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Oxidation state involves a counting if valence electrons, so you have to look at whether and how electrons are transferred between atoms, not how the atoms themselves are transferred. It is true that the transfer of neutral hydrogen atoms provides a way to transfer electrons because each hydrogen atom picks up an electron, or perhaps leaves one behind if it ...


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If oxidation and reduction is taken to be the loss of hydrogen and the gain of hydrogen respectively... That is true only in specific cases, if it leads to change of the formal oxidation number/state. E.g. $\ce{CH4 + O2 -> C + 2H2O}$ is oxidation, as carbon is oxidized from the oxidation state -IV to 0. Exchanging an acidic hydrogen and a ion of an ...


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Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$. Now coming to your question, $\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$. Similarly, $\ce{Mn^{2+}}$ has 5 ...


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This misconception arrises because most of the textbooks and websites starts pairing electrons from left to right in a series. But in general we can pair an electron in any of the degenerate orbitals if all degenerate orbitals are filled with an electron having same spin either up or down ( which is called Hunds rule). Note: The orbitals which are having ...


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Because periodic waves can be represented using exponential functions using imaginary arguments I'm going to attempt a non-rigorous explanation that might give some intuition as to why imaginary numbers appear in the mathematics of quantum stuff. Most of this doesn't require a knowledge of what a Hermitian operator is or about vector spaces or eigenvalues. ...


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Why is the momentum operator imaginary? The simplest explanation hinges on the fact that observables are represented by Hermitian operators in quantum mechanics. Once we accept this, then we can show that the momentum operator $\hat{p} = -\mathrm{i}\hbar\nabla$ is Hermitian precisely because of the factor of $\mathrm{i}$. We need to show that for any two ...


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