Question

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

• A. $1$
• B. $1.49$
• C. $2.98$
• D. None of the above

Answer 1

The correct option is B $1.49$

As we know,
distance between $N{a}^{+}$ and $H^{-}$ $=a/2=2.44A$
So
$r^{+}=2.44-0.95=1.49A$