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Chapter 4: Problem 50

Give the oxidation number of the underlined atoms in the following molecules and ions: (a) \(\mathrm{Mg}_{3} \underline{\mathrm{N}}_{2}\) (b) \(\mathrm{CsO}_{2},\) (c) \(\mathrm{CaC}_{2}\), (d) \(\underline{\mathrm{CO}}_{3}^{2-}\), (e) \(\underline{\mathrm{C}}_{2} \mathrm{O}_{4}^{2-}\) (f) \(\mathrm{Zn} \underline{\mathrm{O}}_{2}^{2-}\) \((\mathrm{g}) \mathrm{NaBH}_{4},(\mathrm{~h}) \underline{\mathrm{W}} \mathrm{O}_{4}^{2-}\)

Short Answer

Expert verified
The oxidation numbers of the underlined atoms are: (a) \(\mathrm{N}\) in \(\mathrm{Mg}_{3} \mathrm{N}_{2}\) is -3, (b) \(\mathrm{O}\) in \(\mathrm{CsO}_{2}\) is -1, (c) \(\mathrm{C}\) in \(\mathrm{CaC}_{2}\) is -1, (d) \(\mathrm{C}\) in \(\mathrm{CO}_{3}^{2-}\) is +4, (e) \(\mathrm{C}\) in \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is +3, (f) \(\mathrm{O}\) in \(\mathrm{Zn} \mathrm{O}_{2}^{2-}\) is -2, (g) \(\mathrm{B}\) in \(\mathrm{NaBH}_{4}\) is -4 and (h) \(\mathrm{W}\) in \(\mathrm{WO}_{4}^{2-}\) is +6.

Step by step solution

01

Calculate for Mg3N2

Considering the molecule \(\mathrm{Mg}_{3} \underline{\mathrm{N}}_{2}\). Here, the magnesium atom \(\mathrm{Mg}\) is likely to have an oxidation state of +2 because it is a group 2 metal. Since the compound is neutral overall, the nitrogen atom \(\mathrm{N}\) will have a total charge of -6 (to balance the +6 from three Mg). Therefore, each nitrogen atom (\(\mathrm{N}\)) would have an oxidation state of -3 (because there are two nitrogen atoms in the molecule).
02

Calculate for CsO2, CaC2, CO32-, C2O42-, ZnO22-

Similarly, in \(\mathrm{CsO}_{2}\), cesium (\(\mathrm{Cs}\)) has an oxidation state of +1 (as alkali metals always have +1 in compounds), and for the total -2 charge from oxygen, \(\underline{\mathrm{O}}\) has an oxidation state of -1 each. \nIn \(\mathrm{CaC}_{2}\), calcium (\(\mathrm{Ca}\)) has +2, and each carbon atom \(\underline{\mathrm{C}}\) has -1 to balance it. \nIn \(\underline{\mathrm{CO}}_{3}^{2-}\), to make a sum of -2 (the charge of the ion), each oxygen atom has -2 and carbon (\(\underline{\mathrm{C}}\)) has +4. \nIn \(\underline{\mathrm{C}}_{2} \mathrm{O}_{4}^{2-}\), each oxygen atom has -2. Therefore to sum up to -2 overall (the charge of the ion), each carbon atom (\(\underline{\mathrm{C}}\)) would have +3. \nIn \(\mathrm{Zn} \underline{\mathrm{O}}_{2}^{2-}\), to make a sum of -2 (the charge of the ion), Zinc (\(\mathrm{Zn}\)) has +2 and each oxygen atom (\(\underline{\mathrm{O}}\)) has -2.
03

Calculate for NaBH4, WO42-

For \(\mathrm{NaBH}_{4}\), sodium (\(\mathrm{Na}\)) has +1, each hydrogen atom has +1 and boron (\(\underline{\mathrm{B}}\)) must have -4 to balance the sum to zero. \nLastly, in the complex ion \(\underline{\mathrm{W}} \mathrm{O}_{4}^{2-}\), each oxygen atom has -2, therefore, tungsten (\(\underline{\mathrm{W}}\)) must have +6 in order to result in an overall charge of -2 (the charge of the ion).

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Key Concepts

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

Redox Chemistry
Redox chemistry, short for reduction-oxidation chemistry, is a branch of chemical science dealing with the transfer of electrons between substances. This process changes the oxidation states of atoms, often leading to observable chemical changes like color shifts, energy release as heat or light, or gas production. In these reactions, one substance is oxidized (loses electrons) and another is reduced (gains electrons). Balancing redox reactions is not just about equalizing the number of atoms on both sides of the equation but also about ensuring that the total charge and the number of electrons lost and gained are balanced.

Understanding redox chemistry is critical, as this concept is central to many processes, including metabolism in living organisms, industrial chemical processes like corrosion and battery operations, and even geological phenomena.
Chemical Compounds Oxidation State
The oxidation state (or oxidation number) is a term used to describe the degree of oxidation of an atom in a chemical compound. It represents the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. Determining oxidation numbers for atoms within molecules or ions allows us to understand the nature of chemical bonding and to balance redox reactions. There are a set of rules to help assign these numbers:
  • The oxidation state of a pure element is always zero.
  • For monoatomic ions, the oxidation state is equal to the charge on the ion.
  • Hydrogen generally has a +1 oxidation state, and oxygen generally has a -2 state.
  • Alkali metals (in group 1 of the periodic table) have an oxidation state of +1, and alkaline earth metals (in group 2) have an oxidation state of +2.
  • For a neutral molecule, the sum of oxidation numbers is zero; for a polyatomic ion, it is equal to the ion charge.
Balancing Oxidation-Reduction Reactions
Balancing oxidation-reduction (redox) reactions requires both mass and charge to be conserved. Typically, the process includes assigning oxidation numbers to all elements in the reaction, determining which are oxidized and reduced, and then using the half-reaction method to balance the electrons gained and lost. First, separate the equation into two half-reactions—one for oxidation and one for reduction. Next, balance each half-reaction for mass (starting with elements other than O and H) and for oxygen by adding water molecules and for hydrogen by adding H+ ions. After that, balance the charges by adding electrons. Lastly, combine the half-reactions while ensuring the number of electrons in both the oxidation and reduction half are equal. Doing so allows us to balance the equation in terms of both mass and charge.
Polyatomic Ions Oxidation Numbers
Polyatomic ions are charged entities composed of two or more atoms covalently bonded, or a metal complex that can be considered as acting as a single unit. The oxidation state of each atom in a polyatomic ion is not always apparent but can be determined using oxidation rules and the overall charge of the ion. The sum of oxidation numbers in a polyatomic ion is equal to its ionic charge. For example, in sulfate ion (SO4^2-), with an overall charge of -2, the sulfur may have an oxidation number of +6, while each oxygen has an oxidation number of -2 to add up to the charge of the ion. Remember, when calculating the oxidation states for polyatomic ions, it's essential to apply the basic rules, taking into account the known oxidation states of certain elements and the overall charge of the ion. This can sometimes involve a bit of problem-solving and logical deduction.

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Most popular questions from this chapter

What volume of a \(0.500 M\) HCl solution is needed to neutralize each of the following? (a) \(10.0 \mathrm{~mL}\) of a \(0.300 \mathrm{M} \mathrm{NaOH}\) solution (b) \(10.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution

The concentration of sulfate in water can be determined by adding a solution of barium chloride to precipitate the sulfate ion. Write the net ionic equation for this reaction. Treating a 145-mL sample of water with excess \(\mathrm{BaCl}_{2}(a q)\) precipitated \(0.330 \mathrm{~g}\) of \(\mathrm{BaSO}_{4} .\) Determine the concentration of sulfate in the original water sample.

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Calculate the molarity of each of the following solutions: (a) \(29.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(545 \mathrm{~mL}\) of solution, (b) \(15.4 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(74.0 \mathrm{~mL}\) of solution, (c) \(9.00 \mathrm{~g}\) of sodium chloride \((\mathrm{NaCl})\) in \(86.4 \mathrm{~mL}\) of solution.

Carbon dioxide in air can be removed by an aqueous metal hydroxide solution such as \(\mathrm{LiOH}\) and \(\mathrm{Ba}(\mathrm{OH})_{2} .\) (a) Write equations for the reactions. (Carbon dioxide reacts with water to form carbonic acid.) (b) Calculate the mass of \(\mathrm{CO}_{2}\) that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.800 \mathrm{M} \mathrm{LiOH}\) and a \(0.800 M \mathrm{Ba}(\mathrm{OH})_{2}\) solution. (c) Which solution would you choose for use in a space capsule and which for use in a submarine?
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