Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
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Chapter 7, Problem 28A
Solution

(a)

Interpretation: The group of nitrogen in the periodic table and the number of valance electrons present in it need to be determined.

Concept Introduction: In a periodic table, the group of an element can be determined from the atomic number and number of valence electrons present in its atom. There are 18 groups in a periodic table. Here, elements are arranged in such a way that group 1 has elements with 1 electron in the outermost shell, group 2 has elements with 2 electrons, and so on. The number of valance electrons can also be calculated from the electronic configuration of an atom of the element. Here, electrons present in the outermost shell are known as valance electrons.

Nitrogen belongs to group 5A and the number of valance electrons is 5.

The given element is nitrogen. The atomic number of nitrogen atom is 7. The electronic configuration can be represented as follows:

  N(7)=1s22s22p3

Here, the number of electrons present in 2s2p are valance electrons; thus, the number of valance electrons is 5. Also, group numbers can be determined from the number of valance electrons. In this case, the number of valance electrons is 5; thus, the group number is 5A.

Therefore, nitrogen belongs to group 5A, and the number of valance electrons is 5.

(b)

Interpretation: The group of lithium in the periodic table and the number of valance electrons present in it need to be determined.

Concept Introduction: In a periodic table, the group of an element can be determined from the atomic number and number of valence electrons present in its atom. There are 18 groups in a periodic table. Here, elements are arranged in such a way that group 1 has elements with 1 electron in the outermost shell, group 2 has elements with 2 electrons, and so on. The number of valance electrons can also be calculated from the electronic configuration of an atom of the element. Here, electrons present in the outermost shell are known as valance electrons.

Lithium belongs to group 1A and the number of valance electrons is 1.

The given element is lithium. The atomic number lithium is 3. The electronic configuration can be represented as follows:

  Li(3)=1s22s1

Here, the number of electrons present in 2s are valance electrons. Thus, the number of valance electrons is 1. Also, group numbers can be determined from the number of valance electrons. In this case, the number of valance electrons is 1; thus, the group number is 1.

Therefore, lithium belongs to group 1 and the number of valance electrons is 1.

(c)

Interpretation: The group of phosphorus in the periodic table and the number of valance electrons present in it need to be determined.

Concept Introduction: In a periodic table, the group of an element can be determined from the atomic number and number of valence electrons present in its atom. There are 18 groups in a periodic table. Here, elements are arranged in such a way that group 1 has elements with 1 electron in the outermost shell, group 2 has elements with 2 electrons, and so on. The number of valance electrons can also be calculated from the electronic configuration of an atom of the element. Here, electrons present in the outermost shell are known as valance electrons.

Phosphorus belongs to group 5A and the number of valance electrons is 5.

The given element is phosphorus. The atomic number of phosphorus is 15. The electronic configuration can be represented as follows:

  P(15)=1s22s22p63s23p3

Here, the number of electrons present in 3s3p are valance electrons; thus, the number of valance electrons is 5. Also, group numbers can be determined from the number of valance electrons. In this case, the number of valance electrons is 5; thus, the group number is 5A.

Therefore, phosphorus belongs to group 5A and the number of valance electrons is 5.

(d)

Interpretation: The group of barium in the periodic table and the number of valance electrons present in it need to be determined.

Concept Introduction: In a periodic table, the group of an element can be determined from the atomic number and number of valence electrons present in its atom. There are 18 groups in a periodic table. Here, elements are arranged in such a way that group 1 has elements with 1 electron in the outermost shell, group 2 has elements with 2 electrons, and so on. The number of valance electrons can also be calculated from the electronic configuration of an atom of the element. Here, electrons present in the outermost shell are known as valance electrons.

Barium belongs to group 2A and the number of valance electrons is 2.

(e)

Interpretation: The group of bromine in the periodic table and the number of valance electrons present in it need to be determined.

Concept Introduction: In a periodic table, the group of an element can be determined from the atomic number and number of valence electrons present in its atom. There are 18 groups in a periodic table. Here, elements are arranged in such a way that group 1 has elements with 1 electron in the outermost shell, group 2 has elements with 2 electrons, and so on. The number of valance electrons can also be calculated from the electronic configuration of an atom of the element. Here, electrons present in the outermost shell are known as valance electrons.

Bromine belongs to group 7A and the number of valance electrons is 7.

The given element is bromine. The atomic number of the bromine atom is 35. The electronic configuration can be represented as follows:

  Br(35)=[Ar]4s23d104p5

Here, the number of electrons present in 4s24p5 are valance electrons. Thus, the number of valance electrons is 7. Also, group numbers can be determined from the number of valance electrons. In this case, the group number is 17.

Therefore, bromine belongs to group 17 and the number of valance electrons is 7.

(f)

Interpretation: The group of carbon in the periodic table and the number of valance electrons present in it need to be determined.

Concept Introduction: In a periodic table, the group of an element can be determined from the atomic number and number of valence electrons present in its atom. There are 18 groups in a periodic table. Here, elements are arranged in such a way that group 1 has elements with 1 electron in the outermost shell, group 2 has elements with 2 electrons, and so on. The number of valance electrons can also be calculated from the electronic configuration of an atom of the element. Here, electrons present in the outermost shell are known as valance electrons.

Carbon belongs to group 4A and the number of valance electrons is 4.

The given element is carbon. The atomic number of carbon atom is 6. The electronic configuration can be represented as follows:

  C(6)=1s22s22p2

Here, the number of electrons present in 2s2p are valance electrons. Thus, the number of valance electrons is 4. Also, group number can be determined from the number of valance electrons. In this case, the group number is 4A.

Therefore, carbon belongs to group 4A and the number of valance electrons is 4.

Chapter 7 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 7.2 - Prob. 11SPCh. 7.2 - Prob. 12LCCh. 7.2 - Prob. 13LCCh. 7.2 - Prob. 14LCCh. 7.2 - Prob. 15LCCh. 7.2 - Prob. 16LCCh. 7.2 - Prob. 17LCCh. 7.2 - Prob. 18LCCh. 7.2 - Prob. 19LCCh. 7.3 - Prob. 20LCCh. 7.3 - Prob. 21LCCh. 7.3 - Prob. 22LCCh. 7.3 - Prob. 23LCCh. 7.3 - Prob. 24LCCh. 7.3 - Prob. 25LCCh. 7.3 - Prob. 26LCCh. 7 - Prob. 27ACh. 7 - Prob. 28ACh. 7 - Prob. 29ACh. 7 - Prob. 30ACh. 7 - Prob. 31ACh. 7 - Prob. 32ACh. 7 - Prob. 33ACh. 7 - Prob. 34ACh. 7 - Prob. 35ACh. 7 - Prob. 36ACh. 7 - Prob. 37ACh. 7 - Prob. 38ACh. 7 - Prob. 39ACh. 7 - Prob. 40ACh. 7 - Prob. 41ACh. 7 - Prob. 42ACh. 7 - Prob. 43ACh. 7 - Prob. 44ACh. 7 - Prob. 45ACh. 7 - Prob. 46ACh. 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. 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. 12STP
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