Final Practice 2012

[leilarishwain_3H]

Can anyone explain how to do number 3 on the 2012 final practice?

[Greta Tobin 2F]

For 3A:
The "n" determines the size, the "l" determines the shape, and the "ml" determines the orbital.
For the first problem,( n=3, l=1) the quantum number is 3p, which can hold 6 electrons (3 because its given, p because l=0 refers to the s orbital, l=1 refers to the p orbital, l=2 refers to the d orbital, and l=3 refers to the f orbital)
For the second problem (n=5, l=3, ml=-1) the quantum number is 5f, and since they specified a specific orbital and an orbital can only hold 2 electrons, there are 2 electrons with this specific quantum number.
For the third problem (n=2, l=1, ml=0) the quantum number is 2p and similar to the second problem, there are only 2 electrons that can be in that specific orbital (0).

[Greta Tobin 2F]

For 3B:
When an electron is removed from an atom, it is always taken from the highest energy orbital (or the last quantum number listed in the electron configuration).
For Zn, the electron configuration is [Ar}3d(^10)4s(^2), and the electron will be taken from 4s, the highest energy orbital.
Cl: [Ne]3s(^2)3p(^5) --> will be taken from 3p
Al: [Ne]3s(^2)3p(^1) --> will be taken from 3p
Cu: [Ar]3d(^10)4s(^1) --> will be taken from 4s

[Tyler_Honrada_1L]

For some reason I remember that an electron was taken from the 3d orbital before it was taken from 4s. Am I just imagining this, or are there acceptions?