Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U1.18, Problem 8E

(a)

Interpretation Introduction

Interpretation:

Interpret the name, symbol and group number of element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(a)

Expert Solution
Check Mark

Answer to Problem 8E

Atomic symbol = Sn

Group number = 14

Name = Tin or stannum

Explanation of Solution

According to long form of the periodic table, the element with atomic number 50 is tin. The atomic symbol of tin is Sn as the Latin name of tin is stannum. It is placed in group-14 and period-5 in the periodic table.

The group-14 is called as carbon family as the first member of group is carbon. Thus, Sn is the member of carbon family with atomic number 50, group number 14.

(b)

Interpretation Introduction

Interpretation:

Interpret the number of protons in the element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(b)

Expert Solution
Check Mark

Answer to Problem 8E

Atomic number = 50

Explanation of Solution

According to long form of the periodic table, the element with atomic number 50 is tin. The atomic symbol of tin is Sn as the Latin name of tin is stannum. It is placed in group-14 and period-5 in the periodic table.

The atomic number of an element can be defined as the number of protons in it. Since the atomic number of Sn is 50 thus the number of protons must be 50.

(c)

Interpretation Introduction

Interpretation:

Interpret the number of neutrons in the element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(c)

Expert Solution
Check Mark

Answer to Problem 8E

Number of neutron = 69

Explanation of Solution

According to long form of the periodic table, the element with atomic number 50 is tin. The atomic symbol of tin is Sn as the Latin name of tin is stannum. It is placed in group-14 and period-5 in the periodic table.

The atomic number of an element can be defined as the number of protons in it. The number of neutrons can be calculated with the help of mass number and atomic number. The mass number of tin is 119. Thus, the number of neutrons must be:

Number of neutrons = Mass number − atomic number

Number of neutrons = 119 − 50 = 69

(d)

Interpretation Introduction

Interpretation:

Interpret the number of electrons in the neutral atom of the element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(d)

Expert Solution
Check Mark

Answer to Problem 8E

Number of electrons = 50

Explanation of Solution

According to long form of the periodic table, the element with atomic number 50 is tin. The atomic symbol of tin is Sn as the Latin name of tin is stannum. It is placed in group-14 and period-5 in the periodic table.

The atomic number of an element can be defined as the number of protons in it. In a neutral atom, the number of protons is always equal to the number of electrons. This is because, protons are positively charge particles and electrons are negatively charged particles. Same number of both sub-atomic particles makes the atom neutral.

Number of protons in tin = number of electrons = 50

(e)

Interpretation Introduction

Interpretation:

Interpret the number of valence electrons in the neutral atom of the element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(e)

Expert Solution
Check Mark

Answer to Problem 8E

Sn = [Kr] 4d10 5s2 5p2 ; 4 valence electrons

Explanation of Solution

The atomic number of an element can be defined as the number of protons in it. In a neutral atom, the number of protons is always equal to the number of electrons. The valence shell electrons are the electrons in the outer most shell of an atom.

According to long form of the periodic table, the element with atomic number 50 is tin. The electronic configuration of tin is Sn = [Kr] 4d10 5s2 5p2.

Thus, there are total 4 valence electrons in Sn.

(f)

Interpretation Introduction

Interpretation:

Interpret the number of core electrons in the neutral atom of the element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(f)

Expert Solution
Check Mark

Answer to Problem 8E

46 core electrons in Sn atom.

Explanation of Solution

In a neutral atom, the number of protons is always equal to the number of electrons. The valence shell electrons are the electrons in the outer most shell of an atom.

According to long form of the periodic table, the element with atomic number 50 is tin. The electronic configuration of tin is Sn = [Kr] 4d10 5s2 5p2.

There are 4 electrons in the valence shell of Sn Si atom. Thus, the remaining electrons must be in innermost shell which are called as core electrons. There are 50-4 = 46 core electrons in Sn atom.

(g)

Interpretation Introduction

Interpretation:

Interpret the names of three other elements with same number of valence electrons line in the element number 50.

Concept introduction:

An atom is considered as the basic unit of matter. It is composed of three sub-atomic particles; electrons, protons and neutrons.

Electrons are placed in certain energy levels which are called as shell. The outermost shell of an atom is called as valence shell and the electrons are called as valence shell electrons. These valence shell electrons take part in all chemical reactions and also involve in chemical bonding.

(g)

Expert Solution
Check Mark

Answer to Problem 8E

  • Carbon
  • Silicon
  • Germanium

Explanation of Solution

According to long form of the periodic table, the element with atomic number 50 is tin. The electronic configuration of tin is Sn = [Kr] 4d10 5s2 5p2.

There are 4 electrons in the valence shell of Sn Si atom. The elements of same group have same number of valence electrons. Since Sn belongs to group-14 thus the elements of group-14 will have same number of valence electrons as in tin. Other three elements are:

  • Carbon
  • Silicon
  • Germanium

Chapter U1 Solutions

Living By Chemistry: First Edition Textbook

Ch. U1.2 - Prob. 1ECh. U1.2 - Prob. 2ECh. U1.2 - Prob. 5ECh. U1.3 - Prob. 1TAICh. U1.3 - Prob. 1ECh. U1.3 - Prob. 2ECh. U1.3 - Prob. 3ECh. U1.3 - Prob. 4ECh. U1.3 - Prob. 5ECh. U1.3 - Prob. 6ECh. U1.4 - Prob. 1TAICh. U1.4 - Prob. 1ECh. U1.4 - Prob. 2ECh. U1.4 - Prob. 3ECh. U1.4 - Prob. 4ECh. U1.4 - Prob. 5ECh. U1.4 - Prob. 6ECh. U1.4 - Prob. 7ECh. U1.4 - Prob. 8ECh. U1.4 - Prob. 9ECh. U1.4 - Prob. 10ECh. U1.4 - Prob. 11ECh. U1.5 - Prob. 1TAICh. U1.5 - Prob. 1ECh. U1.5 - Prob. 2ECh. U1.5 - Prob. 3ECh. U1.5 - Prob. 4ECh. U1.5 - Prob. 5ECh. U1.5 - Prob. 6ECh. U1.5 - Prob. 7ECh. U1.6 - Prob. 1TAICh. U1.6 - Prob. 1ECh. U1.6 - Prob. 2ECh. U1.6 - Prob. 3ECh. U1.6 - Prob. 4ECh. U1.6 - Prob. 5ECh. U1.6 - Prob. 6ECh. U1.7 - Prob. 1TAICh. U1.7 - Prob. 1ECh. U1.7 - Prob. 2ECh. U1.7 - Prob. 3ECh. U1.7 - Prob. 4ECh. U1.7 - Prob. 5ECh. U1.8 - Prob. 1TAICh. U1.8 - Prob. 1ECh. U1.8 - Prob. 2ECh. U1.8 - Prob. 4ECh. U1.8 - Prob. 5ECh. U1.8 - Prob. 6ECh. U1.8 - Prob. 7ECh. U1.9 - Prob. 1TAICh. U1.9 - Prob. 1ECh. U1.9 - Prob. 2ECh. U1.9 - Prob. 5ECh. U1.9 - Prob. 7ECh. U1.10 - Prob. 1TAICh. U1.10 - Prob. 1ECh. U1.10 - Prob. 2ECh. U1.10 - Prob. 3ECh. U1.10 - Prob. 4ECh. U1.10 - Prob. 5ECh. U1.10 - Prob. 6ECh. U1.10 - Prob. 7ECh. U1.10 - Prob. 8ECh. U1.11 - Prob. 1TAICh. U1.11 - Prob. 1ECh. U1.11 - Prob. 2ECh. U1.11 - Prob. 3ECh. U1.11 - Prob. 4ECh. U1.11 - Prob. 5ECh. U1.11 - Prob. 6ECh. U1.11 - Prob. 7ECh. U1.11 - Prob. 9ECh. U1.11 - Prob. 11ECh. U1.11 - Prob. 12ECh. U1.12 - Prob. 1TAICh. U1.12 - Prob. 1ECh. U1.12 - Prob. 2ECh. U1.12 - Prob. 3ECh. U1.12 - Prob. 4ECh. U1.12 - Prob. 5ECh. U1.12 - Prob. 6ECh. U1.12 - Prob. 7ECh. U1.12 - Prob. 8ECh. U1.13 - Prob. 1TAICh. U1.13 - Prob. 1ECh. U1.13 - Prob. 2ECh. U1.13 - Prob. 3ECh. U1.13 - Prob. 4ECh. U1.13 - Prob. 5ECh. U1.13 - Prob. 6ECh. U1.13 - Prob. 7ECh. U1.13 - Prob. 8ECh. U1.13 - Prob. 9ECh. U1.14 - Prob. 1TAICh. U1.14 - Prob. 1ECh. U1.14 - Prob. 2ECh. U1.14 - Prob. 3ECh. U1.14 - Prob. 4ECh. U1.14 - Prob. 5ECh. U1.14 - Prob. 6ECh. U1.14 - Prob. 7ECh. U1.14 - Prob. 8ECh. U1.14 - Prob. 9ECh. U1.14 - Prob. 10ECh. U1.14 - Prob. 11ECh. U1.14 - Prob. 12ECh. U1.14 - Prob. 13ECh. U1.14 - Prob. 14ECh. U1.15 - Prob. 1TAICh. U1.15 - Prob. 1ECh. U1.15 - Prob. 2ECh. U1.15 - Prob. 3ECh. U1.15 - Prob. 4ECh. U1.15 - Prob. 5ECh. U1.15 - Prob. 6ECh. U1.15 - Prob. 7ECh. U1.15 - Prob. 8ECh. U1.15 - Prob. 9ECh. U1.15 - Prob. 10ECh. U1.15 - Prob. 11ECh. U1.15 - Prob. 12ECh. U1.16 - Prob. 1TAICh. U1.16 - Prob. 1ECh. U1.16 - Prob. 2ECh. U1.16 - Prob. 3ECh. U1.16 - Prob. 4ECh. U1.16 - Prob. 5ECh. U1.16 - Prob. 6ECh. U1.17 - Prob. 1TAICh. U1.17 - Prob. 1ECh. U1.17 - Prob. 2ECh. U1.17 - Prob. 3ECh. U1.17 - Prob. 4ECh. U1.17 - Prob. 5ECh. U1.17 - Prob. 6ECh. U1.17 - Prob. 7ECh. U1.17 - Prob. 8ECh. U1.17 - Prob. 9ECh. U1.17 - Prob. 10ECh. U1.17 - Prob. 11ECh. U1.18 - Prob. 1TAICh. U1.18 - Prob. 1ECh. U1.18 - Prob. 2ECh. U1.18 - Prob. 3ECh. U1.18 - Prob. 4ECh. U1.18 - Prob. 5ECh. U1.18 - Prob. 6ECh. U1.18 - Prob. 7ECh. U1.18 - Prob. 8ECh. U1.18 - Prob. 9ECh. U1.18 - Prob. 10ECh. U1.19 - Prob. 1TAICh. U1.19 - Prob. 1ECh. U1.19 - Prob. 2ECh. U1.19 - Prob. 3ECh. U1.19 - Prob. 4ECh. U1.19 - Prob. 5ECh. U1.19 - Prob. 6ECh. U1.19 - Prob. 7ECh. U1.19 - Prob. 8ECh. U1.19 - Prob. 9ECh. U1.19 - Prob. 10ECh. U1.19 - Prob. 11ECh. U1.19 - Prob. 12ECh. U1.19 - Prob. 13ECh. U1.19 - Prob. 14ECh. U1.19 - Prob. 15ECh. U1.19 - Prob. 16ECh. U1.20 - Prob. 1TAICh. U1.20 - Prob. 1ECh. U1.20 - Prob. 2ECh. U1.20 - Prob. 3ECh. U1.20 - Prob. 4ECh. U1.20 - Prob. 5ECh. U1.20 - Prob. 6ECh. U1.20 - Prob. 7ECh. U1.21 - Prob. 1TAICh. U1.21 - Prob. 1ECh. U1.21 - Prob. 2ECh. U1.21 - Prob. 3ECh. U1.21 - Prob. 4ECh. U1.21 - Prob. 5ECh. U1.21 - Prob. 6ECh. U1.21 - Prob. 7ECh. U1.21 - Prob. 8ECh. U1.22 - Prob. 1TAICh. U1.22 - Prob. 1ECh. U1.22 - Prob. 2ECh. U1.22 - Prob. 3ECh. U1.22 - Prob. 4ECh. U1.22 - Prob. 5ECh. U1.22 - Prob. 6ECh. U1.22 - Prob. 7ECh. U1.23 - Prob. 1TAICh. U1.23 - Prob. 1ECh. U1.23 - Prob. 2ECh. U1.23 - Prob. 3ECh. U1.23 - Prob. 4ECh. U1.23 - Prob. 5ECh. U1.24 - Prob. 1TAICh. U1.24 - Prob. 1ECh. U1.24 - Prob. 2ECh. U1.24 - Prob. 3ECh. U1.24 - Prob. 4ECh. U1.24 - Prob. 5ECh. U1.24 - Prob. 6ECh. U1.24 - Prob. 7ECh. U1.24 - Prob. 8ECh. U1.24 - Prob. 9ECh. U1.24 - Prob. 10ECh. U1.24 - Prob. 11ECh. U1.24 - Prob. 12ECh. U1.24 - Prob. 13ECh. U1.25 - Prob. 1TAICh. U1.25 - Prob. 1ECh. U1.25 - Prob. 2ECh. U1.25 - Prob. 3ECh. U1.25 - Prob. 4ECh. U1.25 - Prob. 5ECh. U1.25 - Prob. 6ECh. U1.26 - Prob. 1TAICh. U1.26 - Prob. 1ECh. U1.26 - Prob. 2ECh. U1.26 - Prob. 3ECh. U1.26 - Prob. 4ECh. U1.26 - Prob. 5ECh. U1.26 - Prob. 6ECh. U1.26 - Prob. 7ECh. U1.26 - Prob. 8ECh. U1.26 - Prob. 9ECh. U1.26 - Prob. 10ECh. U1.27 - Prob. 1TAICh. U1.27 - Prob. 1ECh. U1.27 - Prob. 2ECh. U1.27 - Prob. 3ECh. U1.27 - Prob. 4ECh. U1.27 - Prob. 5ECh. U1.27 - Prob. 6ECh. U1.27 - Prob. 7ECh. U1 - Prob. SI1RECh. U1 - Prob. SI2RECh. U1 - Prob. SI3RECh. U1 - Prob. SI4RECh. U1 - Prob. SI5RECh. U1 - Prob. SI6RECh. U1 - Prob. SII1RQCh. U1 - Prob. SII2RQCh. U1 - Prob. SII3RQCh. U1 - Prob. SII4RQCh. U1 - Prob. SIII1RQCh. U1 - Prob. SIII2RQCh. U1 - Prob. SIII3RQCh. U1 - Prob. SIII4RQCh. U1 - Prob. SIV1RQCh. U1 - Prob. SIV2RQCh. U1 - Prob. SIV3RQCh. U1 - Prob. SIV4RQCh. U1 - Prob. SIV5RQCh. U1 - Prob. SIV6RQCh. U1 - Prob. SV1RQCh. U1 - Prob. SV2RQCh. U1 - Prob. SV3RQCh. U1 - Prob. SV4RQCh. U1 - Prob. 1RECh. U1 - Prob. 2RECh. U1 - Prob. 3RECh. U1 - Prob. 4RECh. U1 - Prob. 5RECh. U1 - Prob. 6RECh. U1 - Prob. 7RECh. U1 - Prob. 8RECh. U1 - Prob. 9RECh. U1 - Prob. 10RECh. U1 - Prob. 11RECh. U1 - Prob. 12RE
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