Chapter 18: Problem 93
Assign oxidation states to all of the atoms in each of the following: a. \(\mathrm{PBr}_{3}\) b. \(\mathrm{C}_{3} \mathrm{H}_{8}\) c. \(\mathrm{KMnO}_{4}\) d. \(\mathrm{CH}_{3} \mathrm{COOH}\)
Short Answer
Expert verified
For \(\mathrm{PBr}_{3}\), the oxidation states are: P: +3, Br: -1.
For \(\mathrm{C}_{3} \mathrm{H}_{8}\), the oxidation states are: C: -2.67, H: +1.
For \(\mathrm{KMnO}_4\), the oxidation states are: K: +1, Mn: +7, O: -2.
For \(\mathrm{CH}_{3} \mathrm{COOH}\), the oxidation states are: C: -3, C: 0, O: -2, H: +1.
Step by step solution
01
Identify the elements in the compound
In this compound, we have phosphorus (P) and bromine (Br).
02
Apply the general rules for assigning oxidation states
Using rule 6, we know that the oxidation state of bromine (Br) in this compound is -1. Since there are three bromine atoms, the total charge from the bromine atoms is -3. According to rule 3, the sum of the oxidation states of all the atoms in the compound must be zero, so the oxidation state of phosphorus (P) must be +3.
So for \(\mathrm{PBr}_{3}\), the oxidation states are:
P: +3, Br: -1
For \(\mathrm{C}_{3} \mathrm{H}_{8}\):
03
Identify the elements in the compound
In this compound, we have carbon (C) and hydrogen (H).
04
Apply the general rules for assigning oxidation states
We'll use rule 5 here, stating that hydrogen usually has an oxidation state of +1 in compounds. So each hydrogen has a +1 oxidation state. Since there are 8 hydrogen atoms, the total charge from the hydrogen atoms is +8. Again, rule 3 states that the sum of oxidation states should be zero (since the compound is neutral), the total charge from the carbon atoms must be -8. As there are three carbon atoms, each carbon atom will have an oxidation state of -8/3 = -2.67 (approximately).
So for \(\mathrm{C}_{3} \mathrm{H}_{8}\), the oxidation states are:
C: -2.67, H: +1
For \(\mathrm{KMnO}_4\):
05
Identify the elements in the compound
In this compound, we have potassium (K), manganese (Mn), and oxygen (O).
06
Apply the general rules for assigning oxidation states
Using rule 2, the oxidation state of potassium (K) in this compound is +1. Rule 4 tells us that the oxidation state of oxygen in its compounds is usually -2. So, each oxygen atom has an oxidation state of -2. Since there are four oxygen atoms, the total charge from the oxygen atoms is -8. Finally, the sum of the oxidation states must be zero (rule 3), so the oxidation state of manganese (Mn) must be +7 (as -8 + 1 = -7 and Mn must compensate this charge).
So for \(\mathrm{KMnO}_4\), the oxidation states are:
K: +1, Mn: +7, O: -2
For \(\mathrm{CH}_{3} \mathrm{COOH}\):
07
Identify the elements in the compound
In this compound, we have carbon (C), hydrogen (H), and oxygen (O).
08
Apply the general rules for assigning oxidation states
Again, rule 5 states that hydrogen atoms have an oxidation state of +1 in compounds. So, each hydrogen atom has an oxidation state of +1 (and there are four in total, giving a +4 charge). Rule 4 tells us that the oxidation state of oxygen in its compounds is usually -2. So, the oxygen atoms in the compound have an oxidation state of -2 (and there are 2 oxygen atoms, giving a -4 charge). Now we can use rule 3 to calculate the oxidation state of the carbon atoms. In the \(\mathrm{CH}_{3}\) part of the compound, the carbon oxidation state can be found as follows: \(+1 * 3 + C = 0\) ⟹ C = -3. In the \(\mathrm{COOH}\) part of the compound, the carbon oxidation state can be found as follows: \(-2 * 2 + C + 2*1 = 0\) ⟹ C=0.
So for \(\mathrm{CH}_{3} \mathrm{COOH}\), the oxidation states are:
C(-3), C(0), O: -2, H: +1
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Redox Chemistry
Redox chemistry is a fundamental aspect of chemistry that involves the transfer of electrons between two substances. It's a short form for 'reduction-oxidation'. In these processes, one substance is oxidized, losing electrons, and another is reduced, gaining electrons. The substance that donates electrons is called the reducing agent, and the one that accepts electrons is the reducing agent.
Understanding redox chemistry is key to grasping concepts like energy production in batteries and biological systems, corrosion, and the functioning of antioxidants. When assigning oxidation states in a redox reaction, we are essentially keeping track of the electrons as they move from one atom to another, providing vital clues about how these reactions proceed.
Understanding redox chemistry is key to grasping concepts like energy production in batteries and biological systems, corrosion, and the functioning of antioxidants. When assigning oxidation states in a redox reaction, we are essentially keeping track of the electrons as they move from one atom to another, providing vital clues about how these reactions proceed.
Chemical Compound Oxidation
Chemical compound oxidation involves determining the increase in the oxidation state of an atom within a compound as it undergoes a chemical reaction. Oxidation doesn't always mean oxygen is involved. Rather, it indicates the loss of electrons, which can happen with or without oxygen. Conversely, when an atom's oxidation state decreases, it's gaining electrons, a process known as reduction.
In the context of the provided exercises like \( \mathrm{PBr}_{3} \), \( \mathrm{KMnO}_4 \), and \( \mathrm{CH}_{3} \mathrm{COOH} \), identifying the oxidation states of each atom enables us to understand which atoms are likely to be oxidized or reduced in reactions. For example, in \( \mathrm{KMnO}_4 \), manganese's high oxidation state of +7 makes it a strong oxidizing agent, capable of accepting electrons and thereby being reduced.
In the context of the provided exercises like \( \mathrm{PBr}_{3} \), \( \mathrm{KMnO}_4 \), and \( \mathrm{CH}_{3} \mathrm{COOH} \), identifying the oxidation states of each atom enables us to understand which atoms are likely to be oxidized or reduced in reactions. For example, in \( \mathrm{KMnO}_4 \), manganese's high oxidation state of +7 makes it a strong oxidizing agent, capable of accepting electrons and thereby being reduced.
Oxidation Number Rules
The rules for assigning oxidation numbers are essential for systematically determining the charge distribution in a chemical compound. Here are core rules often used:
The steps involved in pinpointing oxidation numbers often start with identifying known oxidation states (like those of H and O), then calculating the rest based on the charge balance across the compound. Grasping these rules is crucial for students as they offer a clear path to solving complex assignments on redox chemistry.
- The oxidation number of a free element is always zero.
- For a monoatomic ion, the oxidation number is equivalent to its charge.
- The total of the oxidation numbers for a neutral compound is zero; for a polyatomic ion, it is equal to the ion charge.
- Oxygen mostly has an oxidation number of -2, with notable exceptions in peroxides and superoxides.
- Hydrogen is usually +1 when bonded to nonmetals and -1 when bonded to metals.
The steps involved in pinpointing oxidation numbers often start with identifying known oxidation states (like those of H and O), then calculating the rest based on the charge balance across the compound. Grasping these rules is crucial for students as they offer a clear path to solving complex assignments on redox chemistry.