mg: +2, P: +5, O: -2 g. Na: +1, S: +2, O: -2 h. Hg: +1, Cl: -1 i. Ca: +2, N: +5, O: -2

Step-by-step solution

a. Assigning Oxidation States for \(\mathrm{UO}_{2}^{2+}\)

Step 1: Assign known oxidation states Oxygen has an oxidation state of -2. Step 2: Calculate unknown oxidation state Let the oxidation state of Uranium (U) be x. \(x + 2(-2) = +2\) Solve for x: \(x = +6\) Uranium has an oxidation state of +6 in \(\mathrm{UO}_{2}^{2+}\).

b. Assigning Oxidation States for \(\mathrm{As}_{2} \mathrm{O}_{3}\)

Step 1: Assign known oxidation states Oxygen has an oxidation state of -2. Step 2: Calculate unknown oxidation state Let the oxidation state of Arsenic (As) be x. \(2x + 3(-2) = 0\) Solve for x: \(x = +3\) Arsenic has an oxidation state of +3 in \(\mathrm{As}_{2} \mathrm{O}_{3}\).

c. Assigning Oxidation States for \(\mathrm{NaBiO}_{3}\)

Step 1: Assign known oxidation states Sodium (Na) and Oxygen have oxidation states of +1 and -2, respectively. Step 2: Calculate unknown oxidation state Let the oxidation state of Bismuth (Bi) be x. \(+1 + x + 3(-2) = 0\) Solve for x: \(x = +5\) Bismuth has an oxidation state of +5 in \(\mathrm{NaBiO}_{3}\).

d. Assigning Oxidation States for As

Elemental As has an oxidation state of 0.

e. Assigning Oxidation States for \(\mathrm{HAsO}_{2}\)

Step 1: Assign known oxidation states Hydrogen (H) and Oxygen have oxidation states of +1 and -2, respectively. Step 2: Calculate unknown oxidation state Let the oxidation state of Arsenic (As) be x. \(+1 + x + 2(-2) = 0\) Solve for x: \(x = +3\) Arsenic has an oxidation state of +3 in \(\mathrm{HAsO}_{2}\).

f. Assigning Oxidation States for \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\)

Step 1: Assign known oxidation states Magnesium (Mg) and Oxygen have oxidation states of +2 and -2, respectively. Step 2: Calculate unknown oxidation state Let the oxidation state of Phosphorus (P) be x. \(2(+2) + 2x + 7(-2) = 0\) Solve for x: \(x = +5\) Phosphorus has an oxidation state of +5 in \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\).

g. Assigning Oxidation States for \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)

Step 1: Assign known oxidation states Sodium (Na) and Oxygen have oxidation states of +1 and -2, respectively. Step 2: Calculate unknown oxidation state Let the oxidation state of Sulfur (S) be x. \(2(+1) + 2x + 3(-2) = 0\) Solve for x: \(x = +2\) Sulfur has an oxidation state of +2 in \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\).

h. Assigning Oxidation States for \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)

Step 1: Assign known oxidation states Chlorine (Cl) has an oxidation state of -1. Step 2: Calculate unknown oxidation state Let the oxidation state of Mercury (Hg) be x. \(2x + 2(-1) = 0\) Solve for x: \(x = +1\) Mercury has an oxidation state of +1 in \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\).

i. Assigning Oxidation States for \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

Step 1: Assign known oxidation states Calcium (Ca), Oxygen, and Hydrogen have oxidation states of +2, -2, and +1, respectively. Step 2: Calculate unknown oxidation state Let the oxidation state of Nitrogen (N) be x. \(x + 3(-2) = -1\) Solve for x: \(x = +5\) Nitrogen has an oxidation state of +5 in \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\).

Key concepts

Redox Reactions

Redox reactions are a fascinating cornerstone of chemistry, marking the transfer of electrons between two substances. This involves two different processes happening concurrently: oxidation and reduction. In simple terms, oxidation is the loss of electrons and an increase in oxidation state, while reduction is the gain of electrons and a decrease in oxidation state.
The classic example is the reaction between iron and oxygen to form rust. Here, iron is oxidized as it loses electrons to oxygen, which is reduced as it gains those electrons. This elegant electron dance is crucial in understanding chemical transformations. Often, these reactions are also connected to energy changes and are utilized in processes like combustion and electrochemical cells. Understanding how electrons move in redox reactions allows for a deeper understanding of everyday chemical phenomena.

Chemical Compounds

Chemical compounds are substances formed by the combination of two or more elements in fixed ratios, creating a vast variety of substances with unique properties. Each compound has a specific chemical formula that conveys the proportion of its constituent elements. For example, water (H₂O) contains two hydrogen atoms for every oxygen atom.
Compounds are formed when elements react chemically, sharing or transferring electrons to achieve stability. The nature of these bonds can be ionic, covalent, or metallic, depending on how the electrons are distributed within the compound.
Knowing the composition of a chemical compound is crucial for predicting its behavior and reactivity. Through a compound's chemical formula, it's possible to deduce the oxidation states of elements, aiding in understanding more complex reactions such as redox processes. Hence, a solid understanding of chemical compounds underpins the study of chemistry as a whole.

Oxidation Number Calculation

Calculating oxidation numbers, or oxidation states, is a key skill in mastering chemistry. This is a systematic way of keeping track of electron distribution within a molecule or ion during a chemical reaction.

To start, some elements have common oxidation states based on their positions in the periodic table. For instance:

• Alkali metals like sodium and potassium typically have an oxidation number of +1.
• Alkaline earth metals such as calcium and magnesium usually have an oxidation number of +2.
• Oxygen mostly has an oxidation state of -2, except in peroxides or when bonded to fluorine.
• Hydrogen is generally +1, except when it is bonded with metals in hydrides where it's -1.

Using these rules, you can establish unknown oxidation states by setting up equations that reflect the net charge of a compound or ion. For instance, in \(\mathrm{UO}_{2}^{2+}\), uranium's oxidation state is determined by using oxygen's standard oxidation state and equating the sum to the ion's charge. Hence, it’s fundamental to grasp these rules to efficiently solve and balance redox reactions in your chemistry journey.