Chapter 7, Problem 16QAP

Interpretation Introduction

(a)

Interpretation:

The formula of the polyatomic ion needs to be determined having similar Lewis structure as the ${\text{BrO}}^{-}$ molecule.

Concept introduction:

The Lewis structures are used to write a shorthand configuration of number of available valence electrons in an atom for bonding. This structure deals with the magic number 8, and hence, the octet completion is shown by the bonded electrons between the atoms.

This structure is made on basis of octet rule understanding, and it is made sure that the number of electrons surrounding an atom must not divert from the octet.

Answer to Problem 16QAP

${\text{Br}}_{\text{2}}$ molecule

Explanation of Solution

Given Information:

The ion is BrO^-.

Here, the total valence electrons are: $\text{7(Br)+6(O)+1(charge)=14}$.

Bond pair electrons are obtained as:

$\text{1(bond)}\times \text{2= 2}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{14-2}}{2}\text{=6}$

A similar type of Lewis structure is possible in only such molecule, which has nearly matching electronegativities and same number of valence electrons.

In this pattern, the bromine molecule ${\mathrm{Br}}_2$ is best suitable as:

Here, the valence electrons are: $2\times 7(Br)=14$

Bond pair electrons are obtained as:

$\text{1(bond)}\times \text{2= 2}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{14-2}}{2}\text{=6}$

The Lewis structure will be drawn as:

Interpretation Introduction

(b)

Interpretation:

The formula of the polyatomic ion needs to be determined having similar Lewis structure as the ${\text{NH}}_4{}^{+}$ molecule.

Concept introduction:

The Lewis structures are used to write a shorthand configuration of number of available valence electrons in an atom for bonding. This structure deals with the magic number 8, and hence, the octet completion is shown by the bonded electrons between the atoms.

This structure is made on basis of octet rule understanding, and it is made sure that the number of electrons surrounding an atom must not divert from the octet.

Answer to Problem 16QAP

CH₄ molecule

Explanation of Solution

Given Information:

The ion is ${\text{NH}}_4^{+}$.

Here the total valence electrons are: $\text{1×5(N)+4×1(H)-1(Charge)=8}$

Bond pair electrons are obtained as:

$\text{4(bond)}\times \text{2= 8}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{8-8}}{2}\text{=0}$

A similar type of Lewis structure is possible in only such molecule, which has nearly matching electro negativities and same number of valance electrons.

In this pattern, the methane molecule ${\text{CH}}_{\text{4}}$ is best suitable as:

Here the valence electrons are: $\text{4×1(H)+4(C)=8}$

Bond pair electrons are obtained as:

$\text{4(bond)}\times \text{2= 8}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{8-8}}{2}\text{=6}$

The Lewis structure will be drawn as:

Interpretation Introduction

(c)

Interpretation:

The formula of the polyatomic ion needs to be determined having similar Lewis structure as the ${\text{CN}}^{-}$ molecule.

Concept introduction:

The Lewis structures are used to write a shorthand configuration of number of available valence electrons in an atom for bonding. This structure deals with the magic number 8, and hence, the octet completion is shown by the bonded electrons between the atoms.

This structure is made on basis of octet rule understanding, and it is made sure that the number of electrons surrounding an atom must not divert from the octet.

Answer to Problem 16QAP

CO molecule.

Explanation of Solution

Given Information:

The ion is ${\text{CN}}^{-}$.

Here the total valence electrons are: $\text{5(N)+4(C)+1(Charge)=10}$

Bond pair electrons are obtained as:

$\text{1(triple bond)= 6}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{10-6}}{2}\text{=2}$

A similar type of Lewis structure is possible in only such molecule which has nearly matching electronegativities and same number of valence electrons.

In this pattern, the carbon monoxide (CO) is best suitable as:

Here, the valence electrons are: $4(C)+6(O)=10$

Bond pair electrons are obtained as:

$\text{1(triple bond)= 6}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{10-6}}{2}\text{=2}$

The Lewis structure will be drawn as:

Interpretation Introduction

(d)

Interpretation:

The formula of the polyatomic ion needs to be determined having similar Lewis structure as the ${\text{SO}}_4{}^{2-}$ molecule.

Concept introduction:

The Lewis structures are used to write a shorthand configuration of number of available valence electrons in an atom for bonding. This structure deals with the magic number 8, and hence, the octet completion is shown by the bonded electrons between the atoms.

This structure is made on basis of octet rule understanding, and it is made sure that the number of electrons surrounding an atom must not divert from the octet.

Answer to Problem 16QAP

${\text{SO}}_{\text{2}}{\text{F}}_{\text{2}}$ molecule.

Explanation of Solution

Given Information:

The ion is ${\text{SO}}_4^{2-}$.

Here the total valence electrons are: $\text{4×6(O)+6(S)+2(charge)=32}$

Bond pair electrons are obtained as:

$\text{2(bonds)}\times 2+\text{2(double bond)= 12}$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{32-12}}{2}\text{=10}$

A similar type of Lewis structure is possible in only such molecule which has nearly matching structure.

Now, in this pattern, the anion ${\text{SO}}_{\text{2}}{\text{F}}_{\text{2}}$ is best suitable as:

Here the valence electrons are: $\text{6(S)+2×7(F)+2×6(O)=32}$

Looking at the structure,

Bond pair electrons are obtained as:

$\text{4(bonds)}\times \text{2+2 double bonds= }16$

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{32-16}}{2}\text{=8}$

The Lewis structure will be drawn as: