In the cubic crystal of CsCl (d = 3.97g" cm "^(-3) )the eight corners are occupiedbyCl^(-)ions with aCs^(+)ionthe centre and vice -versa. Calculate the distnacebetween the neighbouring Cs^(+) and Cl^(-) ions . What is the radius ratio of the two ions ?[ At masses of Cs = 132.91 and Cl = 35 . 45]

In the cubic crystal of CsCl (d = 3.97g" cm "^(-3) )the eight corners

1.	In the cubic crystal of CsCl (d = 3.97g" cm "^(-3) )the eight corners are occupiedbyCl^(-)ions with aCs^(+)ionthe centre and vice -versa. Calculate the distnacebetween the neighbouring Cs^(+) and Cl^(-) ions . What is the radius ratio of the two ions ?[ At masses of Cs = 132.91 and Cl = 35 . 45]
Answer» Solution : In CsCl , ` Cl^(-)`IONS per UNIT cell ` = 8 xx 1/8 =1` ` Cs^(+)` ions per unit cell = 1 Thus, the unit cell of CsCl contains one CsCl molecule, i.e, Z =1 ` p = ( Zxx M)/(a^(3) xx N_(0))` ` 3.97 = ( 1 xx 168.36)/ (a^(3) xx 6.023 xx 10^(23)) ` ora =`4.13 xx 10^(-18)m ora = 4.13 Å` For a cube of side length (a) equal to 4.13 Å , body diagonal AE =` SQRT3 a = sqrt3 xx 4.13 Å = 7.15 Å` As `Cl^(-)` ions A and E touch `Cs^(+) and Cl^(-) = 3.57 Å` ` 2r^(+) + 2r^(-) = 7.15 Å or r_(+) + r_(-)= 3.15 Å` i.e, Distance between neighboring `Cs^(+) and Cl^(-) = 3.57 Å` Assuming that the two ` Cl^(-)`ions toucheach other. Lengthof the unit cell = ` 2r^(-)=a= 4.13 Å` ` r_(-) = 2.06 Å therefore r_(+)= 3.57 -2.06 = 1.51 Åthereforer_(+)//r_(-) = 1.51// 2.06 = 0.73`