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Old Dec 30th 2014, 01:21 PM   #1
elt2112
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Cubic perovskite bond distances and tolerance factor

If an ideal cubic perovskite ABO3 has cell parameter 3.905 angstroms, how would I work out the A-O and B-O bond distances for this compound?

I thought maybe halving the cell parameter would give the B-O distance, and finding out the diagonal of the cube and halving it would give the A-O bond distance.

The question then asks me to calculate the tolerance factor, which I know is t = (rA + rO)/sqrt.2(rB + rO) but how would I work out the radii having only been given the unit cell parameter?
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Old Jan 9th 2015, 04:46 AM   #2
MGW1248
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inter atomic bond distance in cubic system = a * SQRT(deltax^2 + deltay^2 + deltaz^2) where x y and z are the co ordinates of the atoms in the unit cell

ABO3 perovskite will have
8 A cations on each corner (8 x 1/8 = 1)
1 B cation in the centre (0.5, 0.5, 0.5)
6 O anions, one on each face (6 x1/2 =3)

A-O distance is therefore 3.905A * SQRT(0.5^2 + 0.5^2 + 0^2) = 2.761A (assuming you use A = (0,0,0) and O =(0.5,0.5,0)

B-O distance is 3.905A * SQRT(0^2 + 0^2 + 0.5^2) = 1.9525A (assuming you use B = (0.5,0.5,0.5) and O = (0.5,0.5,0)
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Old Jan 9th 2015, 05:11 AM   #3
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Originally Posted by elt2112 View Post
The question then asks me to calculate the tolerance factor, which I know is t = (rA + rO)/sqrt.2(rB + rO) but how would I work out the radii having only been given the unit cell parameter?
For ideal cubic perovskites:
a = SQRT(2)(rA + rO) = 2(rB + rO)
rearrange these to get
(rA + rO) = a/(SQRT(2)
SQRT(2)(rB + rO) = a*SQRT(2)/2

Sub into your equation, and the tolerance factor should equal 1, expected value for ideal cubic perovskite
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