Chapter 10, Problem 10.4P

Interpretation Introduction

Interpretation:

The magnitude and direction of the electric dipole moment of charges in given xy plane has to be given with the given information.

Concept introduction:

Polar molecules are the molecules having a positive and negative end and so there will be a charge separation.

A polar molecule is a molecule where the polar bonds are asymmetrically arranged (the dipoles do not cancel)

A nonpolar molecule is a molecule with no polar bonds or a molecule where the polar bonds are symmetrically arranged.

In polar molecule the charge separation occurred with respect to the difference in electronegativity of atoms in the molecule.

Calculation of the energy dipole moment for polyatomic molecules:

$\begin{array}{c}\text{μ}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{x}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{y}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{z}}\right)}^{\text{1}/\text{2}}\\ \\ {\text{μ}}_{\text{x}}\text{-}\text{Dipole}\text{moment}\text{vector}\text{in}\text{x}\text{direction}\text{.}\\ \\ {\text{μ}}_{\text{y}}\text{-}\text{Dipole}\text{moment}\text{vector}\text{in}\text{y}\text{direction}\text{.}\\ \\ {\text{μ}}_{\text{z}}\text{-}\text{Dipole}\text{moment}\text{vector}\text{in}\text{z}\text{direction}\\ \end{array}$

The dipole moment vector in x direction can find out by the equation given below:

$\begin{array}{c}{\text{μ}}_{\text{x}}\text{=}\text{å}{\text{Q}}_{\text{J}}{\text{x}}_{\text{J}}\\ \\ {\text{Q}}_{\text{J}}\text{-}\text{Partial}\text{charge}\text{of}\text{the}\text{atom}\text{J}\text{.}\\ \\ {\text{x}}_{\text{J}}\text{-}\text{x}\text{coordinate}\text{of}\text{the}\text{atom}\text{J}\end{array}$

Explanation of Solution

Given data is as follows:

  $\begin{array}{l}{\text{Q}}_{\text{1}}\text{=}\text{3e}\\ {\text{Q}}_{\text{2}}\text{=}\text{-e}\\ {\text{Q}}_{\text{3}}\text{=}\text{-2e}\\ \\ {\text{x}}_{\text{1}}{\text{,y}}_{\text{1}}\text{=}\left(\text{0,0}\right)\\ {\text{x}}_{\text{2}}{\text{,y}}_{\text{2}}\text{=}\left(\text{0}\text{.32}\text{nm,0}\right)\end{array}$

Given that the charge -2e is at an angle of ${\text{20}}^{\text{°}}$ from the x-axis and a distance of $\text{0}\text{.23}\text{nm}$ from the origin.

By writing equations for $\text{sin20}\text{andcos20}$, the x coordinate and y coordinate can be determined.

The coordinates at the charge $-2e$ is given below:

$\begin{array}{l}\text{x}\text{=}\text{0}\text{.23}\text{sin}\text{20}\text{=}\text{7}{\text{.86×10}}^{\text{-11}}\text{m}\\ \\ \text{y}\text{=}\text{0}\text{.23}\text{cos}\text{20}\text{=}\text{2}{\text{.16×10}}^{\text{-10}}\text{m}\end{array}$

Therefore,

${\text{x}}_{\text{3}}{\text{,y}}_{\text{3}}\text{=}\left(\text{7}{\text{.86×10}}^{\text{-11}}\text{m,2}{\text{.16×10}}^{\text{-10}}\text{m}\right)$

The equation for calculating ${\text{μ}}_{\text{x}}$ is given below:

  $\begin{array}{c}{\text{μ}}_{\text{x}}\text{=}\text{å}{\text{Q}}_{\text{J}}{\text{x}}_{\text{J}}\\ \\ {\text{Q}}_{\text{J}}\text{-}\text{Partial}\text{charge}\text{of}\text{the}\text{atom}\text{J}\text{.}\\ \\ {\text{x}}_{\text{J}}\text{-}\text{x}\text{coordinate}\text{of}\text{the}\text{atom}\text{J}\end{array}$

$\begin{array}{c}{\text{μ}}_{\text{x}}\text{=}\left({\text{Q}}_{\text{1}}{\text{×x}}_{\text{1}}\right)\text{+}\left({\text{Q}}_{\text{2}}{\text{×x}}_{\text{2}}\right)\text{+}\left({\text{Q}}_{\text{3}}{\text{×x}}_{\text{3}}\right)\\ \\ \text{=}\left(\text{3e×0}\text{pm}\right)\text{+}\left(\text{-e×0}{\text{.32×10}}^{\text{-9}}\text{m}\right)\text{+}\left(\text{-2e×7}{\text{.86×10}}^{\text{-11}}\text{m}\right)\\ \\ \text{=-4}{\text{.77×10}}^{\text{-10}}\text{e}\text{m}\\ \\ \text{=}\text{-7}{\text{.63×10}}^{\text{-29}}\text{C}\text{.m}\end{array}$

$\begin{array}{c}{\text{μ}}_{\text{y}}\text{=}\left({\text{Q}}_{\text{1}}{\text{×y}}_{\text{1}}\right)\text{+}\left({\text{Q}}_{\text{2}}{\text{×y}}_{\text{2}}\right)\text{+}\left({\text{Q}}_{\text{3}}{\text{×y}}_{\text{3}}\right)\text{+}\left({\text{Q}}_{\text{4}}{\text{×y}}_{\text{4}}\right)\frac{\text{Opposite}}{\text{Hypotenuse}}\\ \\ \text{=}\left(\text{-0}\text{.38e×118}\text{pm}\right)\text{+}\left(\text{0}\text{.45e×0}\text{pm}\right)\text{+}\left(\text{0}\text{.02e×-61pm}\right)\text{+}\left(\text{0}\text{.02e×-61pm}\right)\\ \\ \text{=}\text{-47}\text{.28}\text{e}\text{pm}\\ \\ \text{=}\text{-7}{\text{.57×10}}^{\text{-30}}\text{C}\text{.m}\end{array}$

$\begin{array}{c}{\text{μ}}_{\text{y}}\text{=}\left({\text{Q}}_{\text{1}}{\text{×y}}_{\text{1}}\right)\text{+}\left({\text{Q}}_{\text{2}}{\text{×y}}_{\text{2}}\right)\text{+}\left({\text{Q}}_{\text{3}}{\text{×y}}_{\text{3}}\right)\text{+}\left({\text{Q}}_{\text{4}}{\text{×y}}_{\text{4}}\right)\frac{\text{Opposite}}{\text{Hypotenuse}}\\ \\ \text{=}\left(\text{-0}\text{.38e×118}\text{pm}\right)\text{+}\left(\text{0}\text{.45e×0}\text{pm}\right)\text{+}\left(\text{0}\text{.02e×-61pm}\right)\text{+}\left(\text{0}\text{.02e×-61pm}\right)\\ \\ \text{=}\text{-47}\text{.28}\text{e}\text{pm}\\ \\ \text{=}\text{-7}{\text{.57×10}}^{\text{-30}}\text{C}\text{.m}\end{array}$

The value can be converted into Debye by the relation $\text{1}\text{D}\text{=}\text{3}{\text{.336×10}}^{\text{-30}}\text{C}\text{.m}$

$\text{-7}{\text{.57×10}}^{\text{-30}}\text{C}\text{.m}\text{=}\text{-2}\text{.27}\text{D}$

$\begin{array}{c}{\text{μ}}_{\text{z}}\text{=}\left({\text{Q}}_{\text{1}}{\text{×z}}_{\text{1}}\right)\text{+}\left({\text{Q}}_{\text{2}}{\text{×z}}_{\text{2}}\right)\text{+}\left({\text{Q}}_{\text{3}}{\text{×z}}_{\text{3}}\right)\text{+}\left({\text{Q}}_{\text{4}}{\text{×z}}_{\text{4}}\right)\\ \\ \text{=}\left(\text{-0}\text{.38e×0pm}\right)\text{+}\left(\text{0}\text{.45e×0}\text{pm}\right)\text{+}\left(\text{0}\text{.02e×0pm}\right)\text{+}\left(\text{0}\text{.02e×0pm}\right)\\ \\ \text{=}\text{0}\end{array}$

$\begin{array}{l}\text{μ}\text{=}{\left({\text{μ}}^{\text{2}}{}_{\text{x}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{y}}\text{+}{\text{μ}}^{\text{2}}{}_{\text{z}}\right)}^{\text{1}/\text{2}}\\ \text{=}{\left(\text{0}\text{+}{\left(\text{-2}\text{.27}\text{D}\right)}^{\text{2}}\text{+}\text{0}\right)}^{\text{1}/\text{2}}\\ \text{=}\text{2}\text{.27}\text{D}\end{array}$

The magnitude of the electric dipole moment is $\text{2}\text{.27}\text{D}$.