Chapter 3, Problem 82E

Write Lewis structures that obey the octet rule for each of the following molecules and ions. (In each case the first atom listed is the central atom.)

a. POCl₃, SO₄²⁻, XeO₄, PO₄³⁻, ClO₄⁻

b. NF₃, SO₃²⁻, PO₃³⁻. ClO₃⁻

c. ClO₂⁻, SCl₂, PCl₂⁻

d. Considering your answers to parts a, b, and c, what conclusions can you draw concerning the structures of species containing the same number of atoms and the same number of valence electrons?

Interpretation Introduction

Interpretation: The Lewis structure is to be drawn for the given molecules.

Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.

The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.

Explanation of Solution

(a)

To determine: The Lewis structure of the molecule ${\text{POCl}}_3,{\text{SO}}_4{}^{2-},{\text{XeO}}_4,{\text{PO}}_4{}^{3-}$. and ${\text{ClO}}_{\text{4}}{}^{-}$

The Lewis structures of the given molecules are as follows.

• To determine: The Lewis structure of the molecule ${\text{POCl}}_3$.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of phosphorous $\left(\text{P}\right)$ is $\text{15}$ and its electronic configuration is,

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

The valence electron of phosphorous is $\text{5}$

The atomic number of chlorine $\left(\text{Cl}\right)$ is $\text{17}$ and its electronic configuration is,

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

The valence electron of chlorine is $\text{7}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{POCl}}_3$ is made of three chlorine atoms and one phosphorus and oxygen atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{P}\text{+}\text{O}+3\text{Cl}\text{=}5\text{+}6+3\times 7\\ \text{=}\text{32}\end{array}$

Each chlorine atom requires one electron to complete the octet, whereas oxygen and phosphorus atoms require 2 and 3 electrons to complete their octet. In the molecule ${\text{POCl}}_3$, phosphorus is central atom which is placed at the center whereas oxygen and hydrogen atoms are arranged on the sides Hence, the mutual sharing of eight electrons takes place. The 24 valence electrons present are placed as lone pairs in which each chlorine and oxygen atoms get three lone pairs.

The Lewis structure of ${\text{POCl}}_3$ is

Figure 1

• To determine: The Lewis structure of the molecule ${\text{SO}}_4{}^{2-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of sulfur $\left(\text{S}\right)$ is $\text{16}$ and its electronic configuration is,

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

The valence electron of sulfur is $\text{6}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{SO}}_4{}^{2-}$ is made of four oxygen atoms and one sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S+}4\text{O}+2{\text{e}}^{-}\text{=}6\text{+}4\times 6+2\\ \text{=}\text{32}\end{array}$

Each oxygen atom and sulfur requires two electrons to complete the octet In the molecule ${\text{SO}}_4{}^{2-}$, sulfur is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of eight electrons takes place. The 24 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs.

The Lewis structure of ${\text{SO}}_4{}^{2-}$ is

Figure 2

• To determine: The Lewis structure of the molecule ${\text{XeO}}_4$.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of xenon $\left(\text{Xe}\right)$ is $\text{54}$ and its electronic configuration is,

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

The valence electron of xenon is $\text{8}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{XeO}}_4$ is made of four oxygen atoms and xenon atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Xe+}4\text{O}+\text{=}8\text{+}4\times 6+2\\ \text{=}\text{32}\end{array}$

Each oxygen atom requires two electrons to complete the octet In the molecule ${\text{XeO}}_4$, xenon is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of eight electrons takes place. The 24 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs.

The Lewis structure of ${\text{XeO}}_4$ is

Figure 3

• To determine: The Lewis structure of the molecule ${\text{PO}}_4{}^{3-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. . The atomic number of phosphorous is $\text{15}$ and its electronic configuration is,

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

The valence electron of phosphorous is $\text{5}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{PO}}_4{}^{3-}$ is made of four oxygen atoms and phosphorous atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{P+}4\text{O}+3\text{e}\text{=}5\text{+}4\times 6+3\\ \text{=}\text{32}\end{array}$

Each oxygen atom requires two electrons to complete the octet whereas phosphorous atom requires three electrons to complete the octet. In the molecule ${\text{PO}}_4{}^{3-}$, phosphorous is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of eight electrons takes place. The 24 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs.

The Lewis structure of ${\text{PO}}_4{}^{3-}$ is

Figure 4

• To determine: The Lewis structure of the molecule ${\text{ClO}}_4{}^{-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. . The atomic number of chlorine is $\text{17}$  and its electronic configuration is,

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

The valence electron of chlorine is $\text{7}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{ClO}}_4{}^{-}$ is made of four oxygen atoms and chlorine atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Cl+}4\text{O}+1\text{e}\text{=}7\text{+}4\times 6+1\\ \text{=}\text{32}\end{array}$

Each oxygen atom requires two electrons to complete the octet whereas chlorine atom requires single electron to complete the octet. In the molecule ${\text{ClO}}_4{}^{-}$, chlorine is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of eight electrons takes place. The 24 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs.

The Lewis structure of ${\text{ClO}}_4{}^{-}$ is

Figure 5

(b)

To determine: The Lewis structure of the molecule ${\text{NF}}_3,{\text{SO}}_3{}^{2-},{\text{PO}}_3{}^{3-}$. and ${\text{ClO}}_3{}^{-}$

The Lewis structures of the given molecules are as follows.

• To determine: The Lewis structure of the molecule ${\text{NF}}_3$.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of nitrogen $\left(\text{N}\right)$ is $\text{7}$ and its electronic configuration is,

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

The valence electron of nitrogen is $\text{5}$

The atomic number of fluorine $\left(\text{F}\right)$ is $\text{9}$ and its electronic configuration is,

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

The valence electron of fluorine is $\text{7}$

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

$\begin{array}{c}3\text{F}\text{+}\text{1N}\text{=}\text{3}\times \text{7}\text{+}5\\ \text{=}\text{26}\end{array}$

Each fluorine atom requires one electron to complete the octet whereas nitrogen atom requires three valence electrons to complete the octet. Hence, the mutual sharing of six electrons takes place. The 20 valence electrons left are placed as lone pairs in such a way that each fluorine atom gets three lone pairs and nitrogen atom gets single lone pair.

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

Figure 6

• To determine: The Lewis structure of the molecule ${\text{SO}}_3{}^{2-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of sulfur is $\text{16}$ and its electronic configuration is,

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

The valence electron of sulfur is $\text{6}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{SO}}_3{}^{2-}$ is made of three oxygen atoms and one sulfur atom. Also oxygen atom contains two negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{S+}3\text{O}+2{\text{e}}^{-}\text{=}6\text{+}3\times 6+2\\ \text{=}26\end{array}$

Each oxygen atom and sulfur requires two electrons to complete the octet In the molecule ${\text{SO}}_3{}^{2-}$, sulfur is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of six electrons takes place. The 20 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs and sulfur gets one lone pair.

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

Figure 7

• To determine: The Lewis structure of the molecule ${\text{PO}}_3{}^{3-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. . The atomic number of phosphorous is $\text{15}$ and its electronic configuration is,

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

The valence electron of phosphorous is $\text{5}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{PO}}_3{}^{3-}$ is made of three oxygen atoms and phosphorous atom. Also oxygen atoms contain three negative charges; hence, the total number of valence electrons is,

$\begin{array}{c}\text{P+}3\text{O}+3{\text{e}}^{-}\text{=}5\text{+}3\times 6+3\\ \text{=}26\end{array}$

Each oxygen atom requires two electrons to complete the octet whereas phosphorous atom requires three electrons to complete the octet. In the molecule ${\text{PO}}_3{}^{3-}$, phosphorous is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of six electrons takes place. The 20 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs and phosphorous atom gets one lone pair.

The Lewis structure of ${\text{PO}}_3{}^{3-}$ is

Figure 8

• To determine: The Lewis structure of the molecule ${\text{ClO}}_3{}^{-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. . The atomic number of chlorine is $\text{17}$ and its electronic configuration is,

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

The valence electron of chlorine is $\text{7}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{ClO}}_3{}^{-}$ is made of three oxygen atoms and chlorine atom. Also oxygen atoms contain three negative charges; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Cl+}3\text{O}+1{\text{e}}^{-}\text{=}7\text{+}3\times 6+1\\ \text{=}26\end{array}$

Each oxygen atom requires two electrons to complete the octet whereas chlorine atom requires single electron to complete the octet. In the molecule ${\text{ClO}}_3{}^{-}$, chlorine is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of six electrons takes place. The 20 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs and chlorine atom gets single lone pair.

The Lewis structure of ${\text{ClO}}_3{}^{-}$ is

Figure 9

(c)

To determine: The Lewis structure of the molecule ${\text{ClO}}_2{}^{-},{\text{SCl}}_2$. and ${\text{PCl}}_2{}^{-}$

The Lewis structures of the given molecules are as follows.

• To determine: The Lewis structure of the molecule ${\text{ClO}}_2{}^{-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. . The atomic number of chlorine is $\text{17}$ and its electronic configuration is,

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

The valence electron of chlorine is $\text{7}$

The atomic number of oxygen is $\text{8}$ and its electronic configuration is,

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

The valence electron of oxygen is $\text{6}$

The molecule ${\text{ClO}}_2{}^{-}$ is made of two oxygen atoms and chlorine atom. Also oxygen atoms contain one negative charges; hence, the total number of valence electrons is,

$\begin{array}{c}\text{Cl+}2\text{O}+1{\text{e}}^{-}\text{=}7\text{+}2\times 6+1\\ \text{=}20\end{array}$

Each oxygen atom requires two electrons to complete the octet whereas chlorine atom requires single electron to complete the octet. In the molecule ${\text{ClO}}_2{}^{-}$, chlorine is central atom which is placed at the center whereas oxygen atoms are arranged on the sides Hence, the mutual sharing of four electrons takes place. The 16 valence electrons present are placed as lone pairs in which each oxygen atoms get three lone pairs and chlorine atom gets two lone pair.

The Lewis structure of ${\text{ClO}}_2{}^{-}$ is

Figure 10

• To determine: The Lewis structure of the molecule ${\text{SCl}}_2$.

The first step in determining the Lewis structure is to determine the number of valence electrons. . The atomic number of chlorine is $\text{17}$ and its electronic configuration is,

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

The valence electron of chlorine is $\text{7}$

The atomic number of sulfur is $\text{16}$ 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 electron of sulfur is $\text{6}$

The molecule ${\text{SCl}}_2$ is made of two chlorine atoms and sulfur atom; hence, the total number of valence electrons is,

$\begin{array}{c}\text{2Cl+}\text{S}\text{=}2\times 7\text{+}6\\ \text{=}20\end{array}$

Each sulfur atom requires two electrons to complete the octet whereas chlorine atom requires single electron to complete the octet. In the molecule ${\text{SCl}}_2$, sulfur is central atom which is placed at the center whereas chlorine atoms are arranged on the sides Hence, the mutual sharing of four electrons takes place. The 16 valence electrons present are placed as lone pairs in which each chlorine atoms get three lone pairs and sulfur atom gets two lone pair.

The Lewis structure of ${\text{SCl}}_2$ is

Figure 11

• To determine: The Lewis structure of the molecule ${\text{PCl}}_2{}^{-}$.

The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of chlorine is $\text{17}$ and its electronic configuration is,

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

The valence electron of chlorine is $\text{7}$

The atomic number of phosphorous is $\text{15}$ and its electronic configuration is,

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

The valence electron of phosphorous is $\text{5}$

The molecule ${\text{PCl}}_2{}^{-}$ is made of two chlorine atoms and phosphorous atom. Also chlorine atom contains single negative charge; hence, the total number of valence electrons is,

$\begin{array}{c}\text{2Cl+}\text{P}+1{\text{e}}^{-}\text{=}2\times 7\text{+5}+1\\ \text{=}20\end{array}$

Each phosphorous atom requires three electrons to complete the octet whereas chlorine atom requires single electron to complete the octet. In the molecule ${\text{PCl}}_2{}^{-}$, phosphorous is central atom which is placed at the center whereas chlorine atoms are arranged on the sides Hence, the mutual sharing of four electrons takes place. The 16 valence electrons present are placed as lone pairs in which each chlorine atoms get three lone pairs and phosphorous atom gets two lone pair.

The Lewis structure of ${\text{PCl}}_2{}^{-}$ is

Figure 12

(d)

To determine: The conclusion obtained from the answers from part a, b and c.

All compounds and ions in option (a), (b), and (c) have same number of valence electrons. These compounds contain same geometry. The atoms or compounds having same valence electrons are called isoelectronic.

Conclusion

The isoelectronic atoms contain same geometrical structure.