Chapter 9, Problem 33E

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals of the central atom, and predict the overall polarity.

a. CF₄

b. NF₃

c. OF₂

d. BF₃

e. BeH₂

f. TeF₄

g. AsF₅

h. KrF₂

i. KrF₄

j. SeF₆

k. IF₅

l. IF₃

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{CF}}_{\text{4}}$.

Explanation of Solution

The atomic number of carbon is $6$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence electrons present in carbon are four.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is 7

The molecule ${\text{CF}}_{\text{4}}$ is made of four fluorine atoms and one carbon atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{C}+4\text{F}\text{=}\left(4+\left(4\times 7\right)\right){\text{e}}^{-}\\ \text{=}{\text{32e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{4+4}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a tetrahedral geometry.

The Lewis structure of ${\text{CF}}_{\text{4}}$ is,

Figure 1

The Lewis structure of the given molecule corresponds to a tetrahedral molecular structure.

The resultant of any three $\text{C}-\text{F}$ bonds is equal in magnitude but opposite in direction to the fourth $\text{C}-\text{F}$ bond. Hence, the molecule is non-polar.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{NF}}_3$.

Explanation of Solution

The atomic number of nitrogen is $7$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^3$

The valence electrons present in nitrogen are five.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{NF}}_3$ is made of three fluorine atoms and one nitrogen atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{N}+3\text{F}\text{=}\left(5+\left(3\times 7\right)\right){\text{e}}^{-}\\ \text{=}{\text{26e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+3}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a tetrahedral geometry.

The Lewis structure of ${\text{NF}}_3$ is,

Figure 2

The Lewis structure of the given molecule corresponds to a triangular pyramidal molecular structure.

The given molecule is polar and the bond angle present is $109.5°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{OF}}_2$.

Explanation of Solution

The atomic number of oxygen is $8$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

The valence electrons present in nitrogen are six.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{OF}}_2$ is made of two fluorine atoms and one oxygen atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{O}+2\text{F}\text{=}\left(6+\left(2\times 7\right)\right){\text{e}}^{-}\\ \text{=}{\text{20e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{6+2}{2}\\ =4\end{array}$

This means that the central atom shows ${\text{sp}}^3$ hybridization and should have a tetrahedral geometry.

The Lewis structure of ${\text{OF}}_2$ is,

Figure 3

The Lewis structure of the given molecule corresponds to a V-shaped molecular structure.

The given molecule is polar and the bond angle present is $109.5°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{BF}}_3$.

Explanation of Solution

The atomic number of boron is $5$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^1$

The valence electrons present in boron are three.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{BF}}_3$ is made of three fluorine atoms and one boron atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{B}+3\text{F}\text{=}\left(3+\left(3\times 7\right)\right){\text{e}}^{-}\\ \text{=}{\text{24e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{3+3}{2}\\ =3\end{array}$

This means that the central atom shows ${\text{sp}}^2$ hybridization and should have a triangular planar geometry.

The Lewis structure of ${\text{BF}}_3$ is,

Figure 4

The Lewis structure of the given molecule corresponds to a triangular planar molecular structure.

The given molecule is non-polar and the bond angle present is $120°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{BeH}}_2$.

Explanation of Solution

The atomic number of beryllium is $4$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}$

The valence electrons present in beryllium are two.

The atomic number of hydrogen is $1$ and its electronic configuration is,

${\text{1s}}^1$

Hydrogen has only one electron.

The molecule ${\text{BeH}}_2$ is made of two hydrogen atoms and one beryllium atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{Be}+2\text{H}\text{=}\left(2+\left(2\times 1\right)\right){\text{e}}^{-}\\ \text{=}{\text{4e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{2+2}{2}\\ =2\end{array}$

This means that the central atom shows $\text{sp}$ hybridization and should have a linear geometry.

The Lewis structure of ${\text{BeH}}_2$ is,

Figure 5

The Lewis structure of the given molecule corresponds to a triangular planar molecular structure.

The given molecule is non-polar (because of its linear structure) and the bond angle present is $180°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{TeF}}_4$.

Explanation of Solution

The atomic number of tellurium is $52$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{4}}$

The valence electrons present in tellurium are six.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{TeF}}_4$ is made of four fluorine atoms and one tellurium atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{Te}+4\text{F}\text{=}\left(6+\left(4\times 7\right)\right){\text{e}}^{-}\\ \text{=}34{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{6+4}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3\text{d}$ hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of ${\text{TeF}}_4$ is,

Figure 6

The Lewis structure of the given molecule corresponds to a see-saw shaped molecular structure.

The given molecule is polar (as the equatorial dipoles do not get cancelled) and the bond angle for $\text{F}-\text{Te}-\text{F}$ is $90°$ and $120°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{AsF}}_5$.

Explanation of Solution

The atomic number of arsenic is $33$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^3$

The valence electrons present in arsenic are five.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{AsF}}_5$ is made of five fluorine atoms and one arsenic atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{As}+5\text{F}\text{=}\left(5+\left(5\times 7\right)\right){\text{e}}^{-}\\ \text{=}40{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+5}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3\text{d}$ hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of ${\text{AsF}}_5$ is,

Figure 7

The Lewis structure of the given molecule corresponds to a trigonal bipyramidal molecular structure.

The given molecule is non-polar (as the axial dipoles being opposite in direction get cancelled; and the resultant of any two equatorial $\text{As}-\text{F}$ bond is equal in magnitude but opposite in direction to the third) and the bond angle for $\text{F}-\text{As}-\text{F}$ is $90°$ and $120°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{KrF}}_2$.

Explanation of Solution

The atomic number of krypton is $36$ and its electronic configuration is,

$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6$

The valence electrons present in krypton are eight.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{KrF}}_2$ is made of two fluorine atoms and one krypton atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{Kr}+2\text{F}\text{=}\left(8+\left(2\times 7\right)\right){\text{e}}^{-}\\ \text{=}22{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8+2}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3\text{d}$ hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of ${\text{KrF}}_2$ is,

Figure 8

The Lewis structure of the given molecule corresponds to a linear molecular structure.

The given molecule is non-polar (because of its linear structure) and the bond angle present is $180°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{KrF}}_4$.

Explanation of Solution

The atomic number of krypton is $36$ and its electronic configuration is,

$1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6$

The valence electrons present in krypton are eight.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{KrF}}_4$ is made of two fluorine atoms and one krypton atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{Kr}+4\text{F}\text{=}\left(8+\left(4\times 7\right)\right){\text{e}}^{-}\\ \text{=}36{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{8+4}{2}\\ =6\end{array}$

This means that the central atom shows ${\text{sp}}^3{\text{d}}^2$ hybridization and should have a square bipyramidal geometry.

The Lewis structure of ${\text{KrF}}_4$ is,

Figure 9

The Lewis structure of the given molecule corresponds to a square planar molecular structure.

The given molecule is non-polar (the dipoles get cancelled) and the bond angle present is $90°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{SeF}}_6$.

Explanation of Solution

The atomic number of selenium is $34$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^4$

The valence electrons present in selenium are six.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{SeF}}_6$ is made of six fluorine atoms and one selenium atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{Se}+6\text{F}\text{=}\left(6+\left(6\times 7\right)\right){\text{e}}^{-}\\ \text{=}48{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{6+6}{2}\\ =6\end{array}$

This means that the central atom shows ${\text{sp}}^3{\text{d}}^2$ hybridization and should have a square bipyramidal geometry.

The Lewis structure of ${\text{SeF}}_6$ is,

Figure 10

The Lewis structure of the given molecule corresponds to a square bipyramidal molecular structure.

The given molecule is non-polar (the dipoles get cancelled) and the bond angle present is $90°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{IF}}_5$.

Explanation of Solution

The atomic number of iodine is $53$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{5}}$

The valence electrons present in iodine are seven.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{IF}}_5$ is made of five fluorine atoms and one iodine atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{I}+5\text{F}\text{=}\left(7+\left(5\times 7\right)\right){\text{e}}^{-}\\ \text{=}42{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{7+5}{2}\\ =6\end{array}$

This means that the central atom shows ${\text{sp}}^3{\text{d}}^2$ hybridization and should have a square bipyramidal geometry.

The Lewis structure of ${\text{IF}}_5$ is,

Figure 11

The Lewis structure of the given molecule corresponds to a square pyramidal molecular structure.

The given molecule is polar (the dipoles do not get cancelled, it is polar because of the axial dipole) and the bond angle present is $90°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

• The central atom is identified.
• Its valence electrons are determined.
• $1{\text{e}}^{-}$ is added to the total number of valence electrons for each monovalent atom present.
• The total number obtained is divided by $2$, to find the number of electron pairs.
• This further gives us the hybridization of the given compound.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for ${\text{IF}}_3$.

Explanation of Solution

The atomic number of iodine is $53$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{5}}$

The valence electrons present in iodine are seven.

The atomic number of fluorine is $9$ and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

The valence electron of fluorine is $7$.

The molecule ${\text{IF}}_3$ is made of five fluorine atoms and one iodine atom. Hence, the total number of valence electrons is,

$\begin{array}{c}\text{I}+3\text{F}\text{=}\left(7+\left(3\times 7\right)\right){\text{e}}^{-}\\ \text{=}28{\text{e}}^{-}\end{array}$

The formula of number of electron pairs is,

$X=\frac{V+M\pm C}{2}$

Where,

• $X$ is number of electron pairs.
• $V$ is valence electrons of central atom.
• $M$ is number of monovalent atoms.
• $C$ is charge on compound.

The number of electron pairs is,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{7+3}{2}\\ =5\end{array}$

This means that the central atom shows ${\text{sp}}^3\text{d}$ hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of ${\text{IF}}_3$ is,

Figure 12

The Lewis structure of the given molecule corresponds to a trigonal bipyramidal molecular structure.

The given molecule is non-polar (as the axial dipoles being opposite in direction get cancelled; and the resultant of any two equatorial $\text{I}-\text{F}$ bond is equal in magnitude but opposite in direction to the third) and the bond angle for $\text{F}-\text{I}-\text{F}$ is $90°$ and $120°$.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.