South Pasadena • Honors Chemistry
Name
9 • Atomic Structure
Period
9.2
PROBLEMS
–
Date
ELECTRON CONFIGURATION
1. Using a Periodic Table, write the long form and short form ground state electron configuration for the
following species. Finally, state the number of valence electrons in each.
(a) Co
Long Form:
2
(b) O
2
2
6
2
7
[Ar] 4s 3d
Short Form:
2
4
2
6
2
6
2
10
6
2
10
6
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s
(i) Br
(j) Si
(k) Ga
(l) Sr
(m) U
(n) Ni2+ ion
(o) Fe ion
(p) Se ion
(q) Pb ion
(r) Sn ion
2
10
6
2
6
2
1
6
2
2
14
10
Valence electrons:
2
14
10
3
1
[Ar] 4s 3d
2
Short Form:
Valence electrons:
2
6
2
3
2
3
1s 2s 2p 3s 3p
[Ne] 3s 3p
Long Form:
Short Form:
2
6
2
6
2
10
6
2
10
6
2
14
8
5
Valence electrons:
2
14
8
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d
[Xe] 6s 4f 5d
2
Long Form:
Short Form:
Valence electrons:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
[Kr] 5s2 4d10 5p6
8
Long Form:
Short Form:
Valence electrons:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
[Ar] 4s2 3d10 4p5
7
Long Form:
Short Form:
Valence electrons:
1s2 2s2 2p6 3s2 3p2
[Ne] 3s2 3p2
4
Long Form:
Short Form:
Valence electrons:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1
[Ar] 4s2 3d10 4p1
3
Long Form:
Short Form:
Valence electrons:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2
[Kr] 5s2
2
Long Form:
Short Form:
Valence electrons:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10
6p6 7s2 5f4
[Rn] 7s2 5f4
2
Long Form:
Short Form:
Valence electrons:
2
6
2
6
8
1s 2s 2p 3s 3p 3d
[Ar] 3d
Long Form:
Short Form:
2
6
2
6
5
1s 2s 2p 3s 3p 3d
[Ar] 3d
Long Form:
Short Form:
2
6
2
6
2
10
6
8
0
Valence electrons:
5
0
Valence electrons:
2
10
6
1s 2s 2p 3s 3p 4s 3d 4p
[Ar] 4s 3d 4p
8
Long Form:
Short Form:
Valence electrons:
2
6
2
6
2
10
6
2
10
6
2
14
10
2
14
10
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d
[Xe] 6s 4f 5d
2
Long Form:
Short Form:
Valence electrons:
2
6
2
6
2
10
6
10
[Kr] 4d
Long Form:
2
(t) I ion
2
2
1s 2s 2p 3s 3p 4s 3d 4p 4d
(s) P ion
6
Long Form:
2
3‒
10
1s 2s 2p 3s 3p 4s 3d
2
4+
6
Valence electrons:
2
2+
2
Short Form:
2
2–
6
Long Form:
2
3+
2
Short Form:
5
2
(h) Xe
Valence electrons:
2
[Xe] 6s 4f 5d 6p
2
(g) Pt
6
Short Form:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d
6p3
2
(f) P
2
4
[He] 2s 2p
[Xe] 6s
Long Form:
2
‒
2
2
Valence electrons:
2
Long Form:
2
(e) Sc
7
Long Form:
1s 2s 2p
(d) Bi
Valence electrons:
2
1s 2s 2p 3s 3p 4s 3d
2
(c) Ba
Short Form:
6
2
10
0
Short Form:
6
2
6
2
Valence electrons:
6
1s 2s 2p 3s 3p
[Ne] 3s 3p
Long Form:
Short Form:
2
2
6
2
6
2
10
6
2
10
6
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p
2
8
Valence electrons:
10
6
[Kr] 5s 4d 5p
8
2. State the number of unpaired electrons in each of
the following:
(a) Fe
Transitions metals in general have 2 valence
electrons, which are ns2, so they tend to form +2
ions.
4 unpaired electrons
(b) Cd
0 unpaired electrons
(c) Te
5. An atom can be in an excited stated if the aufbau
principle is not held. For which elements do these
excited state electron configurations represent?
2‒
0 unpaired electrons
(a) 1s2 2s2 3s2 3d5
(d) Ti
Na
2 unpaired electrons
(b) 3p1
(e) S
H
2 unpaired electrons
(c) 1s2 2s2 3p1
(f) Ga
B
1 unpaired electron
(d) 1s2 2s2 2p4 3s1
(g) P
F
3 unpaired electrons
(e) 1s1 2p2 3p3 4d4
(h) Rb
Ne
1 unpaired electron
(f) [Ar] 4s2 3d5 4p1
(i) Ba
Fe
0 unpaired electrons
(j) Mn2+
5 unpaired electrons
(k) Sn
2 unpaired electrons
(l) Br
1 unpaired electron
3. Arsenic (As) can form both a cation and an anion.
State these expected ions and write their short form
electron configurations.
Cation:
Anion:
As3+
As5+
As3‒
[Ar] 4s2 3d10
[Ar] 3d10
[Ar] 4s2 3d10 4p6
4. In general, how many valence electrons do
transition metals have? What ion is likely to form
with transition metals? Explain briefly.
or