# Tag Info

15

The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]: \begin{array}{cc} \hline \text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \...

13

You are mixing apples and oranges. Or, to be more precise, an ionic radius for $\ce{Be^2+}$ with coordination number (C.N.) 6 and van der Waals radius of $\ce{He}$. To make it clear I compiled data for van der Waals and covalent radii [1, p. 9-58] as well as ionic radii [1, p. 12-13]: $$\begin{array}{lccccc} \hline \text{Element} & R_\mathrm{vdW}/\pu{Å}... 12 Since OP is still in the high school, I'll try to explain it simply as possible using mathematical manipulation (hoping OP is more familiar with mathematics than chemistry). Both gold and platinum consist of same crystal packing called body-centered cubic, which is illustrated in following diagram: Crystal studies of both gold and platinum has revealed ... 10 As you may know, atomic orbitals are wave functions, solutions of the Schrödinger equation for an atomic system. In a perfectly spherical system you may express an orbital as a function depending on the distance from the nucleus (r) and two angles (\phi and \theta). [If you pick a particular r, the angles \phi and \theta work in a way similar to ... 9 Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions, Yes, that is a general trend, but as you can see there are many exceptions. The trend exists for the same reason that covalent radii generally ... 8 The whole concept of ionic radius is not sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical ... 8 The van der Waals radius of nitrogen is larger than that of oxygen, and has been calculated as such for quite a long time.$$ \begin{array}{lll} \hline \text{Reference} & R_\ce{O} & R_\ce{N} \\ \hline \text{Pauling, 1939} & 1.40 & 1.5 \\ \text{Bondi, 1964} & 1.52 & 1.55 \\ \text{Zefirov, 1974} & 1.29 & 1.50 \\ \text{Gavezzotti,...

7

As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element. This internuclear distance can be determined by two methods: X-rays method Spectroscopy method Note: Atomic radius is not a set ...

7

There is no such thing as unified atomic radius. An atomic radius is a class consisting of van der Waals radii $R_\mathrm{vdW}$ (steric interactions), covalent radii $R_\mathrm{cov}$, and ionic radii $R_\mathrm{i}$ (and some other as well). From the recent edition of CRC Handbook [1, p. 9-57]: $$\begin{array}{llrr} \hline \text{Element} & \text{... 6 The short answer: no, there is no such a general distance. The long answer: atoms and molecules are not just hanging is the vacuum and waiting someone to grab them. They actually have rather high energy in translation, rotation and vibration modes and particles actually collide during a reaction some of these energies can be involved. The science that study ... 6 Source: http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Atomic_%26_ionic_radii.svg/551px-Atomic_%26_ionic_radii.svg.png Ca2+ is bigger than Mg2+ 6 I completely agree with Wildcat's answer. Excellent stuff. This part, though, got me thinking: Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions ... 6 Iron and sulfur ions bond by electrostatic attraction. This is the right answer. I see a hard time maneuvering around it. Breaking an ionic bond between \ce{Fe^2+} and \ce{S^2-} releases energy. Around the basic definition of forces and bonds, there are usually two big beliefs. One is a big misconception, and the other one is right. The ... 5 Atomic radii cannot really be uniquely nor accurately defined. However, there is generally a very big difference between covalent, van der Waals, and ionic radii. So, by most any definition of the respective terms the covalent radius of fluorine will be smaller than the van der Waals radius of neon. Similarly, the ionic radius of fluoride ion will be ... 5 We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near 10^{-33} centimeters and 10^{-43} seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about 10^{-5} GRAMS or 10^{19} Billion Electron Volts (BeV or GeV). It is a simple matter to ... 5 To answer the second part: We know M=208m_e, Z=3, \hbar=\frac{h}{2\pi}. Part one has a mistake, as it is$$ \begin{align} &&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\ &&Mvr&=n\hbar\\ \implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2} \end{align} $$We also know ... 5 You are right: the outer boundary is quite arbitrary. There is no any intrinsic threshold; the probability just gradually decreases lower and lower, but never reaches 0. You may draw a sphere so as to have the electron inside with 90% probability, or 95%, or 99%, or any other value as you see fit. This, in particular, is the reason why atomic radii are ... 5 The size of the 1s orbital in \ce{Li+} and \ce{Al^3+} is not necessarily the same. In fact, they are quite different because of the much larger effective nuclear charge in \ce{Al^3+}. One can easily look up the wavefunction for the 1s orbital and see the radial dependence on Z_\mathrm{eff}. Therefore, merely looking at the electronic configuration ... 5 This is something students often forget when comparing radii: shells are not simply additive. If you move along eight elements to go from lithium to sodium, you are not only filling the second shell and putting one electron into the third, you are also adding a proton to the nucleus with every additional electron. The larger the nuclear charge is, the ... 5 First, you have to look at the definition of metallic radius, which is the half distance between two atoms in a lattice. It has a significant dependence on crystal structure. Tanget relevant to the question and other answers: Gallium has an orthorhombic crystal structure (CN = 6) whereas as aluminum has a face-centered cubic crystal structure (CN = 12).... 5 The Bohr model is a comparatively simple model, whereas the quantum description is part of the huge theory of quantum mechanics, so to list all the differences including implications would need to explain quantum mechanics itself. But I think quite a good starting point would be to say, the Bohr model has electrons travel on certain, specific, classical ... 5 In multi-electron atoms the energy of the s, p and d orbitals for a given n is determined primarily by shielding by inner electrons and penetration of the orbital towards the nucleus. As s orbitals have some probability of being at the nucleus their energy is lowest, next is p then d orbitals in increasing energy. Also there is spin-orbit coupling which ... 5 The theory for a muon-antimoun "atom" is essentially the same as any hydrogen-like atom, just with masses changed. See this Wikipedia page for the relevant formulae for relativistic (Dirac) and non-relativistic treatments. The non-relativistic energy levels are given by :$$E_n = \frac { \mu c^2 \alpha^2}{2n^2} where $alpha$ is the fine structure ...

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The Rutherford formula, as derived, assumes purely elastic scattering from the Coulomb force. No formation of a compound nucleus is considered. Generally, for most of the initial experiments, the available alpha particle energies from various decays (in the range of a few MeV) were not high enough for large deviations from pure Rutherford scattering. Of ...

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The question is misguided by the obsolete idea of shells and subshells as separated exclusive regions of the space around the nucleus. All orbitals largely overlaps and statistical distribution of electron density, (aside of being in the particular orbital) depends on the nucleus charge and overall electron configuration. It can be said, that distribution of ...

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The series you cited belongs to so known 'metallic' radius, and it depends on crystal structure of the element, which changes through the row. In short, you cited series, that is not suited for consideration of isolated tendencies. There are, indeed, several types of atomic radii (covalent with different valur for bonds of different order, van-der-waals ...

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There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal. This notion is very sensitive to the number of electrons per atom involved in bonding. Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4. These all participate, in some extent, in ...

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definitely $\ce{O-}$ In $\ce{O-}$ 7 outer electrons are tied to nuclei with 6 positive charges (number 8 minus 2 electons of inner shell), while in $\ce{F-}$ 8 outer electrons are tied to nuclei with 7 charges. The difference, however, should be quite small.

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I think the point of this question is for you to realise that options 1 and 2 can't be the correct answer. As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong. Effective nuclear charge increases across a period because the nuclear charge increases but ...

4

There are several definitions of atomic radius. Guessing from context of your answer, I assume you are talking about van der Waals atomic radius. By definition, it is a radius of atom in respect of equilibrium position with other atoms when only van der Waals forces are acting between the atoms. Since atoms of inert gases do not tend to make covalent bonds, ...

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