NEET 2020 Chemistry Question 97 Solution

In a body-centered cubic (BCC) crystal lattice, the central atom touches all eight corner atoms. Therefore, sphere contact occurs along the body diagonal of the cube.

Let the edge length of the unit cell be $$a$$. The face diagonal of any face is $$b = \sqrt{a^2 + a^2} = \sqrt{2}a$$. The body diagonal $$c$$ running from one corner through the central sphere to the opposite corner is: $$c = \sqrt{b^2 + a^2} = \sqrt{(\sqrt{2}a)^2 + a^2} = \sqrt{3a^2} = \sqrt{3}a$$

Along the body diagonal, there is one radius from the starting corner atom, two radii (diameter) of the central atom, and one radius from the ending corner atom: $$\text{Body Diagonal} = r + 2r + r = 4r$$ Equating the two expressions: $$4r = \sqrt{3}a \implies r = \frac{\sqrt{3}}{4}a$$

With $$a = 288\text{ pm}$$: $$r = \frac{\sqrt{3}}{4} \times 288\text{ pm}$$ This corresponds precisely to Option (1).