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3

Octahedral complexes are somewhat of a ‘standard’ for transition-metal complexes. Not only can they happen if there are too few electrons to satisfy the 18-electron rule (e.g. $\ce{[Ti(H2O)6]^3+}$, a $\mathrm d^1$ complex) but also if there are too many electrons to satisfy the 18-electron rule (Jahn-Teller distorted $\ce{[Cu(H2O)6]^2+}$, a $\mathrm d^9$ ...


1

The article text itself does go into details. The first thing to note: while aromaticity on the larger scale is retained in your structures 3 and 4, this is only true for the 10-π-electron system as a whole. In many of these types of 10 π systems made up of a six-membered and a five-membered ring, the two rings are not equal with the double bond ...


5

The product $xyz$ is an odd function and as such should produce the same selection rules as 1 photon (electric dipole) transitions. Thus CO has a 1 photon transition at 146.7 nm in the vacuum uv that can be more easily reached with a 3-photon transition with a pulsed laser at 440 nm. ( Three photon transitions are also used in bio-imaging where a ir photon ...


10

That's betanin, from which beets derive much of their deep purple color, and which may have neuro-protective (Ref. 1) features. The following is the structure retrieved from the wikipedia (on 2019-10-5), which closely matches that in the mosaic: I used the editor available on the Sigma-Aldritch website to generate a structure and submit a search: The ...


8

Resonance structures are one model that explains bonding, but not a very good one Resonance structures are widely misunderstood. Individual resonance structures are not real compounds that exist, but an approximation that assumes all structures are made from bonds consisting of two electrons. The true structure of a compound is thought to be an average ...


1

I was offered the standard explanation that lone pairs are larger back in high school but in the first year of uni that was overturned by a pretty simple counterexample: the heavier homologues $\ce{H2S, PH3}$. While $\ce{SiH4}$ remains tetrahedral and has a similar orbital structure to $\ce{CH4}$, the bond angles in $\ce{H2S}$ and $\ce{PH3}$ are very close ...


0

The Cl atom is electronegative relative to the C atom which polarizes the electrons to Cl. This puts p character in the C-Cl bond at the C atom and shifts the HCH bond angle from sp3 109.5° toward sp2, toward 120°. Remember that if the C-Cl bond at the C atom were pure p the σ bonds would have to be coplanar and the bond angle would be 120°. Obviously the ...


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