In a body-centered cubic (BCC) crystal structure, atoms are located at the corners and at the center of the cube. To find the atomic radius, we need to relate it to the side length (a) of the unit cell.

In a BCC lattice, the atoms touch along the body diagonal. The body diagonal length is $\sqrt{3}a$. Along this diagonal, there are two atomic radii from the corner atom to the center atom, and then another two radii from the center atom to the opposite corner atom, but actually, the entire body diagonal spans 4 atomic radii. This is because the body diagonal passes through two atoms: one at a corner and one at the center. The distance from a corner to the center is 2r (since the center atom is touching the corner atoms), and from the center to the opposite corner is another 2r, making the total body diagonal length 4r. However, wait, let's clarify: in a BCC unit cell, the body diagonal has atoms at both ends (corners) and one in the center. The distance between two adjacent atoms along the diagonal (e.g., from a corner to the center) is 2r (sum of their radii). The entire body diagonal, from one corner to the opposite corner, passes through the center atom, so it consists of two such segments: from corner to center (2r) and from center to opposite corner (2r), totaling 4r. Therefore, the body diagonal length equals 4r.

So, we have:

$\text{Body diagonal}=4r=\sqrt{3}a$

Solving for the atomic radius r:

$r=\frac{\sqrt{3}a}{4}$

Given the side length a = 351 pm, we substitute:

$r=\frac{\sqrt{3}\times 351\text{pm}}{4}$

First, calculate $\sqrt{3}\approx 1.732$:

$r=\frac{1.732\times 351}{4}\text{pm}=\frac{608.132}{4}\text{pm}\approx 152.033\text{pm}$

So, the atomic radius is approximately 152 pm.

Therefore, the correct option is 152 pm.

Related Topics and Formulae

Body-Centered Cubic (BCC) Structure: In a BCC unit cell, atoms are at all corners and one at the center. The relationship between atomic radius (r) and unit cell edge length (a) is given by $r=\frac{\sqrt{3}a}{4}$. This is derived from the body diagonal being equal to 4r.

Other Common Crystal Structures:

• Face-Centered Cubic (FCC): $r=\frac{\sqrt{2}a}{4}$
• Simple Cubic: $r=\frac{a}{2}$

These formulae are crucial for calculating atomic radii, packing efficiency, and density in solid state chemistry.