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I will start off by addressing Jon's comment above. Yes, the Bohr model is flawed. I think it is still worth learning about it just from a historical standpoint, to see how we discovered the quantum m …

If a molecular orbital $\psi$ (in this case, a hybrid orbital) is constructed from an orthonormal basis set of atomic orbitals $\{\phi_n\}$ via a linear combination $$\psi = \sum_n c_n \phi_n$$ then …

For s-orbitals (i.e. $l = m = 0$), it turns out that the angular component is simply a constant: that is to say, there is no angular dependency, and hence we have $\psi \propto R$. … But for other orbitals this is not permissible. I wrote more about this before, so for further information, please see: What is the exact definition of the radial distribution function? …

This won't get you anywhere with atomic orbitals, though. Orbitals are frequently expressed in terms of spherical coordinates. …

Analogously, an angular node is simply a region where the angular wavefunction is zero.* In the case of the p-orbitals, this is a plane. … However, radial and angular nodes are most commonly discussed in the context of real atomic orbitals, obtained by linear combination of the spherical harmonics. …

What you can say is that 2s and 2p orbitals belong to the same principal quantum shell. … The $\text{s}$ orbitals are spherical in nature, and the centre of the sphere corresponds to the nucleus. And the same goes for $\text{p}$ orbitals, $\text{d}$ orbitals, and so on. …

The traditional "orbitals" that introductory chemistry teaches resemble what theoreticians would call spatial orbitals. … Also, orbitals in DFT have an entirely different meaning from orbitals in HF, but that's a story I'm not qualified to tell. …

Consider the case of the ground-state carbon atom; the $\mathrm{1s}$ and $\mathrm{2s}$ orbitals can be neglected since those are closed-shell, so we only need the two highest-energy electrons, i.e. a $ …

This clearly contradicts the supposed trend given in the book of $\mathrm{sp}$ orbitals forming slightly weaker bonds than $\mathrm{sp^3}$. …

No. The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many acc …

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 sho …

The complex d-orbitals are obtained by solution of the Schrödinger equation. In general, these d-orbitals are made up of a radial part $R(r)$ and an angular part $Y(\theta,\phi)$. … are only five valid solutions with $l = 2$, i.e. d-orbitals). …

TL;DR Excitation of an electron typically conserves spin; that is to say, the spin must be left unchanged by the process. If the initial state has no net spin (one spin-up and one spin-down electron), …

$\mathbf{c}_n$ is a vector with $N$ entries (I switched to boldface to indicate - hopefully slightly more clearly - that it is a vector): $$\mathbf{c}_n = \begin{pmatrix}c_{1n} \\ c_{2n} \\ \vdots \\ …

Formally, this is expressed in terms of integrals involving atomic orbitals $\phi_i$: $$\langle \phi_1 | \hat{H} | \phi_2\rangle \begin{cases} \neq 0 & \text{if }\phi_1 \text{ and } \phi_2 \text{ transform …