Chapter 7, Problem 5QAP

Follow the directions of Question 1 for

(a) OCl₂ (b) PF₃ (c) SbCl₆^- (d) ICl₄^-

Interpretation Introduction

(a)

Interpretation:

Lewis structure for ${\text{OCl}}_{\text{2}}$ should be drawn.

Concept introduction:

• Lewis structures are the diagrams that show the bonding between the atoms of the molecules and existing lone pairs of electrons.
• Bonding electrons are those electrons which are shared between the atoms resulting in the formation of bond.
• Non-bonding electrons are the valence electrons of the atom which are not shared with another atom.

Answer to Problem 5QAP

Explanation of Solution

1. Calculate the total number of valence electrons in molecule ${\text{OCl}}_{\text{2}}$ :

Total number of valence electrons = valence electrons of O + 2(valence electrons of Cl)

$=6+2(7)=20{\text{ e}}^{-}$

2. Arrange the atoms in such a way that the least electronegative atom is at the center. Then put the valence electrons around them such that each atom contributes at least 1 electron to single bond and the octet rule for each atom is followed.
3. To complete the octet each chlorine, Cl atom forms single bond with oxygen, O. The rest number of electrons are present as lone pair on oxygen and chlorine atoms, the number of lone pair of electrons = Total number of valence electrons − Total number of electrons involved in bond formation.

The number of lone pair of electrons = 20 - 4 = 16 electrons

4. The Lewis structure of ${\text{OCl}}_{\text{2}}$ as:

Interpretation Introduction

(b)

Interpretation:

Lewis structure for ${\text{PF}}_3$ should be drawn.

Concept introduction:

• Lewis structures are the diagrams that show the bonding between the atoms of the molecules and existing lone pairs of electrons.
• Bonding electrons are those electrons which are shared between the atoms resulting in the formation of bond.
• Non-bonding electrons are the valence electrons of the atom which are not shared with another atom.

Answer to Problem 5QAP

Explanation of Solution

1. Calculate the total number of valence electrons in molecule ${\text{PF}}_3$ :

Total number of valence electrons = valence electrons of P + 3(valence electrons of F)

$=5+3(7)=26{\text{ e}}^{-}$

2. Arrange the atoms in such a way that the least electronegative atom is at the center. Then put the valence electrons around them such that each atom contributes at least 1 electron to single bond and the octet rule for each atom is followed.
3. To complete the octet each fluorine, F atom forms single bond with phosphorus, P. The rest number of electrons are present as lone pair on fluorine and phosphorus atom, the number of lone pair of electrons = Total number of valence electrons − Total number of electrons involved in bond formation.

The number of lone pair of electrons = 26 − 6 = 20 electrons

4. Thus, as P is the least electronegative, it would be the central atom and on arranging all the valence electrons, we get the Lewis structure of ${\text{PF}}_3$ as:

Interpretation Introduction

(c)

Interpretation:

Lewis structure for ${\text{SbCl}}_6^{-}$ should be drawn.

Concept introduction:

• Lewis structures are the diagrams that show the bonding between the atoms of the molecules and existing lone pairs of electrons.
• Bonding electrons are those electrons which are shared between the atoms resulting in the formation of bond.
• Non-bonding electrons are the valence electrons of the atom which are not shared with another atom.
• For a molecule possessing negative charge, the charge is added to the total number of valence electron count.

Answer to Problem 5QAP

Explanation of Solution

1. Calculate the total number of valence electrons in molecule ${\text{SbCl}}_6^{-}$ :

Total number of valence electrons = valence electrons of Sb + 6(valence electron of Cl) + 1

$=5+6(7)+1=48{\text{e}}^{-}$

2. Arrange the atoms in such a way that the least electronegative atom is at the center. Then put the valence electrons around them such that each atom contributes at least 1 electron to single bond and the octet rule for each atom is followed.
3. To complete the octet the each chlorine, Cl atom will form a single bond with chlorine, Cl.

The rest number of electrons are present as lone pair on chlorine atoms, the number of lone pair of electrons = Total number of valence electrons − Total number of electrons involved in bond formation.

The number of lone pair of electrons = 48 − 12 = 36 electrons

4. Thus, as Sb is the least electronegative, it would be the central atom and on arranging all the valence electrons, the Lewis structure of ${\text{SbCl}}_6^{-}$ as:

Interpretation Introduction

(d)

Interpretation:

Lewis structure for ${\text{ICl}}_4^{-}$ should be drawn.

Concept introduction:

• Lewis structures are the diagrams that show the bonding between the atoms of the molecules and existing lone pairs of electrons.
• Bonding electrons are those electrons which are shared between the atoms resulting in the formation of bond.
• Non-bonding electrons are the valence electrons of the atom which are not shared with another atom.
• For a molecule possessing negative charge, the charge is added to the total number of valence electron count.

Answer to Problem 5QAP

Explanation of Solution

1. Calculate the total number of valence electrons in molecule ${\text{ICl}}_4^{-}$ :

Total number of valence electrons = valence electrons of I + 4(valence electrons of Cl) + 1

$=7+4(7)+1={\text{ 36e}}^{-}$

2. Arrange the atoms in such a way that the least electronegative atom is at the center. Then put the valence electrons around them such that each atom contributes at least 1 electron to single bond and the octet rule for each atom is followed.
3. To complete the octet each chlorine, Cl atoms forms single bond with iodine, I. The rest number of electrons are present as lone pair on chlorine atoms and iodine atom, the number of lone pair of electrons = Total number of valence electrons − Total number of electrons involved in bond formation.

The number of lone pair of electrons = 36 − 8 = 28 electrons

4. Thus, as Iis the least electronegative, it would be the central atom and on arranging all the valence electrons, we get the Lewis structure of ${\text{ICl}}_4^{-}$ as: