Question

Metallic nickel crystallizes in a face-centered cubic lattice. If the edge length of the unit cell is found to be 352 pm, what is the metallic radius of Ni in pm? (1 pm = 10¹² m)

Answer 1

Answer:

124.45 pm

Explanation:

Given that:

Metallic nickel crystallizes in a face-centered cubic lattice;

The diagram below shows an illustration of that;

and the edge length of the unit cell is found to be 352 pm;

we are tasked to find the  metallic radius of Ni in pm?

In order to do that; from the diagram; we can use the Pythagoras theorem to determine the value for the radius.

So; we have:

(4r)² = d² + d²

16r²  = 2d²

r² = $\frac{2d^2}{16}$

$r^2= \frac{d^2}{8}$

$r= \sqrt{\frac{d^2}{8} }$

$r=\frac{\sqrt{d^2} }{\sqrt{8} }$

$r =\frac{d}{\sqrt{2*4} }$

$r= \frac{d}{2\sqrt{2} }$

So since d (the edge length of the unit cell) = 352 pm

then  metallic radius of Ni in pm will be:

$r= \frac{d}{2\sqrt{2} }$

$r= \frac{352}{2\sqrt{2} }$

r = 124.4507935 pm

r ≅ 124.45 pm

∴   the metallic radius of Ni in pm = 124.45 pm