Q. What is the radius of iodine atom (at. no. 53, mass number 126)
A) $2.5 \times 10^{- 11}m$
B) $2.5 \times 10^{- 9}m$
C) $7 \times 10^{- 9}m$
D) $7 \times 10^{- 6}m$
Explanation
(a) Electronic configuration of iodine is 2, 8, 18, 18, 7
Here $r_{n} = \left( 0.053 \times 10^{- 9}m \right)\frac{n^{2}}{Z}$
Here $n = 5$ and $Z = 53$
Hence $r_{n} = 2.5 \times 10^{- 11}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.