Chapter 10, Problem 10.57QE

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{CF}}_{\text{4}}$ 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 10.57QE

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

Explanation of Solution

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

${\text{CF}}_{\text{4}}$ has four electron domains and are all in bonding which means that the geometry of ${\text{CF}}_{\text{4}}$ is tetrahedral.  The tetrahedral geometry around the carbon means that it has ${\text{sp}}^{\text{3}}.$ hybrid orbitals on carbon 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

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals on the central atom that form bond in ${\text{SbCl}}_{\text{6}}{}^{-}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.57QE

The hybrid orbitals on the central atom that form bond in ${\text{SbCl}}_{\text{6}}{}^{-}$ is ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}.$

Explanation of Solution

The structure of ${\text{SbCl}}_{\text{6}}{}^{-}$ is,

${\text{SbCl}}_{\text{6}}{}^{-}$ have six electron domains and are all in bonding which means that the geometry of ${\text{SbCl}}_{\text{6}}{}^{-}$ is octahedral.  The octahedral geometry around the $\text{Sb}$ means that it has ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}.$ hybrid orbitals on $\text{Sb}$ 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 six orbitals are directed towards the corners of an octahedron.

Figure 2

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.57QE

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

Explanation of Solution

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

${\text{AsF}}_{\text{5}}$ have five electron domains and are all in bonding which means that the geometry of ${\text{AsF}}_{\text{5}}$ is trigonal bipyramidal.  Trigonal bipyramidal is a molecular geometry that results when there are five bonds and no lone pairs on the central atom in the molecule.  Three of the bonds are arranged along the atoms equator with ${\text{120}}^{\text{o}}$ angles between them; the other two are placed at the atoms axis.  Axial bonds are at right angles to the equatorial bonds.  Molecules with trigonal bipyramidal electron pair geometries have ${\text{sp}}^{\text{3}}\text{d}$ hybridization at the central atom.

Figure 3

(d)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.57QE

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

Explanation of Solution

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

${\text{SiH}}_{\text{4}}$ have four electron domains and are all in bonding which means that the geometry of ${\text{SiH}}_{\text{4}}$ is tetrahedral.  The tetrahedral geometry around the silicon means that it has ${\text{sp}}^{\text{3}}.$ hybrid orbitals on silicon 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.

(e)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 10.57QE

The hybrid orbitals on the central atom that form bond in ${\text{NH}}_4{}^{\text{+}}$ is ${\text{sp}}^{\text{3}}.$

Explanation of Solution

The structure of ${\text{NH}}_4{}^{\text{+}}$ is,

${\text{NH}}_4{}^{\text{+}}$ have four electron domains and are all in bonding which means that the geometry of ${\text{NH}}_4{}^{\text{+}}$ is tetrahedral.  The tetrahedral geometry around the nitrogen means that it has ${\text{sp}}^{\text{3}}.$ hybrid orbitals on nitrogen 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.