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Chapter 7: Problem 107

In April 2010, a research team reported that it had made Element 117. This discovery was confirmed in 2012 by additional experiments. Write the ground- state electron configuration for Element 117 and estimate values for its first ionization energy, electron afnity, atomic size, and common oxidation state based on its position in the periodic table.

Short Answer

Expert verified
Element 117, Tennessine (Ts), has a ground-state electron configuration of \([Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}\). Its first ionization energy is expected to be high, slightly lower than Astatine. Tennessine's electron affinity should be high, also similar to that of other halogens. Its atomic size is likely larger than Astatine, and Ts's common oxidation state is predicted to be -1.

Step by step solution

01

Ground-state electron configuration

To determine the ground-state electron configuration of Tennessine, follow the periodic table and fill the electron orbitals as per the Aufbau principle until you reach Element 117. The electron configuration will be: \[ [Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5} \] This configuration shows that Element 117 has filled 5f and 6d orbitals, two electrons in the 7s orbital, and five electrons in the 7p orbital.
02

Estimate the first ionization energy

The first ionization energy is the energy required to remove one electron from an atom to form a positive ion. As we move across a period, the ionization energy generally increases due to the increased effective nuclear charge. Since Tennessine is in Group 17, it should have a higher ionization energy compared to the elements to its left, but lower than that of the noble gases in Group 18. We can estimate that the first ionization energy of Tennessine will be slightly less than that of Astatine (At), the halogen in Period 6.
03

Estimate the electron affinity

Electron affinity is the energy change when an electron is added to an atom to form a negative ion. Electron affinity generally increases across a period due to the increased effective nuclear charge. Halogens have high electron affinity since their addition of an electron results in a complete outer energy level. We can expect Tennessine to have a high electron affinity, similar to other halogens, and possibly slightly lower than Astatine's electron affinity.
04

Estimate the atomic size

The atomic size generally increases as we move down a group due to the addition of electron shells, leading to increased shielding and a larger atomic radius. Tennessine is the heaviest halogen, and we can expect that its atomic size will be larger than that of Astatine, the halogen in Period 6.
05

Determine the common oxidation state

The common oxidation state for an element can be predicted based on its group number and the number of valence electrons it possesses. Halogens, including Tennessine, typically have an oxidation state of -1 due to their high affinity for electrons. They can gain one electron in chemical reactions to achieve a complete outer energy level resembling the noble gases. In summary, Element 117 (Tennessine) has the ground-state electron configuration of [Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}, high ionization energy, high electron affinity, a large atomic size compared to the other halogens, and a common oxidation state of -1.

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Most popular questions from this chapter

Use electron configurations to explain the following observa tions: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron afnity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

Which of the following statements about effective nuclear charge for the outermost valence electron of an atom is incorrect? (i) The effective nuclear charge can be thought of as the true nuclear charge minus a screening constant due to the other electrons in the atom. (ii) Effective nuclear charge increases going left to right across a row of the periodic table. (iii) Valence electrons screen the nuclear charge more effectively than do core electrons. (iv) The effective nuclear charge shows a sudden decrease when we go from the end of one row to the beginning of the next row of the periodic table. (v) The change in effective nuclear charge going down a column of the periodic table is generally less than that going across a row of the periodic table.

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}^{-} .\) Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

In Table 7.8 , the bonding atomic radius of neon is listed as 0.58 A, whereas that for xenon is listed as 1.40 A. A classmate of yours states that the value for Xe is more realistic than the one for Ne. Is she correct? If so, what is the basis for her statement?

Consider the stable elements through lead (Z = 82). In how many instances are the atomic weights of the elements out of order relative to the atomic numbers of the elements?
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