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)

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 =

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

then  metallic radius of Ni in pm will be:

r = 124.4507935 pm

r ≅ 124.45 pm

∴   the metallic radius of Ni in pm = 124.45 pm

answer:

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answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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