$$ \begin{aligned} &\text { Point out the oxidation number of } \mathrm{C} \text { in the following : }\\\ &\mathrm{CH}_{4}, \mathrm{C}_{3} \mathrm{H}_{8}, \mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{C}_{4} \mathrm{H}_{10}, \mathrm{CO}, \mathrm{CO}_{2} \text { and } \mathrm{HCO}_{3}^{-}, \mathrm{CO}_{3}^{2-} \end{aligned} $$

Short Answer

Expert verified
The oxidation numbers of carbon in the given compounds are: CH4: -4, C3H8: -2.67, C2H6: -3, C4H10: -2.5, CO: +2, CO2: +4, HCO3-: +4, CO32-: +4.

Step by step solution

01

Understanding Oxidation Numbers

Oxidation number is a concept that provides a measure of the degree of oxidation of an atom in a substance. It is often represented as a charge that an atom would have if all bonds to atoms of different elements were completely ionic.
02

Assigning Oxidation Numbers for Hydrogen and Oxygen

Hydrogen generally is assigned an oxidation number of +1, and oxygen is generally assigned an oxidation number of -2, except in peroxides where it is -1, or when bonded to fluorine where it is positive. We will use these rules as a starting point for assigning oxidation numbers to carbon in the given compounds.
03

Determining the Oxidation Number of Carbon in CH4

In methane (CH4), hydrogen has an oxidation number of +1. With four hydrogens, the total is +4. To balance this in a neutral molecule, carbon must have an oxidation number of -4.
04

Determining the Oxidation Number of Carbon in C3H8

In propane (C3H8), like in methane, hydrogen has an oxidation number of +1. With eight hydrogens, the total is +8. To balance the total charge to zero for the entire molecule, each carbon must have an oxidation number that sums to -8 when multiplied by 3. Therefore, the oxidation number of carbon is -8/3 or approximately -2.67.
05

Determining the Oxidation Number of Carbon in C2H6

In ethane (C2H6), hydrogen has an oxidation number of +1. With six hydrogens, the total is +6. To balance the total charge to zero for the entire molecule, each carbon must have an oxidation number of -3.
06

Determining the Oxidation Number of Carbon in C4H10

In butane (C4H10), hydrogen has an oxidation number of +1. With ten hydrogens, the total is +10. To balance the total charge to zero for the molecule, each carbon must have an oxidation number that sums to -10 when multiplied by 4. Therefore, the oxidation number of carbon is -10/4 or -2.5.
07

Determining the Oxidation Number of Carbon in CO

In carbon monoxide (CO), the total charge is zero. Oxygen has an oxidation number of -2, therefore carbon must have an oxidation number of +2 to balance the oxygen.
08

Determining the Oxidation Number of Carbon in CO2

In carbon dioxide (CO2), the total charge is zero. Oxygen has an oxidation number of -2, and since there are two oxygens, the total oxygen has a charge of -4. To balance this, carbon must have an oxidation number of +4.
09

Determining the Oxidation Number of Carbon in HCO3-

In the bicarbonate ion (HCO3-), hydrogen has an oxidation number of +1, and oxygen has an oxidation number of -2 for a total of -6 from oxygen. The overall charge of the ion is -1. Therefore, to balance the charges, carbon must have an oxidation number of +4.
10

Determining the Oxidation Number of Carbon in CO32-

In the carbonate ion (CO32-), oxygen has an oxidation number of -2. With three oxygens, the total is -6. The overall charge of the ion is -2. Therefore, to balance the charges, carbon must have an oxidation number of +4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
Oxidation states, also known as oxidation numbers, serve as a bookkeeping tool in chemistry, giving us insight into the electron distribution within compounds. Essentially, an oxidation state indicates the hypothetical charge an atom would possess if all its bonds were ionic in nature, disregarding the actual bond type—it's essentially a measure of electron 'ownership' within a molecule.

When determining the oxidation state of an atom, certain rules are applied consistently. For instance, the oxidation number of a pure element is always 0, and for monoatomic ions, it is equal to the ion's charge. In compounds, hydrogen typically has an oxidation number of +1, and oxygen normally has an oxidation number of -2, with some exceptions, such as in peroxides. The sum of oxidation states for all atoms in a molecule or ion must equal the overall charge of that species.
Compound Nomenclature
Understanding compound nomenclature is vital for identifying and communicating chemical substances systematically. It follows specific rules, such as prefixes for the number of atoms (mono-, di-, tri-, etc.), and suffixes for the type of compound, like '-ide' for simple anions or binary compounds. For example, CO is known as carbon monoxide, and CO2 is carbon dioxide, reflecting the number of oxygen atoms.

Naming ions involves similar methods: anions typically end in '-ide', while polyatomic ions have unique names, such as the bicarbonate (HCO3-) and carbonate (CO32-) ions mentioned in the provided exercise. The prefixes and suffixes are essential clues to the composition and structure of the compounds, an essential aspect of effective chemical communication.
Balancing Charges
Balancing charges is a crucial aspect of writing correct chemical formulas and equations. It's a concept based on the law of conservation of charge, which states that the net charge in an isolated system remains constant. In chemical compounds, the total positive charge (from cations) must balance the total negative charge (from anions) resulting in a neutral entity. For polyatomic ions, the sum of charges of all atoms included must result in the ion's net charge.

For instance, in the bicarbonate ion (HCO3-), the positive and negative charges must add up to -1. Similarly, in methane (CH4), carbon has a -4 oxidation state to neutralize the +1 oxidation states of the four hydrogen atoms. Understanding how to balance charges enables chemists to predict the formulas of compounds formed and to balance chemical equations for reactions.
Redox Chemistry
Redox chemistry, which stands for reduction-oxidation chemistry, involves reactions in which the oxidation states of atoms change. These processes are fundamental to numerous chemical reactions, such as combustion, respiration, corrosion, and energy generation within batteries.

Redox reactions consist of two complementary processes: reduction (gain of electrons, leading to a decrease in oxidation state) and oxidation (loss of electrons, leading to an increase in oxidation state). For a redox process to occur, there must be a transfer of electrons from the reducing agent to the oxidizing agent. The action of assigning oxidation numbers, as demonstrated in the provided exercise, is a preliminary step to identifying redox reactions and understanding the electron transfer that occurs during these chemical processes.

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Most popular questions from this chapter

Select the nature or type of redox change in the following reactions: (ii) \(2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}^{0}\) (b) \(\quad \mathrm{Cl}_{2} \longrightarrow \mathrm{ClO}^{-}+\mathrm{Cl}^{-}\) (c) \(2 \mathrm{KClO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}+3 \mathrm{O}_{2}\) (d) \(\quad\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \longrightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}\) (e) \(10 \mathrm{FeSO}_{4}+2 \mathrm{KMnO}_{4}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{MnSO}_{4}+5 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\) \(\mathrm{K}_{2} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}\) (f) \(5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+\) \(10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)

Write the half reactions for the following redox reactions: (a) \(2 \mathrm{Fe}^{3+}{ }_{\text {(aq. })}+2 \mathrm{I}_{(\mathrm{aq})} \longrightarrow 2 \mathrm{Fe}^{2+}{ }_{\text {(aq.) }}+\mathrm{I}_{2(\mathrm{aq} .)}\) (b) \(\mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{H}_{(\text {aq. })}^{+} \longrightarrow \mathrm{Zn}^{2+}\) (aq.) \(+\mathrm{H}_{2(\mathrm{~g})}\) (c) \(\mathrm{Al}_{(\mathrm{s})}+3 \mathrm{Ag}^{+}{ }_{\text {(aq.) }} \longrightarrow \mathrm{Al}_{\text {(aq.) }}^{3+}+3 \mathrm{Ag}_{(\mathrm{s})}\)

Identify the substance acting as oxidunt or reductant if any in the following: (i) \(\mathrm{AICl}_{3}+3 \mathrm{~K} \longrightarrow \mathrm{Al}+3 \mathrm{KCI}\) (ii) \(\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{~S} \longrightarrow 3 \mathrm{~S}+\mathrm{H}_{2} \mathrm{O}\) (iii) \(\mathrm{BaCl}_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{BaSO}_{4}+2 \mathrm{NaCl}\) (iv) \(3 \mathrm{I}_{2}+6 \mathrm{NaOH} \longrightarrow \mathrm{NalO}_{3}+5 \mathrm{NaI}+3 \mathrm{H}_{2} \mathrm{O}\)

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