Chapter 9.3, Problem 38PS

Solution

a.

To determine the coordinates $\left(x,y\right)$ of the centre of the oxygen atom in Quadrant $\text{II}\text{.}$

The coordinates $\left(x,y\right)$ are $\left(-58,114\right).$

Given:

The diagram is shown below:

  

Calculation:

Let us draw the line $BC$ for the given diagram as:

  

Now, from the diagram find the angle of $\angle BAC$ as:

  $\begin{array}{l}\angle BAC={180}^{\circ }-{117}^{\circ }\\ \angle BAC={63}^{°}\end{array}$

Using the trigonometric function, the values of $x$ and $y$ can be calculated as:

  $\begin{array}{l}\sin \theta =\frac{\text{Height}}{\text{Hypotenuse}}\\ \sin {63}^{°}=\frac{y}{128}\\ y=128\times 0.891\\ y=114.04\\ y\approx 114\end{array}$

Now,

  $\begin{array}{l}\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}\\ \cos {63}^{°}=\frac{x}{128}\\ x=128\times 0.454\\ x=58.11\\ x\approx 58\end{array}$

As the point lies in the second quadrant, the coordinates of $\left(x,y\right)$ are $\left(-58,114\right).$

b.

To determine the distance $d$ between the centres of the two unbounded oxygen atoms.

The distance $d$ is $218\text{ pm}\text{.}$

Given:

The diagram is shown below:

  

Formula:

The distance $d$ between two coordinates can be calculated as:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Here, the coordinates are:

  $\left(x_1{y}_1\right)=\left(-58,114\right)\text{ and }\left(x_2{y}_2\right)=\left(128,0\right)$

Now, the distance between these two coordinates are:

  $\begin{array}{l}d=\sqrt{{\left(128+58\right)}^2+{\left(0-114\right)}^2}\\ d=\sqrt{{\left(186\right)}^2+{\left(114\right)}^2}\\ d=\sqrt{34596+12996}\\ d=\sqrt{47592}\\ d=218.15\\ d\approx 218\text{ pm}\end{array}$

Hence, the distance $d$ is $218\text{ pm}\text{.}$