Home
Class 11
CHEMISTRY
Assertion. The sum of the radii of Na^(...

Assertion. The sum of the radii of ` Na^(+) and Cl^(-)` ions in NaCl cystal is 281 pm Hence, edge of the unit cell is 281 pm.
Reason. Edge of the unit cell is the distance between the centres of ` Na^(+) and Cl^(-)` ions touching each other.

Text Solution

Verified by Experts

The correct Answer is:
a
Correct A. The edge of NaCl unit cell is the distance from `Na^(+)` to next `Na^(+)` ion or from ` Cl^(-) and Cl^(-)` ions touching each other .
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

If the distance between Na^+ and Cl^(-) ions in NaCl crystal is 'a' pm, the edge length of the unit cell is

If the distance between Na^(+) and Cl^(-) ions in NaCl crystals is 265 pm, then edge length of the unit cell will be ?

If a is the edge length of unit cell of sodium chloride, the distance between the nearest Na^(+) and Cl^(-) ions is

The radii of Na^+ and Cl^- ions are 95 pm and 181 pm respectively. The edge length of NaCl unit cell is

In the distance between Na^(+) and CI^(-) ions in sodium chliride crystal is X pm , the length of the edge of the unit cell is

The distance between Na^(+) and CI^(-) ion ion NaCI is 281 pm. What is the edge length of the cube