Question

If the edge length of a $NaH$ unit cell is 488 pm, what is the length of $Na-H$ bond if it crystallises in the fcc structure:

• A. 122 pm
• B. 346 pm
• C. 488 pm
• D. 976 pm

Answer 1

The correct option is B 346 pm

In a FCC crystal the atom touches at the centre, not at edge centre.

(Refer to Image)

Thus the distance of face centre is $a\sqrt{2}$, where $a$ is the edge length of unit cell.

$2(r_{+}+r_{-})=a\sqrt{2}$ $r_{+}=$ radius of cation; $r_{-}=$ radius of anion

$(r_{+}+r_{-})=\frac{a}{\sqrt{2}}=\frac{488}{\sqrt{2}}pm$

$Na-H(r_{+}+r_{-})=346pm$

This is the $Na-H$ bond length.