Question bank/JEE Main/Chemistry

JEE MainChemistryEasy

Iron has body centerd cubic lattice structure. The edge length of
the unit cell is found to be 286 pm. What is the radius of an iron atom?

Show answer and solution
For bcc structure
3a=4r,r=3a4=1.732×2864=123.8≈124pm\sqrt{3}a = 4r,r = \frac{\sqrt{3}a}{4} = \frac{1.732 \times 286}{4} = 123.8 \approx 124pm

Source: Grass-G / CH-19 Chemistry Paper 28 Nov. Fourth.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans