An fcc lattice has a lattice parameter a = 400 pm. Calculate the molar volume of the lattice including all the empty space.
Volume = a3 = (400 × 10-12m)3 = 64 × 10-24 cm3
Vtotal = VNA = 64 ×10-24 × 6.02 × 1023
For cubic coordination, the value of radius ratio is
The ratio of cationic radius of anionic radius in an ionic crystal is greater than 0.732. its coordination number is
The vacant space in the bcc unit cell is
Calculate the ionic radius of a Cs+ ion, assuming that the cell edge length for CsCl is 0.4123 nm and that the ionic radius of a Cl- ion 0.181 nm.
If the value of ionic radius ratio is 0.52 in an ionic compound, the geometrical arrangement of ions in crystal is
A metal has bcc structure and the edge length of its unit cell is 3.04 . The volume of the unit cell in cm3 will be
For an ionic crystal of the general formula AX and coordination number 6, the value of radius ratio will be
In A+ B- ionic compound, radii of A+ and B- ions are 180 pm and 187 pm respectively. The crystal structure of this compound will be
Sodium metal crystallizes as a body centred cubic lattice with the cell edge 4.29 . What is the radius of sodium atom?
The structure of Na2O crystal is