[Cicada]

Hello, heres my question:
a)write a ground state electron configuration of for the arsenic atom
 b)give a possible set of quantum numbers for each of the three outermost electrons of the arsenic atom

i know how to get the ground state, but i have no idea how to get the quantum number for each of the three electrons.  I search google also, all i've found is articles regarding the ground state.  At this point I am 100% stuck.  thanks for *delete me*

[Yggdrasil]

There are four quantum number: n, l, m_l, and m_s

n is the principle quantum number.  It specifies the energy level of the electron.  This can be any natural number (1,2,3,...)
l is the azimuthal quantum number.  It specifies the type of orbital (s, p, d, f, etc.).  l ranges from 0 to n-1.
m_l is the magnetic quantum number, it specifies the orientation of your orbital (e.g. p_x v, p_y v. p_z.  m_l ranges from -l to l.
m_s is the spin quantum number.  It specifies the spin of the electron (±1/2).

For more information see your chemistry text or wikipedia (the article isn't too well written though):
http://en.wikipedia.org/wiki/Quantum_number

[Ψ×Ψ]

Alrightie.  Which are the three outermost electrons?  (What's the electron configuration of arsenic?  You're looking for the last three electrons listed.)

[Cicada]

okay, for arsenic its this: 3d¹⁰4s²4p³, but thats the ground state, and the quantum numbers, from what i understand are distinct, so how can there be more than one configuration?  I already read the wikipedia article, it didnt give me any help.

[Yggdrasil]

So, the three outermost electrons are the three electron in the 4p orbital.  Here you can already tell two of the quantum numbers, since the 4 denotes that n=4 and the p denotes that l = 1 (s orbitals have l=0, p means l=1, d means l=3, and f means l=4).  Now, all you have to do is figure out the m_l and m_s values for these electrons.  Remember that there are three distinct p orbitals, each with a different value of m_l.

[Cicada]

okay, thanks, but there are 4 "variables" there, n,l,m_l and m_s, i thought there were only 3 quantum numbers?

[Cicada]

oh, i see, it because these are describing electrons, which can go in to direction in the same orbital, thanks again.

[Cicada]

so three possible quantum numbers for the outermost electrons could be:

n=4,l=1,m_l=-1,m_s=1/2
n=4,l=1,m_l=0,m_s=1/2
n=4,l=1,m_l=+1,m_s=1/2

... is this right?

[Yggdrasil]

Yup, this is a correct possibility.  You correctly remembered that each p orbital will have one electron each and that all electrons will have the same spin.  The other correct possibility would be:

n=4,l=1,ml=-1,ms=-1/2
n=4,l=1,ml=0,ms=-1/2
n=4,l=1,ml=+1,ms=-1/2