[Borek]

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.

[Rutherford]

Edit: result noted. Let's wait for others.

[Rutherford]

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]

No idea how you understood the problem then, and how you understand it now. Wording has not changed.

[Rutherford]

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]

No more takers?

[Mr. Inquisitive]

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?

[Big-Daddy]

Do we go for a "simple cubic" lattice structure here? If not, which do we need ...

[Borek]

Go for diamond.

[Big-Daddy]

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 

[Borek]

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).

[Rutherford]

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]

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. 

[Borek]

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?

[Rutherford]

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·n_C/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 n_C is the number of species in the unit cell (8 for diamond).
I get the equation ρ=n_C*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. r_C is the covalent radius of C and x=2r_C, therefore r_c=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=r_C+r_Si, and finally r_Si=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.