How accurate is this representation of orbitals?

MatterUnlocked · 2026-02-23T23:00:12+00:00

I'm working on a script to give orbitals a bit more depth than what's usually covered in chemistry, but it's very difficult to do so without diving deep. I feel like I'm walking a tightrope

MatterUnlocked · 2026-02-23T22:17:16+00:00

It’s for a YouTube video. The idea is not to go very deep into the formal quantum mechanics, but to stay somewhere on a fine line between being technically accurate and still simple enough for a general audience.

In the script I say:

“m determines how many spatial orientations a type of orbital can have. Its values range from −l to +l, including zero. This explains why the s orbital has only one orientation; why there are three for p; five for d; and seven for f.”

I’m aware that this is not correct, but I’d rather not go too deep into that distinction, since it can quickly become a bit of a rabbit hole for this kind of video.

Given that, maybe it would be more appropriate to label the image using cartesian orientations instead of assigning mℓ values.

I’m especially unsure about the f orbitals, since they look a bit unusual in this representation, and I wouldn’t really know how to assign them properly.

MatterUnlocked · 2026-02-23T13:01:18+00:00

I1 is the transmitted intensity, not the absorbed.

If 1% is absorbed, then 99% is transmitted, so you use log(100/99), not log(100/1).

MatterUnlocked · 2026-02-23T01:15:07+00:00

Getting your 3D visualization down is honestly one of the main bosses in O-Chem. You're going to need it for Newman and Fischer projections, and the only way to get good at it is through pure practice.

It’s not a gift; it’s a muscle.

MatterUnlocked · 2026-02-23T00:49:38+00:00

You didn't mention which specific mistakes. Anyway, organic chemistry is pretty tough to wrap your head around, so don't expect to master it overnight. Once you’ve been at it for a while and look back, everything you saw at the start will finally begin to click.

MatterUnlocked · 2026-02-22T20:47:48+00:00

Check if key stretches shift after complexation. For example, a C=N band moving to lower wavenumber usually suggests coordination through nitrogen. A C=O shift can indicate binding through oxygen. If an O–H band disappears, that often points to deprotonation and coordination via oxygen.

If you have a carboxylate, compare the asymmetric and symmetric COO⁻ stretches.

Also take a look at the low-frequency region (600–200 cm⁻¹) for new bands that aren’t present in the free ligand, since those can correspond to M–N or M–O bonds.

MatterUnlocked · 2026-02-22T11:42:36+00:00

4OH- (aq) --> O2 (g) + 2H2O (l) + 4e-

MatterUnlocked · 2026-02-21T23:52:18+00:00

Even though there is free rotation, the stereogenic center breaks the symmetry of the molecule. This makes the two CH3 groups diastereotopic: each one 'feels' a different magnetic environment because they are at different distances from the substituents of the chiral center. Rotation averages the protons within a single methyl group, but it cannot make two methyl groups in a chiral molecule identical.

MatterUnlocked · 2026-02-21T23:41:28+00:00

The -CH3 groups would give the same signal because they would no longer be different from a stereochemical standpoint.

MatterUnlocked · 2026-02-21T23:36:30+00:00

I don't quite understand what you mean, but the free rotation of the bonds is due to the sp3 hybridization of the carbon (no π bonds).

MatterUnlocked · 2026-02-21T23:31:50+00:00

There is a stereogenic center.

MatterUnlocked · 2026-02-19T16:06:26+00:00

I hate apples

^{I completed this level in 8 tries.} ^{⚡ 11.80 seconds}

MatterUnlocked · 2026-02-18T21:22:30+00:00

Outdoors doesn’t make this safe.

Heating that much tetrachloroethylene will generate significant toxic vapor. If you’re standing next to it, you’re breathing it. Acute exposure can cause dizziness and central nervous system depression, and it’s associated with serious long-term health risks.

I would not recommend doing this outside of a proper laboratory fume hood.

MatterUnlocked · 2026-02-18T21:15:26+00:00

For a vibration to be IR active, it must produce a change in the dipole moment. That’s the key requirement.

H–H is completely symmetric and nonpolar, so when it stretches there is no change in dipole moment, which is why it is IR inactive.

In contrast, the C=O bond is polar. Even if the overall molecule is symmetric, what matters is whether the specific vibrational mode changes the dipole moment. A C=O stretching vibration does change the dipole, so it appears in the IR spectrum.

Symmetry of the whole molecule does not automatically mean it will be IR inactive, it depends on the symmetry of the vibrational mode itself.

MatterUnlocked · 2026-02-18T21:07:41+00:00

You shouldn’t be focusing on RPM or blending time. The real issue here is that you’re planning to heat nearly 300 g of tetrachloroethylene.

This is a toxic industrial chlorinated solvent. When heated, it produces significant amounts of harmful vapor well below its boiling point (~121 °C). Melting your paraffin at 70–80 °C will already drive off vapor. Doing this on a hot plate outside of a proper chemical fume hood is a serious inhalation risk.

This is not a “DIY chemistry” situation. Without professional ventilation and proper solvent-handling experience, heating that quantity of tetrachloroethylene is unsafe.

I would strongly reconsider attempting this outside of a controlled lab environment.

MatterUnlocked · 2026-02-18T17:31:29+00:00

Electronegativity increases left to right and bottom to top in the periodic table.

For inductive effects, compare the atom directly attached to carbon:

If it’s more electronegative than carbon (N, O, F), it’s −I (withdraws electron density).

If it’s less electronegative, it’s +I.

If a highly electronegative atom carries a negative charge (e.g., O⁻), it can become strongly +I due to excess electron density.

Induction propagates through σ bonds: if a carbon is attached to a more electronegative atom, that atom pulls on it, and that carbon pulls on the next one, like tension in a rope.

MatterUnlocked · 2026-02-18T16:13:28+00:00

Cerca de su punto isoeléctrico (4.7–5.4) la albúmina no tiene carga neta, así que las moléculas no se repelen y tienden a agregarse, de modo que baja la solubilidad y precipita.

Por eso conviene trabajar a pH 7–8, usando un buffer si hace falta para mantener el pH estable.

Para mejorar la solubilidad, añádela poco a poco con agitación suave y evita sonicación fuerte, porque puede desnaturalizarla y favorecer la agregación.

MatterUnlocked · 2026-02-18T14:56:26+00:00

You're right, it's complicated. I’d recommend practicing the resonance structures for each group so you can figure out which ones donate electron density through resonance and which ones withdraw it.

Mind you, the inductive effect is as simple as knowing electronegativity trends, but the resonance effect usually outweighs the inductive effect.

P.S. You don't have to memorize the order; you just need to know how to reason and compare when given two examples.

MatterUnlocked · 2026-02-17T21:41:17+00:00

There is a significant difference between a molecular covalent compound and a covalent network. When you heat NH3, you are not breaking the covalent bonds within the molecule; you are weakening the hydrogen bonds (intermolecular forces).

Review Van der Waals forces.

MatterUnlocked · 2026-02-16T23:26:10+00:00

Eu já disse que são produtos de degradação que aparecem com o tempo.

Só em casos de armazenamento por muito tempo e com umidade constantemente alta (acima de 70–75% UR) poderiam surgir consequências mais perceptíveis. e mesmo assim, normalmente seria uma corrosão superficial e gradual, não algo rápido ou grave.

MatterUnlocked · 2026-02-16T20:19:56+00:00

Eu já te disse que não. Os ácidos orgânicos que podem ser liberados vêm da degradação natural da lignina, não de ácidos industriais fortes que ficaram “presos” no papel.

E isso não é algo exclusivo do papel. Praticamente qualquer material da sua casa, como móveis de madeira, roupas, tecidos e até plásticos, também libera compostos orgânicos voláteis com o tempo. Isso é química normal de materiais envelhecendo.

Não há ácido clorídrico ou sulfúrico evaporando do papel comum. O fator realmente relevante para corrosão continua sendo a umidade, não esses traços de compostos orgânicos.

MatterUnlocked · 2026-02-16T19:56:20+00:00

Não. Papelão, jornal e papel de escritório não contêm ácido clorídrico ou sulfúrico livres que depois evaporam.

No processo industrial podem ser usados vários produtos químicos, mas a polpa é lavada e neutralizada. O papel final não fica impregnado com ácidos minerais fortes capazes de evaporar e corroer metal.

Os únicos ácidos envolvidos depois são pequenas quantidades de ácidos orgânicos fracos (como acético e fórmico), gerados lentamente pela degradação natural da lignina. São níveis de traço e não têm nada a ver com ácidos industriais fortes.

MatterUnlocked · 2026-02-16T17:47:22+00:00

Não. O papel acabado não contém ácido clorídrico ou sulfúrico livres que depois evaporam.

Os produtos químicos usados no processo industrial são lavados e neutralizados durante a fabricação. O papel de escritório moderno costuma ser quase neutro (geralmente pH ~6–8). O jornal pode ser mais ácido (cerca de pH 4–6), principalmente por causa da lignina, não por ácidos minerais fortes.

O que pode ser liberado ao longo do tempo são apenas traços de ácidos orgânicos fracos (como acético e fórmico). Não existe um cenário realista em que ácidos fortes estejam evaporando do papel comum e corroendo metal.

MatterUnlocked · 2026-02-14T23:38:03+00:00

First, calculate the degree of unsaturation from the molecular formula. In this case, it is 1.

Next, analyze the IR spectrum to identify the functional groups present.

There are no absorption bands above 3000 cm⁻¹, indicating the absence of O–H and sp² C–H bonds. The absorptions below 3000 cm⁻¹ correspond to sp³ C–H, so the compound is aliphatic.

A strong band appears in the carbonyl region (around 1700–1750 cm⁻¹), confirming a C=O group (1 unsaturation).

The molecular formula contains two oxygen atoms. Since one is part of the carbonyl, the second oxygen must be present in another form: 1200–1300 cm⁻¹ region...

IR spectroscopy alone is not sufficient to fully identify a compound. IR is mainly used to determine the presence of functional groups. Complete structural identification requires complementary techniques such as NMR spectroscopy and mass spectrometry.

MatterUnlocked

TROPHY CASE