Why is it impossible for any metal to have a simple cubic crystal structure? In other words, why must all metals be either BCC, FCC, or HCP?
The packing of metal atoms is a function of the shape of that atom. All atoms are not perfectly spherical. Those that are like d-5 and d-10 atoms tend to form closest packing structures like ccp and hcp, cubic closest packing and hexagonal closest packing. Note that face centred cubic fcp is the same thing as ccp. It's just a second name to represent the same packing structure.To see what closest packing looks like, drop spherical balls into a box and see that they line up spontaneously into layers where the balls form little triangles and the subsequent layers sit in the hollows formed in the previous layers. This is closest packing and is what you get spontaneously with spherical balls.Now drop footballs into a large tub and see what they do. It'll be impossible for them to adopt a closest packing structure and will choose something different that is less efficient but is the best they can do.Now drop wooden child's blocks into a box. With a little jiggling they will settle into a more cubic arrangement.Atoms are only perfectly spherical if their subshells are filled or half filled. This includes s atoms d-5 and d-10 atoms. Other atoms with partially filled subshells have different shapes and the best they can do will be something that may not be a “closest packing” structure. Not many atoms are cubic in shape and so you don't find simple cubic crystal structures in metals.Many p atoms are not metallic and form more directional covalent bonds rather than metallic bonds and so they are a different category all together. Po is one such atom, which forms covalent directional bonds and happens to form a cubic crystal structure.
How many atoms are there in 111 plane of simple cubic lattice?
There is a difference between how many atoms are there on a specific plane and how many atoms actually contributing for the same plane. For a simple cubic lattice, if we carefully visualize the (111) plane, it appears that the plane contains 3 atoms. But the actual contribution from each atom to a single (111) plane 1/6. Because one atoms is shared by 6 equivalent (111) planes. Therefore the contribution from each atom is 1/6 and there are 3 atoms of the same kind. Therefore the actual number of atoms there on (111) plane is 3*1/6 = 1/2.
How many tetrahedral and octahedral voids are present in HCP lattice?
First note that there is no HCP lattice, only HCP crystal structure which is built by adding two atom motif to a simple hexagonal lattice. Even with this correction the question as asked is not clear. It can be made precise by either askingHow many tetrahedral and octahedral voids per atom are there in an HCP crystal? oathHow many tetrahedral and octahedral voids per unit cell are there in an HCP crystal structure.The answer to the first question is that the number of tetrahedral voids per atom is two and the number of octahedral voids per atom is one. These are the same as those for Cubic Close-Packed (CCP) structure.To answer the 2nd question we should note that the primitive unit cell of the HCP structure contains two atoms. Thus the previous answers per atom have to be multiplied by 2 to get answers per unit cell. So there are 4 tetrahedral voids and two octahedral voids per primitive unit cell of the HCP structure.
How many number of carbon atoms are there in a unit cell of diamond?
Diamond unit cell is F.C.C type with four additional C atoms tetrahedrally bonded(black ones). Corner aoms contributes to 1/8th of the atom to the unit cell , Face centred atoms contributes 1/2 of the atom .Now, Total atom in one cell can be summed up to :(1/8 X 8) +(1/2 X 6) + 4(tetrahedrally bonded atoms in the cell) = 8.
If all the atoms along any one body diagonal are removed, what is the formula of the compound?
A type of atoms are at 8 corners, B type of atoms are at 6 face centres, C type of atoms occupy all tetrahedral voids and D type atoms occupy all octahedral voids.Here, A and B together make a FCC(or CCP) unit cell structure.Each Corner and face-centered atom contribute 1/8 and 1/2 respectively towards one unit cell.For one FCC unit cell, there are total 8 tetrahedral voids (2 along each body diagonal at 1/4 from eah corners,, all completely inside FCC unit cell so each contribute complete 1).For one FCC unit cell, there are total 4 octahedral voids (12 along each edge centre which contribute 1/3 each towards one unit cell and one complete octahedral void at the body centre of FCC).So, original formula will be AB3C8D4.Now along a body diagonal, 2 corners, 2 tetrahedral voids and one octahedral void are present.If all the atoms along any one body diagonal are removed, remaining atoms are 6 corners, 6 face centers, 6 total tetrahedral voids and 3 total octahedral voids.Contribution by A atoms at corners = 6 x (1/8) = 3/4Contribution by B atoms at face-centre positions = 6 x (1/2) = 3So, new formula is A(3/4) B3 C6 D3 = A3 B12 C24 D12 = A B4 C8 D4