Chapter 4, Problem 86AE

(a)

Interpretation Introduction

Interpretation: The common properties in the given compound are to be stated.

Concept introduction: The following steps are to be followed to determine 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.

To determine: The common properties in ${\text{SO}}_3,{\text{NO}}_3{}^{-}$ and ${\text{CO}}_3{}^{2-}$

Explanation of Solution

For ${\text{SO}}_3$ compound

The central atom in ${\text{SO}}_3$ is sulfur $\left(\text{S}\right)$ and its electronic configuration is,

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

The valence electrons in sulfur are six.

The electronic configuration of oxygen is,

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

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{S}+3\text{O}=\left(6+6\times 3\right){\text{e}}^{-}\\ =\left(24\right){\text{e}}^{-}\end{array}$

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,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{6}{2}\\ =3\end{array}$

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

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

For ${\text{NO}}_3{}^{-}$ ion

The central atom in ${\text{NO}}_3{}^{-}$ is nitrogen $\left(\text{N}\right)$ and its electronic configuration is,

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

The valence electrons in nitrogen are five.

The electronic configuration of oxygen is,

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

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{N}+3{\text{O+1e}}^{-}=\left(5+6\times 3+1\right){\text{e}}^{-}\\ =\left(24\right){\text{e}}^{-}\end{array}$

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,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+1}{2}\\ =3\end{array}$

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

The Lewis structure of ${\text{NO}}_3{}^{-}$ is,

For ${\text{CO}}_3{}^{2-}$ ion

The central atom in ${\text{CO}}_3{}^{2-}$ is carbon $\left(\text{C}\right)$ and its electronic configuration is,

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

The valence electrons in carbon are four.

The electronic configuration of oxygen is,

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

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{C}+3{\text{O+2e}}^{-}=\left(4+6\times 3+2\right){\text{e}}^{-}\\ =\left(24\right){\text{e}}^{-}\end{array}$

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,

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

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

The Lewis structure of ${\text{CO}}_3{}^{2-}$ is,

Hence, it is clear from above calculation that molecules ${\text{SO}}_3,{\text{NO}}_3{}^{-}$ and ${\text{CO}}_3{}^{2-}$ have same hybridization and molecular shape. Moreover these entire compounds contain resonance structure.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.

The given compounds are common in molecular shape and hybridization.

(b)

Interpretation Introduction

Interpretation: The common properties in the given compound are to be stated.

Concept introduction: The following steps are to be followed to determine 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.

To determine: The common properties in ${\text{O}}_3,{\text{SO}}_2$ and ${\text{NO}}_2{}^{-}$

Explanation of Solution

For ${\text{SO}}_2$ compound

The central atom in ${\text{SO}}_2$ is sulfur $\left(\text{S}\right)$ and its electronic configuration is,

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

The valence electrons in sulfur are six.

The electronic configuration of oxygen is,

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

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{S}+2\text{O}=\left(6+6\times 2\right){\text{e}}^{-}\\ =\left(18\right){\text{e}}^{-}\end{array}$

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,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{6}{2}\\ =3\end{array}$

This means that the central atom shows ${\text{sp}}^2$ hybridization and should have a trigonal planar geometry. But the lone pair present on sulfur atom the structure is transformed into bent shape.

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

For ${\text{NO}}_2{}^{-}$ ion

The central atom in ${\text{NO}}_2{}^{-}$ is nitrogen $\left(\text{N}\right)$ and its electronic configuration is,

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

The valence electrons in nitrogen are five.

The electronic configuration of oxygen is,

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

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}\text{N}+2{\text{O+1e}}^{-}=\left(5+6\times 2+1\right){\text{e}}^{-}\\ =\left(18\right){\text{e}}^{-}\end{array}$

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,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{5+1}{2}\\ =3\end{array}$

This means that the central atom shows ${\text{sp}}^2$ hybridization and should have a trigonal planar geometry. But the lone pair present on sulfur atom the structure is transformed into bent shape.

The Lewis structure of ${\text{NO}}_2{}^{-}$ is,

For ${\text{O}}_3$ compound

The electronic configuration of oxygen $\left(\text{O}\right)$ is,

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

The valence elctrons in oxygen is six.

The total number of valence electrons are,

$\begin{array}{c}3\text{O}=\left(6\times 3\right){\text{e}}^{-}\\ =\left(18\right){\text{e}}^{-}\end{array}$

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,

$\begin{array}{c}X=\frac{V+M\pm C}{2}\\ X=\frac{6}{2}\\ =3\end{array}$

This means that the central atom shows ${\text{sp}}^2$ hybridization and should have a trigonal planar geometry. But the lone pair present on sulfur atom the structure is transformed into bent shape.

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

Hence, it is clear from above calculation that molecules ${\text{O}}_3,{\text{SO}}_2$ and ${\text{NO}}_2{}^{-}$ have same hybridization and molecular shape. Moreover these entire compounds contain resonance structure.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The molecular shape can be predicted on the basis of the Lewis structure of the given compound. The lone pairs are depicted in the Lewis structure diagrams.

The given compounds are common in molecular shape and hybridization.