Chapter 36
Problem 28
†
n=7
n=6
n=5
n=4
ml= -3 -2 -1 0 +1 +2 +3
n=3
ml= -2 -1 0 +1 +2
n=2
ml= -1 0 +1
n=1
ml= 0
l=0
s
l=1
p
l=2
d
l=3
f
Solution
Use shell notation to characterize the outermost electron of rubidium.
Since rubidium has an atomic number of 37, it has 37 protons and 37 electrons. Using the shell filling
diagram place two electrons (spin-up, spin-down) in each subshell.
n=7
n=6
n=5
n=4
ml= -3 -2 -1 0 +1 +2 +3
n=3
ml= -2 -1 0 +1 +2
n=2
ml= -1 0 +1
n=1
ml= 0
l=0
s
l=1
p
l=2
d
l=3
f
When this is done, the last electron goes into the 5s subshell. The full electronic structure of rubidium is
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1
†
Problem from Essential University Physics, Wolfson