1.

The unit cube Length for LICL (NACL structure) is 5.14 Å. Assuming anion-anion contact, calculate the ionic radius for chloride ion.

Answer»


Solution :Interionic DISTANCE of LICL=`5.14/2`=2.57 Å
`THEREFORE BC=sqrt(AB^2+AC^2)=sqrt((2.57)^2+(2.57)^2)=3.63`
RADIUS of `Cl^-` ion=`1/2` x 3.63 Å =1.81 Å


Discussion

No Comment Found