# Tag Info

170

This is a nice question, as it confronts a very replicable and common experience with a well established yet seemingly contradictory fact. As you expected, the smell of metal has nothing to do with the metal actually getting into your nose, as most metals have far too low of a vapor pressure at ordinary temperatures to allow direct detection. The ...

54

There are many types of magnetic properties, including ferromagnetism, paramagnetism, diamagnetism, antiferromagnetism, ferrimagnetism, superparamagnetism, metamagnetism, spin glasses, and helimagnetism. Many of these are too weak to cause any noticeable interaction with a magnet. The type of everyday magnetism you're thinking of, which nickel has, is ...

46

Let's consider, for example, a tetrahedral Ni(II) complex ($\mathrm{d^8}$), like $\ce{[NiCl4]^2-}$. According to hybridisation theory, the central nickel ion has $\mathrm{sp^3}$ hybridisation, the four $\mathrm{sp^3}$-type orbitals are filled by electrons from the chloride ligands, and the 3d orbitals are not involved in bonding. Already there are several ...

38

There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, ...

37

In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $\mathrm{d}$-block (a.k.a. the transition metals) obey one special rule: In general, electrons are removed from the valence-shell $\mathrm{s}$-orbitals before they are removed from valence $\mathrm{d}$-orbitals when transition metals are ...

27

The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference. Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. If it were to adopt a ...

24

As I understand this, there are basically two effects at work here. When you populate an $\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between ...

23

You are absolutely correct, it all about the metal's electrons and also about their d orbitals. Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level. However when the metal starts bonding with other ligands, ...

23

On negative oxidation states, in general Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite alright. On the Wikipedia list of oxidation states, there are quite a number of negative oxidation states. Some textbooks have tables which only show positive ...

22

This is just a confirmation to Aesin's answer... Say, we take copper. The expected electronic configuration (as we blindly fill the $\mathrm{d}$-orbitals along the period) is $\ce{[Ar]}\mathrm{3d^9 4s^2}$, whereas the real configuration is $\ce{[Ar]}\mathrm{3d^{10} 4s^1}$. There is a famous interpretation for this, that $\mathrm{d}$-orbitals are more stable ...

22

You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes. Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes ...

22

Disclaimer: I now believe this answer to be fully incorrect. Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22. However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in ...

22

None of the US coins are magnetic (ferromagnetic), except for the 1943 Lincoln penny (Steel Cents, made in steel and zinc to save copper for ammunition during wartime), which are considered magnetic. Almost all of those coins other than Steel Cents are made with higher percentage of copper ($\ce{Cu}$) and lower percentages of other metals such as nickel ($\... 21 Selection rules The intensity of the transition from a state$\mathrm{i}$to a state$\mathrm{f}$is governed by the transition dipole moment$\mu_{\mathrm{fi}}$(strictly, it is proportional to$|\mu_{\mathrm{fi}}|^2$): $$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$ where$\mathrm{d}\tau$is the usual ... 20 These species usually do not exist in nature, but they can be synthesized. Silver has been reduced in liquid ammonia to give$\ce{Ag-}$. A lot of anionic metal carbonyl complexes$\ce{M(CO)_{n}^{m-}}$have been synthesized: -1$\ce{[V(CO)6]-}$,$\ce{[Nb(CO)6]-}$,$\ce{[Ta(CO)6]-}$,$\ce{[Mn(CO)5]-}$,$\ce{[Ir(CO)4]-}$,$\ce{[Co(CO)4]-}$,$\ce{[Rh(CO)4]-}$... 19 The answer simply has to do with the accessibility of the high +6 oxidation state. In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another; the only Cr(VI) compounds that we know of are paired with extremely hard bases like the oxide ion, viz. CrO3, CrO42−, Cr2O72−... 18 Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity. $$\ce{S_0 ->[h\nu_\mathrm{ex}] S_1}$$ In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the ... 18 The electronic configuration has nothing to do with it. The reduction potentials of$\ce{Ni^3+}/\ce{Ni^2+}$,$\ce{Cu^3+}/\ce{Cu^2+}$and$\ce{Zn^3+}/\ce{Zn^2+}$, if they have been/could be measured, would be even greater. The reduction potential for$\ce{M^3+}/\ce{M^2+}$is most dependent upon the third ionisation energy. If$I_3$is large then it will be ... 17 The partially full d-orbitals in transition metals have energy splittings that happen to lie in the visible range. Depending on the arrangement of substituents (known as ligands) that attach to them, the electron energies split according to crystal field theory. Similar splitting in the s or p orbitals produce gaps in the ultraviolet, and any visible light ... 17 Yes, it is all about the absorption of light at specific wavelength. Azobenzene, the parent compound has an absorption maximum around$\lambda$= 430 nm in the visible spectrum. The interesting part is: The absorption can be tuned by substitution of the arenes. This is done before the azo coupling. Some examples are Allura Red (1), Chrysoine Resorcinol (2),... 16 Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann.[1] He explains the apparent stability of half-filled and filled subshells by invoking exchange energy (actually more of a decrease in destabilization due to smaller-than-expected electron-electron repulsions). According to him, there is a ... 16 Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$,$\mathrm{4s}$and$\mathrm{4p}$) and on the right the twelve degenerate ligand p-orbitals (transform as$\mathrm{a_1 + e + t_1 + 2t_2}$). Only orbitals of ... 16 Yes,$^{56}\ce{Fe}$has the most stable nucleus, and$\ce{He}$is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of ... 15 What is the structure of$\ce{FeSO4 \cdot NO}$that is formed when$\ce{NO}$is passed through ferrous sulphate solution? The structure is octahedral. The Fe ion is at the center of the octahedron. Five water molecules and the NO molecule occupy the vertices of the octahedron. Sulfate is a separate spectator ion. The overall charge of the iron ... 14 The question of anomalous electronic configurations, meaning$\mathrm{s^1}$or$\mathrm{s^0}$in one case (Pd) is very badly explained in textbooks. For example, the anomalous configuration of Cr ($\mathrm{3d^5~4s^1}$) is typically explained as being due to "half-filled subshell stability". This is wrong for several reasons. First of all there is nothing ... 14 It is very convenient to use crystal field theory to discuss this. It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$and$d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy. The first six electrons ... 14 Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from$\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$to$\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not only an electron but also a proton to your atom. When you remove electrons to get to a cation, you only remove electrons. Thus it is a different situation, with ... 13 You have to think about the whole process. When a metal loses electrons to make a metal ion the following happens: The metallic bonds holding the metal atoms together are broken. The metal atom loses the electrons. The resulting metal ion is hydrated. In your analysis you are only focusing on step 2. The enthalpy and entropy of the entire process factor ... 13 It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the$5d^{10}$orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3. 13 Although the question is a bit old, I think it still hasn't been answered yet. And by looking at the given answers it seems like the discussion went into a different direction at some point. So let's compare the two compounds, here I plotted the$\ce{Cu(II)}$centers of$\ce{CuSO4.5H2O}$and$\ce{CuSO4}\$ from their crystal structure data. As you can see, ...

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