The edge length of unit cell of a metal having molecular weight 75 g/mol is 5Å which crystallises in cubic lattice. If the density is 2 g/c.c., then the radius of the metal atom in pm is

Explanation

`rho=(ZxxM)/(a^3xxN_0)` or `Z=(rhoxxa^3xxN_0)/M=(2 g cm^(-3) xx(5xx10^(-8) cm)^3xx6.02xx10^23 mol^(-1))/(75 g mol^(-1))=2`
This shows that the metal has body-centred cubic lattice. For BCC lattice,
`r=sqrt3/4a=1.732/4xx5Å=2.165 Å`