Question

NaH crystallizes in the same structure as that of NaCl. The edge length of the cubic unit cell of NaH is $4.88\mathring{A}$ and the ionic radius of $N{a}^{+}$ is $0.95\mathring{A}$. Calculate the ionic radius of $H^{-}$ (in $\mathring{A}$).

• A. $1.0$
• B. $1.5$
• C. $2.5$
• D. None of the above

Answer 1

Answer: (B)

$1.5$

Given that:

Radius of $N{a}^{+},r^{+}=0.95\mathring{A}$

Edge length, $a=4.88\mathring{A}$

For NaCl type structure-
$r^{+}+r^{-}=\frac{a}{2}$

$r^{-}=\frac{4.88}{2}-0.95=2.44-0.95=1.5\mathring{A}$

Hence, the correct option is B.