The hybrid orbitals on the central atom that form bond in NF 3 have to be given. Concept Introduction: Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory .

Question

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Answer 1

Question

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

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Answer 2

Question

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

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Answer 3

Question

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

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Answer 4

Question

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

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Answer 5

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

Answer 6

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Answer 7

Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $NF3$ have to be given.

Concept Introduction:

Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory.

Answer 8

(a)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $NF3$ is $sp3.$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The structure of $NF3$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 1

The $NF3$ have four electron domains and three of them are in bonding which means that the geometry of $NF3$ is trigonal pyramidal. The trigonal pyramidal geometry around the N means that it has $sp3.$ hybrid orbitals on N atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl2$ is $sp3.$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The structure of $SCl2$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 2

$SCl2$ have four electron domains and two of them are in bonding which means that the geometry of $SCl2$ is bent shape. The bent shape geometry around the S means that it has $sp3$ hybrid orbitals on S atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Answer 20

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

Answer 21

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

Answer 22

(c)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $H3O+$ is $sp3.$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The structure of $H3O+$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 3

The $H3O+$ have four electron domains and three of them are in bonding which means that the geometry of $H3O+$ is trigonal pyramidal. The trigonal pyramidal geometry around the O means that it has $sp3$ hybrid orbitals on O atom.

The $sp3$ hybridization:

The process of mixing and rearrangement of one s and three p orbitals of valence shell of same atom to form new four hybrid orbitals having maximum symmetry and definite orientation is known as $sp3$ hybridization.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 4

Figure 1

Answer 27

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

Answer 28

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

Answer 29

(d)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $IF5$ is $sp3d2.$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The structure of $IF5$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 5

The $IF5$ have six electron domains and five of them are in bonding which means that the geometry of $IF5$ is square pyramidal. The Square pyramidal geometry around the iodine means that it has $sp3d2.$ hybrid orbitals on iodine atom.

The $sp3d2$ hybridization:

The $sp3d2$ hybridization has 1s, 3p and 2d orbitals they undergo intermixing to form 6 identical $sp3d2$ hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron.

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 6

Figure 2

Answer 34

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

Answer 35

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

Answer 36

(f)

Expert Solution

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in $SCl4$ is $sp3d.$

Answer 37

(f)

Expert Solution

Answer 38

(f)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

The structure of $SCl4$ is,

Chemistry: Principles and Practice, Chapter 10, Problem 10.58QE , additional homework tip 7

$SCl4$ have five electron domains and four of them are in bonding which means that the geometry of $SCl4$ is see-saw. The see-saw geometry around the sulphur means that it has $sp3d$ hybrid orbitals on chlorine atom.

The $sp3d$ hybridization:

The combination of 1s, 3p and 1d orbital result in the formation of $sp3d$ orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions.

Answer 41

Ch. 10 - Prob. 10.1QE Ch. 10 - Prob. 10.2QE Ch. 10 - Prob. 10.3QE Ch. 10 - Prob. 10.4QE Ch. 10 - Prob. 10.5QE Ch. 10 - Prob. 10.6QE Ch. 10 - Prob. 10.7QE Ch. 10 - Prob. 10.8QE Ch. 10 - Prob. 10.9QE Ch. 10 - Prob. 10.10QE

Concept explainers

Videos

Answer to Problem 10.58QE

Explanation of Solution

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions

Concept explainers

Videos

Answer to Problem 10.58QE

Explanation of Solution

Interpretation: The hybrid orbitals on the central atom that form bond in SCl2<m:math><m:mrow><m:msub><m:mrow><m:mtext>SCl</m:mtext></m:mrow><m:mtext>2</m:mtext></m:msub></m:mrow></m:math> have to be given. Concept Introduction: Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Interpretation: The hybrid orbitals on the central atom that form bond in H3O+<m:math><m:mrow><m:msub><m:mtext>H</m:mtext><m:mtext>3</m:mtext></m:msub><m:msup><m:mtext>O</m:mtext><m:mtext>+</m:mtext></m:msup></m:mrow></m:math> have to be given. Concept Introduction: Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Interpretation: The hybrid orbitals on the central atom that form bond in IF5<m:math><m:mrow><m:msub><m:mrow><m:mtext>IF</m:mtext></m:mrow><m:mtext>5</m:mtext></m:msub></m:mrow></m:math> have to be given. Concept Introduction: Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Interpretation: The hybrid orbitals on the central atom that form bond in SCl4<m:math><m:mrow><m:msub><m:mrow><m:mtext>SCl</m:mtext></m:mrow><m:mtext>4</m:mtext></m:msub></m:mrow></m:math> have to be given. Concept Introduction: Refer to part (a).

Answer to Problem 10.58QE

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl2$ have to be given.

Concept Introduction:

Refer to part (a).

Interpretation:

The hybrid orbitals on the central atom that form bond in $H3O+$ have to be given.

Concept Introduction:

Refer to part (a).

Interpretation:

The hybrid orbitals on the central atom that form bond in $IF5$ have to be given.

Concept Introduction:

Refer to part (a).

Interpretation:

The hybrid orbitals on the central atom that form bond in $SCl4$ have to be given.

Concept Introduction:

Refer to part (a).