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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 ?
enter image description here

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.)

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  • $\begingroup$ 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. $\endgroup$
    – Manu
    Commented Jun 10, 2020 at 6:56
  • $\begingroup$ 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. $\endgroup$
    – Maurice
    Commented Jun 10, 2020 at 7:13
  • $\begingroup$ @Maurice $\frac{\sqrt{3}}{2} < 1$, therefore the central atoms are closer than the other corner atoms. $\endgroup$
    – Aniruddha Deb
    Commented Jun 10, 2020 at 7:43
  • $\begingroup$ @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 ? $\endgroup$
    – Nimantha
    Commented Jun 10, 2020 at 9:17

2 Answers 2

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Let's just take a picture!

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.

And then I searched for the closest neighbours, those on the vertices of the unit cells (in blue) forming an octahedron and those in the center of the former unit cells (in red) forming a cube.

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.

stacked unit cells of CsI

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BCC is a body-centered lattice. The origin and the center of the cell are crystallographically equivalent, i.e. there is a translation operation relating them and the entire crystal. So if the atom in the center has 8 nearest neighbors, the atoms at the corners have 8 nearest neighbors as well.

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