Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
Question
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Chapter 7.1, Problem 5SSC
Interpretation Introduction

(a)

Interpretation:

The electron configuration of Nitrogen and the change to achieve a noble gas configuration needs to be explained.

Concept introduction:

In chemistry, an electron configuration is the arrangement of electrons in a molecule, an atom, or a different physical structure.

Expert Solution
Check Mark

Answer to Problem 5SSC

The electronic configuration of Nitrogen is [He]2s2, 2p3 and it gains 3 electrons to achieve a noble gas configuration.

Explanation of Solution

The elements have their electrons, which are distributed in its atomic orbital and this distribution is known as that element’s electron configuration. This distribution or filling is done as per standard notation, in which filling is at subshell levels and is placed in sequence.

Nitrogen has 5 valence electrons. The electrons are distributed in orbitals 1s2 2s2 2p3. It is also depicted as [He]2s22p3. Thus, nitrogen needs to gain 3 electrons to reach the nearest noble gas configuration of Neon.

(b)

Interpretation Introduction

Interpretation:

The electronic configuration of Sulphur and the change occur to achieve a noble gas configuration needs to be explained.

Concept introduction:

In chemistry, an electron configuration is the arrangement of electrons in a molecule, an atom, or a different physical structure.

(b)

Expert Solution
Check Mark

Answer to Problem 5SSC

The electronic configuration of Sulphur is [Ne]3s23p4 and needs 2 electrons to achieve a noble gas configuration.

Explanation of Solution

The elements have their electrons, which are distributed in its atomic orbital and this distribution is known as that element’s electron configuration. This distribution or filling is done as per standard notation, in which filling is at subshell levels and is placed in sequence.

Sulfur has 6 valence electrons. Initially, the electrons will distribute in orbital 2p and then the next 2 electrons will fill inan orbital 3s. After 3s, the next filling will be in orbital 3p where the last 4 electronsfilled. Hence, the electronic configuration is 1s2 2s2 2p63s2 3p4. It is also depicted as [Ne]3s23p4.

So, sulfur needs to gain 2 electrons to reach the nearest noble gas configuration of Argon.

(c)

Interpretation Introduction

Interpretation:

The electronic configuration of Barium and the change that occur to achieve a noble gas configuration needs to be explained.

Concept introduction:

In chemistry, an electron configuration is the arrangement of electrons in a molecule, an atom, or a different physical structure.

(c)

Expert Solution
Check Mark

Answer to Problem 5SSC

The electronic configuration of Barium is [Xe]6s2 and needs to lose 2 electrons to achieve a noble gas configuration.

Explanation of Solution

The elements have their electrons, which are distributed in its atomic orbital and this distribution is known as that element’s electron configuration. This distribution or filling is done as per standard notation, in which filling is at subshell levels and is placed in sequence.

Atomic Barium has 2 valence electrons. The electronic configuration of Barium is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2. It is also depicted as [Xe]6s2. As it has 2 electrons in ‘p’ shell, indicates it can lose 2 electrons and forms a cation.

So, barium needs to lose 2 electrons to reach the nearest noble gas configuration of xenon.

(d)

Interpretation Introduction

Interpretation:

The electronic configuration of Lithium and the change that occur to achieve a noble gas configuration needs to be determined.

Concept introduction:

In chemistry, an electron configuration is the arrangement of electrons in a molecule, an atom, or a different physical structure.

(d)

Expert Solution
Check Mark

Answer to Problem 5SSC

The electronic configuration of Lithium is [Xe]6s2 and it can lose 2 electrons.

Explanation of Solution

The elements have their electrons, which are distributed in its atomic orbital and this distribution is known as that element’s electron configuration. This distribution or filling is done as per standard notation, in which filling is at subshell levels and is placed in sequence.

Atomic Lithium has 1 valence electrons. The electronic configuration of Lithium is 1s2 2s1.

It is also depicted as [He]2s1. As it has 1 an electron in ‘p’ shell, it can easily lose one electron and become a positively charged ion.

Hence, lithium needs to lose an electron to reach the nearest noble gas configuration of helium.

Chapter 7 Solutions

Chemistry: Matter and Change

Ch. 7.2 - Prob. 11PPCh. 7.2 - Prob. 12SSCCh. 7.2 - Prob. 13SSCCh. 7.2 - Prob. 14SSCCh. 7.2 - Prob. 15SSCCh. 7.2 - Prob. 16SSCCh. 7.2 - Prob. 17SSCCh. 7.2 - Prob. 18SSCCh. 7.3 - Prob. 19PPCh. 7.3 - Prob. 20PPCh. 7.3 - Prob. 21PPCh. 7.3 - Prob. 22PPCh. 7.3 - Prob. 23PPCh. 7.3 - Prob. 24PPCh. 7.3 - Prob. 25PPCh. 7.3 - Prob. 26PPCh. 7.3 - Prob. 27PPCh. 7.3 - Prob. 28PPCh. 7.3 - Prob. 29PPCh. 7.3 - Prob. 30PPCh. 7.3 - Prob. 31PPCh. 7.3 - Prob. 32PPCh. 7.3 - Prob. 33PPCh. 7.3 - Prob. 34SSCCh. 7.3 - Prob. 35SSCCh. 7.3 - Prob. 36SSCCh. 7.3 - Prob. 37SSCCh. 7.3 - Prob. 38SSCCh. 7.3 - Prob. 39SSCCh. 7.4 - Prob. 40SSCCh. 7.4 - Prob. 41SSCCh. 7.4 - Prob. 42SSCCh. 7.4 - Prob. 43SSCCh. 7.4 - Prob. 44SSCCh. 7.4 - Prob. 45SSCCh. 7 - How do positive ions and negative ions form?Ch. 7 - Prob. 47ACh. 7 - Prob. 48ACh. 7 - Prob. 49ACh. 7 - Prob. 50ACh. 7 - Prob. 51ACh. 7 - Prob. 52ACh. 7 - Prob. 53ACh. 7 - Prob. 54ACh. 7 - Prob. 55ACh. 7 - Prob. 56ACh. 7 - Prob. 57ACh. 7 - Prob. 58ACh. 7 - Prob. 59ACh. 7 - Prob. 60ACh. 7 - Prob. 61ACh. 7 - Prob. 62ACh. 7 - Prob. 63ACh. 7 - Prob. 64ACh. 7 - Prob. 65ACh. 7 - Prob. 66ACh. 7 - Prob. 67ACh. 7 - Prob. 68ACh. 7 - Prob. 69ACh. 7 - Prob. 70ACh. 7 - Prob. 71ACh. 7 - Prob. 72ACh. 7 - Prob. 73ACh. 7 - Prob. 74ACh. 7 - Prob. 75ACh. 7 - Prob. 76ACh. 7 - Prob. 77ACh. 7 - Prob. 78ACh. 7 - Prob. 79ACh. 7 - Prob. 80ACh. 7 - Prob. 81ACh. 7 - Prob. 82ACh. 7 - Prob. 83ACh. 7 - Prob. 84ACh. 7 - Prob. 85ACh. 7 - Prob. 86ACh. 7 - Prob. 87ACh. 7 - Prob. 88ACh. 7 - Prob. 89ACh. 7 - Prob. 90ACh. 7 - Prob. 91ACh. 7 - Prob. 92ACh. 7 - Prob. 93ACh. 7 - Prob. 94ACh. 7 - Prob. 95ACh. 7 - Prob. 96ACh. 7 - Prob. 97ACh. 7 - Prob. 98ACh. 7 - Prob. 99ACh. 7 - Prob. 100ACh. 7 - Prob. 101ACh. 7 - Prob. 102ACh. 7 - Prob. 103ACh. 7 - Prob. 104ACh. 7 - Prob. 105ACh. 7 - Prob. 106ACh. 7 - Prob. 107ACh. 7 - Prob. 108ACh. 7 - Prob. 109ACh. 7 - Prob. 110ACh. 7 - Prob. 111ACh. 7 - Prob. 112ACh. 7 - Prob. 113ACh. 7 - Prob. 114ACh. 7 - Prob. 115ACh. 7 - Prob. 116ACh. 7 - Prob. 117ACh. 7 - Prob. 118ACh. 7 - Prob. 119ACh. 7 - Prob. 120ACh. 7 - Prob. 121ACh. 7 - Prob. 122ACh. 7 - Prob. 123ACh. 7 - Prob. 124ACh. 7 - Prob. 125ACh. 7 - Prob. 126ACh. 7 - Prob. 127ACh. 7 - Prob. 128ACh. 7 - Prob. 129ACh. 7 - Prob. 130ACh. 7 - Prob. 131ACh. 7 - Prob. 132ACh. 7 - Prob. 133ACh. 7 - Prob. 134ACh. 7 - Prob. 1STPCh. 7 - Prob. 2STPCh. 7 - Prob. 3STPCh. 7 - Prob. 4STPCh. 7 - Prob. 5STPCh. 7 - Prob. 6STPCh. 7 - Prob. 7STPCh. 7 - Prob. 8STPCh. 7 - Prob. 9STPCh. 7 - Prob. 10STPCh. 7 - Prob. 11STPCh. 7 - Prob. 12STPCh. 7 - Prob. 13STPCh. 7 - Prob. 14STPCh. 7 - Prob. 15STPCh. 7 - Prob. 16STPCh. 7 - Prob. 17STPCh. 7 - Prob. 18STPCh. 7 - Prob. 19STPCh. 7 - Prob. 20STP
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