## Sapling question 21

Chloe Shamtob 2H
Posts: 74
Joined: Wed Sep 30, 2020 9:55 pm

### Sapling question 21

I'm confused on how to find the number of electrons for an atom when only given the principle quatum number n. On the homework it asked how many electrosn an atom could have at n=2. How would I begin this problem?

Rohit Srinivas 2D
Posts: 67
Joined: Wed Sep 30, 2020 9:52 pm

### Re: Sapling question 21

I looked at the periodic table. How many electrons are possible at the second energy level are all of the atoms in the second period. this would be Li-Ne. Or aka 2s^22p^6. This is 2+6=8

Leo Naylor 2F
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### Re: Sapling question 21

At energy level n=2, an atom could have the s and p states. Therefore, there could be two atoms in the 2s state, and six atoms in the 2p state. In total, this atom could have eight electrons. This is represented on the periodic table, as Neon is the atom with the highest number of electrons (8) contained in the n=2 state.

Blake Ballew 1H
Posts: 64
Joined: Wed Sep 30, 2020 9:35 pm

### Re: Sapling question 21

I would draw out the electron configuration for an element at n=2 and then simply count the number of electrons. After doing so it is clear that 2s^2 and 2p^6 gives you a total of 8 possible electrons.

Justin Zhang_1A
Posts: 65
Joined: Wed Sep 30, 2020 9:48 pm

### Re: Sapling question 21

So because n = 2, the question is referring to elements in the second row. It's referring to the last electron of the atom, and since there are 8 elements in the row, there can be 8 electrons with n=2.

Posts: 66
Joined: Wed Sep 30, 2020 9:35 pm

### Re: Sapling question 21

In the n=2 level we know that there are s and p subshells and we know that the s subshell can hold 2 electrons and the p subshell can hold 6 electrons. So the n=2 level can hold a total of 8 electrons.

Farah Abumeri 3F
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### Re: Sapling question 21

Hi Chloe! To start this problem, I would reference the periodic table. The shell n=1 is represented by the first group. For n=1, 2 electrons can occupy this shell as seen by the 1s orbital that is represented by the 2 elements, H and He. For n=2, 8 electrons can occupy this shell because 2 electrons can occupy 2s orbital and 6 electrons can occupy the 2p orbital. Group 2 has 2 elements in the s-block and 6 elements in the p-block. Also, remember that there is 1 s-orbital shape and 3 p-orbital shapes, and 2 electrons can occupy each orbital. I hope that helps!

Ralph Zhang 2L
Posts: 53
Joined: Wed Sep 30, 2020 9:55 pm

### Re: Sapling question 21

If you think of the atom as the classic Borh model (with circles of electrons) each n level corresponds to one circle. If it asks for n=2 then ignore all other circles, focus on what can exist in that one circle, which is 2s2p.

Charmaine Ng 2D
Posts: 71
Joined: Wed Sep 30, 2020 9:37 pm

### Re: Sapling question 21

At energy level n=2, the configuration would be 2s^2 2p^6, since two electrons fit in the s subshell and 6 fit in the p subshell, so 2+6 = 8 electrons! :))

905579227
Posts: 44
Joined: Wed Sep 30, 2020 10:01 pm

### Re: Sapling question 21

I was having the same problem as you but it was because of my calculator not working because it was low on charge, try turning it off and on.

Fernanda Olvera 3D
Posts: 50
Joined: Wed Sep 30, 2020 9:41 pm

### Re: Sapling question 21

I agree with Charmaine ! :)

Michelle Magana 2B
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Joined: Tue Oct 06, 2020 12:17 am

### Re: Sapling question 21

i was also confused so this thread helped