Chapter 7, Problem 7.66PAE

a.

Interpretation Introduction

Interpretation:

The shape of ${\text{PH}}_4^{+}$ should be determined.

Concept Introduction:

Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.

The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.

Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.

Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-

lp-lp>lp-bp>bp-bp

Answer to Problem 7.66PAE

Solution:

The geometry of ${\text{PH}}_4{}^{+}$ is trigonal Planar as the hybridization of P is sp³

Explanation of Solution

The electronic configuration of P is $3s^{2}3p^{3}3d^0$. In the excited state, the configuration becomes $3s^{1}3p^{3}3d^1$. Thus, there is mixing of the s and p orbitals to form four half filledsp³ hybrid orbitals. These orbitals combine with hydrogen to form their respective sigma bonds. The arrangement of the bond pair is such that each of them suffers the minimum repulsion. Hence, the shape is Trigonal Planar.

Structure of ${\text{PH}}_4{}^{+}$

The geometry of ${\text{PH}}_4{}^{+}$ is Trigonal Planar

b.

Interpretation Introduction

Interpretation:

The shape of ${\text{OSF}}_{\text{4}}$ should be determined.

Concept Introduction:

Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.

The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.

Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.

Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-

lp-lp>lp-bp>bp-bp.

Answer to Problem 7.66PAE

Solution:

The geometry of ${\text{OSF}}_4$ is trigonal−bipyramidal as the hybridization of S is sp³d

Explanation of Solution

The electronic configuration of S is $3s^{2}3p^4$. In the excited state, the configuration becomes $3s^{1}3p^{3}3d^2$. Thus, there is mixing of the s, p and d orbitals to form five half filledsp³d hybrid orbitals. These orbitals combine with fluorine to form their respective sigma bonds and pi bonds. The arrangement of the bond pairs is such that each of them suffers the minimum repulsion. Hence, the shape is trigonal-bipyramidal. Structure of ${\text{OSF}}_4$

The geometry of ${\text{OSF}}_4$ is trigonal-bipyramidal.

c.

Interpretation Introduction

Interpretation:

The geometry of ${\text{ClO}}_{\text{2}}^{\text{-}}$ should be determined.

Concept Introduction

Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.

The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.

Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.

Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-

Answer to Problem 7.66PAE

lp-lp>lp-bp>bp-bp

Solution: The geometry of ${\text{ClO}}_2^{-}$ is tetrahedral as the hybridization of Cl is sp²

Explanation of Solution

The electronic configuration of Cl is $3s^{2}3p^5$. Thus there is mixing of the s and p orbitals in the exciting form to form three half filledsp² hybrid orbitals. These orbitals combine with oxygen to form their respective sigma bonds and pi bonds. The arrangement of the bond pairs is such that each of them suffers the minimum repulsion. Hence, the shape is tetrahedral Structure of ${\text{ClO}}_2^{-}$

The geometry of ${\text{ClO}}_2^{-}$ is tetrahedral

d.

Interpretation Introduction

Interpretation:

The geometry of ${\text{I}}_{\text{3}}^{\text{-}}$ should be determined.

Concept Introduction

Hybridization: It relates to the mixing of atomic orbitals into new hybrid orbitals that have varied energies and shapes. It is appropriate for the pairing of the electrons for forming chemical bonds in the Valence Bond Theory. We can predict the shape of a particular molecule by the knowledge of their atomic numbers and VSEPR theory according to which the atoms take such a position that there is a minimum possible repulsion between the bonded atoms and the lone pair of electrons if any.

The main concept behind this theory is that the electron pairs are always present in the outermost shell i.e. valence shell of an atom of a molecule and they repel each other due to which they try to attain the best possible position so that the value of their repulsion is the least. Hence, the electrons occupy such positions around the atom that reduces their repulsion and provides a molecule to their shape.

Here the electrons that take part in the bonding of a molecule are known as the bonding pair and the electrons that do not take part in the bonding are known as the lone pairs. The bond pairs are in the influence of the two bonding atoms whereas the lone pairs are in the influence of only of the atom.

Due to the presence of lone pairs, there is more space occupied between the atoms of the molecules. Now they suffer the repulsion between the lone pair-lone pair and bond pair-lone pair. Their repulsion can be represented as:-

lp-lp>lp-bp>bp-bp

Answer to Problem 7.66PAE

Solution:

The geometry of ${\text{I}}_3{}^{-}$ is linear as the hybridization of I is sp³

Explanation of Solution

The electronic configuration of I is $4d^{10}5s^{2}5p^5$. In the excited state, the configuration becomes $4d^{10}5s^{1}5p^{3}6d^3$. Thus, there is mixing of the s, p and d orbitals in the exciting form to form four half filledsp³ hybrid orbitals. These orbitals combine with iodine to form their respective sigma bonds. The arrangement of the bond pair and lone pair is such that each of them suffers the minimum repulsion. Hence, the shape is linear Structure of ${\text{I}}_3{}^{-}$

The geometry of ${\text{I}}_3{}^{-}$ is linear.