Chapter 14, Problem 20E

(a)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{XeCl}}_{\text{2}}$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{XeCl}}_{\text{2}}$ is sp³d.

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom , present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom .

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{XeCl}}_{\text{2}}$ , central an atom is xenon ( $\text{Xe}$ ). The an atom ic number of xenon is 54 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}{\text{4s}}^{2}4{\text{p}}^{6}4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^6$

Number of valence electrons = 8

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{XeCl}}_{\text{2}}$ , V = 8, M =2 and C = 0

  $X=\frac{8+2}{2}$

  $X=5$

Number of lone pairs = 0

Steric number 5 shows that the hybridization of central an atom is sp³d and geometry is Trigonal bipyramidal.

(b)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{ICl}}_3$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{ICl}}_3$ is sp³d.

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom , present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom .

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{ICl}}_3$ , central an atom is iodine ( $\text{I}$ ). The an atom ic number of iodine is 53 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}{\text{4s}}^{2}4{\text{p}}^{6}4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^5$

Number of valence electrons = 7

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{ICl}}_3$ , V = 7, M =3 and C = 0

  $X=\frac{7+3}{2}$

  $X=5$

Number of lone pairs = 0

Steric number 5 shows that the hybridization of central an atom is sp³d and geometry is Trigonal bipyramidal.

(c)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{TeF}}_{\text{4}}$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{TeF}}_{\text{4}}$ is sp³d.

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom, present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom .

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{TeF}}_{\text{4}}$ , central an atom is tellurium ( $\text{Te}$ ). The atom ic number of tellurium is 52 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}{\text{4s}}^{2}4{\text{p}}^{6}4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^4$

Number of valence electrons = 6

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{TeF}}_{\text{4}}$ , V = 6, M =4 and C = 0

  $X=\frac{6+4}{2}$

  $X=5$

Number of lone pairs = 0

Steric number 5 shows that the hybridization of central an atom is sp³d and geometry is Trigonal bipyramidal.

(d)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{PCl}}_5$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{PCl}}_5$ is sp³d.

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom , present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom.

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{PCl}}_5$ , central an atom is phosphorus ( $\text{P}$ ). The an atom ic number of phosphorus is 15 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^3$

Number of valence electrons = 5

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{PCl}}_5$ , V = 5, M =5 and C = 0

  $X=\frac{5+5}{2}$

  $X=5$

Number of lone pairs = 0

Steric number 5 shows that the hybridization of central an atom is sp³d and geometry is Trigonal bipyramidal.

(a)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{ICl}}_5$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{ICl}}_5$ is sp³d².

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom , present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom .

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{ICl}}_5$ , central an atom is iodine ( $\text{I}$ ).The an atom ic number of iodine is 53 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}{\text{4s}}^{2}4{\text{p}}^{6}4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^5$

Number of valence electrons = 7

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{ICl}}_5$ , V = 7, M =5 and C = 0

  $X=\frac{7+5}{2}$

  $X=6$

Number of lone pairs = 0

Steric number 6 shows that the hybridization of central an atom is sp³d² and geometry is octahedral.

(b)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{XeCl}}_{\text{4}}$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{XeCl}}_{\text{4}}$ is sp³d².

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom , present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom .

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{XeCl}}_4$ , central an atom is xenon ( $\text{Xe}$ ). The an atom ic number of xenon is 54 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}{\text{4s}}^{2}4{\text{p}}^{6}4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^6$

Number of valence electrons = 8

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{XeCl}}_4$ , V = 8, M =4 and C = 0

  $X=\frac{8+4}{2}$

  $X=6$

Number of lone pairs = 0

Steric number 6 shows that the hybridization of central an atom is sp³d² and geometry is octahedral.

(c)

Interpretation Introduction

Interpretation:

The expected hybridization of the central an atom in ${\text{SeCl}}_{\text{6}}$ should be determined.

Concept Introduction:

When two atomic orbitals combine with each other to produce hybrid orbitals, redistribution of energy of orbitals of distinct an atom s to form orbitals with equal energy occurs. This process is known as hybridization and the formed new orbitals are known as hybrid orbitals.

Answer to Problem 20E

The expected hybridization of the central an atom in ${\text{SeCl}}_{\text{6}}$ is sp³d².

Explanation of Solution

On the basis of types of orbitals involved in mixing, different hybridization is classified as sp, sp², sp³, sp³d, sp³d², sp³d³.

sp-hybridization: It is formed when one s and one p orbital mix with each other in same shell of an an atom to produce two new equal orbitals. This hybridization takes place in linear molecules.

  

sp²-hybridization: It is formed when one s and two p orbital mix with each other in same shell of an an atom to produce three new equal orbitals. This hybridization takes place in molecules exhibits trigonal planar geometry.

  

sp³-hybridization: It is formed when one s and three p orbital mix with each other in same shell of an atom to produce four new equal orbitals. This hybridization takes place in molecules exhibits tetrahedral geometry. In this hybridization, no p-unhybridized orbital is present as all are hybridized and form four sigma bonds.

Now, for an atom , present in the structure compound, the sum of the number of bonded an atom s and number of lone pairs present on it results its steric number. The value of steric number tells about the hybridization of central an atom .

Steric Number	Hybridization	Structure
2	sp	Linear
3	sp2	Trigonal Planar
4	sp3	Tetrahedral
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
7	sp3d3	Pentagonal bipyramidal

In ${\text{SeCl}}_{\text{6}}$ , central an atom is selenium ( $\text{Se}$ ).The an atom ic number of selenium is 34 and its electronic configuration is ${\text{1s}}^{\text{2}}{\text{2s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}3{\text{d}}^{10}{\text{4s}}^{2}4{\text{p}}^4$

Number of valence electrons =6

The expression for calculating number of electron pairs is:

  $X=\frac{V+M+C}{2}$

Where, X = number of electron pairs

V = valence electrons of central an atom

M = number of monovalent an atom s

C = charge on compound.

In ${\text{SeCl}}_{\text{6}}$ , V = 6, M =6 and C = 0

  $X=\frac{6+6}{2}$

  $X=6$

Number of lone pairs = 0

Steric number 6 shows that the hybridization of central an atom is sp³d² and geometry is octahedral.