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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on March 11, 2013, 01:09:38 PM

Title: Problem of the week - 11/03/2013
Post by: Borek 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.
Title: Re: Problem of the week - 11/03/2013
Post by: Rutherford on March 11, 2013, 02:12:31 PM
Edit: result noted. Let's wait for others.
Title: Re: Problem of the week - 11/03/2013
Post by: Rutherford 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.
Title: Re: Problem of the week - 11/03/2013
Post by: Borek 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.
Title: Re: Problem of the week - 11/03/2013
Post by: Rutherford 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.
Title: Re: Problem of the week - 11/03/2013
Post by: Borek on March 14, 2013, 11:32:12 AM
No more takers?
Title: Re: Problem of the week - 11/03/2013
Post by: Mr. Inquisitive 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?
Title: Re: Problem of the week - 11/03/2013
Post by: Big-Daddy on March 17, 2013, 08:01:07 AM
Do we go for a "simple cubic" lattice structure here? If not, which do we need ...
Title: Re: Problem of the week - 11/03/2013
Post by: Borek on March 17, 2013, 08:06:05 AM
Go for diamond.
Title: Re: Problem of the week - 11/03/2013
Post by: Big-Daddy 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  :P
Title: Re: Problem of the week - 11/03/2013
Post by: Borek 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).
Title: Re: Problem of the week - 11/03/2013
Post by: Rutherford 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.
Title: Re: Problem of the week - 11/03/2013
Post by: Corribus 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. 
Title: Re: Problem of the week - 11/03/2013
Post by: Borek 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?
Title: Re: Problem of the week - 11/03/2013
Post by: Rutherford 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.
Title: Re: Problem of the week - 11/03/2013
Post by: Borek on March 18, 2013, 04:24:00 PM
Don't worry about these small differences - for example, I used 6.02×1023 for Avogadro's number, if you have added fourth digit, your results will be different in the fourth significant digit (which they are!). And we are given density with 3 SD only, so any difference after 111 is laughable.