## Sunday, June 6, 2010

I haven't written for a really long time and that's mainly because I've been busy trying to adapt to my "new life". Yeah, it's been about 6 months since I entered Junior College (JC) and it's really been a blast! I've met many new and amazing people and learnt many new things. The past 6 months practically flew past.

I have noticed a few odd things about the local education system as well. Students are not taught to understand concepts but are taught to memorize concepts instead and apply them in examinations. This totally obscures the beauty behind the subjects and it encourages indifference towards understanding the world.

Let me share an experience where I tried to overcome this barrier created by the education system. I learnt a lot from this experience and it strengthened my love for knowledge and understanding of principles.

In chemistry, we were learning a chapter called chemical bonding. We were shown some images on how molecules would look like when the central atoms of the molecules had different number of electron pairs around them. According to my notes, a molecule with 4 bond pairs around it's central atom only should spread out in 3 dimensions to form a tetrahedral shape. It should have a bond angle of 109.5 degrees and should look like this:
We were not provided with any explanations whatsoever as to why this phenomenon occurred. We were just expected to memorize it... But, I was wondering why this happens. I imagined that the molecule would spread out in 2 dimensions and that it would have a bond angle of 90 degrees. The picture below provides an illustration of my perception. The T represents the terminal atoms while the C represents the central atom. I also assumed that the bond lengths are of the same value, X.
$\large K_{1}\; \textup{can be found using the cosine rule}$

\large \begin{align*} & \therefore K_{1}=\sqrt{2x^{2}-(2x^{2}\cdot \cos\frac{\pi}{2})} \\ &\: \: \: \: \; \; \; \; \;\, = \sqrt{2x^{2}} \end{align*}

By the end of the day, there were three questions perplexing me.

1. Why and how is it that a tetrahedral molecular geometry can form?

2. If the tetrahedral geometry does exist, how does one determine the specific bond angle of 109.5 degrees?

3. How can one claim that one form of molecular geometry exists and not another?

I did some research and found intriguing results that solved all of my problems. Basically, there is a theory known as the Valence shell electron pair repulsion (VSEPR) theory that states that the electron pairs around atoms in a molecule will arrange themselves as far apart as possible in space so as to minimise their mutual repulsion.

1.

To solve my first problem, I had to make use of a molecule whose central atom had three bond pairs of electrons only. It would acquire a trigonal planar shape. It is basically a 2 dimensional molecule and it would like this:
I think this is quite logical as it is the only orientation in which the electron pairs can spread out to achieve minimal repulsion. So imagine the molecule above is lying on a flat piece of paper. The molecule will lie flat on the paper since it has a planar geometry. Then, imagine attaching another terminal atom vertically above A. This terminal atom will be sticking out of your screen. There will be some vertical repulsion by the newly attached terminal atom on the other 3 terminal atoms. This will cause them to bend downwards slightly, until all of the mutual repulsions cancel off. So, if you think about it, a tetrahedral structure will form. This answered my first question as to why does a tetrahedral structure form.

I was really happy when I thought of this. I did some more research and found an amazing derivation that solved my second problem.

2.

Firstly, imagine I have a cube and each of it's sides has a length of 1 cm. Imagine I place a tetrahedral pyramid inside it. It should look like this:
P and Q are two vertices's of tetrahedral pyramid while O is the centre. O is equidistant from all the vertices's of the tetrahedral pyramid and it represents the central atom. Now if we isolate the triangle OPQ, it should look like this:
The length PQ can be found using Pythagoras Theorem.

$\large PQ= \sqrt{1^{2}+1^{2}}= \sqrt{2}$

The length from O to PQ must be half of the vertical length of the cube, since O is the centre. Thus, it is 0.5cm.

To find the bond angle, I need to find the value of 2A. I can find the value of A using trigonometry.

\large \begin{align*} & \\ A= \tan^{-1}(\frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}})\: \, & \\=\tan^{-1}(\sqrt{2})\, \, & \\\\\textup{Bond angle,}\: 2A & \\=2\, \tan^{-1}(\sqrt{2}) &\\ \approx 109.5^{\circ}\; \; \; \; \; \; \;\, \, \,\, \end{align*}

An amazing result! This solved my second problem as to why the bond angle was 109.5 degrees.

3.

Now, to solve the third problem. As proposed by the VSEPR theory, the terminal atoms in a molecule will spread apart as far as possible to achieve minimum mutual repulsion. So, the molecular geometry that will exist must have the greatest inter-terminal atomic distance, K. I had to check which form of geometry had a greater value of K; the tetrahedral geometry or my predicted square planar geometry.

The distance, K2, between the terminal atoms in the tetrahedral geometry can be found using the cosine rule.
\large \begin{align*} & K_{2}=\sqrt{x^{2}+x^{2}-(2x^{2}\cdot\cos109.5^{\circ})}\\\\ & \cos (109.5^{\circ})\\ &=-\cos(180^{\circ}-109.5^{\circ})\\ &=-\cos(70.5^{\circ})< 0 \\&\textup{Since} \, \cos(70.5^{\circ})> 0 \end{align*}

\large \begin{align*} &\\ K_{2}=\sqrt{2x^{2}+(2x^{2}\cdot \cos70.5^{\circ})}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \, \, \, \, \, &\\= \sqrt{2x^{2}}\cdot\sqrt{1+\cos70.5^{\circ}}> \sqrt{2x^{2}}=K_{1} &\\\\ \therefore K_{2}> K_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\: \end{align*}

This value of K2 is greater than the value of K1. This mathematical proof shows why the molecule would spread out in 3 dimensions to form a tetrahedral structure rather than forming the square planar structure that I predicted. This solved my third problem!

I was amazed at the beauty and intricacy of the model. I felt uplifted to a greater state of life after solving this problem. It made me appreciate and love my studies a lot more. It showed me the hard work scientists and mathematicians had to put in to create the models that we take for granted in daily life. I learnt that all problems can be solved by looking deep inside yourself and through thinking very logically and patiently.

Because of this experience, I'm even more determined never to give in the to the intellectual and emotional indifference encouraged by the education system.

References:
http://mathforum.org/library/drmath/view/55023.html

This is what I found on Economist:
WHAT is the most efficient shape for randomly packing things into a container? Physicists at New York University and Virginia Tech have carried out a series of experiments and, among all the shapes they have tested so far, the tetrahedron (a pyramid with four triangular sides) takes some beating. The researchers poured tetrahedral dice into containers which they shook until completely full. A magnetic-resonance imaging scanner was then used to see how tightly the dice packed themselves (pictured above). To quantify the packing efficiency, the researchers poured water into the containers to measure the amount of space between the dice. They report in the latest issue of Physical Review Letters that the tetrahedra filled roughly 76% of the available space in a large container, whereas experiments with spheres typically filled only 64% of the available volume. The physicists say the data will be useful when calculating, for example, how liquids seep through soils of different densities. They think the work might also have some use in packaging consumer products. Watch out for tetrahedral tomatoes.

Keep writing....Puja

2. i think we are both driven by the same curiosity, i had the very same problems (a few minutes ago), and the very same opinion about the educational sytem (and believe me it's not only a local problem) it seems like everyone is happy to learn by heart without any understanding...i'm not.
You helped me a lot on this one, thank you !

greetings from France

3. Hi Sam!

Thank you for writing!I'm glad to know we think alike. Hopefully we'll be able to do something about this and end all the apathy.

Nice to hear from you :)

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