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Chemistry help please? :3

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I so don't get the electronic configuration thing.. The 1s^1 and 2p thingy. Has anyone got proper complete notes that would help me solve these kind of problems? Like forexample, for Boron the electrons are 5,right? Why is the electronic configuration like 1s^2 , 2s^2 and then 2p^1? Why cant it be like 1s^1 , 1s^2 , 2p^1? I've got this problem in every friggin atom.
 
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I so don't get the electronic configuration thing.. The 1s^1 and 2p thingy. Has anyone got proper complete notes that would help me solve these kind of problems? Like forexample, for Boron the electrons are 5,right? Why is the electronic configuration like 1s^2 , 2s^2 and then 2p^1? Why cant it be like 1s^1 , 1s^2 , 2p^1? I've got this problem in every friggin atom.

Every shell has different orbitals in it.

The first shell only has the s orbital.
Second shell has s and p.
Third shell has s, p, d.

So, you know the electronic configuration of Boron from O-Level, it's (2, 3)

That means 2 electrons in the first shell.
3 electrons in the second shell.

In A-levels, it's more complex. You have to state the electrons in the sub-shells, too.

1s^2 -- Means that the first shell has 2 electrons.
Then, 2s^2, means the s-orbital of second shell has 2 electrons.
Then 2p^1, the third electron of the second shell goes into the p orbital, because every orbital can have a max of 2 electrons.

Also, s orbital is always filled before p, and p before d.
 
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Every shell has different orbitals in it.

The first shell only has the s orbital.
Second shell has s and p.
Third shell has s, p, d.

So, you know the electronic configuration of Boron from O-Level, it's (2, 3)

That means 2 electrons in the first shell.
3 electrons in the second shell.

In A-levels, it's more complex. You have to state the electrons in the sub-shells, too.

1s^2 -- Means that the first shell has 2 electrons.
Then, 2s^2, means the s-orbital of second shell has 2 electrons.
Then 2p^1, the third electron of the second shell goes into the p orbital, because every orbital can have a max of 2 electrons.

Also, s orbital is always filled before p, and p before d.
Why is it
2p^1
Since it is the first p orbital, shouldn't it be:
1p^1
 
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Dark, If each shell has the s orbital why don't we simply continue on with the s orbital? Like 1s then 2s then 3s? Why do we have to go forward to the p and d orbitals?

In O-Levels, you studied that the electrons revolve around the nucleus in shells.

You used to draw dot-and-cross diagrams, right?

You drew circles, and put dots on them, or crosses to represent electrons.
The first shell had a max of 2 electrons, the 2nd shell 8.. and so on..

Now in A-Levels, you get to know that the shells are further divided into sub-shells/orbitals.

Electrons don't just revolve around the nucleus in one path. They have different paths, and those different paths are known as orbitals.

The first shell has only one path. Only one orbital. It's spherical, and 2 electrons can fit into it, and orbit around the nucleus. This is the s-orbital. All spherical subshells are called s-orbitals.
But the second shell has more than one orbital. It has s-orbital, as well as p-orbital. That means, first it's s-orbital will be filled with 2 electrons, and then the p orbital will be filled with 6 electrons. 2+6 makes 8 right? You studied in O-Level that the second shell can hold 8 electrons.

Now it's just that, that second shell has two parts. s and p, and s has 2, p has 6.
It has one spherical path. [S-orbital]
And one other shaped path, which has 6 electrons. [p-orbital]

When electrons are being filled into the atom, they go into the shell which is closest to the nucleus, first. That is the first shell.
And it has one orbital only. That shell itself is the s-orbital. A spherical path. [Imagine it in 3-D]

So, 2 electrons revolve around the nucleus in the first shell, at high speed.

Then the next 2 electrons are added to the second shell, but the s-orbital of the second shell, first. It's always filled before p.
Then 6 electrons in the p-orbital of 2nd shell.

Only when 2 have been added in 2s, and 6 have been aded in 2p.. we can say that 8 electrons have been added in the second shell.

And now, we'll move onto the third shell, which has s orbital, p orbital, and now another orbital called d-orbital.

Same as before, the s orbital fills up first, then p, then d.
First 3s will be filled up with 2 electrons, then 3p is filled up with 6.. and so on!

The s and p orbitals are part of the shell. Unless you fill all orbitals, you can not move onto the next shell.
 
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In O-Levels, you studied that the electrons revolve around the nucleus in shells.

You used to draw dot-and-cross diagrams, right?

You drew circles, and put dots on them, or crosses to represent electrons.
The first shell had a max of 2 electrons, the 2nd shell 8.. and so on..

Now in A-Levels, you get to know that the shells are further divided into sub-shells/orbitals.

Electrons don't just revolve around the nucleus in one path. They have different paths, and those different paths are known as orbitals.

The first shell has only one path. Only one orbital. It's spherical, and 2 electrons can fit into it, and orbit around the nucleus. This is the s-orbital. All spherical subshells are called s-orbitals.
But the second shell has more than one orbital. It has s-orbital, as well as p-orbital. That means, first it's s-orbital will be filled with 2 electrons, and then the p orbital will be filled with 6 electrons. 2+6 makes 8 right? You studied in O-Level that the second shell can hold 8 electrons.

Now it's just that, that second shell has two parts. s and p, and s has 2, p has 6.
It has one spherical path. [S-orbital]
And one other shaped path, which has 6 electrons. [p-orbital]

When electrons are being filled into the atom, they go into the shell which is closest to the nucleus, first. That is the first shell.
And it has one orbital only. That shell itself is the s-orbital. A spherical path. [Imagine it in 3-D]

So, 2 electrons revolve around the nucleus in the first shell, at high speed.

Then the next 2 electrons are added to the second shell, but the s-orbital of the second shell, first. It's always filled before p.
Then 6 electrons in the p-orbital of 2nd shell.

Only when 2 have been added in 2s, and 6 have been aded in 2p.. we can say that 8 electrons have been added in the second shell.

And now, we'll move onto the third shell, which has s orbital, p orbital, and now another orbital called d-orbital.

Same as before, the s orbital fills up first, then p, then d.
First 3s will be filled up with 2 electrons, then 3p is filled up with 6.. and so on!

The s and p orbitals are part of the shell. Unless you fill all orbitals, you can not move onto the next shell.
Thanks a bunch! You solved it all for me! (y)
 
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Can someone help me as well??? I dont get the emission spectrum. All I understand is that when an atom is excited, it releases energy, but i dont get all the series and how all this could possibly relate to creating a model of an atom. Pls help
 
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Uhm.. quick guess.. C? :p

Basically, the p-orbital can hold a maximum of 6 electrons.
In the situation where it has 3 electrons, we say it's a half-filled p-orbital.
If it has 6, we say it's fully-filled.

Nitrogen has a half-filled P-Orbital.
It's electronic configuration is (2,5)

Now, in A-Level, the 5 will be further split up.
So 1s^2, 2s^2, 2p^3.

Nitrogen atom has a half-filled p-orbital. But the question says which particle ON LOSING AN ELECTRON has a half-filled p-orbital, so the option B of Nitrogen atom is wrong. Nitrogen negative ion, has four electrons in p-orbital, because it has gained one electron, losing which it will become Nitrogen atom that has half-filled p-orbital.
 
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D?

Just write the electronic configuration of Lithium --> 1s^2, 2s^1
YEAH IT IS D
but you are only half correct coz you didn't see configuration of Cr and sometimes there may be option which may be like this but it will be incorrect answer so it would be better if you see both option configuration :)
 
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YEAH IT IS D
but you are only half correct coz you didn't see configuration of Cr and sometimes there may be option which may be like this but it will be incorrect answer so it would be better if you see both option configuration :)

Well, A and B first options are wrong so we can skip that.
In C, the first is right but the second is wrong. (They are much more easier to calculate:p)

Thus, we don't need to find Cr. Though I guess all Group 1 elements have S^1.
 
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