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Chapter 2: Problem 129

What is the systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5} ?\) If the charge on the metal remained constant and then sulfur was substituted for oxygen, how would the formula change? What is the difference in the total number of protons between \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) and its sulfur analog?

Short Answer

Expert verified
The systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) is "Ditantalum pentoxide". When sulfur is substituted for oxygen, the new formula becomes \(\mathrm{Ta}_{2} \mathrm{S}_{5}\). The difference in the total number of protons between \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) and its sulfur analog is 40 protons.

Step by step solution

01

Identify the systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5}\)

To determine the systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5}\), we look at the elements present in the compound. In this case, we have two Ta (tantalum) atoms and five O (oxygen) atoms. The systematic name of this compound is determined by the stock notation: the metal (tantalum) name, followed by its oxidation state in roman numerals and the polyatomic ion name. The oxidation state of tantalum in this compound is 5+ since each oxygen atom has a 2- charge. The systematic name of \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) is "Ditantalum pentoxide".
02

Determine the new formula when sulfur substitutes oxygen

If sulfur (S) replaces oxygen (O) in the compound while the charge on the metal remains constant, we need to consider the fact that both oxygen and sulfur have a 2- charge (as they are in the same group of the periodic table). Since the tantalum atom has a 5+ charge, we can re-write the formula as \(\mathrm{Ta}_{2} \mathrm{S}_{5}\). In this case, both the oxidation state of tantalum and the overall charge on the formula remain unchanged.
03

Calculate the difference in the total number of protons between \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) and its sulfur analog

To calculate the difference in the total number of protons, we will compare the number of protons in both compounds: For \(\mathrm{Ta}_{2} \mathrm{S}_{5}\): - Ta (tantalum) has an atomic number of 73, so it has 73 protons. - O (oxygen) has an atomic number of 8, so it has 8 protons. Total protons in \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) = (2 脳 73) + (5 脳 8) = 146 + 40 = 186 protons For \(\mathrm{Ta}_{2} \mathrm{S}_{5}\): - Ta (tantalum) has an atomic number of 73, so it has 73 protons. - S (sulfur) has an atomic number of 16, so it has 16 protons. Total protons in \(\mathrm{Ta}_{2} \mathrm{S}_{5}\) = (2 脳 73) + (5 脳 16) = 146 + 80 = 226 protons Difference in total number of protons = Total protons in \(\mathrm{Ta}_{2} \mathrm{S}_{5}\) - Total protons in \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) = 226 - 186 = 40 protons.

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Key Concepts

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

Understanding Oxidation States
Oxidation states, also known as oxidation numbers, are essential for understanding chemical reactions and compound formation. They are numerical values that represent the number of electrons an atom either gains, loses, or shares when forming a compound. In systematic naming of compounds, oxidation states are crucial.
  • In \(\mathrm{Ta}_{2} \mathrm{O}_{5}\), the oxidation state of tantalum (Ta) is determined by the charge of oxygen (O), which is -2 for each atom. Since there are five oxygen atoms \(5 \times (-2) = -10\), each tantalum must carry a +5 charge \((+10/2)\) to balance out the equation, giving its oxidation state as +5.
  • This leads to the systematic name "Ditantalum pentoxide," where the prefix "Di-" indicates two tantalum atoms, "-pent" refers to the five oxygen atoms, and "oxide" signifies the presence of oxygen.
Analogous to balancing a scale, oxidation states ensure that the sum of charges in a compound is zero, reflecting neutrality. Knowing oxidation states helps predict the chemical behavior of compounds, which is especially useful for understanding reactions in chemistry.
Exploring the Periodic Table
The periodic table is a powerful tool for understanding elements and predicting their chemical behavior. It arranges all known elements in a tabular format based on their atomic number, electron configurations, and recurring chemical properties. Each vertical column, known as a group, contains elements with similar properties.
  • For instance, oxygen (O) and sulfur (S) are in Group 16, also known as the chalcogens. They often exhibit a -2 oxidation state, making their substitution in the compound \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) to \(\mathrm{Ta}_{2} \mathrm{S}_{5}\) straightforward without altering the metal's oxidation state.
  • Elements are also organized horizontally in periods, which show the progression of atomic number and increasing energy levels.
  • The position of an element provides information about its reactivity, potential compounds, and chemical properties.
Understanding the periodic table not only explains how and why certain elements might substitute each other in chemical reactions but also demonstrates the repeatability and predictability of chemical properties across periods and groups.
Unveiling Atomic Number
The atomic number of an element is a fundamental property, representing the number of protons in the nucleus of an atom. It uniquely identifies each element and dictates its position in the periodic table.
  • Tantalum (Ta), for example, has an atomic number of 73. This means every tantalum atom contains 73 protons in its nucleus.
  • Oxygen, with an atomic number of 8, contains 8 protons, while sulfur has an atomic number of 16, containing 16 protons in its atom.
  • The atomic number reflects an element's chemical identity and affects its reactivity, electron configuration, and placement in the periodic table.
When calculating proton differences, such as between \(\mathrm{Ta}_{2} \mathrm{O}_{5}\) and \(\mathrm{Ta}_{2} \mathrm{S}_{5}\), the atomic number plays a critical role. It determines how many protons each element contributes to the compound, allowing us to calculate the differences in total proton count accurately. Recognizing the atomic number's significance helps in simplifying and solving complex chemical puzzles by understanding the elemental composition and behavior.

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Most popular questions from this chapter

When mixtures of gaseous \(\mathrm{H}_{2}\) and gaseous \(\mathrm{Cl}_{2}\) react, a product forms that has the same properties regardless of the relative amounts of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) used. a. How is this result interpreted in terms of the law of definite proportion? b. When a volume of \(\mathrm{H}_{2}\) reacts with an equal volume of \(\mathrm{Cl}_{2}\) at the same temperature and pressure, what volume of product having the formula HCl is formed?

Elements in the same family often form oxyanions of the same general formula. The anions are named in a similar fashion. What are the names of the oxyanions of selenium and tellurium: \(\mathrm{SeO}_{4}^{2-}, \mathrm{SeO}_{3}^{2-}, \mathrm{TeO}_{4}^{2-}, \mathrm{TeO}_{3}^{2-} ?\)

What number of protons and neutrons are contained in the nucleus of each of the following atoms? Assuming each atom is uncharged, what number of electrons are present? a. \(^{235}_{92} \mathrm{U}\) b. \(_{13}^{27} \mathrm{Al}\) c. \(_{26}^{57} \mathrm{Fe}\) d. \(_{82}^{208} \mathrm{Pb}\) e. \(_{37}^{86} \mathrm{Rb}\) f. \(_{20}^{41} \mathrm{Ca}\)

Section 2.3 describes the postulates of Dalton鈥檚 atomic theory. With some modifications, these postulates hold up very well regarding how we view elements, compounds, and chemical reactions today. Answer the following questions concerning Dalton鈥檚 atomic theory and the modifications made today. a. The atom can be broken down into smaller parts. What are the smaller parts? b. How are atoms of hydrogen identical to each other, and how can they be different from each other? c. How are atoms of hydrogen different from atoms of helium? How can H atoms be similar to He atoms? d. How is water different from hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\)even though both compounds are composed of only hydrogen and oxygen? e. What happens in a chemical reaction, and why is mass conserved in a chemical reaction?

Early tables of atomic weights (masses) were generated by measuring the mass of a substance that reacts with 1.00 g of oxygen. Given the following data and taking the atomic mass of hydrogen as 1.00, generate a table of relative atomic masses for oxygen, sodium, and magnesium. How do your values compare with those in the periodic table? How do you account for any differences?
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