Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 77

Identify the specific element that corresponds to each of the following electron configurations and indicate the number of unpaired electrons for each: (a) \(1 s^{2} 2 s^{2}\), (b) \(1 s^{2} 2 s^{2} 2 p^{4}\) (d) \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\) (c) \([\mathrm{Ar}] 4 s^{1} 3 d^{5}\)

Short Answer

Expert verified
(a) The element corresponding to \(1s^2 2s^2\) is Beryllium (Be), and it has 0 unpaired electrons. (b) The element corresponding to \(1s^2 2s^2 2p^4\) is Oxygen (O), and it has 2 unpaired electrons. (d) The element corresponding to \([\mathrm{Kr}] 5s^2 4d^{10} 5p^4\) is Tellurium (Te), and it has 2 unpaired electrons. (c) The element corresponding to \([\mathrm{Ar}] 4s^1 3d^5\) is Chromium (Cr), and it has 6 unpaired electrons.

Step by step solution

01

(a) Identify the element corresponding to \(1s^2 2s^2\)

To identify the element, we need to count the total number of electrons based on its electron configuration. In this case, \(1s^2\) indicates that there are 2 electrons in the \(1s\) subshell and \(2s^2\) indicates there are 2 electrons in the \(2s\) subshell. Thus, the total number of electrons is 2 + 2 = 4. Now, we look at the periodic table and find an element with an atomic number (number of protons) of 4, which corresponds to Beryllium (Be).
02

(a) Determine the number of unpaired electrons for \(1s^2 2s^2\)

Since all orbitals in the configuration are fully filled (s-orbitals can contain 2 electrons), there are no unpaired electrons.
03

(b) Identify the element corresponding to \(1s^2 2s^2 2p^4\)

Count the total number of electrons: 2 (from \(1s^2\)) + 2 (from \(2s^2\)) + 4 (from \(2p^4\)) = 8. Looking at the periodic table, an atomic number of 8 corresponds to Oxygen (O).
04

(b) Determine the number of unpaired electrons for \(1s^2 2s^2 2p^4\)

Here, the p-orbital has a maximum capacity of 6 electrons, and there are 4 in the configuration \(2p^4\). Two of the electrons are unpaired (since each p-orbital can hold 2 electrons). Thus, there are 2 unpaired electrons.
05

(d) Identify the element corresponding to \([\mathrm{Kr}] 5s^2 4d^{10} 5p^4\)

First, we need to determine the electron configuration of the noble gas Kr. Kr stands for Krypton, which has an atomic number of 36. Now, we add the electrons in the rest of the configuration: 2 (from \(5s^2\)) + 10 (from \(4d^{10}\)) + 4 (from \(5p^4\)) = 16 electrons. Adding these to the 36 electrons of Kr, we get 52 electrons in total. Referencing the periodic table, an atomic number of 52 corresponds to Tellurium (Te).
06

(d) Determine the number of unpaired electrons for \([\mathrm{Kr}] 5s^2 4d^{10} 5p^4\)

All the electrons in subshells \(5s\) and \(4d\) are paired, but there are 2 unpaired electrons in the \(5p\) subshell (out of a maximum of 6 electrons, 4 are present).
07

(c) Identify the element corresponding to \([\mathrm{Ar}] 4s^1 3d^5\)

Determine the electron configuration of the noble gas Ar. Ar stands for Argon, which has an atomic number of 18. Now, we add the electrons in the rest of the configuration: 1 (from \(4s^1\)) + 5 (from \(3d^5\)) = 6 electrons. Adding these to the 18 electrons of Ar, we get 24 electrons in total. The periodic table shows that an atomic number of 24 corresponds to Chromium (Cr).
08

(c) Determine the number of unpaired electrons for \([\mathrm{Ar}] 4s^1 3d^5\)

There is 1 unpaired electron in the \(4s\) subshell and 5 unpaired electrons in the \(3d\) subshell. Therefore, a total of 6 unpaired electrons exist for this element. Summary of the results: (a) Beryllium (Be), 0 unpaired electrons (b) Oxygen (O), 2 unpaired electrons (d) Tellurium (Te), 2 unpaired electrons (c) Chromium (Cr), 6 unpaired electrons

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

Determine which of the following statements are false and correct them. (a) The frequency of radiation increases as the wavelength increases. (b) Electromagnetic radiation travels through a vacuum at a constant speed, regardless of wavelength. (c) Infrared light has higher frequencies than visible light. (d) The glow from a fireplace, the energy within a microwave oven, and a foghorn blast are all forms of electromagnetic radiation.

The wavenumber \(\bar{\lambda}\) is the number of waves that exist over a specified distance, very often \(1 \mathrm{~cm}\). The wavenumber can easily be calculated by taking the reciprocal of the wavelength. Give typical wavenumbers for (a) X-rays \((\lambda=1 \mathrm{nm})\) (b) visible light \((\lambda=500 \mathrm{nm})\) (c) microwaves $(\lambda=1 \mathrm{~mm})$.

One of the emission lines of the hydrogen atom has a wavelength of $94.974 \mathrm{nm}$. (a) In what region of the electromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

Label each of the following statements as true or false. For those that are false, correct the statement. (a) Visible light is a form of electromagnetic radiation. (b) Ultraviolet light has longer wavelengths than visible light. (c) X rays travel faster than microwaves. (d) Electromagnetic radiation and sound waves travel at the same speed.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

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