Chapter 10, Problem 10.57P

(a)

Interpretation Introduction

Interpretation:

The molecule that has greater dipole moment among ${\text{SO}}_3$ and ${\text{SO}}_2$ is to be identified.

Concept introduction:

The dipole moment arises when there is a separation of charges between two ions or atoms involved in the bond. The dipole moment is a vector quantity and its direction towards the most electronegative atom.

The direction of the dipole moment is represented as follows:

The polar and non-polar molecule can be identified on the basis of the net dipole moment. Polar molecules have non zero value of net dipole moment and the non polar molecules have zero net dipole moment.

Answer to Problem 10.57P

${\text{SO}}_2$ molecule has a greater dipole moment than ${\text{SO}}_3$ molecule.

Explanation of Solution

The shape of ${\text{SO}}_3$ is bent. In ${\text{SO}}_3$, the oxygen atom is more electronegative than sulfur atom so the direction of dipole moment vector is towards oxygen atom.

The dipole moment in ${\text{SO}}_3$ is represented as follows:

All the $\text{S}-\text{O}$ polar bonds cancel each other’s polarity and thus result in a zero net dipole moment.

The shape of ${\text{SO}}_{\text{2}}$ is bent. In ${\text{SO}}_{\text{2}}$, the oxygen atom is more electronegative than sulfur atom so the direction of dipole moment vector is towards oxygen atom.

The dipole moment in ${\text{SO}}_{\text{2}}$ is represented as follows:

The individual dipole moment does not cancel each other so the value of net dipole moment is not zero and therefore ${\text{SO}}_{\text{2}}$ is polar in nature.

Hence, ${\text{SO}}_2$ molecule is more polar than ${\text{SO}}_3$ molecule.

Conclusion

Molecules in which individual dipole moment does not cancel each other result in net non-zero dipole moment and the overall molecule is said to be polar. In ${\text{SO}}_2$ individual dipole moment adds up and ${\text{SO}}_2$ is a polar molecule.

(b)

Interpretation Introduction

Interpretation:

The molecule that has greater dipole moment among $\text{ICl}$ and $\text{IF}$ is to be identified.

Concept introduction:

Electronegativity is the tendency of an atom to attract the shared electrons in the bond towards itself. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative atom. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge and the electropositive atom will acquire a partial positive charge.

Here, B is the electronegative atom and A is the electropositive atom.

Bond polarity can be estimated by $\Delta \text{EN}$. $\Delta \text{EN}$ is the electronegativity difference between the atoms bonded to each other. The formula to calculate $\Delta \text{EN}$ in AB bond is as follows:

  $\Delta \text{EN}=\left(\text{electronegativity of B}\right)-\left(\text{electronegativity of A}\right)$

Here, B is the electronegative atom and A is the electropositive atom.

The greater electronegativity difference between the atoms induces the greater polarity in the bond thus it generates higher dipole moment.

Answer to Problem 10.57P

$\text{IF}$ has greater dipole moment then the $\text{ICl}$ molecule.

Explanation of Solution

The formula to calculate $\Delta \text{EN}$ in $\text{I}-\text{Cl}$ bond is as follows:

  ${\left(\Delta \text{EN}\right)}_{\text{I}-\text{Cl}}=\left(\text{electronegativity of Cl}\right)-\left(\text{electronegativity of I}\right)$        (1)

Substitute $3.0$ for electro negativity of $\text{Cl}$ and $2.5$ for electro negativity of $\text{I}$ in the equation (1).

  $\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{I}-\text{Cl}}=\left(3.0\right)-\left(2.5\right)\\ =0.5\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{I}-\text{F}$ bond is as follows:

  ${\left(\Delta \text{EN}\right)}_{\text{I}-\text{F}}=\left(\text{electronegativity of F}\right)-\left(\text{electronegativity of I}\right)$        (2)

Substitute $4.0$ for electro negativity of $\text{F}$ and $2.5$ for electro negativity of $\text{I}$ in the equation (2).

  $\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{I}-\text{F}}=\left(4.0\right)-\left(2.5\right)\\ =\mathrm{1.5.}\end{array}$

Bond polarity and dipole moment are directly related to the electronegativity difference.  Therefore, $\text{I}-\text{F}$ molecule has greater dipole moment than $\text{I}-\text{Cl}$ molecule.

Conclusion

Greater the electronegativity difference between the two bonding atoms more polar will be the bond.

(c)

Interpretation Introduction

Interpretation:

The molecule that has greater dipole moment among ${\text{SiF}}_4$ and ${\text{SF}}_4$ is to be identified.

Concept introduction:

The dipole moment arises when there is a separation of charges between two ions or atoms involved in the bond. The dipole moment is a vector quantity and its direction towards the most electronegative atom.

The direction of the dipole moment is represented as follows:

The polar and non-polar molecule can be identified on the basis of the net dipole moment. Polar molecules have non zero value of net dipole moment and the non polar molecules have zero net dipole moment.

Answer to Problem 10.57P

${\text{SF}}_4$ molecule has a greater dipole moment than ${\text{SiF}}_4$ molecule.

Explanation of Solution

The shape of ${\text{SiF}}_4$ is tetrahedral. In ${\text{SiF}}_4$, the $\text{F}$  atom is more electronegative than $\text{Si}$ atom so the direction of dipole moment vector is towards oxygen atom.

The dipole moment in ${\text{SiF}}_4$ is represented as follows:

All the $\text{Si}-\text{F}$ polar bonds canceled each other’s polarity with net dipole moment Zero.

The shape of ${\text{SF}}_{\text{4}}$ is see-saw. In ${\text{SF}}_{\text{4}}$, the fluorine atom is more electronegative than sulfur atom so the direction of dipole moment vector is towards fluorine atom.

The dipole moment in ${\text{SF}}_{\text{4}}$ is represented as follows:

The individual dipole moment does not cancel each other so the value of net dipole moment is not zero and therefore ${\text{SF}}_{\text{4}}$ is polar in nature.

Hence, ${\text{SF}}_4$ molecule has greater dipole moment than ${\text{SiF}}_4$ molecule.

Conclusion

Molecules in which individual dipole moment does not cancel each other result in net non-zero dipole moment and the overall molecule is said to be polar. In ${\text{SF}}_{\text{4}}$ individual dipole moment adds up and ${\text{SF}}_{\text{4}}$ is a polar molecule.

(d)

Interpretation Introduction

Interpretation:

The molecule that has greater dipole moment among ${\text{H}}_2\text{O}$ and ${\text{H}}_2\text{S}$ is to be identified.

Concept introduction:

Electronegativity is the tendency of an atom to attract the shared electrons in the bond towards itself. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative atom. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge and the electropositive atom will acquire a partial positive charge.

Here, B is the electronegative atom and A is the electropositive atom.

Bond polarity can be estimated by $\Delta \text{EN}$. $\Delta \text{EN}$ is the electro negativity difference between the atoms bonded to each other. The formula to calculate $\Delta \text{EN}$ in AB bond is as follows:

  $\Delta \text{EN}=\left(\text{electronegativity of B}\right)-\left(\text{electronegativity of A}\right)$

Here, B is the electronegative atom and A is the electropositive atom.

The greater electronegativity difference between the atoms induces the greater polarity in the bond thus it generates higher dipole moment.

Answer to Problem 10.57P

${\text{H}}_2\text{O}$ has greater dipole moment then the ${\text{H}}_2\text{S}$ molecule.

Explanation of Solution

The formula to calculate $\Delta \text{EN}$ in $\text{O}-\text{H}$ bond is as follows:

  ${\left(\Delta \text{EN}\right)}_{\text{O}-\text{H}}=\left(\text{electronegativity of O}\right)-\left(\text{electronegativity of H}\right)$        (3)

Substitute $3.5$ for electro negativity of $\text{O}$ and $2.1$ for electro negativity of $\text{H}$ in the equation (3).

  $\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{O}-\text{H}}=\left(3.5\right)-\left(2.1\right)\\ =1.4\end{array}$

The formula to calculate $\Delta \text{EN}$ in $\text{S}-\text{H}$ bond is as follows:

  ${\left(\Delta \text{EN}\right)}_{\text{S}-\text{H}}=\left(\text{electronegativity of S}\right)-\left(\text{electronegativity of H}\right)$        (4)

Substitute $3.5$ for electro negativity of $\text{O}$ and $2.1$ for electro negativity of $\text{H}$ in the equation (4).

  $\begin{array}{c}{\left(\Delta \text{EN}\right)}_{\text{S}-\text{H}}=\left(2.5\right)-\left(2.1\right)\\ =\mathrm{0.4.}\end{array}$

Both ${\text{H}}_{\text{2}}\text{O}$ and ${\text{H}}_{\text{2}}\text{S}$ has polar bonds and have a similar shape but $\text{H}-\text{O}$ has a larger electronegative difference than ${\text{H}}_{\text{2}}\text{S}$. Therefore ${\text{H}}_{\text{2}}\text{O}$ has a greater dipole moment.

Conclusion

Greater the electronegativity difference between the two bonding atoms more polar will be the bond.