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Luis Avalos 1D

Lewis acids are lone pair acceptors while bases are donors. This means that if a molecule has a lone pair on the central atom, it will be a lewis base, just like ammonia.

Luis Avalos 1D

As people have stated, it has everything to do with lone pairs on the central atom

Luis Avalos 1D

Bronsted acids donate protons (H+) while bronsted bases accept protons.

Luis Avalos 1D

Since it's trigonal pyramidal, I think you're right that the angles are all the same.

Luis Avalos 1D

I believe form refers to all the electron regions around the central atom (including lone pairs), while shape does not account for the lone pair regions. In the case of water, there are 4 electron regions which would create a tetrahedral form, but since we don't count lone pairs in shape it's just b...

Luis Avalos 1D

I'm pretty sure Lavelle mentioned resonance structures don't affect VSEPR models, but lone pairs do.

Luis Avalos 1D

It's generally the least electronegative element that's in the middle.

Luis Avalos 1D

You basically want the structure with the least amount of charge which is found by calculating each element's formal charge. Ideally, there should be a formal charge of 0 for each element present.

Luis Avalos 1D

The website says test 3 covers chapter 3 material so I think it'd be useful to review what we've gone over in lecture.

Luis Avalos 1D

Beryllium has 4 electrons which means that if we remove 3, we will be in the 1s shell (the core). On the other hand, when we remove 3 electrons from Boron (has 5 electrons), we only remove the valence electrons (the 2s and 2p). As stated, it is way harder to remove these innermost electrons than it ...

Luis Avalos 1D

It specifically says "period 3 element" which is Na through Ar, and nitrogen is a period 2 element. On top of this, the central atom does not strictly follow the octet rule which is also an indication that it would not be nitrogen.

Luis Avalos 1D

I like to do the following formula to calculate it:

formal charge = valence electrons of the atom - (lone pairs + .5* bonded electrons).

Luis Avalos 1D

I think molybdenum is part of this list as well, and you just need to remember that those elements will take an e- from the s shell to fill the d shells (singly for Mo and Cr, and doubly for Cu and Ag).

Luis Avalos 1D

In period 4, Chromium and Copper fill their shells differently than you'd expect. Instead of Chromium having the electron configuration [Ar] 3d 4 4s 2 , it's actually more stable with the following configuration [Ar] 3d 5 4s 1 . Likewise, Copper is expected to be [Ar] 3d 9 4s 2 but it fills the d sh...

Luis Avalos 1D

If you're asking about writing them out, it's a personal preference I believe. For example, some people would write scandium as [Ar] 4s² 3d¹ but others write it as [Ar] 3d¹ 4s²

Luis Avalos 1D

Yup! Since only one electron can have that positive spin number, the answer is 1.

Luis Avalos 1D

If it gains electrons, it will be negatively charged which is an anion. If it loses electrons, it will be positively charged which is a cation.

Luis Avalos 1D

If you're referring to m, it corresponds to the magnetic quantum number which tells us the number of orbitals and their orientation, if i'm not mistaken.

Luis Avalos 1D

As others have said, higher values of n mean higher energies. If you go from n=1 to n=5, you need more energy to do so than if you went from n=1 to n=2. This also means that more energy is released when the electron returns to the ground state from a higher energy level.

Luis Avalos 1D

As stated, the coefficient represents the corresponding period of the specific element.

Luis Avalos 1D

Hund's rule is just stating that electrons cannot have paired spins until the other orbitals are all filled first.

Luis Avalos 1D

The number of electrons ejected has to do with the intensity. The photoelectric effect only takes one electron into account without the intensity in mind, I believe.

Luis Avalos 1D

Since you found the energy per photon, to find the energy of a mol of photons all you'd have to do is multiply by avogadro's number which is 6.022 x10^23

Luis Avalos 1D

Divide the total number of Joules emitted by the amount of energy per photon!

Luis Avalos 1D

You can always plug the values of the reactants, one at a time, into the equation and whichever yields less product will be the limiting reactant.

Luis Avalos 1D

Because the wavelength 102.6 nm corresponds to the lyman series, we know that the energy level, n1, will be at the ground state (n=1).

Luis Avalos 1D

As others have said, your answer should be written in terms of nanometers. I got .486 x 10^-6 which translates to 486 x 10^-9 (486 nanometers). This wavelength corresponds to the balmer series, but more specifically, blue visible light.

Luis Avalos 1D

In a vacuum, yes. The only thing that's different is their frequencies & wavelengths I believe.

Luis Avalos 1D

Like everyone else is saying, chemistry strictly follows the law of conservation of mass. What we would ideally observe is the same masses, but sometimes mass is lost when we transfer the products elsewhere.

Luis Avalos 1D

How exactly do you find ionic concentrations when salts are dissolved in water?

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Re: 12.13.A Significance of lone pairs in determining Lewis acid/base?

Re: Trigonal Planar vs Trigonal Pyramidal

Re: How to identify a Bronsted acid or base

Re: Homework 4.7

Re: Form vs. Shape

Re: VSPER MODEL

Re: central atom

Re: Stable Lewis structures

Re: Test 3

Re: 2.6B

Re: Identifying elements in Lewis structures [ENDORSED]

Re: Formal Charge terms

Re: Electron Configurations Exceptions

Re: Exceptions for electron configuration

Re: D and S oribitals

Re: question on a problem

Re: cation or anion?

Re: How are orbitals corresponding to m sub l numbers

Re: Hydrogen Levels and Energy [ENDORSED]

Re: Coefficients in electron configurations [ENDORSED]

Re: Hund's Rule [ENDORSED]

Re: Frequency on Photoelectric Effect

Re: Energy Absorbed in a Mole? - HW Problem #55

Re: calculating the number of photons

Re: moles + limiting reactant

Re: HW 1.15

Re: 1.13 (b)

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Re: Mass of Products VS. Mass of Reactants- Class Assessment

Ionic Concentrations