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Chapter 2: Problem 79

On the basis of the expectod charges of the monatomic ions, give the chemical formula of each of the following compounds: (a) vanadium(V) oxide; (b) lead(IV) oxide; (c) thallium(III) oxide.

Short Answer

Expert verified
The chemical formulas are: (a) V2O5, (b) PbO2, (c) Tl2O3.

Step by step solution

01

Determine the Charge on Vanadium in Vanadium(V) Oxide

The Roman numeral (V) in vanadium(V) oxide indicates the oxidation state of vanadium, which is +5. Since oxygen typically has a charge of -2, we can use this information to balance the charges.
02

Write the Formula for Vanadium(V) Oxide

To neutralize the positive charge of vanadium(V), multiple oxide ions are needed. The formula unit must have a net charge of zero, so we need 5/2 = 2.5, which is not possible with whole numbers. To ensure whole numbers, multiply the charges by the smallest common multiple (2), leading to V2O5.
03

Determine the Charge on Lead in Lead(IV) Oxide

For lead(IV) oxide, the Roman numeral (IV) denotes the oxidation state of lead, which is +4. Oxygen has a -2 charge, so we balance it in the same manner.
04

Write the Formula for Lead(IV) Oxide

The required number of oxide ions to balance the charge of lead(IV) is 4/2 = 2, leading to the formula PbO2.
05

Determine the Charge on Thallium in Thallium(III) Oxide

The Roman numeral (III) in thallium(III) oxide indicates the charge of thallium, which is +3. Oxygen's charge is -2.
06

Write the Formula for Thallium(III) Oxide

The balancing of charges will be done as before. Multiplying the +3 charge of thallium and the -2 charge of oxygen by the smallest common multiple, which is 2, we get Tl2O3.

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Key Concepts

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

Oxidation States
Understanding oxidation states is vital to mastering the art of writing chemical formulas. An oxidation state, often referred to as oxidation number, is a figure that represents the total number of electrons an atom either gains or loses in order to form a chemical bond with another atom.

Take, for example, vanadium(V) oxide. The Roman numeral (V) signifies that vanadium has an oxidation state of +5. This means vanadium can be assumed to have 'lost' five electrons. On the other hand, oxygen typically has an oxidation state of -2, indicating that it 'gains' two electrons when forming compounds. Balancing these oxidation states is like solving a numeric puzzle: the total positive charge must counterbalance the total negative charge for the compound to be stable. Here, the balancing act leads to the formation of V2O5, which ensures the charges are neutralized.

Step-by-Step Analysis

In our step-by-step solution for vanadium(V) oxide, we multiplied the oxidation states by the smallest common multiple to get whole numbers for the eventual chemical formula. For instances where a simple 1:1 ratio doesn't achieve a neutral compound, this method helps us find the most reduced whole number ratio that does.
Ionic Charges
Closely related to oxidation states are ionic charges. Ionic charges are indicative of the potential an atom has to engage in ionic bonding by gaining or losing electrons. Compounds like lead(IV) oxide and thallium(III) oxide are made up of cations (positively charged ions) and anions (negatively charged ions) that combine in a way that their overall ionic charges add up to zero.

Understanding Ionic Balances

For lead(IV) oxide, lead has a charge of +4, and oxygen has a charge of -2. When we combine them, two oxygen ions are needed to balance out the single lead ion, resulting in the formula PbO2. In the case of thallium(III) oxide, the thallium ion has a +3 charge, while oxygen remains consistent with a -2 charge. Here, we achieve a neutral compound with the formula Tl2O3, where two thallium ions balance three oxide ions.

Students often struggle with understanding how these charges interact, but remembering that the end goal is to achieve a net zero charge can simplify the process. The charges must always balance out, much like how a scale must be balanced by equal weights on both sides.
Compound Nomenclature
The naming of compounds, otherwise known as compound nomenclature, is a standardized method of assigning names to chemical substances. In inorganic chemistry, the system of nomenclature can tell us a lot about the compound, including the elements present and their proportions or oxidation states.

Deciphering Compound Names

For example, the name vanadium(V) oxide contains the element names and, more importantly, the oxidation state of the metal as indicated by the Roman numeral. The (V) after vanadium tells us that vanadium has a +5 oxidation state in the compound. Similarly, lead(IV) oxide and thallium(III) oxide reveal that lead and thallium have +4 and +3 oxidation states, respectively.

The correct use of nomenclature not only helps in writing and understanding chemical formulas, but also facilitates communication within the scientific community, ensuring clarity and precision. For students to excel in nomenclature, practice in both writing and interpreting chemical names is essential for the application in exercises and exams.

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Most popular questions from this chapter

An important principle in chemistry is the isolobal analogy. This very simple principle states that chemical fragments with similar valence orbital structures can replace each other in molecules. For example, \(\cdot \mathrm{C}-\mathrm{H}\) and \(\cdot \mathrm{S}-\mathrm{H}\) are isolobal fragments, cach having three electrons with which to form bonds besides the bond to \(\mathrm{H}\). An isolobal series of molecules would be HCCH, HCSiH, and HSiSiH. Similarly, a lone pair of clectrons can be used to replace a bond so that - \(\mathrm{N}\) : is isolobal with ' \(\mathrm{C}-\mathrm{H}\), with the lone pair taking the place of the \(\mathrm{C}-\mathrm{H}\) bond. The isolobal set here is HOCH, HCN, and NN. (a) Draw the Lewis structures for the molecules HCCH, HCSiH, HSiSiH, \(\mathrm{HCN}\), and NN. (b) Using the isolobal principle, draw Lewis structures for molecules based on the structure of benzenc, \(\mathrm{C}_{6} \mathrm{H}_{6}\), in which one or more \(\mathrm{CH}\) groups are replaced with \(\mathrm{N}\) atoms.

Which \(\mathrm{M}^{2}\) ions are predicted to have the following ground-state electron configurations: (a) \([\mathrm{Ar}] 3 d^{7} ;\) (b) \([\mathrm{Kr}] 4 d^{7} ;\) (c) \([\mathrm{Kr}] 4 d^{10} 5 s^{2} ;\) (d) [Xe] \(5 d^{10}\) ?

lodine can exist in both positive and negative oxidation states. What is the maximum (a) positive and (b) negative oxidation number that iodine may have? (c) Determine the clectron configurations for each of these states. (d) Explain how you arrived at these values.

2.7 Give the ground-state clectron configuration expected for cach of the following ions: (a) \(\mathrm{Sb}^{3+}\); (b) \(\mathrm{Sn}^{2+}\); (c) \(\mathrm{W}^{2+}\); (d) \(\mathrm{S}^{2-}\).

Which of the following members of cach pair is the stronger Lewis base? Explain your reasoning. (a) \(\mathrm{CH}_{3}^{-}\)or \(\mathrm{CH}_{4}\); (b) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S}\); (c) \(\mathrm{NH}_{3}\) or \(\mathrm{NH}_{2}^{-}\).
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