Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
Question
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Chapter U2.18, Problem 3E

(a)

Interpretation Introduction

Interpretation: A Lewis dot structure for H2Se needs to be drawn and the shape needs to be indicated. This is to be determined if H2Se has smell or not.

Concept Introduction: Lewis dot structure of H2Se is similar to H2O, it is because Oxygen and selenium both have same number of valence electrons as both are of same group.

(a)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot structure of different molecules with different shapes can be formed with the help of valence electrons.

Explanation of Solution

The Lewis dot structure of H2Se is as follows:

Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 3E , additional homework tip  1

The molecule is polar, bond angle is 1050 and it hassp3 hybridization. The overall molecule has bent shape. It has horseradish like smell.

(b)

Interpretation Introduction

Interpretation: A Lewis dot structure for H2 needs to be drawn and the shape needs to be indicated. This is to be determined if H2 has smell or not.

Concept Introduction: In Lewis dot structure of H2, Hydrogen has one valence electron.Both hydrogen atoms will share one electron to form one covalent bond.

(b)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot structure of different molecules with different shapes can be formed with the help of valence electrons.

Explanation of Solution

The Lewis dot structure of H2molecule is as follows:

Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 3E , additional homework tip  2

Hydrogen-hydrogen bond is formed after sharing of 1 electron by each H atom. Since, there is no lone pair of electrons on H atom thus, it has linear shape. The hydrogen gas is non-toxic, it does not have any smell or color.

(c)

Interpretation Introduction

Interpretation: A Lewis dot structure for Ar needs to be drawn and the shape needs to be indicated. This is to be determined if Ar has smell or not.

Concept Introduction: In Lewis dot structure of Ar, Argon has eight valence electrons. Due to presence of 8 electrons, it will not share even single electron with any other atom.

(c)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot structure of different molecules with different shapes can be formed with the help of valence electrons.

Explanation of Solution

The Lewis dot structure is represented as follows:

Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 3E , additional homework tip  3

Argon is an inert gas with valence electrons 8, there will be no sharing of electrons. The element does not have any shape as it does not form bond with any other atom of element. Argon gas being inert has no smell and color. It is non-toxic in nature.

(d)

Interpretation Introduction

Interpretation: A Lewis dot structure for HOF needs to be drawn and the shape needs to be indicated. This is to be determined if HOF has smell or not.

Concept Introduction: In Lewis dot structure of HOF, hydrogen has one,oxygen has six and fluorine has seven valence electrons. Hydrogen and fluorine both will share one electron to form single covalent bond. Oxygen will share one electron each with hydrogen as well as with fluorine.

(d)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot structure of different molecules with different shapes can be formed with the help of valence electrons.

Explanation of Solution

The Lewis dot structure of HOF molecule is represented as follows:

Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 3E , additional homework tip  4

It has bent shape and molecule has sp3hybridization.

(e)

Interpretation Introduction

Interpretation: A Lewis dot structure for CHClF2needs to be drawn and the shape needs to be indicated. This is to be determined if CHClF2 has smell or not.

Concept Introduction: In Lewis dot structure of CHClF2, hydrogen has one, carbon has four and chlorine and fluorinehave seven valence electrons. Hydrogen, chlorine and fluorine both will share one electron with carbon atom to form single covalent bond.

(e)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot structure of different molecules with different shapes can be formed with the help of valence electrons.

Explanation of Solution

The Lewis dot structure is represented as follows:

Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 3E , additional homework tip  5

The shape of the molecule is tetrahedral and molecule has sp3 hybridization. It is a colorless gas with ethereal smell.

(f)

Interpretation Introduction

Interpretation: A Lewis dot structure for HCHO needs to be drawn and the shape needs to be indicated. This is to be determined if HCHO has smell or not.

Concept Introduction: In Lewis dot structure of HCHO, hydrogen has one, oxygen has six valence electrons. Hydrogen will share one electron with carbon atom to form single covalent bond and oxygen will share two electrons with carbon to complete its octet.

(f)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot structure of different molecules with different shapes can be formed with the help of valence electrons.

Explanation of Solution

The given Lewis dot structure is as follows:

Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 3E , additional homework tip  6

The shape is trigonal planar and molecule is sp2 hybridized. It has a strong pickle like smell.

Chapter U2 Solutions

Living By Chemistry: First Edition Textbook

Ch. U2.2 - Prob. 3ECh. U2.2 - Prob. 4ECh. U2.2 - Prob. 5ECh. U2.2 - Prob. 6ECh. U2.2 - Prob. 7ECh. U2.3 - Prob. 1TAICh. U2.3 - Prob. 1ECh. U2.3 - Prob. 2ECh. U2.3 - Prob. 3ECh. U2.3 - Prob. 4ECh. U2.3 - Prob. 5ECh. U2.4 - Prob. 1TAICh. U2.4 - Prob. 1ECh. U2.4 - Prob. 2ECh. U2.4 - Prob. 3ECh. U2.4 - Prob. 4ECh. U2.4 - Prob. 5ECh. U2.4 - Prob. 6ECh. U2.4 - Prob. 7ECh. U2.4 - Prob. 8ECh. U2.5 - Prob. 1TAICh. U2.5 - Prob. 1ECh. U2.5 - Prob. 2ECh. U2.5 - Prob. 3ECh. U2.5 - Prob. 4ECh. U2.5 - Prob. 5ECh. U2.5 - Prob. 6ECh. U2.5 - Prob. 7ECh. U2.5 - Prob. 8ECh. U2.5 - Prob. 9ECh. U2.6 - Prob. 1TAICh. U2.6 - Prob. 1ECh. U2.6 - Prob. 2ECh. U2.6 - Prob. 3ECh. U2.6 - Prob. 4ECh. U2.6 - Prob. 5ECh. U2.6 - Prob. 6ECh. U2.6 - Prob. 7ECh. U2.6 - Prob. 8ECh. U2.6 - Prob. 9ECh. U2.6 - Prob. 10ECh. U2.6 - Prob. 11ECh. U2.7 - Prob. 1TAICh. U2.7 - Prob. 1ECh. U2.7 - Prob. 2ECh. U2.7 - Prob. 3ECh. U2.7 - Prob. 4ECh. U2.7 - Prob. 5ECh. U2.8 - Prob. 1TAICh. U2.8 - Prob. 1ECh. U2.8 - Prob. 2ECh. U2.8 - Prob. 3ECh. U2.8 - Prob. 4ECh. U2.8 - Prob. 5ECh. U2.8 - Prob. 6ECh. U2.9 - Prob. 1TAICh. U2.9 - Prob. 1ECh. U2.9 - Prob. 2ECh. U2.9 - Prob. 3ECh. U2.9 - Prob. 4ECh. U2.9 - Prob. 5ECh. U2.9 - Prob. 6ECh. U2.9 - Prob. 7ECh. U2.10 - Prob. 1TAICh. U2.10 - Prob. 1ECh. U2.10 - Prob. 2ECh. U2.10 - Prob. 3ECh. U2.10 - Prob. 4ECh. U2.10 - Prob. 5ECh. U2.10 - Prob. 6ECh. U2.10 - Prob. 7ECh. U2.10 - Prob. 8ECh. U2.10 - Prob. 9ECh. U2.11 - Prob. 1TAICh. U2.11 - Prob. 1ECh. U2.11 - Prob. 2ECh. U2.11 - Prob. 3ECh. U2.11 - Prob. 4ECh. U2.11 - Prob. 5ECh. U2.11 - Prob. 6ECh. U2.11 - Prob. 7ECh. U2.12 - Prob. 1TAICh. U2.12 - Prob. 1ECh. U2.12 - Prob. 2ECh. U2.12 - Prob. 3ECh. U2.12 - Prob. 4ECh. U2.12 - Prob. 5ECh. U2.12 - Prob. 6ECh. U2.12 - Prob. 7ECh. U2.13 - Prob. 1TAICh. U2.13 - Prob. 1ECh. U2.13 - Prob. 2ECh. U2.13 - Prob. 3ECh. U2.13 - Prob. 4ECh. U2.13 - Prob. 5ECh. U2.14 - Prob. 1TAICh. U2.14 - Prob. 1ECh. U2.14 - Prob. 2ECh. U2.14 - Prob. 3ECh. U2.14 - Prob. 4ECh. U2.14 - Prob. 5ECh. U2.14 - Prob. 6ECh. U2.14 - Prob. 7ECh. U2.14 - Prob. 8ECh. U2.15 - Prob. 1TAICh. U2.15 - Prob. 1ECh. U2.15 - Prob. 2ECh. U2.15 - Prob. 4ECh. U2.15 - Prob. 5ECh. U2.15 - Prob. 6ECh. U2.16 - Prob. 1TAICh. U2.16 - Prob. 1ECh. U2.16 - Prob. 2ECh. U2.16 - Prob. 3ECh. U2.16 - Prob. 4ECh. U2.16 - Prob. 6ECh. U2.16 - Prob. 7ECh. U2.17 - Prob. 1TAICh. U2.17 - Prob. 1ECh. U2.17 - Prob. 2ECh. U2.17 - Prob. 3ECh. U2.17 - Prob. 4ECh. U2.17 - Prob. 5ECh. U2.17 - Prob. 6ECh. U2.17 - Prob. 7ECh. U2.17 - Prob. 8ECh. U2.17 - Prob. 9ECh. U2.18 - Prob. 1TAICh. U2.18 - Prob. 1ECh. U2.18 - Prob. 2ECh. U2.18 - Prob. 3ECh. U2.18 - Prob. 4ECh. U2.18 - Prob. 5ECh. U2.18 - Prob. 6ECh. U2.18 - Prob. 7ECh. U2.19 - Prob. 1TAICh. U2.19 - Prob. 1ECh. U2.19 - Prob. 2ECh. U2.19 - Prob. 3ECh. U2.19 - Prob. 4ECh. U2.19 - Prob. 5ECh. U2.19 - Prob. 6ECh. U2.19 - Prob. 7ECh. U2.19 - Prob. 8ECh. U2.19 - Prob. 9ECh. U2.20 - Prob. 1TAICh. U2.20 - Prob. 1ECh. U2.20 - Prob. 2ECh. U2.20 - Prob. 3ECh. U2.20 - Prob. 4ECh. U2.20 - Prob. 5ECh. U2.20 - Prob. 6ECh. U2.20 - Prob. 7ECh. U2.20 - Prob. 8ECh. U2.20 - Prob. 9ECh. U2.21 - Prob. 1TAICh. U2.21 - Prob. 1ECh. U2.21 - Prob. 2ECh. U2.21 - Prob. 3ECh. U2.21 - Prob. 4ECh. U2.21 - Prob. 6ECh. U2 - Prob. SI1RECh. U2 - Prob. SI2RECh. U2 - Prob. SI3RECh. U2 - Prob. SI4RECh. U2 - Prob. SI5RECh. U2 - Prob. SII1RECh. U2 - Prob. SII2RECh. U2 - Prob. SII3RECh. U2 - Prob. SII4RECh. U2 - Prob. SII5RECh. U2 - Prob. SIII1RECh. U2 - Prob. SIII2RECh. U2 - Prob. SIII3RECh. U2 - Prob. SIII4RECh. U2 - Prob. SIII5RECh. U2 - Prob. SIII6RECh. U2 - Prob. SIV1ECh. U2 - Prob. SIV2ECh. U2 - Prob. SIV3ECh. U2 - Prob. SIV4ECh. U2 - Prob. SIV5ECh. U2 - Prob. 1RECh. U2 - Prob. 2RECh. U2 - Prob. 3RECh. U2 - Prob. 4RECh. U2 - Prob. 5RECh. U2 - Prob. 6RECh. U2 - Prob. 7RECh. U2 - Prob. 8RE
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