electron configuration

[valkener]

Could anyone give me a quick summary of how to assign electron configuration by looking at the periodic table? Or lets say a question gives the atomic number and asks how many electrons are in the p orbital - how would I figure that out??
Kaplan sucks. 👎

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[Phantastic]

The periodic table is aligned in such a way that similarly behaving atoms (atoms with the same number of valence electrons, which are the outer shell electrons) are aligned into families or groups (columns). Atoms with the same shell # are aligned into periods (rows).

Exception: subtract 1 from d orbital rows in order to find the shell number of that orbital, and subtract 2 from f orbital rows.

The first 2 columns make up the s orbital. The final 6 make up the p orbital. The columns connecting the s and p orbitals are the d orbital, and, finally, the section that is superimposed on the bottom of the periodic table is the f orbital. The s orbital holds 2 e-, p holds 6 e-, d holds 10 e-, and f holds 14 e-.

Exception: He (Helium) is "right justified" in the table, if you will. This just shows that He has a full valence shell, like the other noble gases. The difference here is that shell 1 only has an s orbital.

The reason that the above exceptions exist is that they allows us to "fill in" electrons into the next lowest energy level by reading the chart like a book--left to right, top to bottom.

So, for carbon, which has an atomic number of 6 (6 protons in its nucleus):

The first electron goes into the 1s orbital. I know this, because it is the top-left (first atom), and therefore the lowest energy level. My electron configuration at this point is the same as that for Hydrogen: 1s^(1). The coefficient of 1 is the shell number, and the superscript (power) of 1 represents the number of electrons in that orbital. I keep filling just like I'm reading a book.

The second electron goes into the second spot in the first s orbital. We now have the e- configuration of helium: 1s^(2). From here, I keep filling them in like I'm reading a book, because that is the order of stability (i.e. the order that the electrons are most likely to fill) until I get to: 1s^(2)2s^(2)2p^(2).

Other exceptions exist, such as half-filled and full d orbitals, but try to focus on the basics above.

Note: The above description is crude and based on application of concepts. For a better understanding, look into electron quantum numbers, and try to reconciliate that information with what was provided here. I really could go on and on with the valuable information contained within the periodic table, but I think this should suffice for now.

 

[valkener]

The periodic table is aligned in such a way that similarly behaving atoms (atoms with the same number of valence electrons, which are the outer shell electrons) are aligned into families or groups (columns). Atoms with the same shell # are aligned into periods (rows).

Exception: subtract 1 from d orbital rows in order to find the shell number of that orbital, and subtract 2 from f orbital rows.

The first 2 columns make up the s orbital. The final 6 make up the p orbital. The columns connecting the s and p orbitals are the d orbital, and, finally, the section that is superimposed on the bottom of the periodic table is the f orbital. The s orbital holds 2 e-, p holds 6 e-, d holds 10 e-, and f holds 14 e-.

Exception: He (Helium) is "right justified" in the table, if you will. This just shows that He has a full valence shell, like the other noble gases. The difference here is that shell 1 only has an s orbital.

The reason that the above exceptions exist is that they allows us to "fill in" electrons into the next lowest energy level by reading the chart like a book--left to right, top to bottom.

So, for carbon, which has an atomic number of 6 (6 protons in its nucleus):

The first electron goes into the 1s orbital. I know this, because it is the top-left (first atom), and therefore the lowest energy level. My electron configuration at this point is the same as that for Hydrogen: 1s^(1). The coefficient of 1 is the shell number, and the superscript (power) of 1 represents the number of electrons in that orbital. I keep filling just like I'm reading a book.

The second electron goes into the second spot in the first s orbital. We now have the e- configuration of helium: 1s^(2). From here, I keep filling them in like I'm reading a book, because that is the order of stability (i.e. the order that the electrons are most likely to fill) until I get to: 1s^(2)2s^(2)2p^(2).

Other exceptions exist, such as half-filled and full d orbitals, but try to focus on the basics above.

Note: The above description is crude and based on application of concepts. For a better understanding, look into electron quantum numbers, and try to reconciliate that information with what was provided here. I really could go on and on with the valuable information contained within the periodic table, but I think this should suffice for now.

Wow thank you so much. Ok so let me try to apply this:

Li: 1s^(2)2s^(1)
Be: 1s^(2)2s^(2)
K: 1s^(2)2s^(2)2p^(6)4s^(1)?
Ca: 1s^(2)2s^(2)2p^(6)4s^(2)??
Cr: 1s^(2)2s^(2)2p^(6)4s^(2)1d^(4?)
Rb: 1s^(2)2s^(2)3p^(6)4s^(2)1d^(10)3p^(6)5s^(1) ?

Am I doing it right? Could you give me an example of the f orbital config.? Like Ce and U, too? I tried learning it the quantum way first, but I don't understand the relationship between the element and n. Nor do I quite understand anything other than how to get from n to l.. lol 😳

You rock. 👍

 

[valkener]

Wow thank you so much. Ok so let me try to apply this:

Li: 1s^(2)2s^(1)
Be: 1s^(2)2s^(2)
K: 1s^(2)2s^(2)2p^(6)4s^(1)?
Ca: 1s^(2)2s^(2)2p^(6)4s^(2)??
Cr: 1s^(2)2s^(2)2p^(6)4s^(2)1d^(4?)
Rb: 1s^(2)2s^(2)3p^(6)4s^(2)1d^(10)3p^(6)5s^(1) ?

Am I doing it right? Could you give me an example of the f orbital config.? Like Ce and U, too? I tried learning it the quantum way first, but I don't understand the relationship between the element and n. Nor do I quite understand anything other than how to get from n to l.. lol 😳

You rock. 👍

I think I know what I messed up . Is this right now? I don't understand why the d orbital isn't on the same line as the rest. Why isn't the first row (3d) actually 4d since it's sandwiched between 4s and 4p?? 😕

Li: 1s^(2)2s^(1)
Be: 1s^(2)2s^(2)
K: 1s^(2)2s^(2)3p^(6)4s^(1)
Ca: 1s^(2)2s^(2)3p^(6)4s^(2)
Cr: 1s^(2)2s^(2)3p^(6)4s^(2)3d^(4) ??
Rb: 1s^(2)2s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(1) ??

 

[Phantastic]

Li: 1s^(2)2s^(1)

Good.

Be: 1s^(2)2s^(2)

Good.

K: 1s^(2)2s^(2)2p^(6)4s^(1)?)

Change the 4 to a 3 and you're solid. 3s comes before 3p, then 4s, then 3d, then 4p...just like reading a book.

Ca: 1s^(2)2s^(2)2p^(6)4s^(2)??

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2) would be the correct e- config.

No need to skip the 3rd row.

Cr: 1s^(2)2s^(2)2p^(6)4s^(2)1d^(4?)

You had to choose an exception! lol

****Notice that for d, I subtracted 1 even though it was in the 4th row.******

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(4) is what we would expect...but since it is a d^(4), it's actually slightly more stable by promoting one of its valence electrons (from the 4s orbital) into the d orbital to make it half-filled. The configuration looks like:

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5)

This also applies to any d^(9) elements.

Rb: 1s^(2)2s^(2)3p^(6)4s^(2)1d^(10)3p^(6)5s^(1) ?

Where did the 1d come from? Remember, the first row to introduce the d orbital is row 4, and since we subtract 1 from all d orbitals, then the first is 3d.

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(1) is the correct configuration.

Am I doing it right?

Practice, practice, practice.

Could you give me an example of the f orbital config.? Like Ce and U, too?

These are very unlikely to be on the MCAT and have some weird cases, but I'll go a step further here:

First, the configurations:

Ce:
1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(2)4d^(10)5p^(6)6s^(2)4f^(2)

Notice how I built that just by following the chart. Also notice that I subtracted 2 from the f orbital's row number.

U:
1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(2)4d^(10)5p^(6)6s^(2)4f^(14)6p^(6)7s^(2)5f^(4)

These are actually incorrect e- configs for these elements. The reason is, to maintain stability, some f orbital electrons occupy similar energy d levels. So you might get 4f^(1)5d^(1) for Ce at the end.

Note also that I can abbreviate most of this using Noble gases.

e.g.
[He] = 1s^(2)

so, I can write Lithium as [He]2s^(1) or simply [He]2s.

Applying it to later elements...

[Xe] = 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(2)4d^(10)5p^(6)

Ce becomes: [Xe]6s^(2)4f^(2), or you may see it as [Xe]6s^(2)4f^(1)5d^(1) as this is also energetically favorable. e- won't be this ambiguous on the MCAT, because you likely won't be dealing with the f orbital.

I tried learning it the quantum way first, but I don't understand the relationship between the element and n. Nor do I quite understand anything other than how to get from n to l.. lol 😳

Remember, elements are defined by the number of protons (atomic number). We're working with electrons here, which should equal the number of protons, unless we're dealing with ions. Ooooo, nice transition. Try working out the electron configuration for Li+ cation!

As for the quantum, it's really just a set of rules that defines all of the applied concepts above. It does however help you understand what's really going on...which isn't always necessary for MCAT, lol.

You rock. 👍

Thanks, I do love the basics.

 

[valkener]

Good.


1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(4) is what we would expect...but since it is a d^(4), it's actually slightly more stable by promoting one of its valence electrons (from the 4s orbital) into the d orbital to make it half-filled. The configuration looks like:

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(5)

This also applies to any d^(9) elements.

___

Where did the 1d come from? Remember, the first row to introduce the d orbital is row 4, and since we subtract 1 from all d orbitals, then the first is 3d.

1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)4p^(6)5s^(1) is the correct configuration.

Why do we subtract one? If it's just a rule than don't bother trying to explain 🙄.
Thanks so much for your replies, btw. Could I still use these rules if I were asked what the e- configuration is for let's say Calcium at an excited state?

Li+ ion? Lost an electron so is it 1s^(2) and that's it? or 1s^(2)2s^(0)?

Thanks again!!

 

[Phantastic]

Why do we subtract one? If it's just a rule than don't bother trying to explain 🙄.

We subtract one because it's only in the 4th row because we fill electrons in the 3d orbital before the 4s. So, yes, think of it as a rule in order for the periodic table to be useful in that regard.

Thanks so much for your replies, btw. Could I still use these rules if I were asked what the e- configuration is for let's say Calcium at an excited state?

Now you're on the right track! Yes, indeed I could! So, I would just say promote one of its 4s electrons into the 3d orbital. This is an excited Calcium (not a Ca+ cation or Ca- anion, that's only when we ionize by adding or removing electrons). Where did this energy come from? What happens when the excited Ca "drops" back down into its stable e- configuration? Use your intuition. If I excite it with a photon of energy E=hf, then that's the same energy it will release when it returns!

Li+ ion? Lost an electron so is it 1s^(2) and that's it? or 1s^(2)2s^(0)?

Yup! The lithium cation is just [He] or 1s^(2), the helium configuration (but its NOT helium, because it has an extra proton). Don't worry about the zero power stuff.