# 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]: $$\begin{array}{lccccc} \hline \text{Element} & R_\mathrm{vdW}/\pu{Å}... 10 As you may know, atomic orbitals are wave functions, solutions of the Schrödinger equation for an atomic system. In a perfectly spherical system you may express an orbital as a function depending on the distance from the nucleus (r) and two angles (\phi and \theta). [If you pick a particular r, the angles \phi and \theta work in a way similar to ... 9 Let me start by providing a reference to a nice collection of van der Waals' data (actually the site provides a lot of other interesting data about the elements, you might want to poke around). As to your questions, Yes, that is a general trend, but as you can see there are many exceptions. The trend exists for the same reason that covalent radii generally ... 8 The whole concept of ionic radius is not sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical ... 8 As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element. This internuclear distance can be determined by two methods: X-rays method Spectroscopy method Note: Atomic radius is not a set ... 7 There is no such thing as unified atomic radius. An atomic radius is a class consisting of van der Waals radii R_\mathrm{vdW} (steric interactions), covalent radii R_\mathrm{cov}, and ionic radii R_\mathrm{i} (and some other as well). From the recent edition of CRC Handbook [1, p. 9-57]:$$ \begin{array}{llrr} \hline \text{Element} & \text{...

6

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 ...

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

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 ...

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 ...

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 ...

4

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.

4

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 ...

4

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 ...

4

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, ...

4

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 ...

4

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 ...

4

The first thing you should realize is that there are various definitions of the size of the atom. One distinguishes for example the covalent radius, the Van der Waals radius and the ionic radius. The covalent radius is based on the binding of atoms into molecules and on the resulting bond length. The van der Waals radius is half the minimum distance between ...

4

As Paul has pointed out, there are many different ways of defining atomic radii. ron has also pointed out that the trend goes in the opposite direction from what you said. For instance, according to Ptable, |Calculated radii | | |r(pm)| |r(pm)| |H | 53 |He| 31 | |F | 42 |Ne| 38 | |Cl| 79 |Ar| 71 | |Br| 94 |Kr| 88 | I have added hydrogen here as ...

4

You're right about the expected order. From largest to smallest the expected order would be: $\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe] $\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr] $\ce{_{35}Br^-}\text{ }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr] $\ce{_8O^{2-}}\text{ }\text{ }$ = [He] $2s^2\text{ ... 4 The Bohr model is a comparatively simple model, whereas the quantum description is part of the huge theory of quantum mechanics, so to list all the differences including implications would need to explain quantum mechanics itself. But I think quite a good starting point would be to say, the Bohr model has electrons travel on certain, specific, classical ... 4 Covalent radius is measured as the distance ($r_\text{cov} = d/2$) between the nuclei of two bonded atoms (covalent). But, if you try to do the same for noble gases/inert gases (good luck!), as they have fully filled$np$orbitals, they will repel each other, hence the closest distance between the two atoms is taken (high pressure, low temperature) as the ... 4 That is impressive. What you may need to bear in mind is that as the nuclear charge increases, all the electron orbitals move in closer to the nucleus. You may have been assuming, consciously or unconsciously, that the electron shells appear at a relatively fixed distance from the nucleus -- that is, that the n=1 shell is always at about the same distance ... 4 The commonly used method of obtaining ionic radii for higher coordination numbers (C.N.) is to extrapolate values from the Shannon's scale [1] using the relationship between ionic radius and coordination number proposed by Zachariasen [2]: ... the bond lengths$D(N_1)$and$D(N_2)$for cation coordination numbers$N_1$and$N_2\$ were related as follows ...

3

Atomic radius is inversely proportional to the effective nuclear charge. As we move from left to right in a period the effective nuclear charge increases. This will decrease the radius of an atom. At the same time, in transition elements the number of electrons in the 3d sub-shell will increase. This will repel the already present 4s electrons. This ...

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