Jan 10, 2017 · $\ce O$ is a free oxygen atom and $\ce{O2}$ is two oxygen atoms chemically bound to form an oxygen molecule. There is no common analogy for $\ce C$, but $\ce{N2}$ is called nitrogen, $\ce{H2}$ is hydrogen and $\ce{Cl2}$ is chlorine, each having the same name as that of their constituent elements.

$\\ce{O2}$ has a double bond in its normal form. That is $\\ce{O=O}$. There are no unpaired electrons in this case are there since there are 2 lone pairs on each oxygen. However 1 resonance structure

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Dec 11, 2021 · $\begingroup$ In the typical way these terms are used, the LUMO can't be the same as the HOMO. Since each of the pi spatial orbitals already have one electron, they are not unoccupied, so they can't be the LUMO.

Aug 8, 2012 · $\begingroup$ @GregE. generating oxygen is a separate concern from CO2 removal (i.e. wouter dorgelo's 'providing astronauts with fresh oxygen' ... stated in the context of CO2 conversion to C and O2). Yes, you can generate O2 via electrolysis, but it is the concentration of CO2 and its build-up that creates a hazard for humans.

Jul 9, 2015 · Oxygen, the $\ce{O2}$ molecule, is an absolutely lowsy σ donor — especially when compared to $\ce{CO}$. Its molecular orbital scheme is generally that of carbon monoxide except that it is entirely symmetric and two more electrons are included: They populate the 2π orbitals to give a triplet ground state.

$\begingroup$ The reaction $\ce{2H2 + O2 -> 2H2O}$ is a redox reaction (also if you go to atomic recombination), there is no Lewis acid or base, but electron transfer and this results in the contradiction you observed. $\endgroup$ –

Oxidation at anode: $\ce{2 H2O(l) -> O2(g) + 4 H+(aq) + 4e^{−}}$ The same half reactions can also be balanced with base as listed below. Not all half reactions must be balanced with acid or base. Many do, like the oxidation or reduction of water listed here. To add half reactions they must both be balanced with either acid or base.

$$\text{combustible stuff} \ce{+ O2 -> } \text{various oxides} + \text{heat}$$ Note that nitrous oxide ($\ce{N2O}$) is also a supporter of combustion, which means that you can use it to carry out a combustion reaction without oxygen--but this is only because $\ce{N2O}$ generates oxygen itself.

The free energy change of total reaction: $\ce{H2O -> 1/2 O2 + H2}$ is fixed at the experimentally found value of 2.46 eV per water molecule. This is done in order to avoid calculations of $\ce{O2}$, since this molecule has a complicated electronic structure, which is not described accurately with DFT.