Chapter 11, Problem 11.38P

(a)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{BrO}}_3^{-}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{BrO}}_3^{-}$, the hybridization of bromine is $s{p}^3$. Its shape is trigonal pyramidal, the ideal bond angle is $109.5°$ and the deviated bond angle is less than $109.5°$.

Explanation of Solution

The Lewis structure of ${\text{BrO}}_3^{-}$ is as follows:

Boron forms three single bonds with three oxygen atoms and one lone pair is present on it so four hybrid orbitals are required and therefore the hybridization of boron in ${\text{BrO}}_3^{-}$ is $s{p}^3$. According to hybridization, its shape should be tetrahedral and the ideal bond angle should be $109.5°$. But due to the presence of a lone pair on bromine atom, its shape becomes trigonal pyramidal and the bond angle becomes less than $109.5°$.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

(b)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{AsCl}}_4^{-}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{AsCl}}_4^{-}$, the hybridization of bromine is $s{p}^{3}d$. Its shape is seesaw, ideal bond angles are $120°$ and $90°$. But the deviated bond angles are less than $120°$ and $90°$.

Explanation of Solution

The Lewis structure of ${\text{AsCl}}_4^{-}$ is as follows:

Arsenic forms four single bonds with four chlorine atoms and one lone pair is present on it so five hybrid orbitals are required and therefore the hybridization of arsenic in ${\text{AsCl}}_4^{-}$ is $s{p}^{3}d$. According to hybridization, its shape should be trigonal bipyramidal and the ideal bond angles should be $120°$ and $90°$. But due to the presence of a lone pair on the arsenic atom, its shape becomes seesaw and the bond angle becomes less than $120°$ and $90°$.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

(c)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{SeO}}_4^{2-}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{SeO}}_4^{2-}$, the hybridization of selenium is $s{p}^3$. Its shape is tetrahedral and the ideal bond angles is $109.5°$. There is no deviation in the bond angle.

Explanation of Solution

The Lewis structure of ${\text{SeO}}_4^{2-}$ is as follows:

Selenium forms four single bonds with four oxygen atoms so four hybrid orbitals are required and therefore the hybridization of selenium in ${\text{SeO}}_4^{2-}$ is $s{p}^3$. According to hybridization, its shape is tetrahedral and the ideal bond angle should be $109.5°$. There will be no deviation in the bond angle due to the absence of a lone pair of electrons.

Conclusion

The shape of the molecule is determined by the hybridization only if no lone pair is present on the central atom.

(d)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{BiF}}_5^{2-}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{BiF}}_5^{2-}$, the hybridization of bismuth is $s{p}^3{d}^2$. Its shape is square pyramidal and ideal bond angle is $90°$. But the deviated bond angle is less than $90°$.

Explanation of Solution

The Lewis structure of ${\text{BiF}}_5^{2-}$ is as follows:

Bismuth forms five single bonds with five fluorine atoms and one lone pair is present on it so six hybrid orbitals are required and therefore the hybridization of bismuth in ${\text{BiF}}_5^{2-}$ is $s{p}^3{d}^2$. According to hybridization, its shape should be octahedral and the ideal bond angle should be $90°$. But due to the presence of a lone pair on bismuth atom, its shape becomes square pyramidal and the bond angle becomes less than $90°$.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

(e)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{SbF}}_4^{+}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{SbF}}_4^{+}$, the hybridization of selenium is $s{p}^3$. Its shape is tetrahedral and ideal bond angle is $109.5°$. There is no deviation in the bond angle.

Explanation of Solution

The Lewis structure of ${\text{SbF}}_4^{+}$ is as follows:

Antimony forms four single bonds with four fluorine atoms so four hybrid orbitals are required and therefore the hybridization of antimony in ${\text{SbF}}_4^{+}$ is $s{p}^3$. According to hybridization, its shape is tetrahedral and the ideal bond angle is $109.5°$. There will be no deviation in the bond angle due to the absence of a lone pair of electrons.

Conclusion

The shape of the molecule is determined by the hybridization only if no lone pair is present on the central atom.

(f)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{AlF}}_6^{3-}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{AlF}}_6^{3-}$, the hybridization of aluminium is $s{p}^3{d}^2$. Its shape is octahedral, the ideal bond angle is $90°$. There is no deviation in the bond angle.

Explanation of Solution

The Lewis structure of ${\text{AlF}}_6^{3-}$ is as follows:

Aluminium forms six single bonds with six fluorine atoms so six hybrid orbitals are required and therefore the hybridization of aluminium in ${\text{AlF}}_6^{3-}$ is $s{p}^3{d}^2$. According to hybridization, its shape is octahedral and the ideal bond angle is $90°$. There will be no deviation in the bond angle due to the absence of lone pair of electrons.

Conclusion

The shape of the molecule is determined by the hybridization only if no lone pair is present on the central atom.

(g)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in ${\text{IF}}_4^{+}$ are to be determined.

Concept introduction:

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital. Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

Answer to Problem 11.38P

In ${\text{IF}}_4^{+}$, the hybridization of iodine is $s{p}^{3}d$. Its shape is seesaw, ideal bond angles are $120°$ and $90°$. But the deviated bond angles are $120°$ and $90°$.

Explanation of Solution

The Lewis structure of ${\text{IF}}_4^{+}$ is as follows:

Iodine forms four single bonds with four fluorine atoms and one lone pair is present on it so five hybrid orbitals are required and therefore the hybridization of iodine in ${\text{IF}}_4^{+}$ is $s{p}^{3}d$. According to hybridization, its shape should be trigonal bipyramidal and the ideal bond angles should be $120°$ and $90°$. But due to the presence of a lone pair on iodine atom, its shape becomes seesaw and the bond angles become less than $120°$ and $90°$.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.