Question

Assign oxidation states for all atoms in each of the following compounds a. \({KMnO}_{4} \quad\quad\quad f. {Fe}_{3} {O}_{4}\) b. \({NiO}_{2} \quad\quad\quad\quad g. {XeOF}_{4}\) c. \({Na}_{4} {Fe}({OH})_{6} \quad h. {SF}_{4}\) d. \({NH}_{4} {h}_{2} {HPO}_{4} \quad i. {CO}\) e. \({P}_{4} {O}_{6} \quad\quad\quad\quad\quad j. {C}_{6} {H}_{12} {O}_{6}\)

Short answer

The oxidation states for the given compounds are: a. K(+1), Mn(+7), O(-2) b. Ni(+4), O(-2) c. Na(+1), Fe(+2), O(-2), H(+1) d. N(-3), H(+1), P(+5) e. P(+3), O(-2) f. Fe(+2, +3), O(-2) g. Xe(+6), O(-2), F(-1) h. S(+4), F(-1) i. C(+2), O(-2) j. C(0), H(+1), O(-2)

Step-by-step solution

Assign oxidation states for K, Mn, and O

Assign +1 for K, -2 for O. The sum of the oxidation states must be zero since it's a neutral compound. \(+1 + x + 4(-2) = 0\) \(x = +7\) So, the oxidation states are K(+1), Mn(+7), and O(-2). b. NiO2

Assign oxidation states for Ni and O

Assign -2 for O. The sum of oxidation states must be zero since it's a neutral compound. \(x + 2(-2) = 0\) \(x = +4\) So, the oxidation states are Ni(+4) and O(-2). c. Na4Fe(OH)6

Assign oxidation states for Na, Fe, O, and H

Assign +1 for Na, -2 for O, and +1 for H. The sum of oxidation states must be zero since it's a neutral compound. \(4(+1) + x + 6(-2) + 6(+1) = 0\) \(x = +2\) So, the oxidation states are Na(+1), Fe(+2), O(-2), and H(+1). d. NH4H2HPO4

Assign oxidation states for N, H, and P

Assign +1 for H (both in NH4 and H2). Assign -2 for O. The sum of oxidation states must be zero since it's a neutral compound. \(+1 + 4(+1) + 2(+1) + x + 4(-2) = 0\) \(x = +5\) So, the oxidation states are N(-3), H(+1), and P(+5). e. P4O6

Assign oxidation states for P and O

Assign -2 for O. The sum of oxidation states must be zero. \(4x + 6(-2) = 0\) \(x = +3\) So, the oxidation states are P(+3) and O(-2). f. Fe3O4

Assign oxidation states for Fe and O

Assign -2 for O. We know iron can have oxidation states of +2 or +3, and the sum of oxidation states must be zero. \(3x + 4(-2) = 0\) It becomes clear that there are two Fe(+3) and one Fe(+2). So, the oxidation states are Fe(+2, +3) and O(-2). g. XeOF4

Assign oxidation states for Xe, O, and F

Assign -2 for O and -1 for F. The sum of oxidation states must be zero. \(x - 2 + 4(-1) = 0\) \(x = +6\) So, the oxidation states are Xe(+6), O(-2), and F(-1). h. SF4

Assign oxidation states for S and F

Assign -1 for F. The sum of oxidation states must be zero. \(x + 4(-1) = 0\) \(x = +4\) So, the oxidation states are S(+4) and F(-1). i. CO

Assign oxidation states for C and O

Assign -2 for O. The sum of oxidation states must be zero. \(x + (-2) = 0\) \(x = +2\) So, the oxidation states are C(+2) and O(-2). j. C6H12O6

Assign oxidation states for C, H, and O

Assign -2 for O and +1 for H. In most organic compounds, carbon has an oxidation state of -4, 0, or +4. Each carbon atom \(x = \frac{12(+1) + 6(-2)}{6}\); thus, the oxidation state of carbon is 0. So, the oxidation states are C(0), H(+1), and O(-2).

Key concepts

Redox Reactions

Redox reactions are fascinating because they involve the transfer of electrons between substances. These reactions are important as they can produce energy or drive many industrial processes. In a redox reaction, one atom gets oxidized, meaning it loses electrons, and another gets reduced, meaning it gains electrons. This electron transfer results in changes in oxidation states. A useful way to keep track of the electrons in a redox reaction is by assigning oxidation numbers, which indicate the charge of an atom if it were entirely ionic. A practical example is the combustion of hydrogen gas with oxygen, producing water. In this reaction, hydrogen goes from an oxidation state of 0 to +1, losing electrons, while oxygen goes from 0 to -2, gaining electrons.

Chemical Compounds

Chemical compounds are substances formed when two or more elements are chemically bonded together. They can be classified based on the types of bonds, such as ionic, covalent, or metallic. These compounds play crucial roles in forming all substances around us. Ionic compounds, like sodium chloride, consist of charged ions held together by strong electrostatic forces. Covalent compounds, such as water, are formed by atoms sharing electrons. Knowing the formula of a compound helps determine the proportion of atoms, which reflects both its type and the kind of elements involved. This knowledge is essential when determining oxidation states, as seen in the exercise examples where Na, Fe, and O form different kinds of compounds with varying oxidation states.

Chemical Bonding

Chemical bonding is the glue that holds atoms together to form molecules and compounds. It's through bonding that atoms achieve stability, creating the compounds that fill our world. There are primarily three types of bonds: ionic, covalent, and metallic.

• Ionic bonds occur when electrons are transferred from one atom to another, forming charged ions that attract each other, common in metal and non-metal combinations.
• Covalent bonds involve the sharing of electrons between atoms, typical among non-metal atoms.
• Metallic bonds are seen in metals where electrons are shared across a lattice of atoms.

Understanding these bonds can aid students in grasping how elements like oxygen form compounds in the exercise, consistently taking an oxidation state of -2, reflecting its tendency to gain two electrons for stability.

Oxidation Numbers

Oxidation numbers are vital in understanding and writing redox reactions. They represent the hypothetical charge an atom would have if all bonds with atoms of different elements were 100% ionic. Here's the basic rule-of-thumb for assigning these numbers:

• Elements in their pure form (like O₂ or H₂) have an oxidation number of 0.
• The oxidation number for a monoatomic ion is equal to its charge.
• In compounds, fluorine is always -1, oxygen is generally -2 (except in peroxides or when bonded to fluorine), and hydrogen is +1 (except in hydrides).

The sum of oxidation numbers in a neutral compound must be zero, while in a polyatomic ion, it equals the charge of the ion. Practicing this, as done in the solutions for compounds like KMnO₄ and CO, helps in balancing redox reactions and predicting chemical behavior.