Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 5, Problem 80E
Interpretation Introduction

(a)

Interpretation:

The electron configuration for Br ion with the help of periodic table is to be predicted.

Concept introduction:

Ionic charge is defined as the charge possesses by an ion that result by the gain or loss of an electron by an atom or a molecule. Metals are electropositive in nature because they lose electrons from their outermost shells. Nonmetals are electronegative in nature because they gain electrons in their outermost shells.

Expert Solution
Check Mark

Answer to Problem 80E

The predicted electronic configuration of Br is [Ar]4s24d104p6.

Explanation of Solution

The given negative ion is Br.

Bromine belongs to the group 17 of a periodic table. The atomic number of bromine is 35. The atomic number corresponds to the number of electrons present in an element. Electronic configuration tells about the arrangement of an electron in each subshell and orbit of an atom. The electronic configuration of Br is [Ar]4s24d104p5. It gain one electron and act as an Br ion. Therefore, the predicted electronic configuration of Br is [Ar]4s24d104p6.

Conclusion

The predicted electronic configuration of Br is [Ar]4s24d104p6.

Interpretation Introduction

(b)

Interpretation:

The electron configuration for Te2 ion with the help of periodic table is to be predicted.

Concept introduction:

Ionic charge is defined as the charge possesses by an ion that result by the gain or loss of an electron by an atom or a molecule. Metals are electropositive in nature because they lose electrons from their outermost shells. Nonmetals are electronegative in nature because they gain electrons in their outermost shells.

Expert Solution
Check Mark

Answer to Problem 80E

The predicted electronic configuration of Te2 is [Kr]4d105s25p6.

Explanation of Solution

The given negative ion is Te2.

Tellurium belongs to the group 16 of a periodic table. The atomic number of tellurium is 52. The atomic number corresponds to the number of electrons present in an element. Electronic configuration tells about the arrangement of an electron in each subshell and orbit of an atom. The electronic configuration of Te is [Kr]4d105s25p4. It gain two electrons and act as an Te2 ion. Therefore, the predicted electronic configuration of Te2 is [Kr]4d105s25p6.

Conclusion

The predicted electronic configuration of Te2 is [Kr]4d105s25p6.

Interpretation Introduction

(c)

Interpretation:

The electron configuration for As3 ion with the help of periodic table is to be predicted.

Concept introduction:

Ionic charge is defined as the charge possesses by an ion that result by the gain or loss of an electron by an atom or a molecule. Metals are electropositive in nature because they lose electrons from their outermost shells. Nonmetals are electronegative in nature because they gain electrons in their outermost shells.

Expert Solution
Check Mark

Answer to Problem 80E

The predicted electronic configuration of As3 is [Ar]4s24p6.

Explanation of Solution

The given negative ion is As3.

Arsenic belongs to the group 15 of a periodic table. The atomic number of arsenic is 33. The atomic number corresponds to the number of electrons present in an element. Electronic configuration tells about the arrangement of an electron in each subshell and orbit of an atom. The electronic configuration of As is [Ar]3d104s24p3. It gain three electrons and act as an As3 ion. Therefore, the predicted electronic configuration of As3 is [Ar]4s24p6.

Conclusion

The predicted electronic configuration of As3 is [Ar]4s24p6.

Interpretation Introduction

(d)

Interpretation:

The electron configuration for O2 ion with the help of periodic table is to be predicted.

Concept introduction:

Ionic charge is defined as the charge possesses by an ion that result by the gain or loss of an electron by an atom or a molecule. Metals are electropositive in nature because they lose electrons from their outermost shells. Nonmetals are electronegative in nature because they gain electrons in their outermost shells.

Expert Solution
Check Mark

Answer to Problem 80E

The predicted electronic configuration of O2 is [He]2s22p6.

Explanation of Solution

The given negative ion is O2.

Oxygen belongs to the group 16 of a periodic table. The atomic number of oxygen is 8. The atomic number corresponds to the number of electrons present in an element. Electronic configuration tells about the arrangement of an electron in each subshell and orbit of an atom. The electronic configuration of O is [He]2s22p4. It gain two electrons and act as an O2 ion. Therefore, the predicted electronic configuration of O2 is [Kr]5s24d105p6.

Conclusion

The predicted electronic configuration of O2 is [He]2s22p6.

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Chapter 5 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 5 - Prob. 11CECh. 5 - Prob. 12CECh. 5 - Prob. 13CECh. 5 - Prob. 14CECh. 5 - Prob. 1KTCh. 5 - Prob. 2KTCh. 5 - Prob. 3KTCh. 5 - Prob. 4KTCh. 5 - Prob. 5KTCh. 5 - Prob. 6KTCh. 5 - Prob. 7KTCh. 5 - Prob. 8KTCh. 5 - Prob. 9KTCh. 5 - Prob. 10KTCh. 5 - Prob. 11KTCh. 5 - Prob. 12KTCh. 5 - Prob. 13KTCh. 5 - Prob. 14KTCh. 5 - Prob. 15KTCh. 5 - Prob. 16KTCh. 5 - Prob. 17KTCh. 5 - Prob. 18KTCh. 5 - Prob. 19KTCh. 5 - Prob. 20KTCh. 5 - Prob. 21KTCh. 5 - Prob. 22KTCh. 5 - Prob. 23KTCh. 5 - Prob. 1ECh. 5 - Prob. 2ECh. 5 - Prob. 3ECh. 5 - Prob. 4ECh. 5 - Prob. 5ECh. 5 - Prob. 6ECh. 5 - Prob. 7ECh. 5 - Prob. 8ECh. 5 - Prob. 9ECh. 5 - Prob. 10ECh. 5 - Prob. 11ECh. 5 - Prob. 12ECh. 5 - Prob. 13ECh. 5 - Prob. 14ECh. 5 - Prob. 15ECh. 5 - Prob. 16ECh. 5 - Prob. 17ECh. 5 - Prob. 18ECh. 5 - Prob. 19ECh. 5 - Prob. 20ECh. 5 - Prob. 21ECh. 5 - Prob. 22ECh. 5 - Prob. 23ECh. 5 - Prob. 24ECh. 5 - Prob. 25ECh. 5 - Prob. 26ECh. 5 - Prob. 27ECh. 5 - Prob. 28ECh. 5 - Prob. 29ECh. 5 - Prob. 30ECh. 5 - Prob. 31ECh. 5 - Prob. 32ECh. 5 - Prob. 33ECh. 5 - Prob. 34ECh. 5 - Prob. 35ECh. 5 - Prob. 36ECh. 5 - Prob. 37ECh. 5 - Prob. 38ECh. 5 - Prob. 39ECh. 5 - Prob. 40ECh. 5 - Prob. 41ECh. 5 - Prob. 42ECh. 5 - Prob. 43ECh. 5 - Prob. 44ECh. 5 - Prob. 45ECh. 5 - Prob. 46ECh. 5 - Prob. 47ECh. 5 - Prob. 48ECh. 5 - Prob. 49ECh. 5 - Prob. 50ECh. 5 - Prob. 51ECh. 5 - Prob. 52ECh. 5 - Prob. 53ECh. 5 - Prob. 54ECh. 5 - Prob. 55ECh. 5 - Prob. 56ECh. 5 - Prob. 57ECh. 5 - Prob. 58ECh. 5 - Prob. 59ECh. 5 - Prob. 60ECh. 5 - Prob. 61ECh. 5 - Prob. 62ECh. 5 - Prob. 63ECh. 5 - Prob. 64ECh. 5 - Prob. 65ECh. 5 - Prob. 66ECh. 5 - Prob. 67ECh. 5 - Prob. 68ECh. 5 - Prob. 69ECh. 5 - Prob. 70ECh. 5 - Prob. 71ECh. 5 - Prob. 72ECh. 5 - Prob. 73ECh. 5 - Prob. 74ECh. 5 - Prob. 75ECh. 5 - Prob. 76ECh. 5 - Prob. 77ECh. 5 - Prob. 78ECh. 5 - Prob. 79ECh. 5 - Prob. 80ECh. 5 - Prob. 81ECh. 5 - Prob. 82ECh. 5 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90ECh. 5 - Prob. 91ECh. 5 - Prob. 92ECh. 5 - Prob. 1STCh. 5 - Prob. 2STCh. 5 - Prob. 3STCh. 5 - Prob. 4STCh. 5 - Prob. 5STCh. 5 - Prob. 6STCh. 5 - Prob. 7STCh. 5 - Prob. 8STCh. 5 - Prob. 9STCh. 5 - Prob. 10STCh. 5 - Prob. 11STCh. 5 - Prob. 12STCh. 5 - Prob. 13STCh. 5 - Prob. 14STCh. 5 - Prob. 15STCh. 5 - Prob. 16STCh. 5 - Prob. 17STCh. 5 - Prob. 18ST
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