Why Only Five Platonic Solids in Our Geo-Bio-Chemical World?

The DNA-surrounding capsid of the cold virus and the atomic arrangement of  a certain allotrope of  boron consists of  20 triangles arranged in a three dimensional shape known as a icosahedron.

If you form a three-dimensional structure with 8 triangles, you get an octahedron. An example is the molecular structure of the electrical insulator and potent greenhouse gas, sulfur hexafluoride, SF6. It can also appear in the mineral pyrite.

If you limit the number of triangles to four, you get a tetrahedron. The tetrahedral silicate unit,  SiO42- , is the basic component of most silicates in the Earth’s crust. On our planet’s surface, any time carbon makes four single bonds with other carbons or other elements in a wide variety of life’s hydrocarbons, we also get a tetrahedral arrangement. This allows the four bonding pairs of electrons to get as far from each other as possible.

Changing polygon, we can use 6 squares to make a cube. The similarly-sized ions of cesium and chloride can form an arrangement where each whole ion centers 8 vertices occupied by an ion of the opposite charge. But viewed within the lattice, the boundaries of the cube are such that only 1/8 of each ion is at a vertex. Given that there are 8 vertices, this maintains the ratio of one cesium for every one chloride ion. In the mineral halite, which is composed of NaCl, the chloride ion is considerably bigger than the sodium. They don’t pack into a cube in the same manner, but overall they still form the same shape.

Then there is the dodecahedron, consisting of a dozen pentagons. Cubes of pyritohedron form the macro-illusion of a dodecahedron, but otherwise this “perfect” solid is rare in nature.pyrite_really_cubes_molecular

Are there more than these 5 possible perfect or Platonic solids? No. But why?

There is a relationship that holds for all solids of this type. If you let V represent the number of vertices, E = number of edges and F = the number polygonal faces, slide1-l

you will observe that in a tetrahedron V= 4, E = 6 and F = 4.

For a cube, download

Notice that for both solids, the following simple formula(called Euler’s Formula)holds true:

V + F – E = 2.                                 Equation (1)

That doesn’t constitute a proof for why it should apply to all cases, but you can find one here. We can use this formula to prove that there are only a limited number of Platonic solids.

First let’s introduce two other variables, N = the number of sides in a polygonal face and R = number of edges that meet at a vertex.

Since each edge is shared by two vertices, if we multiply R by the number of vertices,V , and divide by two, we will get the total  number of  edges:

RV / 2  = E;   Solving for V we get

V = 2E / R                                        Equation (2)

The number of edges can also be obtained by the number of faces. Each face has one edge for each of the number of sides, N. But each edge is shared with another face, so again not to count things twice:

NF / 2 = E.   Solving for F we get

F = 2E / N                                        Equation (3)

Substituting equations (2) and (3) into equation 1:

2E / R  + 2E / N   – E = 2.     

Now divide each term by 2E:

1/R + 1/N1/2 = 1/E. 

We need at least three edges to get a 3D shape so R ≥ 3. Similarly to get a polygon, N ≥ 3. Interestingly N and R cannot simultaneously be greater than 3,  because as they create progressively smaller fractions, 1/R  and 1/N will add up to a maximum sum of 1/2 ( if R=N=4), which in the formula will yield 0 = 1/E.  

Letting N = 3,  if R = 3  then E = 6. Using this result and equation(3),  F = 2E/N = 2(6)/3= 4: the tetrahedron.

Having no choice due to the restriction we mentioned, we have to keep either N or R at 3 while increasing the other, so

letting N = 3,  if R = 4, E = 12 and F = 8, the octahedron. 

Reversing the values and letting N = 4 and R = 3, E = 12 and F = 2(12)/4= 6,  the cube.

We can then try the combinations of N= 3,  R= 5 and N= 5, R= 3, which will solve for the icosahedron ( E = 30; F = 20) and dodecahedron ( E = 30; F = 12), respectively.

But 5 is the limit because if we try values of 3 and 6:

1/3 + 1/6 – 1/2 = 0, which means the impossibility of no edges. A value larger than 6 yields a negative value for E.

Given that there are only these 5 solutions to Euler’s formula , then only five Platonic solids can exist in three dimensional space.

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From a Pesticide to an Alternate Quadratic Formula

In the English language, changing the suffix of a word like baked to baking might make it conform to the rules of grammar, but it won’t dramatically change the meaning of the word. That’s not the case when it comes to the language of chemistry.

Consider aluminum phosphide versus aluminum phosphate. Aluminum phosphide (AlP) is a common and inexpensive pesticide. But it releases deadly phosphine gas when it comes in contact either with moisture or with hydrochloric acid in the stomach , as shown by the following reactions:

AlP + 3 H2O → Al(OH)3 + PH3

AlP + 3 HCl → AlCl3 + PH3

Aluminum phosphide-poisoning is a serious problem in the Middle East and Asia. It has also led to rare but profound tragedies in the United States.  Four children were killed in Texas  last year (2017) after Fumitoxin pellets near their mobile home were hosed down by their father after neighbors complained about the smell.

Change phosphide’s suffix from ide to ate, and you get a totally different compound known as aluminum phosphate (AlPO4). This substance is formed from the reaction between aluminum sulfate and phosphate in waste water. Since AlPO is poorly soluble, the phosphate is unavailable to algae and eutrophication becomes less likely.

The maximum solubility of aluminum phosphate at a given temperature can be calculated from its equilibrium solubility product constant, known as Ksp.

For AlPO4  the expression is given by:

Ksp= [ Al 3+][ PO43-],

where the square brackets represent the equilibrium concentration of the respective ions in moles/L. Since the stoichiometric ratio of AlPO4  to Al 3+ is one to one,  the latter’s concentration is equal to how many moles of AlPO4 dissolved per liter of solution.

But what if there was already some PO43- present in a solution before AlPO4 appeared? In the same manner that any phosphate naturally present in an individual’s mouth makes his teeth enamel less vulnerable to bacterial acidic attack, phosphate’s present will decrease the solubility of AlPO4 .

Let’s say there are 0.050 M of phosphate already present, then the Ksp expression becomes:

Ksp= [ x ][ x +0.050] . Since Ksp for AlPO4 is 9.84 X 10-21. Expanding we get:

x2 + 0.050x – 9.84 X 10-21 = 0.

But if this is solved with a typical calculator, it will yield an answer of zero. this happens because in the quadratic formula based on ax2+ bx + c = 0 :

quadraticb2 >> 4ac, which means that the square rooted expression yields b to the typical calculator. ± b, when combined with –b from the formula’s numerator yields either the useless negative value and zero.

What’s the solution then?

In such cases we can use an alternate version of the quadratic formula, which can be obtained by multiplying the numerator and denominator by –±√( b2– 4ac). This will yield:

QuadraticAlternate

With this version of the formula, when b2 >> 4ac, we get either an undefined expression (division by zero) which we ignore, or approximately c/-b, which in our example gives us the solubility(x) to be not quite zero (although awfully close!) but 1.97 X 10-19 mol/L. More generally this version increases accuracy anytime the absolute value of one of the roots is much smaller than the other. A calculator can be easily programmed with conditional statements so that it can handle any situation. I wrote this one for a TI-83.

quadprogram

 

Other Sources:
Managing aluminum phosphide poisonings
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3162709/

The Quadratic Equation http://cse.unl.edu/~ylu/raik283/notes/Resoureces/Quadratic_equation.htm

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