Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 7, Problem 7.1VP

(a)

Interpretation Introduction

Interpretation: The highlighted elements are to be stated. The elements that have single s electron, filled sets of s,p and d orbital in the valence shell, half-filled sets of d orbital, and two s electrons in the valence shell are to be identified.

Concept introduction: The atomic number of element helps to find out the electronic configuration of an element. The electronic configuration helps to find out the electronic distribution in each orbital.

To determine: The elements having a single s electron in the valence shell.

(a)

Expert Solution
Check Mark

Answer to Problem 7.1VP

Solution

The elements that have single s electron in its valence shell are chromium, gold and sodium.

Explanation of Solution

Explanation

There are total five elements that are highlighted in the Figure 1. The names of these elements are chromium (Cr) , gold (Au) , sodium (Na) , neon (Ne) and chlorine (Cl) . Out of these five elements, three elements possess single s electron in its valence shell. The atomic number of Cr , Au , Na , Ne and Cl are 24 , 79 , 11 , 10 , and 17 respectively. The presence of single s electron can be determined by its electronic configuration.

Chemistry: The Science in Context (Fifth Edition), Chapter 7, Problem 7.1VP

Figure 1.

The electrons that are present in different shells of an element is shown as,

Cr=[Ar]3d54s1Au=[Xe]4f145d106s1Na=1s22s22p63s1Ne=1s22s22p6Cl=1s22s22p63s23p5

Therefore, on the basis of electronic configuration, it is inferred that chromium, gold and sodium are the elements that have single s electron in its valence shell.

(b)

Interpretation Introduction

To determine: The elements having a filled set of s and p orbital.

(b)

Expert Solution
Check Mark

Answer to Problem 7.1VP

Solution

Neon element has filled sets of s and p orbital.

Explanation of Solution

Explanation

The elements that have filled sets of s and p orbital do not participate in chemical reactions.

Out of five highlighted elements, neon is the only element that possesses filled sets of s and p orbital.

The electronic configuration of neon is shown as,

Ne=1s22s22p6

Therefore, neon element has filled sets of s and p orbital.

(c)

Interpretation Introduction

To determine: The elements having filled sets of d orbital is to be determined.

(c)

Expert Solution
Check Mark

Answer to Problem 7.1VP

Solution

Gold element has filled set of d orbital.

Explanation of Solution

Explanation

The electronic configuration helps to find out the electronic distribution in each orbital. In gold element last electron goes to d orbital instead of s orbital to provide stability to the element.

Out of given five highlighted elements, gold is the only element that possesses filled sets of d orbital. The electronic configuration of gold is shown as,

Au=[Xe]4f145d106s1

Therefore, gold has filled set of d orbital.

(d)

Interpretation Introduction

To determine: The elements having half filled sets of d orbital are to be determined.

(d)

Expert Solution
Check Mark

Answer to Problem 7.1VP

Solution

Chromium element of periodic table consists of half filled sets of d orbital.

Explanation of Solution

Explanation

The electronic configuration helps to find out the electronic distribution in each orbital. As half filled d orbital are stable in nature so, in this element last electron goes to d orbital to provide stability.

Out of given five highlighted elements, chromium is the only element that possesses filled sets of d orbital. The electronic configuration of gold is shown as,

Cr=[Ar]3d54s1

Therefore, chromium has half filled sets of d orbital.

(e)

Interpretation Introduction

To determine: The elements contain two s electrons in valence shell are to be determined.

(e)

Expert Solution
Check Mark

Answer to Problem 7.1VP

Solution

Neon and chlorine elements consist of two s electrons in their valence shell.

Explanation of Solution

Explanation

The electronic configuration helps to find out the electronic distribution in each orbital. Out of given five highlighted elements, neon and chlorine are the elements that possess two s electrons in its external orbital.

The electronic configuration of neon and chlorine is shown as,

Ne=1s22s22p6Cl=1s22s22p63s23p5

Therefore, Neon and chlorine are the only highlighted elements that have two s electrons in valence shell.

Conclusion

  1. a. The elements that have single s electron in its valence shell are chromium, gold and sodium.
  2. b. Neon element has filled sets of s and p orbital.
  3. c. Gold element has filled set of d orbital.
  4. d. Chromium element of periodic table consists of half filled sets of d orbital.
  5. e. Neon and chlorine elements consist of two s electrons in their valence shell.

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Chapter 7 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 7.9 - Prob. 11PECh. 7.10 - Prob. 12PECh. 7.10 - Prob. 13PECh. 7.11 - Prob. 14PECh. 7.12 - Prob. 15PECh. 7 - Prob. 7.1VPCh. 7 - Prob. 7.2VPCh. 7 - Prob. 7.3VPCh. 7 - Prob. 7.4VPCh. 7 - Prob. 7.5VPCh. 7 - Prob. 7.6VPCh. 7 - Prob. 7.7VPCh. 7 - Prob. 7.8VPCh. 7 - Prob. 7.9VPCh. 7 - Prob. 7.10VPCh. 7 - Prob. 7.11QPCh. 7 - Prob. 7.12QPCh. 7 - Prob. 7.13QPCh. 7 - Prob. 7.14QPCh. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - Prob. 7.42QPCh. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Prob. 7.100QPCh. 7 - Prob. 7.101QPCh. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Prob. 7.112QPCh. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - Prob. 7.119QPCh. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Prob. 7.123QPCh. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Prob. 7.126QPCh. 7 - Prob. 7.127APCh. 7 - Prob. 7.128APCh. 7 - Prob. 7.129APCh. 7 - Prob. 7.130APCh. 7 - Prob. 7.131APCh. 7 - Prob. 7.132APCh. 7 - Prob. 7.133APCh. 7 - Prob. 7.134APCh. 7 - Prob. 7.135APCh. 7 - Prob. 7.136APCh. 7 - Prob. 7.137APCh. 7 - Prob. 7.138APCh. 7 - Prob. 7.139APCh. 7 - Prob. 7.140APCh. 7 - Prob. 7.141APCh. 7 - Prob. 7.142AP
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