Chapter 8, Problem 31E

(i)

(a)

Interpretation Introduction

Interpretation: The exercise 31 and 33 is to be repeated using electro-negativity values of the elements. Any difference, if present, is to be accounted for.

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

To determine: The order of increasing electro-negativity and polarity of given elements from exercise 31 and 33; the comparison with figure 3-4.

Answer to Problem 31E

No, there are no differences.

Explanation of Solution

Refer to figure 3-4.

The electro negativity of carbon $\left(\text{C}\right)$, nitrogen $\left(\text{N}\right)$, and oxygen $\left(\text{O}\right)$ elements obtained from figure 3-4 is:

• Carbon: $\text{2}\text{.5}$
• Oxygen: $\text{3}\text{.5}$
• Nitrogen: $\text{3}\text{.0}$

Thus, the increasing order of electro negativity is $\text{C < N < O}$.

Repeat exercise 31 (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 electro negativity increases along the period from left to right.

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

The carbon $\left(\text{C}\right)$, nitrogen $\left(\text{N}\right)$, and oxygen $\left(\text{O}\right)$ are the elements of period 2 and their electronic configurations are as follows:

• Carbon: ${\text{1s}}^2{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}$
• Nitrogen: ${\text{1s}}^2{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$
• oxygen: ${\text{1s}}^2{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$

The electro negativity increases along the period from left to right. Hence the correct order of increasing electro negativity of $\text{C, N, O}$ element is:

• $\text{C < N < O}$

The answer obtained from exercise 31 (a) is $\text{C < N < O}$.

(b)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The order of increasing electro-negativity of $\text{S, Se, Cl}$ elements from exercise 31; the comparison with figure 3-4.

Answer to Problem 31E

No, there is  no difference.

Explanation of Solution

Refer to figure 3-4

The electro negativity of Selenium (Se), sulfur (S) and chlorine (Cl) obtained from figure 3-4 are:

• Selenium: 2.4
• Sulfur: 2.5
• Chlorine: 3.0

The increasing order of electro negativity is $\text{Se < S < Cl}$.

Repeat exercise 31 (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 electro-negativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electro-negativity decreases down the group.

The element Selenium $\left(\text{Se}\right)$ lies in 4^th period, 16th group, sulfur $\left(\text{S}\right)$ lie in 3^rd period, 16th group and chlorine $\left(\text{Cl}\right)$ lie in 3^rd period, 17th group. The electronic configuration of these elements as follows:

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

The electro-negativity decreases down the group. Hence the correct order of increasing electro-negativity of $\text{S, Se, Cl}$ element is:

• $\text{Se < S < Cl}$

The answer obtained from exercise 31(b) is $\text{Se < S < Cl}$.

(c)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The order of increasing electro-negativity of $\text{Si, Ge, Sn}$ elements from exercise 31; the comparison with figure 3-4.

Answer to Problem 31E

Yes, the answer is different.

Explanation of Solution

Refer to figure 3-4

The electro-negativity of Silicon $\left(\text{Si}\right)$, Germanium $\left(\text{Ge}\right)$ and Tin $\left(\text{Sn}\right)$ obtained from figure 3-4 are:

• Silicon: 1.8
• Germanium: 1.8
• Tin: 1.8

The increasing order of electro-negativity is $\text{Si}=\text{Ge}=\text{Sn}$.

Repeat exercise 31 (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 electro-negativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electro-negativity decreases down the group.

The element Silicon lies in 3^rd period, 14th group, Germanium lie in 4^th period, 14th group and Tin lie in 5^th period, 14th group. The electronic configuration of these elements as follows:

• Silicon: ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^{\text{2}}$
• Germanium: ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{2}}$
• Tin: ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^6{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{2}}$

The electro-negativity increases along the period from left to right. Hence the correct order of increasing electro-negativity of $\text{Si, Ge, Sn}$ element is:

• $\text{Sn}<\text{Ge}<\text{Si}$

The answer obtained from exercise 31(c) is $\text{Sn}<\text{Ge}<\text{Si}$.

(d)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The order of increasing electro-negativity of $\text{Tl, S, Ge}$ elements from exercise 31; the comparison with figure 3-4.

Answer to Problem 31E

Yes, the answer is different.

Explanation of Solution

Refer to figure 3-4

The electro-negativity of Thallium $\left(\text{Tl}\right)$, Germanium $\left(\text{Ge}\right)$ and sulfur $\left(\text{S}\right)$ obtained from figure 3-4 are:

• Thallium: 1.8
• Germanium: 1.8
• Sulfur: 2.5

The increasing order of electro-negativity is $\text{Tl}=\text{Ge}<\text{S}$.

Repeat exercise 31 (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 electro-negativity increases along the period from left to right.

The atomic radii increase down the group because nuclear attraction decreases. As a result electro-negativity decreases down the group.

The element Thallium lies in 6^th period, 13th group, Germanium lie in 4^th period, 14th group and sulfur lie in 3^rd period, 16th group. The electronic configuration of these elements as follows:

• Thallium: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{1}}$
• Germanium: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{2}}$
• Sulfur: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}$

The electro-negativity increases along the period from left to right and decreases down the group. Hence the correct order of increasing electro-negativity of $\text{Tl, S, Ge}$ element is:

• $\text{Tl}<\text{Ge}<\text{S}$

The answer obtained from exercise 31(d) is $\text{Tl}<\text{Ge}<\text{S}$.

Conclusion

Both the answer was same for elements $\text{C, N, O}$.

The order of increasing electro-negativity of $\text{C, N, O}$ elements is $\text{C}<\text{N}<\text{O}$.

Both the answer was same for elements $\text{S, Se, Cl}$.

The order of increasing electro-negativity of $\text{S, Se, Cl}$ elements is $\text{Se}<\text{S}<\text{Cl}$.

The answer obtained from figure 3-4 was different from exercise 31 (c).

According to figure 3-4,

The order of increasing electro-negativity of $\text{Si, Ge, Sn}$ elements is $\text{Sn}=\text{Ge}=\text{Si}$.

According to exercise 31 (c),

The order of increasing electro-negativity of $\text{Si, Ge, Sn}$ elements is $\text{Sn}<\text{Ge}<\text{Si}$.

The answer obtained from figure 3-4 was different from exercise 31 (d).

According to figure 3-4,

The order of increasing electro-negativity of $\text{Tl, S, Ge}$ elements is $\text{Tl}=\text{Ge}<\text{S}$.

According to exercise 31 (d),

The order of increasing electro-negativity of $\text{Tl, S, Ge}$ elements is $\text{Tl}<\text{Ge}<\text{S}$.

(ii)

(a)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The most polar bond among $\text{C}-\text{F},\text{ Si}-\text{F},\text{ Ge}-\text{F}$ bonds from exercise 33; the comparison with figure 3-4.

Answer to Problem 31E

No, there are no differences.

Explanation of Solution

Refer to figure 3-4

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

$\begin{array}{c}\text{C}-\text{F}=4.0-2.5\\ =1.5\end{array}$

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

$\begin{array}{c}\text{Si}-\text{F}=4.0-1.8\\ =2.2\end{array}$

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

$\begin{array}{c}\text{Ge}-\text{F}=4.0-1.8\\ =2.2\end{array}$

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

Repeat exercise 33 (a):

Bond polarity is measured by electro-negativity of elements. Higher the difference in electro-negativity lower 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 Germanium $\left(\text{Ge}\right)$ and silicon $\left(\text{Si}\right)$ has larger atomic radii whereas carbon $\left(\text{C}\right)$ has smaller atomic radii.

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

The answer obtained from exercise 33 (a): $\text{C}-\text{F}$ bond is most polar.

(b)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The most polar bond between $\text{P}-\text{Cl or S}-\text{Cl}$ bonds from exercise 33; the comparison with figure 3-4.

Answer to Problem 31E

No, there are  no differences.

Explanation of Solution

Refer to figure 3-4

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

$\begin{array}{c}\text{P}-\text{Cl}=3.0-2.1\\ =0.9\end{array}$

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

$\begin{array}{c}\text{S}-\text{Cl}=3.0-2.5\\ =0.5\end{array}$

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

Repeat exercise 33 (b):

Bond polarity is measured by electro-negativity of elements. Higher the difference in electro-negativity lower 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 Sulfur $\left(\text{S}\right)$ has smaller atomic radii whereas phosphorus $\left(\text{P}\right)$ atom has bigger atomic radii.

Hence, $\text{S}-\text{Cl}$ is most polar bond.

The answer obtained from exercise 33 (b): $\text{S}-\text{Cl}$ is most polar bond.

(c)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The most polar bond among $\text{S}-\text{F},\text{ S}-\text{Cl},\text{ S}-\text{Br}$ bonds from exercise 33; the comparison with figure 3-4.

Answer to Problem 31E

No, there are  no differences.

Explanation of Solution

Refer to figure 3-4

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

$\begin{array}{c}\text{S}-\text{F}=4.0-3.0\\ =1.0\end{array}$

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

$\begin{array}{c}\text{S}-\text{Cl}=3.0-2.5\\ =0.5\end{array}$

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

$\begin{array}{c}\text{S}-\text{Br}=2.8-2.5\\ =0.3\end{array}$

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

Repeat exercise 33 (c):

Bond polarity is measured by electro-negativity of elements. Higher the difference in electro-negativity lower 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 cation is same.

The size of anion such as chlorine $\left(\text{Cl}\right)$ and fluorine $\left(\text{F}\right)$ has smaller atomic radii whereas Bromine $\left(\text{Br}\right)$ has higher atomic radii.

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

The answer obtained from exercise 33 (c) $\text{S}-\text{Br}$ is most polar bond.

(d)

Interpretation Introduction

Interpretation: The exercise 31 and 33 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.

To determine: The most polar bond among $\text{Ti}-\text{Cl},\text{ Si}-\text{Cl},\text{ Ge}-\text{Cl}$ bonds from exercise 33; the comparison with figure 3-4.

Answer to Problem 31E

Yes, the answer is different.

Explanation of Solution

Refer to figure 3-4

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

$\begin{array}{c}\text{Ti}-\text{Cl}=3.0-1.5\\ =1.5\end{array}$

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

$\begin{array}{c}\text{Si}-\text{Cl}=3.0-1.8\\ =1.2\end{array}$

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

$\begin{array}{c}\text{Ge}-\text{Cl}=3.0-1.8\\ =1.2\end{array}$

Higher the difference in electro-negativity Lower will be the bond polarity. Hence, $\text{Si}-\text{Cl}$ and $\text{Ge}-\text{Cl}$ are most polar bonds.

Repeat exercise 33 (d):

Bond polarity is measured by electro-negativity of elements. Higher the difference in electro-negativity lower 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 smaller atomic radii in comparison to Titanium and germanium.

Hence, $\text{Si}-\text{Cl}$ is most polar bond.

The answer obtained from exercise 33 (d): $\text{Si}-\text{Cl}$ is most polar bond.

Conclusion

The answer calculated from figure 3-4 and exercise 33 is:

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

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

Both the answer was same for bonds $\text{P}-\text{Cl or S}-\text{Cl}$.

The most polar bond between $\text{P}-\text{Cl or S}-\text{Cl}$ is $\text{S}-\text{Cl}$.

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

The most polar bond among $\text{S}-\text{F},\text{ S}-\text{Cl},\text{ S}-\text{Br}$ is $\text{S}-\text{Br}$.

The answer obtained from figure 3-4 was different from exercise 33 (d).

According to figure 3-4,

The most polar bond among $\text{Ti}-\text{Cl},\text{ Si}-\text{Cl},\text{ Ge}-\text{Cl}$ is $\text{Si}-\text{Cl}$ and $\text{Ge}-\text{Cl}$.

According to exercise 33 (d),

The most polar bond among $\text{Ti}-\text{Cl},\text{ Si}-\text{Cl},\text{ Ge}-\text{Cl}$ is $\text{Si}-\text{Cl}$.