Silver metal crystallises with a face-centred cubic lattice. The length of the unit cell is found to be 3 × 10 - 8 cm . Calculate the atomic radius and density of silver. (Molar mass of Ag = 108 g mol - 1 , N A = 6 . 02 × 10 23 mol - 1 ).

Given, silver metal crystallises in face-centred cubic lattice and hence the effective umber of atoms present in face-centred cubic lattice is Z 4 . The relation between the edge length of the unit cell and radius of the constituting atom in face-centred cubic unit cell is r = a 2 2 Given, edge length of the unit cell a = 3 × 10 - 8 cm Atomic radius r = a 2 2 = 3 . 0 × 10 - 8 2 × 1 . 414 = 1 . 06 × 10 - 8 cm Formula of density of cubic crystal system is given by d = Z × M N A × a 3 Where Z is the number of atom per unit cell, M is the molar mass of molecule, a is the edge length and N A is the avogadro's number. d = 4 × 108 ( 3 . 0 × 10 - 8 ) 3 × ( 6 . 02 × 10 23 ) = 26 . 6 g cm - 3 Density of silver which has face-centred cubic lattice is 26 . 6 g cm - 3 .

Copper crystallises into a fcc lattice. Its edge length is 3 . 62 × 10 - 8 cm . Calculate the density of copper (atomic mass of Cu = 63 . 5 u , N A = 6 . 022 × 10 23 mol - 1 ).

Given, edge length, a = 3 . 62 × 10 - 8 cm . Copper crystallises into face-centred cubic lattice, number of atom per unit cell Z is four. Gram atomic mass of Cu , M = 63 . 5 g / mol N A = 6 . 022 × 10 23 mol - 1 We know the density, d = ZM a 3 N A Substitute the values in the above formula, d = 4 × 63 . 5 3 . 62 × 10 - 8 3 × 6 . 022 × 10 23 = 8 . 9 g / cm 3

ZnO turns yellow on heating. Why?

In ZnO , Zn + 2 ions occupy the interstitial sites and electrons are trapped in the interstitial sites for neutralization. Crystals that show such type of defects are generally coloured. Zinc oxide is white in colour at room temperature. On heating, it loses oxygen and turns yellow.