Chapter 7, Problem 15QAP

Interpretation Introduction

(a)

Interpretation:

The formula of the polyatomic ion needs to be determinedhaving similar Lewis structure as the ${\text{Cl}}_{\text{2}}$ 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 15QAP

${\text{ClO}}^{-}$ hypochlorite anion.

Explanation of Solution

Given Information:

The molecule is Cl₂.

Here thetotal valence electrons are: $2\times 7\text{(Cl})=14$ and the Cl-Cl atoms are attached by single covalent bond.

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 polyatomic ion which has nearly matching electronegativities and same number of valence electrons.

In this pattern, the hypochlorite anion ${\text{ClO}}^{-}$ is best suitable as:

Here, the valence electrons are: $\text{1×7(Cl)+1×6(O)+1(Charge)=14}$

The bond pair electrons are the number of electrons used in bond formation.

Looking at the structure,

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 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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 15QAP

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$ ion.

Explanation of Solution

Given Information:

The molecule is H₂ SO₄.

Here the total valence electrons are: $\text{2×1(H)+6(S)+4×6(O)=32}$

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:

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

Now, in this pattern, the anion ${\text{H}}_{\text{2}}{\text{PO}}_4^{-}$ is best suitable as:

Here the valence electrons are: $\text{5(P)+2×1(H)+4×6(O)+1(Charge)=32}$

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:

Interpretation Introduction

(c)

Interpretation:

The formula of the polyatomic ion needs to be determined having similar Lewis structure as the ${\text{CH}}_{\text{4}}$ 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 15QAP

${\text{NH}}_4^{+}$ ammonium ion.

Explanation of Solution

Given Information:

The molecule is CH₄.

Here the total 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:

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

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

In this pattern, the ammonium ion ${\text{NH}}_4^{+}$ is best suitable as:

Here, the 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:

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

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 ${\text{GeCl}}_{\text{4}}$ 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 15QAP

${\text{AsCl}}_4^{+}$ ion.

Explanation of Solution

Given Information:

The molecule is GeCl₄.

Here the total valence electrons are: $\text{4×7(H)+4(Ge)=32}$

Bond pair electrons are obtained as:

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

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{32-8}}{2}\text{=12}$

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

In this pattern, the ion ${\text{AsCl}}_4^{+}$ is best suitable as:

Here, the valence electrons are: $\text{1×5(As)+4×7(Cl)-1(Charge)=32}$

Bond pair electrons are obtained as:

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

Similarly, lone pairs are counted as:

$\frac{\text{Valence electrons-bond pair electrons}}{2}\text{=}\frac{\text{32-8}}{2}\text{=12}$

The Lewis structure will be drawn as: