CHEMISTRY COMMUNITY

George_Yin_3I

Valence electrons are basically electrons on the outer shell, which you can easily calculate by subtracting all electrons in inner shells.

George_Yin_3I

Yes, if there is a pi bond present, there will at least a double bond or bond with greater strength present.

George_Yin_3I

Delocalized pi bonds occur when pi bonds extend to more than 2 atoms, so you can see after drawing out the Lewis structure.

George_Yin_3I

Basically, you draw out the Lewis structure first. For H2S, there are only 2 single bonds: pi bonds only. For SO2, there are 2 double bonds, it means that it has 2 sigma bonds and 2 pi bonds.

George_Yin_3I

Also, as the bond strength increases, like in a triple bond, there is still one sigma bonds, and (3-1) pi bonds. It all acts this way.

George_Yin_3I

Yes, a single bond contain one sigma bond, while a double bond contain one single and one pi bonds.

George_Yin_3I

It is the same logic with cations. For example, the ionic radius of Ca2+ is smaller than K+, given that both of them contain 18 electrons.

George_Yin_3I

These three anions are isoelectronic; in this case, the greater the proton number, the smaller the ionic radius is.

George_Yin_3I

If the two atoms/ions are isoelectronic, then the one with greater proton number is smaller because it means greater effective nuclear charge from the central atom.

George_Yin_3I

The atomic radius increases down a group and generally decreases across a period, so Br- is naturally greater in radius than Na.

George_Yin_3I

Thanks professor!

George_Yin_3I

Delocalized pi bonds occur when pi bonds extend to more than 2 atoms. There is an exercise on Sapling about that.

George_Yin_3I

Yes, we have done some exercises on Sapling that require us to deduce the formal charge based on the charge on the molecule. In this case, the sum of the formal charge is equal to the charge of the molecule.

George_Yin_3I

Based on what Prof Lavelle said, single bond contains one sigma bond; double bonds contain one sigma bond and one pi bond; while triple bonds contain one sigma bond and two pi bonds.

George_Yin_3I

So basically, any bond contains a sigma bond, whether it is single, double, or even triple bonds.

George_Yin_3I

It should have three horizontal orbitals each with 120 degrees, and two vertical orbitals with 90 degrees in between.

George_Yin_3I

Electrons usually enter the 4s orbital before the 3d because the 4s is initially lower in energy, while that is the case for elements #19, #20.

George_Yin_3I

Oxygen has a decrease in ionization energy despite the fact that the trend is all increasing generally across the period. That is because atoms have a tendency to either form a full octet or a half-filled octet.

George_Yin_3I

Besides ionic radius, I believe electron affinity, (first) ionization energy, and electronegativity are some other important properties.

George_Yin_3I

The special case in ionic radius is regarding radius comparison among isoelectronic ions, in which cations are smaller than anions due to the larger proton number and greater effective nuclear charge.

George_Yin_3I

First you have to remember the ionic radius, which is similar to atomic radius principle.

George_Yin_3I

The molecule is definitely polar because if you draw out the Lewis structure, the central atom N will have a lone pair of electrons, making the molecule polar.

George_Yin_3I

The central atom is determined by the electronegativity of the atom: the central atom is also usually the least electronegative. Besides, the central atom will be the one that has the most unpaired valence electrons. Thus, in this case, it should be N.

George_Yin_3I

Resonance structures are used when a single Lewis structure cannot fully describe the bonding. Resonance hybrid represents the combination of possible resonance structures.

George_Yin_3I

Resonance structures are used when a single Lewis structure cannot fully describe the bonding. Resonance hybrid represents the combination of possible resonance structures.

George_Yin_3I

No, you do not need to do that. Lewis structure is basically a demonstration of bond types, so there is no need to represent the bond lengths.

George_Yin_3I

No, you do not need to do that. Lewis structure is basically a demonstration of bond types, so there is no need to represent the bond lengths.

George_Yin_3I

Regarding the number of valence number, you just have to consider the outer most shell of the atom and subtract all electrons from inner shells.

George_Yin_3I

Regarding the number of valence number, you just have to consider the outer most shell of the atom and subtract all electrons from inner shells.

George_Yin_3I

When ions have the same electron configuration, you need to look at the proton number. A greater proton number means a greater effective nuclear charge, and therefore smaller ionic radius.

George_Yin_3I

Regarding the excited state, you can see that the electron moved towards a higher state instead of filling the 2p shell first. Therefore, the answer should be c.

George_Yin_3I

As for carbon, its ground state configuration is 1s^2 2s^2 2p^2 instead of 1s^2 2s^2 2p^1 3s^1 because the electrons tend to first fill the p-block in the second shell.

George_Yin_3I

Basically, for the atomic radius, it increases down the group and decreases across the period.

George_Yin_3I

For electrons, you can use De Broglie's equation,
which is wavelength lambda=h divided by p, in which p=mv.

George_Yin_3I

By using c=lambda*v, and E=hv, the answer can be obtained.
It is a two-step calculation.

George_Yin_3I

The function means that the kinetic energy of electrons ejected from the metal is equal to the kinetic energy of the photons in total minus the work function of the energy.

George_Yin_3I

For the second part of this question, all you have to understand is that "maximum" means that there is no kinetic energy ejected as electrons. In this way, all of the energy from the photons are used to satisfy the work function of the electrons. The answer can be obtained from a simple di...

George_Yin_3I

Since photons (particles of light) have no mass, they must obey E = pc and therefore get all of their energy from their momentum. Now there is an interesting additional effect contained in the general equation. If a particle has no mass (m = 0) and is at rest (p = 0), then the total energy is zero (...

George_Yin_3I

I think both ways are fine, it is a matter of units.

George_Yin_3I

Intensity is proportionate to the amplitude, while it does not determine whether or not electrons can be ejected off the metal. Light sources with short wavelengths (high frequencies) can eject electrons even with low intensity light.

George_Yin_3I

Generally, as for wavelength, we use "nm" nanometers.

George_Yin_3I

Basically, intensity can be reflected upon amplitude. However, in the photoelectric model, intensity is not crucial as the light sources with short wavelengths (high frequencies) can eject electrons even with low intensity light.

George_Yin_3I

Elimination is one way to solve this problem, since the other three options are obviously wrong. Regarding option C, to clarify, "the extent of change" can be understood as the slope of the wave. Then it should be obvious after you draw two waves. You'll find that the slope is significantl...

George_Yin_3I

We can either model light as an electromagnetic wave a stream of photons, as demonstrated by the photoelectric effect off metal surfaces. Regarding how we can prove that light is also a wave, I am not familiar with certain experiments (if there are any), but Maxwell’s equations, which describe the r...

George_Yin_3I

If we need an extra camera to record our work space, how can the second device connect to our computer in lockdown browser?

George_Yin_3I

I think you overcomplicate this question. Basically, theoretical yield is just the yield that we can calculate based on the given reactants. Thus, in this case, it will be 0.2mol * 81g/mol= 16.2g. This question directly gives you the amount of products formed so you do not have to worry about the li...

George_Yin_3I

Hey Isis, AgCl is a solid product that precipitates from this reaction while both reactants are in aqueous forms. However, it does not affect the calculation of mass of AgCl produced. The limiting reagent only refers to one of the reactants that determine the amount of reactants that actually underg...

George_Yin_3I

The answer is no. When the relative ratio of atoms cannot be divided further, the molecular formula may be the same as the empirical formula. However, the empirical formula cannot be greater in the number of atoms than molecular formula, given the definition of empirical formula being the simplest p...

George_Yin_3I

If like you said, the grams of all elements present are known, then actually there is no need to calculate the mass percentage. The number of moles for all atoms in the molecule can be deduced, and thereby a ratio can be calculated. Empirical formula can be obtained using the relative ratio.

George_Yin_3I

5 in China.

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