/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Give oxidation numbers for the u... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give oxidation numbers for the underlined atoms in the following molecules and ions: (a) \(\underline{\mathrm{Cs}_{2} \mathrm{O}},\) (b) \(\mathrm{CaI}_{2}\), (c) \(\underline{\mathrm{Al}_{2} \mathrm{O}_{3}}\) (d) \(\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{3},\) (e) \(\underline{\mathrm{Ti}} \mathrm{O}_{2}\) (f) \(\mathrm{MoO}_{4}^{2-}\) \((\mathrm{g}) \underline{\mathrm{Pt}} \mathrm{Cl}_{4}^{2-}\) (h) \(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\) (i) \(\underline{\mathrm{Sn} \mathrm{F}_{2}}\) (j) \(\underline{C l} F_{3}\) (k) \(\underline{\mathrm{Sb}} \mathrm{F}_{6}^{-}\)

Short Answer

Expert verified
(a) O: -2; (c) Al: +3; (e) Ti: +4; (g) Pt: +2; (h) Pt: +4; (i) Sn: +2; (j) Cl: +3; (k) Sb: +5

Step by step solution

01

Determine Oxidation Numbers for Cs2O

In Cs2O, cesium (Cs) is more electropositive and usually has an oxidation number of +1. Since the compound is neutral, the sum of the oxidation numbers must be zero. Given Cs's oxidation number of +1, the oxidation number of oxygen (O) must be -2 to balance the +2 from the two cesium atoms (2\(+1\) + (\(-2\)) = 0).
02

Determine Oxidation Numbers for Al2O3

In Al2O3, aluminum (Al) typically has an oxidation number of +3. There are three oxygen atoms, each with an oxidation number of -2. Therefore, the oxidation number for Al must be +3 to balance the -6 from oxygen ((2\(+3\)) + (3\(-2\)) = 0).
03

Determine Oxidation Numbers for TiO2

In TiO2, titanium (Ti) is the more electropositive element. Oxygen's oxidation number is -2. To balance the two -2 charges from oxygen, the oxidation number of Ti must be +4 ((1\(+4\)) + (2\(-2\)) = 0).
04

Determine Oxidation Number for Pt in PtCl4^2−

The chloride ion (Cl) has an oxidation number of -1. With four chloride ions, the total negative charge contribution is -4. The compound has an overall charge of -2; therefore, Pt must have an oxidation number of +2 to satisfy the overall charge ((1\(+2\)) + (4\(-1\)) = -2).
05

Determine Oxidation Number for Pt in PtCl6^2−

Each chloride ion (Cl) contributes -1. With six Cl ions, the total negative charge is -6. To achieve the overall charge of -2 for the complex, Pt needs an oxidation number of +4 ((1\(+4\)) + (6\(-1\)) = -2).
06

Determine Oxidation Numbers for SnF2

In SnF2, the fluoride ion (F) has an oxidation number of -1. With two fluoride ions, the total negative contribution is -2. Therefore, Sn must have an oxidation number of +2 to balance this ((1\(+2\)) + (2\(-1\)) = 0).
07

Determine Oxidation Number for Cl in ClF3

In ClF3, fluorine (F) is more electronegative and has an oxidation number of -1. With three F ions, the total negative contribution is -3. Cl must have an oxidation number of +3 to maintain neutrality ((1\(+3\)) + (3\(-1\)) = 0).
08

Determine Oxidation Number for Sb in SbF6^-

Each fluoride ion (F) has an oxidation number of -1. With six F ions, the total negative contribution is -6. To balance the charge and obtain the overall ion charge of -1, Sb must have an oxidation number of +5 ((1\(+5\)) + (6\(-1\)) = -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 91Ó°ÊÓ!

Key Concepts

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

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes that involve the transfer of electrons between two substances. In a redox reaction, one substance gains electrons (reduction) while another loses electrons (oxidation). Redox reactions are fundamental in both biological and industrial processes, playing a crucial role in energy production and chemical synthesis.
  • In a redox reaction, the species that donates electrons is called the reducing agent, and the one that accepts electrons is known as the oxidizing agent.
  • The concept of oxidation numbers helps in identifying which atoms are being oxidized and reduced in a chemical reaction.
  • Balancing redox reactions requires ensuring that both mass and charge are conserved, keeping track of the number of electrons transferred.
Understanding the flow of electrons in redox reactions is vital for grasping more complex chemical phenomena such as combustion, corrosion, and even cellular respiration in living organisms.
Electronegativity
Electronegativity is the ability of an atom to attract electrons in a chemical bond. This concept plays a significant role in determining how atoms interact in molecules, affecting their chemical reactivity and the nature of bonds.
  • Elements with high electronegativity, like fluorine and oxygen, are strong oxidizing agents, meaning they tend to gain electrons easily.
  • Electronegativity differences between atoms can lead to unequal sharing of electrons in covalent bonds, resulting in partial charges and polar molecules.
  • The greater the difference in electronegativity between two bonded atoms, the more ionic (as opposed to covalent) the bond will be.
By evaluating electronegativity values, we can predict how atoms in a molecule might behave in chemical reactions, such as those involving oxidation and reduction. This understanding is key in predicting the outcome of chemical reactions, including redox processes.
Oxidation States
Oxidation states, or oxidation numbers, help to keep track of electron distribution in compounds. They signify the hypothetical charges that atoms would have if the compound was composed of ions. This concept is essential for identifying the electron flow in redox reactions.
  • Oxidation states are assigned based on a set of rules, such as the more electronegative element in a bond gets the negative oxidation state.
  • Some elements have fixed oxidation states in most compounds, like alkali metals (+1) and alkaline earth metals (+2).
  • Transition metals, however, can exhibit multiple oxidation states due to the electron configuration of d orbitals, adding versatility to their chemistry.
Determining the oxidation state of an element in a compound allows chemists to deduce how atoms are interacting and the type of chemical changes they undergo in reactions. Especially for redox reactions, explaining changes in oxidation states is crucial for understanding which atoms are oxidized and which are reduced.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is an important ingredient of vinegar. A sample of \(50.0 \mathrm{~mL}\) of a commercial vinegar is titrated against a \(1.00 \mathrm{M} \mathrm{NaOH}\) solution. What is the concentration (in \(M\) ) of acetic acid present in the vinegar if \(5.75 \mathrm{~mL}\) of the base were required for the titration?

Predict and explain which of the following systems are electrically conducting: (a) solid \(\mathrm{NaCl}\) (b) molten \(\mathrm{NaCl},\) (c) an aqueous solution of \(\mathrm{NaCl}\).

Sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) can be obtained in very pure form and can be used to standardize acid solutions. What is the molarity of an HCl solution if 28.3 \(\mathrm{mL}\) of the solution is required to react with \(0.256 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Magnesium is a valuable, lightweight metal. It is used as a structural metal and in alloys, in batteries, and in chemical synthesis. Although magnesium is plentiful in Earth's crust, it is cheaper to "mine" the metal from seawater. Magnesium forms the second most abundant cation in the sea (after sodium); there are about \(1.3 \mathrm{~g}\) of magnesium in \(1 \mathrm{~kg}\) of seawater. The method of obtaining magnesium from seawater employs all three types of reactions discussed in this chapter: precipitation, acid-base, and redox reactions. In the first stage in the recovery of magnesium, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is heated at high temperatures to produce quicklime, or calcium oxide \((\mathrm{CaO})\) : $$ \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ When calcium oxide is treated with seawater, it forms calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right],\) which is slightly soluble and ionizes to give \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions: $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ The surplus hydroxide ions cause the much less soluble magnesium hydroxide to precipitate: $$ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) $$ The solid magnesium hydroxide is filtered and reacted with hydrochloric acid to form magnesium chloride \(\left(\mathrm{MgCl}_{2}\right):\) $$ \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \underset{\mathrm{MgCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)}{\longrightarrow} $$ After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The half-reactions are After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The half-reactions are $$ \begin{aligned} \mathrm{Mg}^{2+}+2 e^{-} \longrightarrow & \mathrm{Mg} \\ 2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-} \end{aligned} $$ The overall reaction is $$ \mathrm{MgCl}_{2}(l) \longrightarrow \mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) $$ This is how magnesium metal is produced. The chlorine gas generated can be converted to hydrochloric acid and recycled through the process. (a) Identify the precipitation, acid-base, and redox processes. (b) Instead of calcium oxide, why don't we simply add sodium hydroxide to precipitate magnesium hydroxide? (c) Sometimes a mineral called dolomite (a combination of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) ) is substituted for limestone \(\left(\mathrm{CaCO}_{3}\right)\) to bring about the precipitation of magnesium hydroxide. What is the advantage of using dolomite? (d) What are the advantages of mining magnesium from the ocean rather than from Earth's crust?

Calculate the mass of precipitate formed when \(2.27 \mathrm{~L}\) of \(0.0820 M \mathrm{Ba}(\mathrm{OH})_{2}\) are mixed with \(3.06 \mathrm{~L}\) of \(0.0664 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.