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### Topic: Problem of the week - 11/03/2013  (Read 34921 times)

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#### Borek

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##### Problem of the week - 11/03/2013
« on: March 11, 2013, 01:09:38 PM »
Given that the density of silicon carbide is 3.21 g/cm3 and that of diamond is 3.51 g/cm3, calculate the covalent radius of a silicon atom in silicon carbide crystal.
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#### Rutherford

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##### Re: Problem of the week - 11/03/2013
« Reply #1 on: March 11, 2013, 02:12:31 PM »
Edit: result noted. Let's wait for others.
« Last Edit: March 11, 2013, 03:56:51 PM by Borek »

#### Rutherford

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##### Re: Problem of the week - 11/03/2013
« Reply #2 on: March 13, 2013, 08:53:22 AM »
Have you changed the wording of this problem? I thought that it is needed to calculate the length of Si-C bond in the carbide. For the covalent radius of silicon I have a different answer.

#### Borek

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##### Re: Problem of the week - 11/03/2013
« Reply #3 on: March 13, 2013, 09:23:39 AM »
No idea how you understood the problem then, and how you understand it now. Wording has not changed.
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#### Rutherford

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##### Re: Problem of the week - 11/03/2013
« Reply #4 on: March 13, 2013, 09:34:34 AM »
Okay, sorry then  . I initially thought that it is asked to calculate the Si-C bond length in silicon carbide crystal, and that's the answer I posted. I saw now that it is asked to calculate the covalent radius of Si, therefore I have a new answer that I will post (with calculations) when you want me to.

#### Borek

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##### Re: Problem of the week - 11/03/2013
« Reply #5 on: March 14, 2013, 11:32:12 AM »
No more takers?
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#### Mr. Inquisitive

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##### Re: Problem of the week - 11/03/2013
« Reply #6 on: March 16, 2013, 03:23:12 AM »
Although silicon carbide have three major polytypes (β)3C-SiC, 4H-SiC, and (α)6H-SiC and 250 crystalline forms I only found from the table that the covalent radius is 111 pm. Is this correct?

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##### Re: Problem of the week - 11/03/2013
« Reply #7 on: March 17, 2013, 08:01:07 AM »
Do we go for a "simple cubic" lattice structure here? If not, which do we need ...

#### Borek

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##### Re: Problem of the week - 11/03/2013
« Reply #8 on: March 17, 2013, 08:06:05 AM »
Go for diamond.
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##### Re: Problem of the week - 11/03/2013
« Reply #9 on: March 17, 2013, 09:10:38 AM »
Go for diamond.

I'm out for the moment, I'll have to wait for your next hint! What I need is a description of the unit cell, without that I have no idea
« Last Edit: March 17, 2013, 11:50:36 AM by Big-Daddy »

#### Borek

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##### Re: Problem of the week - 11/03/2013
« Reply #10 on: March 18, 2013, 12:13:32 PM »
You can easily google the unit cell for diamond, this is an open book problem!

So far we have no correct answer (the one Raderford gave originally - 94 pm - was wrong).
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#### Rutherford

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##### Re: Problem of the week - 11/03/2013
« Reply #11 on: March 18, 2013, 02:46:40 PM »
Yes, but I got a new answer by calculation as I wrote:
Okay, sorry then  . I initially thought that it is asked to calculate the Si-C bond length in silicon carbide crystal, and that's the answer I posted. I saw now that it is asked to calculate the covalent radius of Si, therefore I have a new answer that I will post (with calculations) when you want me to.
and it's in agreement with the data Mr. Inquisitive found. I had to find the covalent radius of carbon to calculate the covalent radius of silicon, as I don't think that it is otherwise possible.

#### Corribus

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##### Re: Problem of the week - 11/03/2013
« Reply #12 on: March 18, 2013, 03:05:06 PM »
111.4 pm is the answer I calculated as well.  The only "outside" information I used to arrive at this answer was the molecular masses of carbon and silicon and a little information about the way the atoms are arranged in the diamond unit cell.  I did make the assumption that the covalent radius of carbon is the same in diamond and silicon carbide.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Borek

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##### Re: Problem of the week - 11/03/2013
« Reply #13 on: March 18, 2013, 03:23:24 PM »
Yes, 111 pm is the answer I was looking for (and 77 pm for carbon - important intermediate result).

Anyone willing to explain how they got the result?
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#### Rutherford

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##### Re: Problem of the week - 11/03/2013
« Reply #14 on: March 18, 2013, 04:06:33 PM »
I've done this again to be more precise before posting it here and I got a slightly different result  , it's 111.61pm.

First to calculate the length of the unit cell of diamond using the density of diamond:
ρ=m/V, m=M·nC/Na, V=a3, where a is the length of the unit cell, M is the molar mass of carbon, Na is Avogadro's number and nC is the number of species in the unit cell (8 for diamond).
I get the equation ρ=nC*M/(Na·a3). From this equation a is equal to 3.5722·10-8cm.

Now, to calculate the covalent radius of carbon. The half of the smaller diagonal of the unit cell is equal to a·sqrt(2)/2, and it is one side of a triangle that 3 C atoms are making, the other two sides are C-C bonds (as shown in the attached picture). The angle between the two C-C bonds are 109°28' (because the C atom that is in the picture connected to two C atoms is in a tetrahedral hole and therefore the tetrahedron angle). I mark the C-C bond length with x, and using the law of cosines: a2/2=2x2-2x2cos109°28' I got that x=1.5469·10-8cm. rC is the covalent radius of C and x=2rC, therefore rc=77.34pm.

The same procedure is for the unit cell of SiC. Using the formula for density I got that a=4.3634·10-8cm. Using the law of cosines (C atoms are in the tetrahedral holes), x=1.8895·10-8cm. x=rC+rSi, and finally rSi=111.61pm

The C atoms in the picture aren't in the same plane. The ones in the tetrahedral holes are a little further from the observer that the other three C atoms.