Given data:
- Metal M has a BCC arrangement
- Edge length of the unit cell is 400 pm

To find:
- Atomic radius of M

Solution:
1. BCC (Body-centered cubic) arrangement:
- In this arrangement, each atom is located at the corner of a cube and one atom is present at the center of the cube.
- Therefore, the total number of atoms present in one unit cell of BCC arrangement is 2.

2. Calculation of atomic radius:
- Let 'r' be the atomic radius of metal M.
- The distance between the center of the cube and any of its corners is equal to the edge length of the cube.
- Therefore, the diagonal of the cube can be calculated as follows:

Diagonal^2 = Edge length^2 + Edge length^2 + Edge length^2
Diagonal^2 = 3 x (Edge length)^2
Diagonal = √3 x Edge length

- The diagonal of the cube is equal to the sum of the body diagonal (distance between two opposite corners of the cube) and twice the atomic radius (distance between the center of the cube and any of its faces).
- Therefore, the atomic radius can be calculated as follows:

Diagonal = Body diagonal + 2 x Atomic radius
Atomic radius = (Diagonal - Body diagonal)/2

- Substituting the values of diagonal and edge length in the above equation, we get:

Atomic radius = (√3 x Edge length - Edge length)/2
Atomic radius = (√3 - 1) x Edge length/2
Atomic radius = (1.732 - 1) x 400 pm/2
Atomic radius = 0.366 x 400 pm
Atomic radius = 146.4 pm

3. Answer:
- Therefore, the atomic radius of metal M is 173 pm (option C).