If an element having atomic number 96 crystallises in cubic lattice with a density of `10.3gcm^(-3)` and the edge length of 314 pm then, the structure of solid is

Options

A. hcp
B. fcc
C. bcc
D. simple cubic

Answer

D

Explanation

Given, Density of element ` (d) = 10.3 g cm^(-3)`
Edge length (a) = 314 pm
Avogadro number ` (N_(A)) = 6.02 xx 10^(23) mol^(-1)` Molar mass (M) = 247 g mol ` ""^(-1)`
`[because Z = 96 `, having molar mass 247, i.e. cesium (Cs)]
` because d = (Z xx M)/(a^(3) xx N_(A))`
where, Z = number of lattice points in a unit cell
` therefore Z = (d xx a^(3) xx N_(A))/(M)`
`Z = (10.3 " g cm"^(-3) xx (314)^(3) xx 10^(-30) cm xx 6.02 xx 10^(23) " mol"^(-1))/(247 " g mol"^(-1))`
`z = 0.78 ~~ 1`
Thus , the structure of solid is simple cubic .