Let Mg TiO_(3) exists in pervoskite structure. In this lattice, all the atoms of one of the face diagonals are removed. Calculate the denstiy of unit cell if the radius of Mg^(2+) is 0.7 Å and the corner ions are touching each other. [Given atomic mass of Mg = 24, Ti = 48]

Let Mg TiO_(3) exists in pervoskite structure. In this lattice, all th

1.	Let Mg TiO_(3) exists in pervoskite structure. In this lattice, all the atoms of one of the face diagonals are removed. Calculate the denstiy of unit cell if the radius of Mg^(2+) is 0.7 Å and the corner ions are touching each other. [Given atomic mass of Mg = 24, Ti = 48]
Answer» Solution :No. of `Mg^(2+)` PER unit cell = 8 [At corners] `xx (1)/(8) = 1` No. of Ti per unit cell = 1 [body center ] `xx (1)/(1) = 1` No of O per unit cell = 6 [Face center] `xx (1)/(2) = 3` So FORMULA `= MgTiO_(3)` Atom are removed along face diagonal No. of `Mg^(2+) = 6`[At CORNER] `xx (1)/(8) = (6)/(8) = (3)/(4)` No. of Ti per unit cell = 1 [Body center] `xx (1)/(1) = 1` No. of O per unit cell = 5[Face center] `xx (1)/(2) = (5)/(2)` So formula of compound `= Mg_((3)/(4)) TiO_((5)/(2))` Formular MASS `= 24 xx (3)/(4) + 48 + 16 xx (5)/(2) = 18 + 48 + 40 = 106` amu As corner ion are touching so `= a = 2 r_(Mg^(2+)) = 2 xx 0.7 = 1.4 Å` `d = ("mass")/("Volume") = (106 xx 1.76 xx 10^(-24))/((1.4)^(3) xx 10^(-24)) g//cm^(3) = 64.5 g//cm^(3) ~~ 65 g//cm^(3)`