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Chapter 5, Problem 5.87EP

Successive substitution of F atoms for H atoms in the molecule CH4 produces the molecules CH3F, CH2F2, CHF3, and CF4.

  1. a. Draw Lewis structures for each of the five molecules.
  2. b. Using VSEPR theory, predict the geometry of each of the five molecules.
  3. c. Specify the polarity (polar or nonpolar) for each of the five molecules.

(a)

Expert Solution
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Interpretation Introduction

Interpretation:

Lewis structures for the five molecules that are formed by successive substitution of F atom instead of H atom in CH4 has to be drawn.

Concept Introduction:

Lewis structure clearly depicts the bonding and nonbonding electrons in the atom.  This is only partially useful for the molecule that contains one or more multiple bonds and when coordinate covalent bond is present in the molecule.  For drawing Lewis structure a systematic procedure is followed.  They are,

  • The total number of valence electrons that is present in molecule is calculated by adding all the valence electrons of the atoms present in the molecule.
  • The chemical symbols for the atoms that is present in the molecule is written in the order that they are bonded.  After this a single covalent bond is placed between each atoms as two electrons.
  • The nonbonding electrons are added to each atom that is bonded to the central atom so that it contains octet of electrons.  For hydrogen alone the “octet” is only two electrons.
  • The remaining electrons has to be placed on the central atom in the structure.
  • If there is no octet of electrons present in the central atom, then use one or more pairs of nonbonding electrons that is bonded to the central atom to form double or triple bonds.
  • The total number of electrons has to be counted and it has to be confirmed whether the count is same as that of the number of valence electrons that is available for bonding.

Explanation of Solution

Given molecules are CH4, CH3F, CH2F2, CHF3, and CF4.

For CH4:

Given molecule is CH4.  Carbon belongs to Group IVA and it contains four valence electrons.  Hydrogen belongs to Group IA and it contains only one valence electron.  The total number of valence electrons present in CH4 molecule is 8.  These atoms are written order they are bonded.  After the atomic symbol is written, two dots are placed between two atoms and this means a single covalent bond is present.  This can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  1

All the atoms present in the above structure contains octet of electrons.  The total number of electron dots present in the above structure is 8 and it is same as the valence electrons of CH4 molecule.  The Lewis structure for CH4 is drawn as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  2

For CH3F:

Given molecule is CH3F.  Carbon belongs to Group IVA and it contains four valence electrons.  Hydrogen belongs to Group IA and it contains only one valence electron.  Fluorine belongs to Group VIIA and it contains seven valence electrons.  The total number of valence electrons present in CH3F molecule is 14.  These atoms are written order they are bonded.  After the atomic symbol is written, two dots are placed between two atoms and this means a single covalent bond is present.  This can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  3

Nonbonding electrons are added to the fluorine atom.  The resulting structure is,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  4

All the atoms present in the above structure contains octet of electrons.  The total number of electron dots present in the above structure is 14 and it is same as the valence electrons of CH3F molecule.  The Lewis structure for CH3F is drawn as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  5

For CH2F2:

Given molecule is CH2F2.  Carbon belongs to Group IVA and it contains four valence electrons.  Hydrogen belongs to Group IA and it contains only one valence electron.  Fluorine belongs to Group VIIA and it contains seven valence electrons.  The total number of valence electrons present in CH2F2 molecule is 20.  These atoms are written order they are bonded.  After the atomic symbol is written, two dots are placed between two atoms and this means a single covalent bond is present.  This can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  6

Nonbonding electrons are added to the fluorine atom.  The resulting structure is,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  7

All the atoms present in the above structure contains octet of electrons.  The total number of electron dots present in the above structure is 20 and it is same as the valence electrons of CH2F2 molecule.  The Lewis structure for CH2F2 is drawn as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  8

For CHF3:

Given molecule is CHF3.  Carbon belongs to Group IVA and it contains four valence electrons.  Hydrogen belongs to Group IA and it contains only one valence electron.  Fluorine belongs to Group VIIA and it contains seven valence electrons.  The total number of valence electrons present in CHF3 molecule is 26.  These atoms are written order they are bonded.  After the atomic symbol is written, two dots are placed between two atoms and this means a single covalent bond is present.  This can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  9

Nonbonding electrons are added to the fluorine atom.  The resulting structure is,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  10

All the atoms present in the above structure contains octet of electrons.  The total number of electron dots present in the above structure is 26 and it is same as the valence electrons of CHF3 molecule.  The Lewis structure for CHF3 is drawn as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  11

For CF4:

Given molecule is CF4.  Carbon belongs to Group IVA and it contains four valence electrons.  Fluorine belongs to Group VIIA and it contains seven valence electrons.  The total number of valence electrons present in CF4 molecule is 32.  These atoms are written order they are bonded.  After the atomic symbol is written, two dots are placed between two atoms and this means a single covalent bond is present.  This can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  12

Nonbonding electrons are added to the fluorine atom.  The resulting structure is,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  13

All the atoms present in the above structure contains octet of electrons.  The total number of electron dots present in the above structure is 32 and it is same as the valence electrons of CF4 molecule.  The Lewis structure for CHF3 is drawn as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  14

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Molecule geometry has to be predicted for the five molecules using VSEPR theory.

Concept Introduction:

Information about the number of bonds and types of bonds can be obtained from Lewis structure but the molecular geometry cannot be obtained.  Three dimensional arrangement of atoms in a molecule can be given by molecular geometry.  Physical and chemical properties are determined by the molecular geometry of the molecule.

Using VSEPR theory and Lewis structure, the molecular geometry of the molecule that contain less number of atoms can be predicted.  VSEPR theory uses the information from Lewis structure of the molecule to predict the molecular geometry of the molecule.  Main concept of VSEPR theory is that electron pairs that are present in the valence shell adopt arrangement in a way that minimize the repulsion between like charges.

If the central atom contains two electron pairs, then it has to be far apart means, it has to be on opposite side of the nucleus.  This means the angle has to be 180° between them.  This type of arrangement for electron pair result in linear.

If the central atom contains three electron pairs, then it has to be far apart means, it has to be on corner of a triangle.  This means the angle has to be 120° between them.  This type of arrangement for electron pair result in trigonal planar.

If the central atom contains four electron pairs, then it has to be far apart means, it has to be in a tetrahedral arrangement.  This means the angle has to be 109° between them.  This type of arrangement for electron pair result in tetrahedral.

The collection of valence electron that is present in localized region about central atom in a molecule is known as VSEPR electron group.  This may contain two electrons, four electrons, or six electrons.  The electron group that contain four and six electrons repel each other.

Tetrahedral VSEPR electron group:

The four electron pairs can be of three VSEPR electron groups.  They are 4 bonding electron groups, 3 bonding and 1 nonbonding electron groups, and 2 bonding and 2 nonbonding electron groups.  The molecular geometry that is associated with 4 bonding electron groups is tetrahedral.  The molecular geometry that is associated with 3 bonding and 1 nonbonding electron groups is trigonal pyramidal.  The molecular geometry that is associated with 2 bonding and 2 nonbonding electron groups is angular.

Trigonal planar VSEPR electron group:

The three electron pairs can be of two VSEPR electron groups.  They are 3 bonding electron groups, and 2 bonding and 1 nonbonding electron groups.  The molecular geometry that is associated with 3 bonding electron groups is trigonal planar.  The molecular geometry that is associated with 2 bonding and 1 nonbonding electron groups is angular.

Linear VSEPR electron group:

The two electron pairs can be of only one VSEPR electron groups.  It is only 2 bonding electron groups and the geometry associated with it is linear geometry.

Explanation of Solution

Given molecule is CH4.  Lewis structure of CH4 can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  15

The central atom in the above molecule is found to be carbon.  This has four bonding electron groups and zero nonbonding electron groups.  The arrangement around the central atom is tetrahedral arrangement.  Looking for molecular geometry, the central atom that contains four bonding electron groups and zero nonbonding electron groups and it has tetrahedral geometry as per VSEPR theory.

Given molecule is CH3F.  Lewis structure of CH3F can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  16

The central atom in the above molecule is found to be carbon.  This has four bonding electron groups and zero nonbonding electron groups.  The arrangement around the central atom is tetrahedral arrangement.  Looking for molecular geometry, the central atom that contains four bonding electron groups and zero nonbonding electron groups and it has tetrahedral geometry as per VSEPR theory.

Given molecule is CH2F2.  Lewis structure of CH2F2 can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  17

The central atom in the above molecule is found to be carbon.  This has four bonding electron groups and zero nonbonding electron groups.  The arrangement around the central atom is tetrahedral arrangement.  Looking for molecular geometry, the central atom that contains four bonding electron groups and zero nonbonding electron groups and it has tetrahedral geometry as per VSEPR theory.

Given molecule is CHF3.  Lewis structure of CHF3 can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  18

The central atom in the above molecule is found to be carbon.  This has four bonding electron groups and zero nonbonding electron groups.  The arrangement around the central atom is tetrahedral arrangement.  Looking for molecular geometry, the central atom that contains four bonding electron groups and zero nonbonding electron groups and it has tetrahedral geometry as per VSEPR theory.

Given molecule is CF4.  Lewis structure of CF4 can be given as,

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card, Chapter 5, Problem 5.87EP , additional homework tip  19

The central atom in the above molecule is found to be carbon.  This has four bonding electron groups and zero nonbonding electron groups.  The arrangement around the central atom is tetrahedral arrangement.  Looking for molecular geometry, the central atom that contains four bonding electron groups and zero nonbonding electron groups and it has tetrahedral geometry as per VSEPR theory.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The five molecules that are given has to be classified as polar and nonpolar.

Concept Introduction:

Measure of the degree of inequality in attraction of the bonding electrons to the various locations present within a molecule is known as molecular polarity.  This can also be said in terms of electron attraction and that is in a molecule one part is favored than the other parts of the molecule.

If in a molecule there is an uneven distribution of electronic charges means it is known as polar molecule.  If there is a symmetrical distribution of electron charge over the molecule means it is known as nonpolar molecule.  Two factors that decide molecular polarity is bond polarity and geometry of molecule.  If a molecule is symmetrical means then there won’t be any molecular polarity because the effect given by the polar bonds may cancel out each other.

The polarity of the bonds, arrangement of the bonds determines the degree of molecular polarity.  If the electronegativity difference is more, then the molecule will be more polar.

Explanation of Solution

Given molecules are CH4, CH3F, CH2F2, CHF3, and CF4.  Molecular geometry of CH4, CH3F, CH2F2, CHF3, and CF4 are tetrahedral.  The molecules that has same atoms in all four corners of the tetrahedron will be nonpolar because the bond polarity cancel out each other.  The nonpolar molecules is found to be CH4, and CF4.  The molecules that do not have same atoms in all four corners of the tetrahedron will be polar as the bond polarity do not cancel out each other.  Hence, CH3F, CH2F2, and CHF3 will be polar.

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Chapter 5 Solutions

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card

Ch. 5.3 - Prob. 1QQCh. 5.3 - Prob. 2QQCh. 5.3 - Prob. 3QQCh. 5.3 - Prob. 4QQCh. 5.3 - Prob. 5QQCh. 5.4 - Prob. 1QQCh. 5.4 - Prob. 2QQCh. 5.4 - Prob. 3QQCh. 5.5 - Which of the following is an incorrect statement...Ch. 5.5 - Prob. 2QQCh. 5.5 - Prob. 3QQCh. 5.6 - Prob. 1QQCh. 5.6 - Prob. 2QQCh. 5.6 - Prob. 3QQCh. 5.6 - Prob. 4QQCh. 5.6 - Prob. 5QQCh. 5.7 - Prob. 1QQCh. 5.7 - Prob. 2QQCh. 5.7 - Prob. 3QQCh. 5.8 - Prob. 1QQCh. 5.8 - In VSEPR theory, an angular molecular geometry is...Ch. 5.8 - Prob. 3QQCh. 5.8 - Prob. 4QQCh. 5.8 - Prob. 5QQCh. 5.9 - Prob. 1QQCh. 5.9 - Prob. 2QQCh. 5.9 - Prob. 3QQCh. 5.9 - Prob. 4QQCh. 5.10 - Prob. 1QQCh. 5.10 - Prob. 2QQCh. 5.10 - Prob. 3QQCh. 5.10 - As the difference in electronegativity between two...Ch. 5.10 - Prob. 5QQCh. 5.10 - Prob. 6QQCh. 5.11 - Prob. 1QQCh. 5.11 - Prob. 2QQCh. 5.11 - Prob. 3QQCh. 5.11 - Prob. 4QQCh. 5.11 - Prob. 5QQCh. 5.12 - Prob. 1QQCh. 5.12 - Prob. 2QQCh. 5.12 - Prob. 3QQCh. 5.12 - Prob. 4QQCh. 5.12 - Prob. 5QQCh. 5.12 - Prob. 6QQCh. 5 - Contrast the types of elements involved in ionic...Ch. 5 - Contrast the mechanisms by which ionic and...Ch. 5 - Prob. 5.3EPCh. 5 - Prob. 5.4EPCh. 5 - Indicate whether or not covalent bond formation is...Ch. 5 - Indicate whether or not covalent bond formation is...Ch. 5 - Draw Lewis structures to illustrate the covalent...Ch. 5 - Draw Lewis structures to illustrate the covalent...Ch. 5 - How many nonbonding electron pairs are present in...Ch. 5 - How many nonbonding electron pairs are present in...Ch. 5 - The component elements for four binary molecular...Ch. 5 - The component elements for four binary molecular...Ch. 5 - Specify the number of single, double, and triple...Ch. 5 - Specify the number of single, double, and triple...Ch. 5 - Convert each of the Lewis structures in Problem...Ch. 5 - Convert each of the Lewis structures in Problem...Ch. 5 - Prob. 5.17EPCh. 5 - Prob. 5.18EPCh. 5 - Prob. 5.19EPCh. 5 - Identify the Period 3 nonmetal that would normally...Ch. 5 - How many valence electrons do atoms possess that...Ch. 5 - Prob. 5.22EPCh. 5 - What aspect of the following Lewis structure...Ch. 5 - What aspect of the following Lewis structure...Ch. 5 - Identify the coordinate covalent bond(s) present,...Ch. 5 - Identify the coordinate covalent bond(s) present,...Ch. 5 - Without actually drawing the Lewis structure,...Ch. 5 - Without actually drawing the Lewis structure,...Ch. 5 - Prob. 5.29EPCh. 5 - Prob. 5.30EPCh. 5 - Draw the Lewis structure for each of the molecules...Ch. 5 - Draw the Lewis structure for each of the molecules...Ch. 5 - Draw Lewis structures to illustrate the bonding in...Ch. 5 - Draw Lewis structures to illustrate the bonding in...Ch. 5 - How many electron dots should appear in the Lewis...Ch. 5 - Prob. 5.36EPCh. 5 - Draw Lewis structures for the following polyatomic...Ch. 5 - Draw Lewis structures for the following polyatomic...Ch. 5 - Draw Lewis structures for the following compounds...Ch. 5 - Draw Lewis structures for the following compounds...Ch. 5 - Draw Lewis structures for the following molecules...Ch. 5 - Draw Lewis structures for the following molecules...Ch. 5 - In which of the following pairs of diatomic...Ch. 5 - In which of the following pairs of diatomic...Ch. 5 - Prob. 5.45EPCh. 5 - What is the molecular geometry associated with...Ch. 5 - Specify the molecular geometry of each of the...Ch. 5 - Specify the molecular geometry of each of the...Ch. 5 - Prob. 5.49EPCh. 5 - Prob. 5.50EPCh. 5 - Prob. 5.51EPCh. 5 - Prob. 5.52EPCh. 5 - Prob. 5.53EPCh. 5 - Prob. 5.54EPCh. 5 - Using VSEPR theory, predict the molecular geometry...Ch. 5 - Using VSEPR theory, predict the molecular geometry...Ch. 5 - Prob. 5.57EPCh. 5 - Specify both the VSEPR electron group geometry...Ch. 5 - Prob. 5.59EPCh. 5 - Prob. 5.60EPCh. 5 - Using a periodic table, but not a table of...Ch. 5 - Using a periodic table, but not a table of...Ch. 5 - Prob. 5.63EPCh. 5 - Prob. 5.64EPCh. 5 - Place + above the atom that is relatively positive...Ch. 5 - Place + above the atom that is relatively positive...Ch. 5 - Rank the following bonds in order of increasing...Ch. 5 - Rank the following bonds in order of increasing...Ch. 5 - Classify each of the following bonds as nonpolar...Ch. 5 - Classify each of the following bonds as nonpolar...Ch. 5 - Prob. 5.71EPCh. 5 - Prob. 5.72EPCh. 5 - Fill in the blanks in each line of the following...Ch. 5 - Fill in the blanks in each line of the following...Ch. 5 - Four hypothetical elements, A, B, C, and D, have...Ch. 5 - Four hypothetical elements, A, B, C, and D, have...Ch. 5 - Indicate whether each of the following...Ch. 5 - Prob. 5.78EPCh. 5 - Indicate whether each of the following triatomic...Ch. 5 - Indicate whether each of the following triatomic...Ch. 5 - Indicate whether each of the following molecules...Ch. 5 - Indicate whether each of the following molecules...Ch. 5 - Prob. 5.83EPCh. 5 - Prob. 5.84EPCh. 5 - Indicate which molecule in each of the following...Ch. 5 - Indicate which molecule in each of the following...Ch. 5 - Successive substitution of F atoms for H atoms in...Ch. 5 - Successive substitution of F atoms for H atoms in...Ch. 5 - Prob. 5.89EPCh. 5 - Indicate whether or not each of the following...Ch. 5 - Prob. 5.91EPCh. 5 - Name the following binary molecular compounds? a....Ch. 5 - Prob. 5.93EPCh. 5 - Prob. 5.94EPCh. 5 - Prob. 5.95EPCh. 5 - Prob. 5.96EPCh. 5 - Prob. 5.97EPCh. 5 - Prob. 5.98EPCh. 5 - Write chemical formulas for the following binary...Ch. 5 - Write chemical formulas for the following binary...Ch. 5 - Prob. 5.101EPCh. 5 - Prob. 5.102EPCh. 5 - The compound whose molecles contain one atom of C...Ch. 5 - Prob. 5.104EPCh. 5 - Prob. 5.105EPCh. 5 - The correct name for the compound K2SO4 is not...Ch. 5 - Prob. 5.107EPCh. 5 - Prob. 5.108EP
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