Chapter 2, Problem 2C.3E

(a)

Interpretation Introduction

Interpretation:

Lewis structure of periodate ion $\left({\text{IO}}_4^{+}\right)$ has to be drawn.

Concept Introduction:

Lewis structure represents covalent bonds and describes valence electrons configuration of atoms. The covalent bonds are depicted by lines and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

• The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
• Total valence electrons are estimated for each atom.
• A single bond is first placed between each atom pair.
• The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
• Add charge on overall structure in case of polyatomic cation or anion.

Explanation of Solution

The molecule ${\text{IO}}_4^{-}$ consists of one $\text{I}$ atom and four $\text{O}$ atoms. Since $\text{I}$ is least electronegative and is drawn as central atom.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

The symbol for iodine is $\text{I}$; with atomic number 53. Its electronic configuration is $\left[\text{K}\text{}\text{r}\right]4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^5$. Thus, it possesses 7 valence electrons.

One negative charge on molecule is added up as one valence electron in total count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{IO}}_4^{-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7+6\left(4\right)+1\\ =32\left(16\text{pairs}\right)\end{array}$

The skeleton structure in ${\text{IO}}_4^{-}$ has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=32-8\\ =24\left(12\text{ pairs}\right)\end{array}$

To complete the valence electrons of iodine, it forms double bond with three oxygen atoms.

  $\begin{array}{c}\text{Total remaining electrons}=24-6\\ =18\left(9\text{ pairs}\right)\end{array}$

Hence, the Lewis structure for ${\text{IO}}_4^{-}$ is as follows:

To complete the octet of iodine atom the negative charge is delocalized on each oxygen atom and since there are four oxygen atoms, therefore, four resonance structures will be formed and possible resonance structures are as follows:

(b)

Interpretation Introduction

Interpretation:

Lewis structure of hydrogen phosphate ion $\left({\text{HPO}}_4^{2-}\right)$ has to be drawn.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Hydrogen phosphate ion $\left({\text{HPO}}_4^{2-}\right)$ consists of one $\text{H}$ atom and four $\text{O}$ atoms.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

The symbol for hydrogen is $\text{H}$; with atomic number 1. Its electronic configuration is $1{\text{s}}^1$. Thus it possesses 1 valence electron.

The symbol for phosphorus is $\text{P}$; with atomic number 15. Its electronic configuration is $\left[\text{Ne}\right]3{\text{s}}^{2}3{\text{p}}^3$. Thus, it possesses 5 valence electrons.

Two negative charges on molecule are added up as one valence electron in total count.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{HPO}}_4^{2-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=1+6\left(4\right)+5+2\\ =32\left(16\text{pairs}\right)\end{array}$

The skeleton structure in ${\text{HPO}}_4^{2-}$ has five bonds that comprise 10 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=32-10\\ =22\left(11\text{ pairs}\right)\end{array}$

To complete the valence electrons of phosphorous, it forms double bond with one oxygen atom.

  $\begin{array}{c}\text{Total remaining electrons}=22-2\\ =20\left(\text{10 pairs}\right)\end{array}$

Hence, 20 electrons are allocated as 10 lone pairs on remaining oxygen atoms to complete their octet. The Lewis structure is as follows:

The negative charge is delocalized on three oxygen atoms and therefore three resonance structures will be formed and possible resonance structures are as follows:

(c)

Interpretation Introduction

Interpretation:

Lewis structure of chloric acid $\left({\text{HClO}}_3\right)$ has to be drawn:

Concept Introduction:

Refer to part (a).

Explanation of Solution

Chloric acid $\left({\text{HClO}}_3\right)$ consists of one $\text{H}$ atom, three $\text{O}$ atoms, and one $\text{Cl}$ atom.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

The symbol for hydrogen is $\text{H}$; with atomic number 1. Its electronic configuration is $1{\text{s}}^1$. Thus it possesses 1 valence electron.

The symbol for chlorine is $\text{Cl}$; with atomic number 17. Its electronic configuration is $\left[\text{Ne}\right]3{\text{s}}^{2}3{\text{p}}^5$. Thus, it possesses 7 valence electrons.

Thus total valence electrons are sum of the valence electrons for each atom in ${\text{HClO}}_3$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=1+6\left(3\right)+7\\ =26\left(\text{13 pairs}\right)\end{array}$

The skeleton structure in ${\text{HClO}}_3$ has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-8\\ =18\left(\text{9 pairs}\right)\end{array}$

To complete the valence electrons of chlorine, it forms double bond with two oxygen atoms.

  $\begin{array}{c}\text{Total remaining electrons}=18-4\\ =14\left(\text{7 pairs}\right)\end{array}$

Hence, 14 electrons are allocated as 6 lone pairs on remaining oxygen atoms and 1 lone pair on chlorine to complete their respective octet. The Lewis structure is as follows:

The lone pair on oxygen atom that is attached to chlorine participates in resonance with that and produces $+1$ charge on chlorine and $-1$ on oxygen atom. Since there are two oxygen atoms, therefore two resonating structures are drawn as follows:

(d)

Interpretation Introduction

Interpretation:

Lewis structure of arsenate ion ${\left({\text{AsO}}_4\right)}^{-3}$ and its resonance structures has to be drawn:

Concept Introduction:

Refer to part (a).

Explanation of Solution

Arsenate ion ${\left({\text{AsO}}_4\right)}^{-3}$ consists of one $\text{As}$ atom and four $\text{O}$ atoms.

The symbol for oxygen is $\text{O}$; with atomic number 8. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^4$. Thus, it possesses 6 valence electrons.

The symbol for arsenic is $\text{As}$; with atomic number 33. Its electronic configuration is $\left[\text{Ar}\right]3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^3$. Thus it possesses 5 valence electrons.

The symbol for chlorine is $\text{Cl}$; with atomic number 17. Its electronic configuration is $\left[\text{Ne}\right]3{\text{s}}^{2}3{\text{p}}^5$. Thus, it possesses 7 valence electrons.

Three negative charges on molecule is added up as three valence electrons in total count. Thus total valence electrons are sum of the valence electrons for each atom in arsenate ion ${\left({\text{AsO}}_4\right)}^{-3}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5+6\left(4\right)+3\\ =32\left(\text{16 pairs}\right)\end{array}$

The skeleton structure in arsenate ion ${\left({\text{AsO}}_4\right)}^{-3}$ has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=32-8\\ =24\left(\text{12 pairs}\right)\end{array}$

To complete the valence electrons of arsenic, it forms double bond with one oxygen atom.

  $\begin{array}{c}\text{Total remaining electrons}=24-2\\ =22\left(\text{11 pairs}\right)\end{array}$

Hence, 22 electrons are allocated as 11 lone pairs on remaining oxygen atoms to complete their octet. The Lewis structure is as follows:

The negative charge is delocalized on each oxygen atom and since there are four oxygen atoms, therefore, four resonance structures will be formed and possible resonance structures are as,