Sapling question 21
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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?
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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
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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.
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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.
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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.
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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.
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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!
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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.
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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! :))
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.
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