Question

For a tetrahedral void the correct relation between r and R is :
Here
$r\text{is radius of the tetrahedral void}$
$R\text{is radius of the atoms in close packing}$

• A. $r=0.5R$
• B. $r=0.732R$
• C. $r=0.225R$
• D. $r=0.156R$

Answer 1

The correct option is C $r=0.225R$

To simplyfy calculations , a tetrahedral void may be represented in a cube as shown in the figure :


Here, three spheres A,E and F form the triangular base, the fourth (B) lies at the top and the black coloured sphere occupies the tetrahedral void.
Length of the side of cube is a

$\text{In right angled}△ABC$:
$AB=\sqrt{(AC{)}^2+(BC{)}^2}=\sqrt{a^2+a^2}AB=\sqrt{2}a$
As, spheres A and B are touvhing each other in close packing,
$AB=2R$
$\Rightarrow 2R=\sqrt{2}aR=\frac{a}{\sqrt{2}}...eqn\left(i\right)$
Again,
$\text{From right angled}△ABD$
Body diagonal
$AD=\sqrt{(AB{)}^2+(BD{)}^2}AD=\sqrt{(\sqrt{2}a{)}^2+a^2}AD=\sqrt{3}a$
Since, black coloured sphere touches the other spheres, so
$AD=2(R+r)$
$\Rightarrow \sqrt{3}a=2(R+r)R+r=\frac{\sqrt{3}a}{2}...eqn\left(ii\right)$
Dividing eqn (ii) by eqn (i), we get :
$\frac{R+r}{R}=\frac{\frac{\sqrt{3}a}{2}}{\frac{a}{\sqrt{2}}}$

$\frac{R+r}{R}=\sqrt{\frac{3}{2}}$

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

$r=0.225R$

Alternatively :

Joining the centre of spheres formed which form tetrahedral void we get :

Tetrahedral void is formed at point O

$AB=2ROA=(R+r)$
$\sin {54}^{\circ }{44}^{\prime }=\frac{R}{R+r}\frac{r}{R}=0.225$

$r=0.225R$