# How many nearest neighbours does a corner atom have? [closed]

Figure shows a unit cell of a cubic lattice. The atom in the middle has 8 nearest neighbours (the atoms in the corner). How many nearest neighbours does a corner atom has ?

I identified the given lattice is a BCC lattice. But I have trouble finding the number of nearest neighbour atoms of the corner atom. Is there a definition for nearest neighbour atoms ? (Please note I don't know much about crystal structures. I am a student in the first semester of an engineering degree. Please can someone explain how to solve this in simple terms.)

• The nearest neighbor of corner atom is at a distance √3a/2 where a is the length of side of unit cell. The no. of nearest neighbor is 8. Nearest neighbor of an atom means those atoms which surround the given atom at the closest distance to that atom. – Manu Jun 10 '20 at 6:56
• A corner atom has 6 neighbours at distance a, two per axis : one before, on behind, one left, one right, one above, one under. The central atoms are too far away from corner. – Maurice Jun 10 '20 at 7:13
• @Maurice $\frac{\sqrt{3}}{2} < 1$, therefore the central atoms are closer than the other corner atoms. – Aniruddha Deb Jun 10 '20 at 7:43
• @Manu So the 8 neighbour atoms that a corner atom have are the atoms in center of each unit cell surrounding the corner atom. Am I correct ? – Nimantha Jun 10 '20 at 9:17

For that I took the crystal structure data (CIF) of CsI from literature and changed the $$1b$$ position from iodide to another cesium, so all atoms are the same, like in your picture above. I then created a $$2 × 2 × 2$$ cell and took the central cesium atom which used to be the corner of one unit cell and colored it yellow.
If you check on the distances you will find that for this case of a $$\ce{CsI}$$ lattice, the distances are $$d_\mathrm{yellow-blue} = \pu{4.56 Å},$$ while $$d_\mathrm{yellow-red} = \pu{3.95 Å}$$. So the answer should be 8 neighbours here.