Question

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{KMnO}_{4}\) f. \(\mathrm{Fe}_{2} \mathrm{O}_{4}\) \(\begin{array}{ll}\text { b. } \mathrm{NiO}_{2} & \text { g. } \mathrm{XeOF}_{4}\end{array}\) c. \(\mathrm{Na}_{4} \mathrm{Fe}(\mathrm{OH})_{6}\) h. \(\mathrm{SF}_{4}\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) i. \(\mathrm{CO}\) e. \(\mathrm{P}_{4} \mathrm{O}_{6}\) j. \(\mathrm{C}_{6} \mathrm{H}_{1}, \mathrm{O}_{e}\)

Short answer

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, O: -2 e. P: +3, O: -2 f. Fe: +4, O: -2 g. Xe: +6, O: -2, F: -1 h. S: +4, F: -1 i. C: +2, O: -2 j. C: +4, H: +1, O: -2

Step-by-step solution

a. KMnO4

In the case of KMnO_4, we first assign +1 for K and -2 for each O atom. Summing up the oxidation states, we get 1 + x - 8 = 0, where x is the oxidation state of Mn. Therefore, the oxidation state of Mn is +7.

b. NiO2

For the compound NiO_2, each O atom has an oxidation state of -2. So, the equation for oxidation states is x - 4 = 0, where x is the oxidation state of Ni. The oxidation state of Ni is +4.

c. Na4Fe(OH)6

For the compound Na_4Fe(OH)_6, we give Na +1, Fe unknown (x), O -2, and H +1. The sum of oxidation states in this compound is 4(1) + x + 6(-2 + 1) = 0. The oxidation state of Fe is +2.

d. (NH4)2HPO4

In the compound (NH_4)_2HPO_4, we assign N unknown (x), H +1, P unknown (y), and O -2. The sum of oxidation states is 2(1x + 4(1)) + 1 + y - 8 = 0. From the NH_4 group, we know hydrogen is +1; therefore, 1x + 4 = 0, meaning x = -3. The oxidation state of N is -3, and P's oxidation state is +5.

e. P4O6

In the compound P_4O_6, each P atom has an unknown oxidation state (x), and each O atom has an oxidation state of -2. The sum of oxidation states in this compound is 4x - 12 = 0. The oxidation state of P is +3.

f. Fe2O4

For the compound Fe_2O_4, each Fe atom has an unknown oxidation state (x), and each O has an oxidation state of -2. The sum of oxidation states is 2x - 8 = 0. The oxidation state of Fe is +4.

g. XeOF4

For the compound XeOF_4, Xe has an unknown oxidation state (x), O has -2, and each F atom has -1. The sum of oxidation states is x - 2 - 4 = 0. The oxidation state of Xe is +6.

h. SF4

In the molecule SF_4, S has an unknown oxidation state (x), and each F atom has -1 oxidation state. The sum of oxidation states is x - 4 = 0. The oxidation state of S is +4.

i. CO

For CO, we assign C an unknown oxidation state (x) and O an oxidation state of -2. The sum of oxidation states is x - 2 = 0. The oxidation state of C is +2.

j. C6H12O6

In the case of C_6H_12O_6, we assign C an unknown oxidation state (x), H has +1, and O has -2. The sum of oxidation states is 6x + 12 - 12 = 0. The oxidation state of each C is +4.

Key concepts

Assigning Oxidation Numbers

Understanding how to assign oxidation numbers to atoms within molecules or ions is foundational in studying redox chemistry. The oxidation number refers to the hypothetical charge an atom would have if all bonds to atoms of different elements were completely ionic. Here are the general rules we follow for assigning oxidation numbers:

• For an atom in its elemental form (e.g., N2, O2, P4), the oxidation number is always zero.
• For a monatomic ion (e.g., Na+, Cl-), the oxidation number equals the charge of the ion.
• Oxygen usually has an oxidation number of -2, except in peroxides where it's -1.
• Hydrogen is generally assigned a +1 oxidation number, except when it is bonded to metals in hydrides, where it is -1.
• The oxidation numbers in a molecule or ion should sum up to the charge of that species.

In our exercise, for instance, the oxidation number of Mn in KMnO4 is determined by considering the known oxidation states of K (+1) and O (-2), then solving for Mn, which gives us +7. These principles guide us through complex compounds and allow us to systematically compute the oxidation numbers for each element.

Balancing Redox Equations

Balancing redox equations involves making sure that both the number of atoms and the total charge are balanced on either side of the reaction. This is essential because it reflects the conservation of mass and charge. In doing so, we need to follow a few steps, which are essentially the underlying principle of the half-reaction method:

• Separate the redox reaction into two half-reactions: one for oxidation and one for reduction.
• Balance all elements in the half-reactions other than O and H.
• Balance the oxygen atoms by adding H2O to the side that requires oxygen.
• Balance the hydrogen atoms by adding H+ to the opposite side from where H2O was added.
• Balance the charges by adding electrons - to the more positive side on each half-reaction.
• Equalize the number of electrons transferred in both half-reactions by multiplying the half-reactions by appropriate factors.
• Combine both half-reactions and simplify if possible.

Once this process is understood and applied, any redox equation can be balanced, ensuring that students can tackle even the most daunting redox challenges.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are chemical reactions where electrons are transferred between atoms, leading to a change in their oxidation states. This transfer fundamentally changes the properties of the reacting substances. Redox reactions are characterized by the movement of electrons from one molecule to another:

• Oxidation is the loss of electrons, resulting in an increase in oxidation state.
• Reduction is the gain of electrons, resulting in a decrease in oxidation state.

A mnemonic to remember this is 'OIL RIG'—Oxidation Is Loss, Reduction Is Gain. In our textbook examples, compounds are analyzed for their oxidation states to understand the electron transfer process during reactions. When balancing redox reactions, it's crucial to consider both atom balance and charge balance to reflect the true nature of the reaction. Grasping the concept of redox reactions allows students to comprehend a broad area of chemistry that includes energy production, metabolism, and many industrial processes.