/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 118 Predict some of the properties o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Predict some of the properties of element 117 (the symbol is Uus, following conventions proposed by the International Union of Pure and Applied Chemistry, or IUPAC). a. What will be its electron configuration? b. What element will it most resemble chemically? c. What will be the formula of the neutral binary compounds it forms with sodium, magnesium, carbon, and oxygen? d. What oxyanions would you expect Uus to form?

Short Answer

Expert verified
a. The electron configuration of Uus (element 117) is: \[1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}, 6p^6, 7s^2, 5f^{14}, 6d^{10}, 7p^5\]. b. Uus will most closely resemble halogens such as fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). c. The neutral binary compounds formed by Uus with sodium, magnesium, carbon, and oxygen are NaUus, MgUus2, CUus4, and UusO respectively. d. The expected oxyanions of Uus include UusO3- (Uus halate), UusO2- (Uus halite), and UusO- (hypohalite).

Step by step solution

01

Electron Configuration for Uus

Uus is element 117, so it has 117 electrons. To find the electron configuration, we need to follow the periodic table and fill in the orbitals accordingly. The electron configuration will be: \[1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}, 6p^6, 7s^2, 5f^{14}, 6d^{10}, 7p^5\]
02

Chemical Resemblance

To find the element that Uus will most resemble chemically, we need to look at its position in the periodic table. Uus is in Group 17 and Period 7, making it a halogen. Therefore, it will most closely resemble the other halogens: fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). Uus will likely exhibit a similar chemistry to these elements.
03

Neutral Binary Compounds

To determine the neutral binary compounds formed by Uus with sodium (Na), magnesium (Mg), carbon (C), and oxygen (O), we will need to know the common charges of these elements. Uus is in Group 17 and is a halogen, so it will have a charge of -1. Sodium (Na) and magnesium (Mg) are both in Group 1 and Group 2 respectively, so they'll have charges of +1 and +2. Carbon (C) is in Group 14, meaning it can have a charge of ±4 (depending on the compound), while Oxygen is in Group 16 and has a charge of -2. Therefore, the neutral binary compounds formed will be: - Uus and Sodium (Na): NaUus (as +1 and -1 charges balance), - Uus and Magnesium (Mg): MgUus2 (as +2 charge of Mg balances with 2Uus with -1 charge), - Uus and Carbon (C): CUus4 (since C has a charge of +4 and 4Uus with -1 charge), - Uus and Oxygen (O): UusO (as -1 and -2 charges balance when they both bond and create an O2- ion).
04

Oxyanions of Uus

Oxyanions are negatively charged polyatomic ions containing an element (in this case, Uus) bonded to oxygen. As a halogen, Uus is expected to form oxyanions similar to those formed by other halogens. Some examples of halogen-based oxyanions follow a general formula: XOm, where X is the halogen (in this case, Uus) and m is the number of oxygen atoms involved. Some common examples include: - Halate ion: XO3-, such as ClO3- (chlorate), BrO3- (bromate), or IO3- (iodate) – UusO3- - Halite ion: XO2-, such as ClO2- (chlorite), BrO2- (bromite), or IO2- (iodite) – UusO2- - Hypohalite ion: XO-, such as ClO- (hypochlorite), BrO- (hypobromite), or IO- (hypoiodite) – UusO-

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle has a velocity that is \(90 . \%\) of the speed of light. If the wavelength of the particle is \(1.5 \times 10^{-15} \mathrm{m},\) calculate the mass of the particle.

The four most abundant elements by mass in the human body are oxygen, carbon, hydrogen, and nitrogen. These four elements make up about 96% of the human body. The next four most abundant elements are calcium, phosphorus, magnesium, and potassium. Write the expected ground-state electron configurations for these eight most abundant elements in the human body.

The ionization energy for a 1\(s\) electron in a silver atom is \(2.462 \times 10^{6} \mathrm{kJ} / \mathrm{mol} .\) a. Determine an approximate value for \(Z_{\text { eff }}\) for the \(A g 1 s\) electron. Assume the Bohr model applies to the 1 s electron. \(Z_{\mathrm{eff}}\) is the apparent nuclear charge experienced by the electrons. b. How does \(Z_{\text { eff }}\) from part a compare to \(Z\) for Ag? Rationalize the relative numbers.

In the second row of the periodic table, Be, N, and Ne all have endothermic (unfavorable) electron affinities, whereas the other second-row elements have exothermic (favorable) electron affinities. Rationalize why Be, N, and Ne have unfavorable electron affinities.

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of 253.4 \(\mathrm{nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.