Chemistry, asked by Nutanc419, 1 year ago

If the radius of the tetrahedral void is r and radius of atom r, derive a relation btw r and r

Answers

Answered by Anonymous

❣❣If the radius of the octahedral void is r and radius of the atoms in close- packing is R relation between r and R.

Answered by Anonymous

SOLUTION

A tetrahedral void may be represented in a cube. In which there spheres from the triangular base, the fourth lies at the top & the sphere occupies the tetrahedral void.

Let the length of the side of the cube=a

Radius of sphere= R

Radius of Void=r

From right angled triangle ABC, face diagonal

$= > AB= \sqrt{AB {}^{2} + BC {}^{2} } = \sqrt{ {a}^{2} + {a}^{2} } = \sqrt{2a}$

As sphere A & B are actually touching each other, face diagonal AB=2R therefore, 2R= √2a

=) R= 1/√2a......(1)

Again from the right angled triangle ABD

$= > AD= \sqrt{ {AB}^{2} + {BD}^{2} } = \sqrt{ \sqrt{2 {a}^{2} } + {a}^{2} } = \sqrt{3a}$

But as small sphere that is void touches other spheres, evidently body diagonal AD= 2(R+r) therefore, 2(R+r)= √3a

or R+r= √3/2a...........(2)

Dividing equation (2) by (1), we get

$= > \frac{(R+ r)}{R} = \frac{ \frac{ \frac{ \sqrt{3a} }{2} }{1} }{ \sqrt{2} } a = \sqrt{ \frac{3}{2} } \\ \\ = > \frac{1 + r}{R} = \sqrt{ \frac{3}{2} } = 1.225 \\ \\ = > \frac{r}{R} = 1.225 - 1 = 0.225 \\ = > r = 0.225R$