Chapter 8, Problem 32E

(i)

(a)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of order of increasing electronegativity of $\text{Na},\text{ K},\text{ Rb}$ elements from exercise 32 and figure 3-4.

Answer to Problem 32E

Yes, the answer is different.

Explanation of Solution

Refer to figure 3-4

The electro negativity of Sodium $\left(\text{Na}\right)$ Potassium $\left(\text{K}\right)$ and Rubidium $\left(\text{Rb}\right)$ elements obtained from figure 3-4 are:

• Sodium: 0.9
• Potassium: 0.8
• Rubidium: 0.8

The increasing order of electro negativity is $\text{Rb}=\text{K}<\text{Na}$.

Repeat exercise 32 (a):

In a periodic table nuclear attraction on incoming electrons increases along the period because the atomic radii decreases from left to right. As a result electronegativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electronegativity decreases down the group.

The Sodium $\left(\text{Na}\right)$ Potassium $\left(\text{K}\right)$ and Rubidium $\left(\text{Rb}\right)$ are the elements of group 1 and their electronic configuration as follows:

• Sodium: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{1}}$
• Potassium: $\begin{array}{l}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{1}}\hfill \end{array}$
• Rubidium: $\begin{array}{l}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{1}}\hfill \end{array}$

The electronegativity decreases down the group. Hence the correct order of increasing electronegativity of $\text{Na},\text{ K},\text{ Rb}$ element is:

• $\text{Rb}<\text{K}<\text{Na}$

The answer obtained from exercise 32 (a) is $\text{Rb}<\text{K}<\text{Na}$.

(b)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of order of increasing electronegativity of $\text{B, O, Ga}$ elements from exercise 32 and figure 3-4.

Answer to Problem 32E

No, there are no differences.

Explanation of Solution

Refer to figure 3-4

The electronegativity of Boron $\left(\text{B}\right)$, Oxygen $\left(\text{O}\right)$ and Gallium $\left(\text{Ga}\right)$ obtained from figure 3-4 are:

• Boron (B): 2.0
• Oxygen (O): 3.5
• Gallium (Ga): 1.6

The increasing order of electronegativity is $\text{Ga < B < O}$.

Repeat exercise 32 (b):

In a periodic table nuclear attraction on incoming electrons increases along the period because the atomic radii decreases from left to right. As a result electronegativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electronegativity decreases down the group.

The element Boron lies in 2^nd period, 13th group, Oxygen lie in 2^nd period, 16th group and Gallium lie in 4^th period, 13th group. The electronic configuration of these elements as follows:

• Boron: ${\text{1s}}^2{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}$
• Oxygen: ${\text{1s}}^2{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$
• Gallium: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{1}}$

The electronegativity decreases down the group and increases from left to right along the period. Hence the correct order of increasing electronegativity of $\text{B, O, Ga}$ element is:

• $\text{Ga < B < O}$

The answer obtained from exercise 32 (b) is $\text{Ga < B < O}$.

(c)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of order of increasing electronegativity of $\text{F, Cl, Br}$ elements from exercise 32 and figure 3-4.

Answer to Problem 32E

No, there are not any difference.

Explanation of Solution

Refer to figure 3-4

The electronegativity of Fluorine $\left(\text{F}\right)$, Chlorine $\left(\text{Cl}\right)$ and Bromine $\left(\text{Br}\right)$ obtained from figure 3-4 are:

• Fluorine: 4.0
• Chlorine: 3.0
• Bromine: 2.8

The increasing order of electronegativity is $\text{Br < Cl < F}$.

Repeat exercise 32 (c):

In a periodic table nuclear attraction on incoming electrons increases along the period because the atomic radii decreases from left to right. As a result electronegativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electronegativity decreases down the group.

The element Fluorine $\left(\text{F}\right)$, Chlorine $\left(\text{Cl}\right)$ and Bromine $\left(\text{Br}\right)$ are the elements of group 17 and their electronic configuration as follows:

• Fluorine: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}$
• Chlorine: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{5}}$
• Bromine: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{5}}$

The electronegativity decreases down the group. Hence the correct order of increasing electronegativity of $\text{F, Cl, Br}$ element is:

• $\text{Br < Cl < F}$

The answer obtained from exercise 32 (c) is $\text{Br < Cl < F}$.

(d)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of order of increasing electronegativity of $\text{S, O, F}$ elements from exercise 32 and figure 3-4.

Answer to Problem 32E

No, there are no differences.

Explanation of Solution

Refer to figure 3-4

The electronegativity of Fluorine $\left(\text{F}\right)$, Oxygen $\left(\text{O}\right)$ and sulfur $\left(\text{S}\right)$ obtained from figure 3-4 are:

• Fluorine: 2.5
• Oxygen: 3.5
• Sulfur: 4.0

The increasing order of electronegativity is $\text{S}<\text{O}<\text{F}$.

Repeat exercise 32 (d):

In a periodic table nuclear attraction on incoming electrons increases along the period because the atomic radii decreases from left to right. As a result electronegativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electronegativity decreases down the group.

The element Fluorine lies in 2^nd period, 17th group, Oxygen lie in 2^nd period, 16th group and sulfur lie in 3^rd period, 16th group. The electronic configuration of these elements as follows:

• Fluorine: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{5}}$
• Oxygen: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$
• Sulfur: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}$

The electronegativity increases along the period from left to right and decreases down the group. Hence the correct order of increasing electronegativity of $\text{S, O, F}$ element is:

• $\text{S}<\text{O}<\text{F}$

The answer obtained from exercise 32 (d) is $\text{S}<\text{O}<\text{F}$.

Conclusion

The answer calculated from figure 3-4 and exercise 32 is:

The answer obtained from figure 3-4 was different from exercise 32 (a).

According to figure 3-4,

The order of increasing electronegativity of $\text{Na},\text{ K},\text{ Rb}$ elements is $\text{Rb}=\text{K}<\text{Na}$.

According to exercise 32 (a),

The order of increasing electronegativity of $\text{Na},\text{ K},\text{ Rb}$ elements is $\text{Rb}<\text{K}<\text{Na}$.

Both the answer was same for elements $\text{B, O, Ga}$.

The order of increasing electro negativity of $\text{B, O, Ga}$ elements is $\text{Ga}<\text{B}<\text{O}$.

Both the answer was same for elements $\text{F, Cl, Br}$.

The order of increasing electro negativity of elements $\text{F, Cl, Br}$ is $\text{Br}<\text{Cl}<\text{F}$.

Both the answer was same for elements $\text{S, O, F}$.

The order of increasing electro negativity of $\text{S, O, F}$ elements is $\text{S}<\text{O}<\text{F}$.

(ii)

(a)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of most polar bond among $\text{C}-\text{H},\text{ Si}-\text{H},\text{ Sn}-\text{H}$ bonds from exercise 34 and figure 3-4.

Answer to Problem 32E

No, there are no differences.

Explanation of Solution

Refer to figure 3-4

The electro negativity difference of $\text{C}-\text{H}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{C}-\text{H}=2.5-2.1\\ =0.4\end{array}$

The electro negativity difference of $\text{Si}-\text{H}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{Si}-\text{H}=2.1-1.8\\ =0.3\end{array}$

The electro negativity difference of $\text{Sn}-\text{H}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{Sn}-\text{H}=2.1-1.8\\ =0.3\end{array}$

Higher the difference in electro negativity Lower will be the bond polarity. Hence, $\text{Sn}-\text{H}$ bond is more polar.

Repeat exercise 34 (a):

When two different atoms involves in a covalent bond to form a molecule, the shared pair of electrons will not be at the midway between the two atoms. This means the electrons are shared unequally by the atoms. This type of covalent bond is called as polar covalent bond.

Bond polarity is measured by electro negativity of elements. Higher the difference in electro negativity higher will be the bond polarity. Also, size of atomic radii plays a great role in formation of polar covalent bond.

According to Fajan’s rule:

“The covalent bond will be more polar if the size of cation is small and size of anion is large.”

If hydrogen is present in bond formation then p-block elements takes up negative charge and acts as anion.

Therefore, the size of anion such as tin has larger atomic radii whereas carbon and silicon have smaller atomic radii.

Hence, $\text{Sn}-\text{H}$ bond is more polar.

The answer obtained from exercise 34 (a): $\text{Sn}-\text{H}$ bond is more polar.

(b)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of most polar bond among $\text{Al}-\text{Br},\text{ Ga}-\text{Br},\text{ In}-\text{Br},\text{ Tl}-\text{Br}$ bonds from exercise 34 and figure 3-4.

Answer to Problem 32E

No, there are no differences.

Explanation of Solution

Refer to figure 3-4

The electro negativity difference of $\text{Al}-\text{Br}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{Al}-\text{Br}=2.8-1.5\\ =1.3\end{array}$

The electro negativity difference of $\text{Ga}-\text{Br}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{Ga}-\text{Br}=2.8-1.6\\ =1.2\end{array}$

Higher the difference in electro negativity Lower will be the bond polarity. Hence, $\text{Al}-\text{Br}$ is most polar bond.

Repeat exercise 34 (b):

Bond polarity is measured by electro negativity of elements. Higher the difference in electro negativity higher will be the bond polarity. Also, size of atomic radii plays a great role in formation of polar covalent bond.

According to Fajan’s rule:

“The covalent bond will be more polar if the size of cation is small and size of anion is large.”

In this question size of anion is same.

The size of cation Aluminium $\left(\text{Al}\right)$ has smaller atomic radii whereas gallium $\left(\text{Ga}\right)$ and thallium $\left(\text{Tl}\right)$ have bigger atomic radii.

Hence, $\text{Al}-\text{Br}$ is most polar bond.

The answer obtained from exercise 34 (b): $\text{Al}-\text{Br}$ is most polar bond.

(c)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of most polar bond among $\text{C}-\text{O and Si}-\text{O}$ bonds from exercise 34 and figure 3-4.

Answer to Problem 32E

No, there are not any difference.

Explanation of Solution

Refer to figure 3-4

The electro negativity difference of $\text{C}-\text{O}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{C}-\text{O}=3.5-2.5\\ =1.0\end{array}$

The electro negativity difference of $\text{Si}-\text{O}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{Si}-\text{O}=3.5-1.8\\ =1.7\end{array}$

Higher the difference in electro negativity Lower will be the bond polarity. Hence, $\text{C}-\text{O}$ is most polar bond.

Repeat exercise 33 (c):

Bond polarity is measured by electro negativity of elements. Higher the difference in electro negativity higher will be the bond polarity. Also, size of atomic radii plays a great role in formation of polar covalent bond.

According to Fajan’s rule:

“The covalent bond will be more polar if the size of cation is small and size of anion is large.”

In this question size of anion is same.

The size of cation such as silicon $\left(\text{Si}\right)$ has larger atomic radii whereas carbon $\left(\text{C}\right)$ has smaller atomic radii.

Hence, $\text{C}-\text{O}$ is most polar bond.

The answer obtained from exercise 34 (c): $\text{C}-\text{O}$ is most polar bond.

(d)

Interpretation Introduction

Interpretation: The exercise 32 and 34 is to be repeated using electronegativity values of the elements. Any difference, if present, is to be accounted for.

Concept introduction: Electronegativity is defined as the measure of how strongly elements attract bonding electrons to themselves. More the electronegativity more will be the degree of attraction.

A bond formed by sharing of electrons by two atoms is known as covalent bond. It is of two types that is polar and non-polar. A polar covalent bond is a type of bond in which electron pairs are shared unequally

To determine: The comparison of most polar bond among $\text{O}-\text{F or O}-\text{C}$ bonds from exercise 34 and figure 3-4.

Answer to Problem 32E

No, there are not any difference.

Explanation of Solution

Refer to figure 3-4

The electro negativity difference of $\text{O}-\text{F}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{O}-\text{F}=4.0-3.5\\ =0.5\end{array}$

The electro negativity difference of $\text{O}-\text{Cl}$ bond obtained from figure 3-4:

$\begin{array}{c}\text{O}-\text{Cl}=4.0-3.0\\ =1.0\end{array}$

Higher the difference in electro negativity Lower will be the bond polarity. Hence, $\text{O}-\text{F}$ is most polar bond.

Repeat exercise 33 (d):

Bond polarity is measured by electro negativity of elements. Higher the difference in electro negativity higher will be the bond polarity. Also, size of atomic radii plays a great role in formation of polar covalent bond.

According to Fajan’s rule:

“The covalent bond will be more polar if the size of cation is small and size of anion is large.”

Halogen elements are less electro negative than oxygen atom. Therefore, halogens will act as cation whereas oxygen will act as anion.

The size of cation such as Fluorine has smaller atomic radii in comparison to chlorine.

Hence, $\text{O}-\text{F}$ is most polar bond.

The answer obtained from exercise 34 (d): $\text{O}-\text{F}$ is most polar bond.

Conclusion

The answer calculated from figure 3-4 and exercise 34 is:

Both the answer was same for bonds $\text{C}-\text{H},\text{ Si}-\text{H},\text{ Sn}-\text{H}$.

The most polar bond among $\text{C}-\text{H},\text{ Si}-\text{H},\text{ Sn}-\text{H}$ is $\text{Sn}-\text{H}$.

Both the answer was same for bonds $\text{Al}-\text{Br},\text{ Ga}-\text{Br},\text{ In}-\text{Br},\text{ Tl}-\text{Br}$.

The most polar bond among $\text{Al}-\text{Br},\text{ Ga}-\text{Br},\text{ In}-\text{Br},\text{ Tl}-\text{Br}$ is $\text{Al}-\text{Br}$.

Both the answer was same for bonds $\text{C}-\text{O and Si}-\text{O}$.

The most polar bond between $\text{C}-\text{O and Si}-\text{O}$ is $\text{C}-\text{O}$.

Both the answer was same for bonds $\text{O}-\text{F and O}-\text{Cl}$.

The most polar bond between $\text{O}-\text{F and O}-\text{Cl}$ is $\text{O}-\text{F}$.