Chapter 10, Problem 10.58QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{NF}}_{\text{3}}$ 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.

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in ${\text{NF}}_{\text{3}}$ is ${\text{sp}}^{\text{3}}.$

Explanation of Solution

The structure of ${\text{NF}}_{\text{3}}$ is,

The ${\text{NF}}_{\text{3}}$ have four electron domains and three of them are in bonding which means that the geometry of ${\text{NF}}_{\text{3}}$ is trigonal pyramidal.  The trigonal pyramidal geometry around the N means that it has ${\text{sp}}^{\text{3}}.$ hybrid orbitals on N atom.

The ${\text{sp}}^{\text{3}}$ 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 ${\text{sp}}^{\text{3}}$ hybridization.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{SCl}}_{\text{2}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in ${\text{SCl}}_{\text{2}}$ is ${\text{sp}}^{\text{3}}.$

Explanation of Solution

The structure of ${\text{SCl}}_{\text{2}}$ is,

${\text{SCl}}_{\text{2}}$ have four electron domains and two of them are in bonding which means that the geometry of ${\text{SCl}}_{\text{2}}$ is bent shape.  The bent shape geometry around the S means that it has ${\text{sp}}^{\text{3}}$ hybrid orbitals on S atom.

The ${\text{sp}}^{\text{3}}$ 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 ${\text{sp}}^{\text{3}}$ hybridization.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is ${\text{sp}}^{\text{3}}.$

Explanation of Solution

The structure of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is,

The ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ have four electron domains and three of them are in bonding which means that the geometry of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is trigonal pyramidal.  The trigonal pyramidal geometry around the O means that it has ${\text{sp}}^{\text{3}}$ hybrid orbitals on O atom.

The ${\text{sp}}^{\text{3}}$ 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 ${\text{sp}}^{\text{3}}$ hybridization.

Figure 1

(d)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{IF}}_{\text{5}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in ${\text{IF}}_{\text{5}}$ is ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}.$

Explanation of Solution

The structure of ${\text{IF}}_{\text{5}}$ is,

The ${\text{IF}}_{\text{5}}$ have six electron domains and five of them are in bonding which means that the geometry of ${\text{IF}}_{\text{5}}$ is square pyramidal.  The Square pyramidal geometry around the iodine means that it has ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}.$ hybrid orbitals on iodine atom.

The ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}$ hybridization:

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

Figure 2

(f)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{SCl}}_{\text{4}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.58QE

The hybrid orbitals on the central atom that form bond in ${\text{SCl}}_{\text{4}}$ is ${\text{sp}}^{\text{3}}\text{d}\text{.}$

Explanation of Solution

The structure of ${\text{SCl}}_{\text{4}}$ is,

${\text{SCl}}_{\text{4}}$ have five electron domains and four of them are in bonding which means that the geometry of ${\text{SCl}}_{\text{4}}$ is see-saw.  The see-saw geometry around the sulphur means that it has ${\text{sp}}^{\text{3}}\text{d}$ hybrid orbitals on chlorine atom.

The ${\text{sp}}^{\text{3}}\text{d}$ hybridization:

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