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### Quantum numbers for the "exception" cases

Posted: Wed Oct 24, 2018 2:12 pm
I'm confused about how we would write the quantum numbers for something like Copper (Cu) whose electron configuration is [Ar] 3d^10 4s^1
How would we differentiate this quantum number from that of Potassium (K) whose electron configuration is [Ar] 4s^1

### Re: Quantum numbers for the "exception" cases

Posted: Wed Oct 24, 2018 2:21 pm
If we were following the regular rules, copper would be [Ar] 3d^9 4s^2. However, some copper is more stable when all of the d orbitals are full (meaning there are 10 electrons in that orbital instead of 9). In order for the d orbital to be full, though, the s orbital has to "lose" an electron. So, copper is [Ar] 3d^10 4s^1, which is more stable.

You'll be able to tell the difference between this and another atom, K for example, because of how many electrons are given. When you look at the notation [Ar] 3d^10 4s^1, you can see that there are 11 more electrons after argon (10+1). So, you can count 11 elements over, and that will be copper. Potassium is written [Ar] 4s^1, so it will have one more electron than argon.

Hope this helps!

### Re: Quantum numbers for the "exception" cases

Posted: Wed Oct 24, 2018 4:29 pm
Hey there, how do you calculate the frequency of light emitted? I know the equation, but how do I apply it? Is it only utilized when hydrogen atoms are available?

### Understanding a textbook problem

Posted: Wed Oct 24, 2018 4:35 pm
Hey there, there was a problem that I didn't understand the concept behind.
A particle (6.644 * 10^-34 kg) and textbook (2.18 kg) travel the same speed. Which will have a smaller wavelength?
I know it is the textbook, but what is the concept behind it?
Is it that the larger the mass, the smaller the wavelength?

### Re: Quantum numbers for the "exception" cases

Posted: Wed Oct 24, 2018 4:44 pm
Right thanks, but I meant in terms of writing the quantum numbers... wouldn't these two atoms have the same quantum numbers (n = 4, l = 0)

or do we write copper as : n= 3, l = 2, ml = +2

### Re: Quantum numbers for the "exception" cases

Posted: Wed Oct 24, 2018 4:56 pm
Quantum numbers are basically used to describe an electrons position in the atom. So it is not the atoms that have quantum numbers but the electrons in the atom.

Electrons in the atoms could have the same 4 quantum numbers for example the electron in the 4s orbital in either copper or potassium with the spin up state would have the quantum numbers n=4, l=0, ml = 0 and ms = +1/2.

### Re: Quantum numbers for the "exception" cases

Posted: Wed Oct 24, 2018 4:58 pm
I think you have a misconception here.

Firstly, the quantum numbers do not describe atoms. They describe individual electrons within an atom. So when you say atoms have quantum numbers, you are actually talking about the outermost electron, in your case 4s1.

Now, to answer your question, the electrons in the 4s1 in K and Cu have the same quantum numbers (assuming the spin is the same). But, this was never impossible. What is impossible is electrons having the same quantum numbers within ONE atom (due to the fact that orbitals can only contain 2 electrons).

Here is an example to illustrate my point. Take H, with its electron in 1s1. Now, focus on He. It also has an electron in 1s, which has the same quantum numbers (n=1, p=0, ml=0) as the one in H. You can do this for ALL the elements on the periodic table as they all have 1s electrons (assuming the spin is the same).

I think you may be confusing it with energy of electrons across atoms, which can not be the same as the nuclear charge is different for all atoms (and proved through the spectrum lines).

### Re: Understanding a textbook problem

Posted: Wed Oct 24, 2018 5:01 pm
Simran Athwal-Dis 3A wrote:Hey there, there was a problem that I didn't understand the concept behind.
A particle (6.644 * 10^-34 kg) and textbook (2.18 kg) travel the same speed. Which will have a smaller wavelength?
I know it is the textbook, but what is the concept behind it?
Is it that the larger the mass, the smaller the wavelength?

For the particle and textbook problem, you just simply apply De Broglie's equation: wavelength = h / (mass * velocity). You see that as mass goes up, the wavelength falls as h and velocity are constants.

### Re: Quantum numbers for the "exception" cases

Posted: Sun Oct 28, 2018 9:42 pm
ohhh got it. thanks!

### Re: Understanding a textbook problem

Posted: Mon Oct 29, 2018 10:06 pm
Te Jung Yang 1G wrote:
For the particle and textbook problem, you just simply apply De Broglie's equation: wavelength = h / (mass * velocity). You see that as mass goes up, the wavelength falls as h and velocity are constants.

Thank You! I understand it now.

### Valence Electrons?

Posted: Mon Oct 29, 2018 10:11 pm
Hello, I had a quick question about figuring out valence electrons. I know that you can figure how many valence electrons there are for each element by electron configurations or just simply figuring out where the element is located in the column. But, for the transition metals, how do you imply the column "theory?"

### Re: Quantum numbers for the "exception" cases

Posted: Sun Nov 18, 2018 9:02 pm
Hey there, can someone explain why copper has an electron configuration that is [Ar] 3d10 4s1 instead of [Ar] 3d9 4s2 and same goes for chromium.

### Determining the Shape of a Molecule

Posted: Sun Nov 18, 2018 9:05 pm
Hey there, when determining the shape of a molecule, do we have to draw the Lewis structure in a certain way ?

### Determining the Shape of a Molecule

Posted: Sun Nov 18, 2018 9:07 pm
Hey there, can someone explain how to draw Lewis structures as polyatomic species?

Posted: Sun Nov 18, 2018 9:09 pm
Hey there, how can you determine the least amount of repulsion for a molecule after drawing the Lewis structure?

### Determining the Shape of a Molecule

Posted: Sun Nov 18, 2018 9:17 pm
Hey there, how can you know the shape of a molecule by just drawing the Lewis structure? I mean, Dr. Lavelle did mention that you could just indicate which shape it is and not draw it out, but how do you know?