Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card  Package
Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card Package
1st Edition
ISBN: 9780134024516
Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus
Publisher: PEARSON
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Chapter 15, Problem 1DE

(a)

Interpretation Introduction

To determine: The similarities of and differences between the 1s and 2s orbitals of the hydrogen atom.

(a)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: They have similar shape but differ in size and number of radial nodes.

Explanation of Solution

In the designation of an orbital the number that is the principal quantum number n gives the shell number and the letter that corresponds to the angular momentum quantum number l gives the subshell.

The shell number decides the size of an orbital and the subshell defines its shape.

Now, s subshell means that the shape of orbital is spherical. So, both 1s and 2s orbitals are spherical in shape. As n increase size of orbital increases and number of radial nodes also increase as number of radial nodes =n1 .

So 1s orbital has 11=0 radial node while 2s orbital has 21=1 radial node. Also, 1s orbital is smaller in size than 2s orbital.

Conclusion

They have similar shape but differ in size and number of radial nodes.

(b)

Interpretation Introduction

To determine: The directional character of 2p orbital and comparison in directional characteristic of the px and dx2-y2 orbitals.

(b)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The directional character of 2p orbital means that electron density is concentrated in two regions on either side of the nucleus. A px orbital is directed along the x-axis while the lobes of a dx2-y2 orbital lie along the x and y-axes.

Explanation of Solution

The directional character of 2p orbital means that electron density in a p orbital is not uniform around the nucleus and is concentrated in two regions on either side of the nucleus.

A px orbital is dumbbell shaped whereas dx2-y2 orbital is four leaf clover shaped because it has four lobes. A px orbital is oriented along the x-axis while the lobes of dx2-y2 orbital lie along the x and y-axes.

Conclusion

The directional character of a 2p orbital means that electron density is concentrated in two regions on either side of the nucleus. A px orbital is directed along the x-axis while the lobes of dx2-y2 orbital lie along the x and y-axes.

(c)

Interpretation Introduction

To determine: The comparison between average distance from nucleus of 2s orbital and 3s orbital.

(c)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The average distance of 2s orbital from the nucleus is less as compared to that of 3s orbital.

Explanation of Solution

The distance from the nucleus of an orbital is defined by principal quantum number n . As n increases size of the orbital increases and number of radial nodes also increase as number of radial nodes =n1 . So, the average distance of 2s orbital from the nucleus is less as compared to that of 3s orbital.

Conclusion

The average distance of 2s orbital from the nucleus is less as compared to that of 3s orbital.

(d)

Interpretation Introduction

To determine: The increasing energy order of the orbitals 4f,6s,3d,1s,2p .

(d)

Expert Solution
Check Mark

Answer to Problem 1DE

Solution: The increasing energy order of the orbitals is given below.

    1s<2p<3d<6s<4f

Explanation of Solution

The energy of orbital increases with increase in principal quantum number n but the subshells s,p,d,f also differ in energy as s<p<d<f . So, theorder of increasing energy of given orbitals is,

    1s<2p<3d<6s<4f
Conclusion

The increasing energy order of the orbitals is 1s<2p<3d<6s<4f .

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Chapter 15 Solutions

Chemistry: The Central Science, Books a la Carte Edition & Solutions to Red Exercises for Chemistry & Mastering Chemistry with Pearson eText -- Access Card Package

Ch. 15.4 - Prob. 15.6.1PECh. 15.4 - The hydrogen atom can absorb light of wavelength...Ch. 15.5 - Prob. 15.7.1PECh. 15.5 - Prob. 15.7.2PECh. 15.5 - Use the de Brogue relationship to determine the...Ch. 15.5 - Prob. 15.8.2PECh. 15.6 - Neutron diffraction is an important technique for...Ch. 15.6 - The electron microscope has been widely used to...Ch. 15.6 - Prob. 15.10.1PECh. 15.6 - An AM radio station broadcasts at 1010 kHz, and...Ch. 15.6 - One type of sunburn occurs on exposure to UV light...Ch. 15.6 - Prob. 15.11.2PECh. 15.7 - Prob. 15.12.1PECh. 15.7 - A stellar object is emitting radiation at 3.55 mm....Ch. 15 - Prob. 1DECh. 15 - Prob. 1ECh. 15 - Identify the group of elements that corresponds to...Ch. 15 - Prob. 3ECh. 15 - Using the periodic table as a guide, write the...Ch. 15 - Arrange Be, C, K, and Ca in order of increasing...Ch. 15 - Prob. 6ECh. 15 - Prob. 7ECh. 15 - Prob. 8ECh. 15 - Consider the isoelectronic ions F- and Na+. (a)...Ch. 15 - Prob. 10ECh. 15 - Prob. 11ECh. 15 - Prob. 12ECh. 15 - Give the values for n, I,and mlfor each orbital in...Ch. 15 - Prob. 14ECh. 15 - Prob. 15ECh. 15 - Which of the following represent impossible...Ch. 15 - For the table that follows, write which orbital...Ch. 15 - Sketch the shape and orientation of the following...Ch. 15 - Prob. 19ECh. 15 - Prob. 20ECh. 15 - Two possible electron configurations for an Li...Ch. 15 - 6.70 An experiment called the Stern—Gerlach...Ch. 15 - Prob. 23ECh. 15 - Prob. 24ECh. 15 - What are "valence electrons"? What are "core...Ch. 15 - For each element, indicate the number of valence...Ch. 15 - Write the condensed electron configurations for...Ch. 15 - Write the condensed electron configurations for...Ch. 15 - Identify the specific element that corresponds to...Ch. 15 - Prob. 30ECh. 15 - Prob. 31ECh. 15 - Prob. 32ECh. 15 - Prob. 33ECh. 15 - Prob. 34ECh. 15 - Prob. 35ECh. 15 - Prob. 36ECh. 15 - Prob. 37ECh. 15 - In an experiment to study the photoelectric...Ch. 15 - Prob. 39ECh. 15 - Prob. 40ECh. 15 - Prob. 41ECh. 15 - Prob. 42ECh. 15 - Prob. 43ECh. 15 - Prob. 44ECh. 15 - Prob. 45ECh. 15 - Prob. 46ECh. 15 - Prob. 47ECh. 15 - [6.100] The Chemistry and Life box in Section 6.7...Ch. 15 - Prob. 49ECh. 15 - [6.104] In the experiment shown schematically...Ch. 15 - Microwave ovens use microwave radiation to heat...Ch. 15 - Prob. 52ECh. 15 - The discovery of hafnium, element number 72,...Ch. 15 - Account for formation of the following series of...Ch. 15 - Prob. 55ECh. 15 - The two most common isotopes of uranium are 235U...Ch. 15 - Hypothetical elements X and Y form a molecule XY2,...Ch. 15 - Prob. 58ECh. 15 - Prob. 59ECh. 15 - Prob. 60ECh. 15 - Prob. 61ECh. 15 - Prob. 62ECh. 15 - Prob. 63ECh. 15 - Prob. 64ECh. 15 - Consider the following statements about first...Ch. 15 - Prob. 66ECh. 15 - Prob. 67ECh. 15 - Write the electron configurations for (a) Ga3+...Ch. 15 - Prob. 69AECh. 15 - Prob. 70AECh. 15 - Prob. 71AECh. 15 - Prob. 72AECh. 15 - Prob. 73AECh. 15 - Prob. 74AECh. 15 - Consider the hypothetical reaction A(g) 2B(g). A...Ch. 15 - 15.76 As shown in Table 15.2, the equilibrium...Ch. 15 - Prob. 77AECh. 15 - Prob. 78AECh. 15 - Prob. 79AECh. 15 - Prob. 80AECh. 15 - Prob. 81AECh. 15 - Prob. 82AECh. 15 - Prob. 83AECh. 15 - Prob. 84AECh. 15 - Prob. 85AECh. 15 - Prob. 86AECh. 15 - Prob. 87AECh. 15 - Prob. 88AECh. 15 - Prob. 89AECh. 15 - Prob. 90AECh. 15 - Prob. 91AECh. 15 - Prob. 92AECh. 15 - Prob. 93IECh. 15 - Prob. 94IECh. 15 - Prob. 95IECh. 15 - Prob. 96IECh. 15 - Write the equilibrium-constant expression for the...Ch. 15 - In Section 11.5, we defined the vapor pressure of...Ch. 15 - Prob. 99IECh. 15 - Prob. 100IE
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