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 4E

(a)

Interpretation Introduction

Interpretation: The dipole in the H2Se molecule needs to be explained.

Concept Introduction: The chemical compounds can be classified as covalent compounds and ionic compounds. Ionic compounds have complete negative and positive charges on it, whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of a molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to the electronegativity difference between bonded atoms, partial charges are induced on the bonded atoms. The partial charges affect the physical properties of the polar molecules.

(a)

Expert Solution
Check Mark

Answer to Problem 4E

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  1

Explanation of Solution

In H2Se molecule, Se is more electronegative element than H atom. This is because Se is a non-metal with 6 valence electrons. Therefore, Se tends to accept electrons to get octet configuration. That makes it more electronegative than H atom. Due to this difference in electronegativity of bonded atoms, Se gets a partial negative charge and H gets a partial positive charge.

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  2

(b)

Interpretation Introduction

Interpretation: The dipole in the H2 molecule needs to be explained.

Concept Introduction: Concept Introduction: The chemical compounds can be classified as covalent compounds and ionic compounds. Ionic compounds have complete negative and positive charges on it, whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of a molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to the electronegativity difference between bonded atoms, partial charges are induced on the bonded atoms. The partial charges affect the physical properties of the polar molecules.

(b)

Expert Solution
Check Mark

Answer to Problem 4E

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

No dipole exists on H2 molecule.

Explanation of Solution

In H2 molecule, both H atoms are the same and have 1 valence electron. They form a covalent bond with another H atom to get the duplet configuration. Due to the presence of the same atom, there will be no partial charge on both H atoms and the molecule will be non-polar overall with zero dipole moment.

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

(c)

Interpretation Introduction

Interpretation: The dipole in the Ar molecule needs to be explained.

Concept Introduction: Concept Introduction: The chemical compounds can be classified as covalent compounds and ionic compounds. Ionic compounds have complete negative and positive charges on it, whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of a molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to the electronegativity difference between bonded atoms, partial charges are induced on the bonded atoms. The partial charges affect the physical properties of the polar molecules.

(c)

Expert Solution
Check Mark

Answer to Problem 4E

No dipole exists on Ar molecule.

Explanation of Solution

  Ar is the monoatomic Noble gas. It has a completely filled octet configuration, therefore, it does not form any chemical bond with other atoms.

Because of no bond formation, the overall dipole on the Ar monoatomic gas is zero.

(d)

Interpretation Introduction

Interpretation: The dipole in the HOF molecule needs to be explained.

Concept Introduction: Concept Introduction: The chemical compounds can be classified as covalent compounds and ionic compounds. Ionic compounds have complete negative and positive charges on it, whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of a molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to the electronegativity difference between bonded atoms, partial charges are induced on the bonded atoms. The partial charges affect the physical properties of the polar molecules.

(d)

Expert Solution
Check Mark

Answer to Problem 4E

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  5

Explanation of Solution

In HOF molecule, H atom has 1 valence electron, whereas O and F are electronegative elements with 6 and 7 valence electrons respectively. Therefore in HOF molecule, O and F will get a partial negative charge, whereas H will get a partial positive charge.

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  6

(e)

Interpretation Introduction

Interpretation: The dipole in the CHClF2 molecule needs to be explained.

Concept Introduction: Concept Introduction: The chemical compounds can be classified as covalent compounds and ionic compounds. Ionic compounds have complete negative and positive charges on it, whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of a molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to the electronegativity difference between bonded atoms, partial charges are induced on the bonded atoms. The partial charges affect the physical properties of the polar molecules.

(e)

Expert Solution
Check Mark

Answer to Problem 4E

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  7

Explanation of Solution

In CHClF2 molecule, Cl and F are more electronegative elements compared to C and H. Therefore, C and H will get a partial positive charge, whereas Cl and F will get a partial negative charge in the molecule as given.

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  8

(f)

Interpretation Introduction

Interpretation: The dipole in the CH2O molecule needs to be explained.

Concept Introduction: Concept Introduction: The chemical compounds can be classified as covalent compounds and ionic compounds. Ionic compounds have complete negative and positive charges on it, whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of a molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to the electronegativity difference between bonded atoms, partial charges are induced on the bonded atoms. The partial charges affect the physical properties of the polar molecules.

(f)

Expert Solution
Check Mark

Answer to Problem 4E

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  9

Explanation of Solution

In CH2O molecule, O is a more electronegative element compared to C and H. Therefore, C and H will get a partial positive charge, whereas O atom will get a partial negative charge in the molecule as given.

  Living By Chemistry: First Edition Textbook, Chapter U2.18, Problem 4E , additional homework tip  10

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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