Quantum Mechanics and
Electron Configuration Problems
Solutions
First: n = 1,2,3,4 ... infinity
l = 0,1,2,3 .. (n-1) (to specifiy l values you have to know
which n you are talking about)
ml = -l, (-l+1) .. 0 .. (l-1), +l (to specifiy ml values you
have to know which l you are talking about)
ms = +1/2 or -1/2
also: if
if
if
if
l
l
l
l
=0 this corresponds to an s-orbital
=1 this is a p-orbital
= 2 this is a d-orbital
= 3 this is an f-orbital
1. For n = 4, l can take four possible values, 0,1,2,3
For l = 2, ml can take five possible values, -2, -1, 0, 1, 2
2. a) Impossible. If n = 1 the only allowed value of l is 0. In other
words, 1s is the only possible orbital with n = 1.
b) ok. The 4s orbital can exist since the rules for combinations of
quantum numbers allow n = 4 and l = 0.
c) ok. The required quantum numbers for a 5f orbital are n=5, l=3
(plus an ml value for each of the seven possible degenerate orbitals)
d) Impossible. If n = 2, l can be 0 or 1. A d-orbital corresponds to
an orbital with an l value of 2.
3. a) this set of quantum numbers is ok and obeys the above rules.
b) this set is not ok because if l =0 the only possible ml value is
0.
c) this set is ok
d) this set is not ok since if n = 3 the highest possible l value is
2.
a) [Xe] 6s1
b) [Ar] 4s23d8
c) [Ar] 4s23d104p4
d) [Kr] 5s24d10
e) [Rn] 7s26d1
f) [Xe] 6s24f145d106p2
5. [Rn] 7s26d105f147p6
This element will be part of the noble gas group of elements.
6.a) This is not a ground-state electron configuration .. the
electron in the 3s orbital should be in a 2p orbital
b)[Ne] implies a completely filled n=2 level. The next few electrons
will fill the 3s and 3p orbitals.
c)Not a ground-state electron configuration. The electrons in the 3d
orbitals should be in the lower energy 3p orbitals.
7.a) Nitrogen, 1s22s22p3
b) Selenium, [Ar] 4s23d104p4
c) Rhenium, [Kr] 5s24d7
8. If there were three possible spin quantum numbers for the electron
then it would be possible to put three electrons into an orbital
rather than two. Thus, the s-block of elements in the table would be
three elements wide instead of two. The p-block would be 9 elements
wide and the d-block would be 15 elements wide. Hydrogen and helium
would be s-block elements and lithium would be at the top of the
noble gas group. Berylium, boron and carbon would be period 2 s-block
elements and the period 2 p-block elements would consist of nitrogen
to phosphorus, phosphorus being the next noble gas. The chemistry of
elements would be utterly different!