NaH crystallizes in the same structure as that of NaCl. The edge length of the cubic unit cell of NaH is 4.88 Å. (a) Calculate the ionic radius of `H^(–)`, provided the ionic radius of `Na^(+)` is 0.95 Å. (b) Calculate the density of NaH.

Explanation

a = 4.88 Å = `4.88 xx 10^(–8)` cm
(a) `r_(Na^(+))+r_(H^(-))=(4.88)/(2)`,
`r_(H^(-))`=2.44-0.95=1.49 Å
(b) `d=(ZM)/(a^(3)N_(A))`=(4xx24)/((4.88xx10^(-8))^(3)xx6.023xx10^(23))`
1.37 `gm//cm^(3)`