Problem 25
Question
Oxidation numbers of carbon in \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{CH}_{4}\) and diamond respectively are (a) \(+3,4\) and \(+4\) (b) \(+3,-4\) and zero (c) \(+6,+4\) and zero (d) \(+6,+4\) and \(+4\)
Step-by-Step Solution
Verified Answer
The correct answer is (b) +3, -4, and zero.
1Step 1: Determine oxidation state in H2C2O4
The total oxidation state of hydrogen is usually +1, and there are two hydrogen atoms, contributing +2. Oxygen usually has an oxidation state of -2, and there are four oxygen atoms, contributing -8. Let the oxidation state of carbon be x for each carbon atom (there are two carbon atoms). The sum of oxidation states should equal zero for neutral compounds.\[ 2(+1) + 2x + 4(-2) = 0\]Simplifying gives:\[ 2 + 2x - 8 = 0 \]\[ 2x - 6 = 0 \]\[ 2x = 6 \]\[ x = +3 \]So the oxidation number of carbon in \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\) is +3.
2Step 2: Determine oxidation state in CH4
In \(\mathrm{CH}_4\), hydrogen has an oxidation state of +1, and there are four hydrogen atoms, contributing a total of +4. The oxidation state of carbon is x. Since the compound is neutral, the sum of the oxidation states must be zero:\[ x + 4(+1) = 0 \]\[ x + 4 = 0 \]\[ x = -4 \]Thus, the oxidation number of carbon in \(\mathrm{CH}_4\) is -4.
3Step 3: Determine oxidation state in diamond
Diamond is a form of elemental carbon where carbon atoms are bonded to each other in a regular pattern. Since the element is in its elemental form, the oxidation state of the carbon atom in diamond is 0.
4Step 4: Compare with answer choices
From the preceding steps, we found the oxidation states: - In \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\), carbon is +3.- In \(\mathrm{CH}_4\), carbon is -4.- In diamond, carbon is 0.Comparing these results to the answer choices, the correct answer is (b) \(+3, -4\) and zero.
Key Concepts
Oxidation StateOrganic CompoundsElementary Chemistry Concepts
Oxidation State
The oxidation state, also known as oxidation number, is a fundamental concept in chemistry. It describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. It is a useful tool in determining electron flow in redox reactions.
Oxidation states are assigned based on a set of rules. For example:
Oxidation states are assigned based on a set of rules. For example:
- The oxidation state of an element in its natural form, like O2 or diamond (which is pure carbon), is always 0.
- Fluorine in compounds is always -1, and oxygen is typically assigned -2 in most compounds.
- Hydrogen usually has +1 unless it's paired with more electropositive elements, like metals, where it is -1.
Organic Compounds
Organic compounds are primarily composed of carbon and hydrogen, often with additional elements such as oxygen, nitrogen, sulfur, and halogens. The structure and arrangement of these atoms define the properties and reactions of the compounds.
In the context of this exercise:
In the context of this exercise:
- \(\text{H}_2\text{C}_2\text{O}_4\) (oxalic acid) is an organic compound where two carbon atoms are bonded with each other and to oxygen.
- \(\text{CH}_4\) (methane) is one of the simplest organic compounds, consisting of one carbon atom bonded to four hydrogen atoms. It's a significant component of natural gas.
- Diamond is a form of pure carbon, structured in a robust three-dimensional lattice.
Elementary Chemistry Concepts
Elementary chemistry concepts lay the foundation for understanding more complex chemical principles. These basics include understanding atoms, molecules, compounds, and how they interact with each other.
Key elements of elementary chemistry involve:
Key elements of elementary chemistry involve:
- Atoms: The basic unit of chemical elements, composed of protons, neutrons, and electrons.
- Molecules: Formed when two or more atoms bond together. The composition determines the chemical properties and reactivity.
- Compounds: Substances formed when atoms of different chemical elements are chemically bonded.
- Chemical Reactions: Processes where substances change into different substances through breaking and forming chemical bonds.
Other exercises in this chapter
Problem 23
A metal ion \(\mathrm{M}^{3+}\) loses three electrons to produce another cation. The oxidation number of the metal in the cation is (a) \(+3\) (b) \(+4\) (c) \(
View solution Problem 24
\(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) have the oxidation numbers of \(+6,-2\) and \(-1\) respectively, the possible molecular formula when these atoms c
View solution Problem 26
Which of the following does not represent redox reactions? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}+\ma
View solution Problem 27
If \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\), oxidation number of \(\mathrm{Cl}\) changes from (a) \(-1\) to 0 (b) 0 to \(-1\) (c) 0 to \(+5\)
View solution