Problem 44 - Introductory Chemistry Atoms First

Question

Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{CO}\) (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{HCOO}^{-}\) (d) \(\mathrm{PtCl}_{6}^{2-}\)

Step-by-Step Solution

Answer

The oxidation states of the atoms in the given molecules/ions are: a) CO: C (+2), O (-2) b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): C (0), H (+1), Cl (-1) c) \(\mathrm{HCOO}^{-}\): C (+1), H (+1), O (terminal) (-2), O (bridging) (-1) d) \(\mathrm{PtCl}_{6}^{2-}\): Pt (+4), Cl (-1)

Step 1: a) CO

To assign the oxidation state in CO, we'll use the following steps: 1. Assign 'O' an oxidation state of -2 (since it is not in a peroxide or bonded to F). 2. Calculate the oxidation state of 'C'. Since \(\mathrm{CO}\) is a neutral molecule, the sum of the oxidation states must be equal to zero. So, \[x-2=0\] Solving this equation, we get \[x=+2\] So the oxidation state of each atom is: C: +2 O: -2

Step 2: b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\)

Assign the oxidation state in \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\): 1. Assign H an oxidation state of +1 2. Assign Cl an oxidation state of -1 (the Chlorine is not bonded to oxygen) 3. Calculate the oxidation state of C. Since \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) is neutral, the sum of the oxidation states must be equal to zero. So, \[x + 2 \cdot (+1) + 2 \cdot (-1)=0\] Solving this equation, we get \[x=0\] So the oxidation state of each atom is: C: 0 H: +1 Cl: -1

Step 3: c) \(\mathrm{HCOO}^{-}\)

Assign the oxidation state in \(\mathrm{HCOO}^{-}\): 1. Assign the 'H' an oxidation state of +1 2. Assign the terminal 'O' (the one bonded to 'C' exclusively) an oxidation state of -2. 3. Assign the bridging 'O' (the one bonded to 'C' and the other 'O')the oxidation state of -1 (because the oxygen atom bonded to `C` are in a polyatomic-ion). 4. Calculate the oxidation state of 'C'. Since \(\mathrm{HCOO}^{-}\) has a charge of -1, the sum of the oxidation states must be equal to -1. So, \[x + 1 + (-2) + (-1)=-1\] Solving this equation, we get \[x=+1\] So the oxidation state of each atom is: C: +1 H: +1 O (terminal): -2 O (bridging): -1

Step 4: d) \(\mathrm{PtCl}_{6}^{2-}\)

Assign the oxidation state in \(\mathrm{PtCl}_{6}^{2-}\): 1. Assign each 'Cl' atom an oxidation state of -1. 2. Calculate the oxidation state of 'Pt'. Since \(\mathrm{PtCl}_{6}^{2-}\) has a charge of -2, the sum of the oxidation states must be equal to -2. So, \[x + 6 \cdot (-1)=-2\] Solving this equation, we get \[x=+4\] So the oxidation state of each atom is: Pt: +4 Cl: -1

Key Concepts

Oxidation State Rules

Understanding the oxidation state is crucial because it helps in determining how electrons are distributed in atoms and ions within compounds. In chemistry, oxidation state or oxidation number is an indicator of the degree of oxidation of an atom within a molecule or ion.

Here are some simplified rules to follow when assigning oxidation states:

• The oxidation state of a pure element is always zero.
• For ions composed of only one atom, the oxidation state is equal to the charge of the ion.
• Hydrogen usually has an oxidation state of +1, while oxygen typically has an oxidation state of -2.
• In compounds or ions, the sum of oxidation states must equal the total charge of the compound or ion.
• Fluorine, being the most electronegative element, always has an oxidation state of -1 in its compounds.

When applying these rules to complex molecules or polyatomic ions, you may encounter exceptions or need additional rules, such as differentiating between bonding and antibonding oxygen atoms, as seen in polyatomic ions like \(\mathrm{HCOO}^{-}\), where the bonding patterns influence the oxidation states.

Oxidation-Reduction Reactions

The movement of electrons from one element to another is what characterizes oxidation-reduction (redox) reactions. In these processes, one species loses electrons (oxidation) and another gains electrons (reduction). These two processes always occur simultaneously - when one substance is oxidized, another is reduced.

When assigning oxidation numbers in a redox reaction, consider:

• The substance that increases its oxidation number is being oxidized.
• The substance that decreases its oxidation number is being reduced.
• Agents promoting oxidation are called oxidizing agents; similarly, those promoting reduction are reducing agents.

The ability to identify changes in oxidation states can aid in classifying the roles substances play in a reaction and in balancing redox equations.

Polyatomic Ions

A polyatomic ion is a charged species (ion) composed of two or more atoms covalently bonded or of a metal complex that can be considered as acting as a single unit. The process of determining oxidation states in these ions is particularly important because they often participate in redox reactions.

When assigning oxidation numbers to polyatomic ions, keep in mind:

• The sum of oxidation states within a polyatomic ion should equal its overall charge.
• The most electronegative atoms are usually assigned their typical oxidation states first.
• Oxygen in polyatomic ions generally has an oxidation state of -2, unless it's part of a peroxide, where it has an oxidation state of -1.
• Different positions of the same element in a polyatomic ion could hold different oxidation numbers, as seen with oxygen in \(\mathrm{HCOO}^{-}\).

Understanding these particulars allows us to compute the oxidation states correctly, even in more complex structures.

Oxidation Number Calculation

Calculating oxidation numbers is a fundamental skill in chemistry, necessary for analyzing redox reactions and understanding the composition and reactivity of molecules or ions. To calculate an oxidation number:

• Apply the oxidation state rules systematically, taking into account the molecule's or ion's overall charge.
• Keep track of each atom's oxidation state by writing an equation where the sum of oxidation states equals the charge of the molecule or ion.
• Solve the equation for the unknown oxidation state.

For instance, in the compound \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), carbon has an oxidation state of 0, since the +2 from the hydrogen atoms cancels out the -2 from the chlorine atoms. These calculations are essential for predicting the behavior of different species in chemical reactions.