Chapter 12.5, Problem 17P

To determine

Find the angle between two of the carbon-chlorine bonds.

Answer to Problem 17P

  $\theta \approx 109.47°$

Explanation of Solution

Given:

In a molecule of carbon tetrachloride, the four chlorine atoms are at the vertices of a regular tetrahedron, with the carbon atom in the center.

Formula Used:

  $\cos \theta =\frac{\text{length of adjacent side to }\theta }{\text{length of hypotenuse }}$

Calculation:

Represent the given situation:

  

Let the angle between two of the carbon-chlorine bonds be $\theta$ and the length of each carbon-chlorine bond be 1 unit.

We need to find $\theta$ .

  

Let h be the height from the carbon atom to the base of the tetrahedron. Above is the 2D diagram of the tetrahedron.

We now write equations :

  $\begin{array}{l}\cos {\theta }_s=\frac{\text{length of adjacent side to }{\theta }_s}{\text{length of hypotenuse }}\\ \Rightarrow \cos {\theta }_s=\frac{h}{1}\\ \Rightarrow \cos {\theta }_s=h\\ \Rightarrow \cos \theta =\cos (180°-{\theta }_s)=-\cos {\theta }_s\\ \Rightarrow \cos \theta =-h\\ \Rightarrow \theta ={\cos }^{-1}(-h)\mathrm{..........................}(1)\end{array}$

It is clear that the big tetrahedron is made up of four smaller tetrahedrons. So, the volume of the big tetrahedron is 4 times the volume of the smaller tetrahedron.

The base of the big and the small tetrahedron is same.

So, the base area B of both the tetrahedrons is same.

Now,

  $\begin{array}{l}{V}_{big}=4\cdot V_{small}\\ \Rightarrow \frac{1}{3}(base\text{ area)(height)}=4\cdot \frac{1}{3}(base\text{ area)(height)}\\ \Rightarrow \frac{1}{3}B\text{(}h+1\text{)=}\frac{4}{3}Bh\\ \Rightarrow h+1=4h\\ \Rightarrow 3h=\mathrm{1....................}[\text{subtract h from each side]}\\ \Rightarrow h=\frac{1}{3}\mathrm{....................}[\text{divide each side by 3]}\end{array}$

  $\begin{array}{l}\theta ={\cos }^{-1}(-h)\\ \Rightarrow \theta ={\cos }^{-1}\left(-\frac{1}{3}\right)\\ \Rightarrow \theta \approx 109.47°\end{array}$