Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 79

Use the basic rules for electron configurations to indicate the number of (a) unpaired electrons in an atom of \(\mathrm{P} ;\) (b) \(3 d\) electrons in an atom of \(\mathrm{Br} ;\) (c) \(4 p\) electrons in an atom of \(\mathrm{Ge} ;\) (d) \(6 \mathrm{s}\) electrons in an atom of \(\mathrm{Ba}\) (e) \(4 f\) electrons in an atom of Au.

Short Answer

Expert verified
(a) 3 unpaired electrons for P. (b) 0 3d electrons for Br. (c) 2 4p electrons for Ge. (d) 2 6s electrons for Ba. (e) 14 4f electrons for Au.

Step by step solution

01

Unpaired Electrons in Atom of P

Phosphorus (P) has 15 electrons. The electron configuration is \(1s^22s^22p^63s^23p^3\). We can see that there are 3 unpaired electrons in p orbital of the third energy level.
02

Number of 3d Electrons in Atom of Br

Bromine (Br) is the 35th element with 35 electrons. The electron configuration for bromine ends with \(4p^5\). There are no d electrons in the third energy level because d orbital starts filling from the 4th period.
03

Number of 4p Electrons in Atom of Ge

Germanium (Ge) is the 32nd element with 32 electrons. The electron configuration ends with \(4p^2\). Therefore, Germanium has 2 electrons in the 4p orbital.
04

Number of 6s Electrons in Atom of Ba

Barium (Ba) has 56 electrons. The electron configuration ends with \(6s^2\). Hence, the Barium atom has 2 electrons in the 6s orbital.
05

Number of 4f Electrons in Atom of Au

Gold (Au) has 79 electrons. Its electron configuration ends with \(6s^14f^{14}5d^{10}\). Thus, there are 14 electrons in the 4f orbital.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The most intense line in the cerium spectrum is at 418.7 nm. (a) Determine the frequency of the radiation producing this line. (b) In what part of the electromagnetic spectrum does this line occur? (c) Is it visible to the eye? If so, what color is it? If not, is this line at higher or lower energy than visible light?

Use the Balmer equation (8.2) to determine (a) the frequency, in \(s^{-1}\), of the radiation corresponding to \(n=5\) (b) the wavelength, in nanometers, of the line in the Balmer series corresponding to \(n=7\) (c) the value of \(n\) corresponding to the Balmer series line at \(380 \mathrm{nm}\)

Without doing detailed calculations, arrange the following electromagnetic radiation sources in order of increasing frequency: (a) a red traffic light, (b) a \(91.9 \mathrm{MHz}\) radio transmitter, (c) light with a frequency of \(3.0 \times 10^{14} \mathrm{s}^{-1}\) (d) light with a wavelength of \(49 \mathrm{nm}\).

Use Planck's equation (8.3) to determine (a) the energy, in joules per photon, of radiation of frequency $7.39 \times 10^{15} \mathrm{s}^{-1}$ (b) the energy, in kilojoules per mole, of radiation of frequency $1.97 \times 10^{14} \mathrm{s}^{-1}$

The following electron configurations correspond to the ground states of certain elements. Name each element. (a) \([\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{3} ;\) (b) \([\mathrm{Ne}] 3 s^{2} 3 p^{4} ;\) (c) \([\mathrm{Ar}] 3 d^{1} 4 s^{2}\) (d) \([\mathrm{Kr}] 4 d^{6} 5 s^{2} ;\) (e) \([\mathrm{Xe}] 4 f^{12} 6 s^{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation
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.

Sign-up for free