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Chapter 7: Problem 164

While Mendeleev predicted the existence of several undiscovered elements, he did not predict the existence of the noble gases, the lanthanides, or the actinides. Propose reasons why Mendeleev was not able to predict the existence of the noble gases.

Short Answer

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Mendeleev was unable to predict the existence of noble gases due to their low reactivity, limited isolation techniques available during his time, and the limited knowledge of atomic structure at the time. The noble gases' full electron shells made them unreactive and difficult to detect, as they did not readily form compounds with other elements. The discovery of noble gases eventually led to the establishment of Group 18 in the periodic table.

Step by step solution

01

Understand Mendeleev's Periodic Table

Mendeleev's periodic table was based on the atomic mass and the properties of elements. He arranged the elements in ascending order according to their atomic mass and noted that elements with similar properties shared the same group. Mendeleev left gaps in his table to make room for undiscovered elements that would have similar properties to the neighboring elements.
02

Characteristics of Noble Gases

The noble gases are a group of chemical elements that, under standard conditions, are odorless, colorless, and tasteless. They are characterized by their low reactivity, which stems from their outer electron shell being completely filled. This makes them reluctant to form bonds with other elements, and as a result, they are found as monatomic gases in nature.
03

Why Mendeleev didn't predict Noble Gases

There are several reasons why Mendeleev did not predict the existence of noble gases: 1. Low reactivity: Noble gases are very unreactive due to their full electron shells. This makes it difficult for them to form compounds with other elements. Thus, it was challenging for Mendeleev to predict their existence since they did not readily form compounds with other elements that he could study in his research. 2. Isolation techniques: Mendeleev relied on the chemical and physical properties of known elements and their compounds to predict the existence of undiscovered elements. Since noble gases did not readily interact with other elements, it was difficult to isolate them using the techniques available during Mendeleev's time. 3. Limited knowledge: Mendeleev's periodic table was an early method of organizing elements, and although it was a groundbreaking achievement, it was limited by the understanding of atomic structure at the time. The existence of the noble gases was not known during Mendeleev's lifetime, and without knowledge of their unique properties, he was unable to predict their existence.
04

Impact of the Discovery of Noble Gases on the Periodic Table

The discovery of noble gases led to the establishment of a new group called Group 18 in the periodic table, also known as the noble gases group. This addition improved the understanding of the periodicity of elements and increased the comprehensiveness of Mendeleev's periodic table.

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Most popular questions from this chapter

Neutron diffraction is used in determining the structures of molecules. a. Calculate the de Broglie wavelength of a neutron moving at \(1.00 \%\) of the speed of light. b. Calculate the velocity of a neutron with a wavelength of \(75 \mathrm{pm}\left(1 \mathrm{pm}=10^{-12} \mathrm{~m}\right)\)

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

In the ground state of element 115, Uup, a. how many electrons have \(n=5\) as one of their quantum numbers? b. how many electrons have \(\ell=3\) as one of their quantum numbers? c. how many electrons have \(m_{\ell}=1\) as one of their quantum numbers? d. how many electrons have \(m_{s}=-\frac{1}{2}\) as one of their quantum numbers?

In each of the following sets, which atom or ion has the smallest ionization energy? a. \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) b. \(\mathrm{K}, \mathrm{Mn}, \mathrm{Ga}\) c. \(\mathrm{N}, \mathrm{O}, \mathrm{F}\) d. \(\mathrm{S}^{2-}, \mathrm{S}, \mathrm{S}^{2+}\) e. \(\mathrm{Cs}\), Ge, Ar

From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let \(V\) represent the principal quantum number for the valence shell of element \(X\). If an electron in the hydrogen atom falls from shell \(V\) to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).
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