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The structure of $\ce{XeF5}$ is approximately described by the $\ce{C_{4v}}$ point group. This means that the four equatorial F atoms may lie in the same plane as the Xe atom and that the fifth, axial …

In a simple molecule such as oxygen, the bond length measured will alway be the same as the interaction between any two oxygen atoms is the same since one oxygen atom is indistinguishable from another …

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 …

The bond angle in NO$_2$ is $\approx 134^\mathrm{o}$ and the NO bond length 120 pm, in $\ce{N2O4}$ the N-N bond is very long 175 pm and the angle $\approx 126^\mathrm{o}$. Nitrogen dioxide is paramag …

First I would calculate the centre of mass (com) then go on and calculate the moment of inertia axis vectors for each structure and use these to align them via a rotation matrix. Make the centre of ma …

Presumably because there is a small amount of steric interaction between the two methyl groups and this opens up the angle from the ideal tetrahedral ones. The HCH angle is less because the other is l …

The molecule is a 1,3-butadiene derivative. The butadiene has a planar structure as conjugation from the pi bonds spreads out along the carbon backbone so the carbon carbon angles would be close to 12 …

Obviously if you have a diatomic then yes, you can put energy in and the bond will break. There are three cases; (a) as long as the photon has enough energy it can excite a dissociative excited state, …

The figure shows the $\psi,\, \phi$ angles on the peptide backbone.

The RKR method determines the classical turning points $r_{\pm}$ of the potential energy based on knowing experimentally determined spectroscopic constants. ( Rydberg, 1931, Z. Physik. v73,376, Klein, …

No, not symmetry per se but generally the lowest energy arrangement of the atoms. What structure this is is not always obvious by just combining atoms according to the valency rules; if it were, exp …

The question asks why water has a larger angle than other hydrides of the form $\ce{XH2}$ in particular $\ce{H2S}$ and $\ce{H2Se}$. There have been other similar questions, so an attempt at a general …

The fluorescence yield is $\varphi =k_k/(k_f+k_n)$ where $k_f$ is the rate constant for fluorescence and $k_n$ that for non-radiative processes. If $k_n \gg k_f$ then the fluorescence yield dis very l …