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chemistry 4s orbital

can someone explain to me the 4s orbital?
i was told when writing out the electronic configs, 4s should come before 3p, but then in most mark schemes it's 3p then 4s, and this applies to when it asks you to write configs for ions too, like 4s loses electrons before 3p? so what was the point in teaching us about 4s coming first since it just confuses me :/
Original post by cherrybunny
can someone explain to me the 4s orbital?
i was told when writing out the electronic configs, 4s should come before 3p, but then in most mark schemes it's 3p then 4s, and this applies to when it asks you to write configs for ions too, like 4s loses electrons before 3p? so what was the point in teaching us about 4s coming first since it just confuses me :/

you are confusing the 3p subshell with the 3d subshell. electrons go in 3s then 3p then 4s then 3d. and then writing out electronic configurations you can write the 3d and 4s in either order eg... 3s23p63d14s2 or 3s23p64s23d1. and electrons go in 4s before 3d as 4s is at lower energy than 3d, however they leave 4s before 3d.
Original post by i_love_dogs
you are confusing the 3p subshell with the 3d subshell. electrons go in 3s then 3p then 4s then 3d. and then writing out electronic configurations you can write the 3d and 4s in either order eg... 3s23p63d14s2 or 3s23p64s23d1. and electrons go in 4s before 3d as 4s is at lower energy than 3d, however they leave 4s before 3d.

OHHH thank you so much
I recommend you check with your teacher/exam board on the electronic configuration notation and whether to write 4s or 3d first.
But just to quickly summarize why this 3d-4s switch happens:
4s orbitals are of a slightly lower energy than 3d orbitals, so electrons occupy 4s first (because the universe is lazy and wants to go for the less-energy-required route)
But once electrons begin to occupy the inner 3d orbitals, they cause a certain shielding effect that repels the outermost 4s electrons. Since the 4s electrons are already being repelled, they require less energy to remove. Going by the universe-is-lazy logic, cations would thus prefer to remove the 4s electrons before removing the 3d electrons.
Original post by ziqzaczuck
i recommend you check with your teacher/exam board on the electronic configuration notation and whether to write 4s or 3d first.
But just to quickly summarize why this 3d-4s switch happens:
4s orbitals are of a slightly lower energy than 3d orbitals, so electrons occupy 4s first (because the universe is lazy and wants to go for the less-energy-required route)
but once electrons begin to occupy the inner 3d orbitals, they cause a certain shielding effect that repels the outermost 4s electrons. Since the 4s electrons are already being repelled, they require less energy to remove. Going by the universe-is-lazy logic, cations would thus prefer to remove the 4s electrons before removing the 3d electrons.

thank you so much
but in a question i came across, it says "State the full electron configuration of a cobalt(II) ion."
answer: 1s2 2s2 2p6 3s2 3p6 3d7
how come the 3d orbital fills up first because shouldn't it be 1s2 2s2 2p6 3s2 3p6 3d5 as 4s would've lost two electrons?
Original post by ziqzaczuck
I recommend you check with your teacher/exam board on the electronic configuration notation and whether to write 4s or 3d first.
But just to quickly summarize why this 3d-4s switch happens:
4s orbitals are of a slightly lower energy than 3d orbitals, so electrons occupy 4s first (because the universe is lazy and wants to go for the less-energy-required route)
But once electrons begin to occupy the inner 3d orbitals, they cause a certain shielding effect that repels the outermost 4s electrons. Since the 4s electrons are already being repelled, they require less energy to remove. Going by the universe-is-lazy logic, cations would thus prefer to remove the 4s electrons before removing the 3d electrons.


Original post by i_love_dogs
you are confusing the 3p subshell with the 3d subshell. electrons go in 3s then 3p then 4s then 3d. and then writing out electronic configurations you can write the 3d and 4s in either order eg... 3s23p63d14s2 or 3s23p64s23d1. and electrons go in 4s before 3d as 4s is at lower energy than 3d, however they leave 4s before 3d.

but in a question i came across, it says "State the full electron configuration of a cobalt(II) ion."
answer: 1s2 2s2 2p6 3s2 3p6 3d7
how come the 3d orbital fills up first because shouldn't it be 1s2 2s2 2p6 3s2 3p6 3d5 as 4s would've lost two electrons?
Original post by cherrybunny
but in a question i came across, it says "State the full electron configuration of a cobalt(II) ion."
answer: 1s2 2s2 2p6 3s2 3p6 3d7
how come the 3d orbital fills up first because shouldn't it be 1s2 2s2 2p6 3s2 3p6 3d5 as 4s would've lost two electrons?


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(edited 3 years ago)
Original post by cherrybunny
but in a question i came across, it says "State the full electron configuration of a cobalt(II) ion."
answer: 1s2 2s2 2p6 3s2 3p6 3d7
how come the 3d orbital fills up first because shouldn't it be 1s2 2s2 2p6 3s2 3p6 3d5 as 4s would've lost two electrons?

Hmm okay let's tackle the question step by step:
After filling up the first 18 electrons of a cobalt atom, we get 1s22s22p63s23p6.

We're left with 9 more electrons to assign (Cobalt has 27 electrons). Since 4s orbitals fill up first, the next 2 electrons go into the 4s orbital and we get 1s22s22p63s23p64s2. We're now left with 7 more electrons to assign, so we put all 7 of them in the 3d orbital (since it can accommodate up to 10 electrons).

Therefore, the electronic configuration of a cobalt atom is 1s22s22p63s23p63d74s2, because as mentioned earlier, the 4s orbital fills up before the 3d orbital. You may have noticed that I wrote the 3d7 before the 4s2. That's just my preference, it doesn't matter if I wrote 4s2 before 3d7. Both mean that the 4s orbital is holding 2 electrons and the 3d orbital has 7 electrons.

When you want to find the electronic configuration of an ion, you need to find the configuration of its atom first, which I've just written above.
As I mentioned, said earlier, the 3d electrons cause a bit of repulsion, causing the 4s electrons to be removed first.
Cobalt (II) has a charge of 2+ because 2 electrons were removed. By removing 2 electrons from the 4s orbital, we get the final answer:

Configuration of a cobalt (II) ion: 1s22s22p63s23p63d7.

Hope this helps!! Feel free to dm or ask further if you're still having trouble :>
Original post by CaptainDuckie
Cobalt is a transition element, it doesn’t follow the general consensus

Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration.

Iirc most transition metals, or at least those in the first row, fill up how you'd expect (assuming you know that 4s fills up before 3d). The only 2 anomalies in the first row of d block elements are Copper and Chromium. I'm not sure if it is in your syllabus but I had to learn why for my (international) A-levels so I'd be happy to explain why if @cherrybunny (or anyone else) wants
Original post by ziqzaczuck
Iirc most transition metals, or at least those in the first row, fill up how you'd expect (assuming you know that 4s fills up before 3d). The only 2 anomalies in the first row of d block elements are Copper and Chromium. I'm not sure if it is in your syllabus but I had to learn why for my (international) A-levels so I'd be happy to explain why if @cherrybunny (or anyone else) wants


Cobalt (III) is AR 3d6 right?
(edited 3 years ago)
Original post by CaptainDuckie
Cobalt (III) is AR 3d6 right?

yep!
Original post by ziqzaczuck
yep!


I got confused in the rules :s

My bad, you are right, they fill up normally but only exceptions are copper & chromium :s

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