# Tag Info

7

X-rays do not allow you to look at the material just like an microscope. You get the spacing information from the diffraction pattern of only crystalline materials and a lot of mathematics. On the other hand, atomic force microscopy may actually be able to tell you where one atomic layer starts and where it ends on a surface. For example, imagine a gold ...

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It's actually a sneakily disguised version of a lattice you already know. Read on ... . Let's try out the staggered square packing. The distance unit used here is the center to center distance between adjacent spheres the diameter of a single sphere. Start with the A layer with spheres A1, A2, A3, A4 in rotational order at the corners ofca unit square. ...

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In the picture, the cubic face-centered packing is shown, with layers of identical atoms colored in red and blue for better contrast. The unit cell shown with thin lines is the conventional face-centered setting. It has red spheres on the corner, and on the center of the bottom face (the faces not normal to the plain of the paper have blue spheres in their ...

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Crystallographic background A unit cell cannot contain parts of atoms from a formula unit, otherwise it would not be a geometric unit of repeatability. The simplest condition arising from translational symmetry is that the total number of atoms present in the unit cell must be either equal to or be a multiple of the number of atoms in the chemical formula, ...

4

You have to use a matrix to convert. This is derived in most textbooks on crystallography, such as McKie & McKie 'Essentials of Crystallography'. The matrix is $$M=\begin{bmatrix} a & 0 & \\b\cos(\gamma)& b\sin(\gamma) & 0\\c\cos(\beta) & cn_2 & c\sqrt{\sin^2(\beta)-n_2^2} \end{bmatrix}$$ where $$n_2=\frac{\cos(\alpha)-\cos(\... 4 Let's just take a picture! For that I took the crystal structure data (CIF) of CsI from literature and changed the 1b position from iodide to another cesium, so all atoms are the same, like in your picture above. I then created a 2 × 2 × 2 cell and took the central cesium atom which used to be the corner of one unit cell and colored it yellow. And then ... 4 Don't think of it so much as putting the center mass of an atom on a vertex but as matching the unit cell symmetry to the crystal symmetry. Take the definition from google below, I've bolded key parts: The smallest group of atoms of a substance that has the overall symmetry of a crystal of that substance, and from which the entire lattice can be built up ... 3 The CRC Handbook of Chemistry and Physics contains a dedicated compilation by H. W. King, titled «Crystal Structures and Lattice Parameters of Allotropes of the Elements». In case your research library is closed, you may access some of its editions freely or borrow them with the library card of archive.org. In case of the 97th edition (by 2016), the section ... 2 Body-centered tetragonal is face-centered cubic only if c/a=\sqrt2. If you try your transformation with a c/a value greater/less than \sqrt2, your "cube" will have lateral edges that are longer/shorter (resp.) than the basal edge and so really remains tetragonal. You can also read this the other way. If you are presented with a body-... 2 Your adjacent plane isn't. Worth a thousand words, they say. 2 There are two equivalent ways to define the meaning of the Miller indices: via a point in the reciprocal lattice, or as the inverse intercepts along the lattice vectors. The reflecting plane are parallel to a plane that includes the following three points:$$\frac{\mathbf{a}_1}{h}; \frac{\mathbf{a}_2}{k}; \frac{\mathbf{a}_3}{l} To show that the plane ...

2

They actually form: ions can be dissolved up till a certain concentration, and beyond that, they form ionic lattices just as you predict. The easiest way to describe the scenario as an equilibrium between ions-in-water vs ions-in-lattice: If the ion-water interaction is very strong, the ions can be very stable in water, even more than in an ionic lattice, ...

2

I looked and there are a lot of questions on the site about units cells, many of which are pertinent. However I couldn't find one that answers your question directly. First there is no reason to guess that $\ce{LiH}$ has a FCC structuree. You can look that up, for instance in Wikipedia. Second $\ce{LiH}$ doesn't really have molecules in the solid. Think ...

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I think the cell after EDIT is correct. Here I added a few neighboring cells: Admittedly, the atom in the right bottom corner is missing. Apparently the author didn't consider it important.

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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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The lattice energy is not directly measurable, and it is difficult to calculate. In a paper about estimating lattice energies, the estimation for $\ce{Na2CO3}$ is 30% off from the reference value (which is from CRC Handbook of Chemistry and Physics). I mention sodium carbonate because ammonium carbonate is not one of their examples. The introduction explains ...

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If we assume that the lattices have the same structure, the lattice enthalpy is dominated by the charge of the ions and the distance between them. Since we are dealing here with cations from the same group, the charges are the same, so the remaining variable is the distance between them. With moderately sized or small anions, a smaller cation can get ...

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Yes of course, if you do the math right, that limes gives the exact same number. Don't forget to look out for atoms that are completely within the unit cell.

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In answer to you first question, no if hkl differ, yes if they do not. I try to explain below. In a crystal the unit cell defines the repeating unit. Inside the unit cell the atoms are arranged as they are in the molecular structure with the molecule being at the same angle and position within each of the unit cells. Each atom scatters the x-ray radiation ...

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The actual, "real" state of the atoms (and then ions) that combine to form an ionic compound is not particularly significant to the question at hand. When we want to calculate the lattice energy of an ionic compound, one method used is a Born-Haber Cycle. In essence, we use Hess' Law to combine a series of elementary steps that results in the overall ...

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Between 273 K and 500 K the intrinsic carrier concentration in Si increases by about 5 orders of magnitude. Compare with the decrease in conductivity of a metal over that range. Increase in carriers far outweighs the increased scattering of carriers.

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Your statement 'For example, Na+ bonds with OH− and Cl− bonds with H+ when NaOH is dissolved in H2O' is not correct if the ions are in solution. In an ionic solid each ion is surrounded by other ions of the same charge and those of opposite charge, in NaCl each Na$^+$ ion is surrounded by 6 Cl$^-$ and vice versa. The interaction between the ions is ...

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