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Element 117 has never been found. but what would we expect its valence to be?

Short Answer

Expert verified
The expected valence of Element 117 is likely to be 1 since it belongs to Group 17 (VII) of the periodic table, just like the halogens.

Step by step solution

01

Understanding the Periodic Table layout

The periodic table is organized into groups and periods. Elements in the same group (also known as family) have the same number of valence electrons and therefore similar chemical behaviors. This means that if an unknown element is mentioned, like Element 117, one could derive its chemical properties by identifying its position in the periodic table relative to familiar elements.
02

Identifying the group of Element 117

Element 117 would be in the same group as the elements in the halogen family because it falls under Group 17 (VII) on the periodic table. This group includes elements like Fluorine (F), Chlorine (Cl), and Iodine (I).
03

Defining the Valence of Element 117

Since the valence is the same within a group in the periodic table and Element 117 belongs to Group 17 (VII) which is represented by halogens known to have a valence of 1, it would be logical to say that Element 117 is likely to have a valence of 1.

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Key Concepts

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

Element 117
Element 117, known as Tennessine with the symbol Ts, is a superheavy artificial chemical element that has stirred considerable intrigue since its discovery. While it doesn't naturally occur on Earth and is synthesized in a laboratory setting, scientists predict certain attributes based on its position in the periodic table.

As part of the 7th period and Group 17, it falls within the series of halogens, a group known for their extreme reactivity and distinctive nonmetallic properties. Understanding Tennessine involves delving into more than its theoretical valence; it includes speculating about its potential reactions, stability, and even its physical state at room temperature, considering the trends in its group.
Valence Electrons
Valence electrons are the outermost electrons of an atom and they play a pivotal role in chemical bonding and reactions. These electrons determine an element's ability to gain, lose, or share electrons with other atoms. Group trends on the periodic table mean that elements within the same group share the same number of valence electrons, thus exhibiting similar reactivity.

For Element 117, or Tennessine, this implies a count of seven valence electrons, aligning with the general characteristic of the halogens. The valence electron count is essential for predicting how an element will engage in chemical bonding. For students, a clear grasp of valence electrons can unlock the understanding of many chemical reactions and the formation of compounds.
Chemical Properties
The chemical properties of an element include its reactivity, the types of bonds it forms, its electronegativity, and its ionization energies. These properties are deeply influenced by an element's valence electrons since they are involved directly in chemical reactions. In the case of hypothetical Element 117, these properties would exhibit traits common to halogens, such as high reactivity and a tendency to form anions by gaining an electron, resulting in a negatively charged state.

Theoretical predictions also suggest that Tennessine would likely form ionic and covalent bonds, engaging in reactions that align with the behavior of other halogen elements. Discussing the chemical properties doesn’t only help in academic exercises but also provides foundational knowledge that underpins much of modern chemistry.
Halogens
Halogens, occupying Group 17 of the periodic table, are five nonmetallic elements: Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), and Astatine (At), with Tennessine (Ts) joining as the heaviest and most recently discovered. These elements are known for their electron affinity and high reactivity, especially with alkali and alkaline earth metals.

Students can observe patterns among the halogens, such as their electron configurations and physical state changes down the group, transitioning from gaseous to solid. By studying halogens, students not only learn about their individual characteristics but also gain insights into how periodic trends can predict the behavior of newly discovered elements like Element 117.

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Most popular questions from this chapter

Using a beam of electrons accelerated in an X-ray tube, we wish to knock an electron out of the \(K\) shell of a given element in a target. Section 7.8 gives the energies in a hydrogenlike atom as \(Z^{2}\left(-13.6 \mathrm{eV} / \mathrm{n}^{2}\right)\). Assume that for \(f\) airly high \(Z\), a \(K\) -shell electron can be treated as orbiting the nucleus alone, (a) A typical accelerating potential in an X-ray tube is \(50 \mathrm{kV}\). In roughly how high a \(Z\) could a hole in the \(K\) -shell be produced? (b) Could a hole be produced in elements of higher \(Z\) ?

Two particles in a box have a total energy \(5 \pi^{2} \hbar^{2} / 2 m L^{2}\) (a) Which states are occupied? (b) Make a sketch of \(P_{S}\left(x_{1}, x_{2}\right)\) versus \(x_{1}\) for points along the line \(x_{2}=x_{1}\) (c) Make a similar sketch of \(P_{A}\left(x_{1}, x_{2}\right)\). (d) Repeat parts (b) and (c) but for points on the line \(x_{2}=L-x_{1}\). (Note: \(\left.\sin [m \pi(L-x) / L]=(-1)^{m+1} \sin \left(m \pi .0^{\prime} L\right) .\right)\)

The Zeeman effect occurs in sodium just as in hydrogen - sodium's lone 3 s valence electron behaves much as hydrogen's 1.5. Suppose sodium atoms are immersed in a \(0.1 \mathrm{~T}\) magnetic field. (a) Into how many levels is the \(3 p_{1 / 2}\) level split? (b) Determine the energy spacing between these states. (c) Into how many lines is the \(3 p_{1 / 2}\) to \(3 s_{1 / 2}\) spectral line split by the field? (d) Describe quantitatively the spacing of these lines. (e) The sodium doublet \((589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm}\) ) is two spectral lines. \(3 p_{3 n} \rightarrow 3 s_{1 / 2}\) and \(3 p_{1 / 2} \rightarrow 3 s_{1 / 2}\) which are split according to the two different possible spin-orbit ener gies in the \(3 p\) state (see Exercise 60 ). Detemine the splitting of the sodium doublet (the energy diff erence between the two photons). How does it compare with the line splitting of part (d), and why?

All other things being equal, should the spin-orbit interaction be a larger or smaller effect in hydrogen as \(n\) increases? Justify your answer.

Slater Determinant: A convenient and compact way of expressing multiparticle states of antisymmetric character for many fermions is the Slater determinant. lt is based on the fact that for \(N\) fermions there must be \(N\) different individual-particle states, or sets of quantum numbers. The ith state has sparial quantum numbers (which might be \(n_{i}, \ell_{i},\) and \(m_{c i}\) ) represented simply by \(n_{t}\) and spin quanturn number \(m_{s i^{i}}\). Were it occupied by the ith particle, the state would be \(\psi_{n}\left(x_{j}\right) m_{s i}\). A column corresponds to a given state and a row to a given particle. For instance, the first column corresponds to individualparticle state \(\psi_{n}(x,) m_{3},\) where \(j\) progresses (through the rows) from particle 1 to particle \(N\). The first row corresponds to particle I. which successively occupies all individual-particle states (progressing through the columns). (a) What property of determinants ensures that the multiparticle state is 0 if any two individualparticle states are identical? (b) What property of deterninants ensures that switching the labels on any two particles switches the sign of the multiparticle state?

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