Home
Class 11
CHEMISTRY
A compound AB crystallises in bcc lattic...

A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate (i) the distance between oppositely charged ions in the lattice ,(ii) radius of `B^-` if the radius of `A^+` is 190 pm

Text Solution

Verified by Experts

The correct Answer is:
299.9 pm
(i) For bcc, distance ` A^(+) an B^(-) = sqrt3/2 ` a, i.e, it is half of body diagonal
Promotional Banner

Topper's Solved these Questions

  • APPENDIX

    PRADEEP|Exercise 6 WHAT HAPPENS WHEN|1 Videos

  • APPENDIX

    PRADEEP|Exercise 7 WHAT HAPPENS WHEN|1 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    PRADEEP|Exercise Curiosity Questions|2 Videos

Similar Questions

Explore conceptually related problems

A compound AB crystallizes in bcc lattice with unit edge length of 380 pm. Calculate (i)The distance between oppositely charged ions in the lattice. (ii)Radius of A^+ if radius of B is 175 pm

A compound AB crystallises in the b.c.c lattice with unit cell edge length of 390 pm. Calculate (a) the distance between the oppsitely charged ions in the lattice. (b) the radius of A^(+) ion if radius of B^(-) ion is 175 pm.

AB crystallises in a bcc lattice with edge length qa equal to 387 pm .The distance between two oppositely chariged ions in the lattice is

NH_(4)Cl crystallizes in a body-centered cubic type lattice with a unit cell edge length of 387 pm. The distance between the oppositively charged ions in the lattice is

Ammonium chloride crystallises in a body-centred cubic lattice with a unit distance of 387 pm. Calculate (a) the distance between oppositely charged ions in the lattice and (b) and radius of the NH_(4)^(+) ion of the radius of Cl^(-) ion is 181 pm.