Can you help me with this Chemistry question?

Question

Write the 14 sets of quantum numbers that describe the 14 electrons of Silicon. How would I do this?

Answer 1

You have to consider the electron configuration of silicon. This would be written as:

1s2 2s2 2p6 3s2 3p2

This information is easily gathered from the periodic table. The first two columns (Li, Be; Na, Mg; K,Ca, etc) are your s orbitals. The 6 columns starting with (B to Ne; Al to Ar), etc are your p orbitals. The 10 columns starting with (Sc to Zn; etc) are your d orbitals and the 14 columns starting with (La to Yb, etc) are your f orbitals. The number of columns is precisely defined by quantum theory.

Lets think about what quantum numbers define the 1s orbital. n is the number that comes before the orbital type, it defines the energy of the orbital. Therefore n=1. l defines the orbital shape. When l =0, it is an s orbital, 1, a p orbital, 2 a d orbital or 3 it is a f orbital. l can have values of 0 to (n-1). Therefore when n=1 you can only have l=0 (s orbital). YOU CAN SEE THIS BY LOOKING AT THE PERIODIC TABLE. ml describes the subshells of an orbital, it can have values of (-l to l), coincidentally, the number of subshells an orbital has is the total number of electrons the orbital can hold divided by two, thus s has 1 value for ml, p has 3 values for ml(px, py, pz), d has 5 values for ml(dx2-y2, dz2, etc), and f has 7 values for ml. For 1s, there is only one value for ml, 0. Finally you have the electronic spin, ms. Usually the first electron in gets spin up(Hund's rule), which corresponds to ms of +1/2, the second electron in gets a spin down, or a value of -1/2.

Thus the quantum numbers for the first two electrons (1s) would be:

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

For 2s you would get:

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

For 2p you would get:

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

Now you try to finish the rest!

Answer 2

N is the principle quantum number, once you get n you should be able to get the rest of the possibilities.

For example
If n = to 4
then l (0, to n-1) can be 0, 1, 2, or 3
then M_(l) ( -l to l ) can be -3, -2, -1, 0, 1, 2, 3
and M_(s) (spin of the electron) is either -1/2 or 1/2 to represent an upward or downward spin.

Find n and you have all the possibilites of L, M_L, and M_S ( which is always going to be 1/2 or -1/2)

Answer 3

start out with:

1s^1 1s^2 2s^1 2s^2 2p1^.....

Usually you would write:
1s^2 2s^2 2p^6 3s^2 3p^2