Chapter 6, Problem 16QRT

(a)

Interpretation Introduction

Interpretation:

The Lewis structure for $C{l}_{2}O$ molecule has to be written.

Concept Introduction:

Lewis structure is also known as Lewis dot diagrams or electron dot structures. The bond between atoms and lone pairs of electrons that is present in the molecule.  Lewis structure represents each atom and their position in structure using the chemical symbol.  Excess electrons forms the lone pair are given by pair of dots, and are located next to the atom.

Explanation of Solution

Oxygen is in Group 6A and Chlorine is in Group 7A and the valence electrons present in the $C{l}_{2}O$ molecule are ${\text{ 6 e}}^{\text{–}}\text{ + 2}\times {\text{(7 e}}^{\text{–}}{\text{) = 20 e}}^{\text{-}}\text{. }$ Oxygen makes two bonds, but chlorine prefers to make only one bond.

The two chlorine atoms connect with one Oxygen atom through single bonds.

  $\left({\text{20 e}}^{\text{–}}\right)\text{ - }\left({\text{4 e}}^{\text{–}}\right){\text{ = 16 e}}^{\text{-}}\text{. }$

Chlorine atoms attain octet by adding six electrons as dots in pairs.

Complete the octet of the two chlorine atom uses $12$ electrons and subtract $16$ electrons form the current electron count, $16e^{–}– 12e^{–}= 4e^{–}$.

Put the last six electrons on Oxygen atom.

The correct Lewis structure of the $C{l}_{2}O$ can be drawn as,

Oxygen has eight electrons four in the bonds and four as dots, hence the structure is complete.

Hence, the total number of electrons can be counted as

  $\text{6 dots + 2 in a bond + 4 dots + 2 in a bond + 6 dots = 20 total}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The Lewis structure for $H_2{O}_2$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Hydrogen atom is from Group one A and Oxygen atom is from group 6A, hence the valence electrons are ${\text{2 ×(1e}}^{\text{-}}{\text{)+2×(6e}}^{\text{-}}{\text{)=14e}}^{\text{-}}$.  So, Oxygen atoms to form two bonds and H atoms can only form one bond, the two Oxygen atoms must be bonded to Hydrogen.

Complete the octet of the two chlorine atom uses $12$ electrons and subtract $16$ electrons form the current electron count, $14e^{–}– 6e^{–}= 8e^{–}$.

The incomplete Lewis structure of $H_2{O}_2$ can be written as

The correct Lewis structure of the $H_2{O}_2$ can be drawn as,

Hence, the total number of electrons can be counted as

  ${\text{6 e}}^{\text{-}}\text{ in three}{\text{bonds + 8e}}^{\text{-}}{\text{ dots = 14e}}^{\text{-}}\text{ total}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The Lewis structure for $B{H}_4^{-}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

The four Hydrogen atoms connect to boron with single bonds uses eight electrons.  Boron atom is the central atom with the hydrogen atoms around it.  So, the valence electron present in the $B{H}_4^{-}$ ion is

  $\left({\text{3 e}}^{\text{–}}\right)\text{ + 4 × }\left({\text{1 e}}^{\text{–}}\right){\text{ + 1e}}^{\text{-}}{\text{ = 8e}}^{\text{–}}\text{. }$

Boron atom must be the central atom with the four Hydrogen atoms bonded to it.  Boron has eight electrons so, the structure is complete.  Boron has eight electrons, and each Hydrogen atom has just two electrons, the structure clockwise, the total number of electrons can be counted ${\text{8 e}}^{\text{–}}$ in four bonds = ${\text{8 e}}^{\text{–}}\text{total}\text{.}$

The correct Lewis structure of the $B{H}_4^{-}$ can be drawn as,

(d)

Interpretation Introduction

Interpretation:

The Lewis structure for $P{H}_4^{+}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

The four Hydrogen atoms connect to Phosphorous with single bonds uses eight electrons.  Phosphorous atom is the central atom with the hydrogen atoms around it.  So, the valence electrons present in the $P{H}_4^{+}$ ion are

  $\left({\text{5 e}}^{\text{–}}\right)\text{ + 4 × }\left({\text{1 e}}^{\text{–}}\right){\text{ - 1e}}^{\text{-}}{\text{ = 8e}}^{\text{–}}\text{. }$

Phosphorous atom must be the central atom with the four Hydrogen atoms bonded to it.  Phosphorous has eight electrons so, the structure is complete.  It has eight electrons, and each Hydrogen atom has just two electrons, the structure clockwise, the total number of electrons can be counted ${\text{8 e}}^{\text{–}}$ in four bonds = ${\text{8 e}}^{\text{–}}\text{total}\text{.}$

The structure is a $1+$ ion, so add brackets and the charge.

The correct Lewis structure of the $P{H}_4^{+}$ can be drawn as,

(e)

Interpretation Introduction

Interpretation:

The Lewis structure for $PC{l}_5$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

The five chlorine atoms connect to Phosphorous with single bonds uses ten electrons.  Phosphorous atom is the central atom with the Chlorine atoms around it.

The number of valence electrons present in $PC{l}_5$ molecule are

  $\text{5+5}\times {\text{(7) = 40e}}^{\text{–}}\text{. }$

Chlorine atoms prefer making only one bond, and Phosphorous prefers to make three and five bonds.  So use Phosphorous atom as central atom with the five Chlorine atoms around it.

Each Chlorine atom has three lone pair and one bond pair so it attains octet.

The correct Lewis structure of the $PC{l}_5$ can be drawn as,

The total number of electrons can be counted as

  ${\text{5×(2 e}}^{\text{–}}{\text{in a bond) + 5×(6 e}}^{\text{–}}{\text{as dots) = 10 e}}^{\text{–}}{\text{+ 30 e}}^{\text{–}}{\text{= 40 e}}^{\text{–}}\text{total}\text{.}$