10/7/2011
How many atomic orbitals are in the subshell
with quantum numbers n = 6 and l = 2?
Only one of the following sets of four QNs (n l ml ms) is possible. Which one? a) 1
b) 2
c) 3
d) 5
e) More information is required to answer this question
a) 2
b) 4
c) 3
d) 3
e) 5
n = 6 and l = 2 is a 6d orbital.
A d subshell has five ml values: –2 –1 0 +1 +2
All d subshells have five orbitals
1
2
a) Be
b) C
c) Si
d) Ge
e) none of the above
3
l cannot equal n, only as high as n–1
ms cannot equal 0
ml: cannot exceed ±l
this set is a 3s orbital
l cannot be negative
n: 1, 2, 3 … l: 0, 1, … n–1
ml: 0, ±1, ±2 … ±l
ms: +½, –½ For Pb (in Group 14), how many electrons are listed in the ground state electronic configuration after the [Xe] closed shell?
Which element has a ground‐state configuration of [Ne]3s23p2 ?
[Ne]3s23p2 = 1s22s22p63s23p2
= 14e, so Z = 14: Si
valence configuration is 3s23p2, np2 is Group 14, n = 3 is third row: Si
2 –2 +½ 1 0 0 1 +2 –½ 0 0 +½ –2 –1 +½ a) 4 note: only 4 of the 28e– are valence electrons
b) 8
c) 14
d) 28
e) none of the above
4
Pb is Z = 82, [Xe]6s24f145d106p2, 28e– past Xe.
Don’t forget the d‐ and f‐block electrons!
Also, Xe is Z = 54, and 82 – 54 = 28.
1
10/7/2011
What general name corresponds to elements with a valence electronic configuration of ns1(n–1)d10?
Which element has the greatest number of valence electrons in its ground‐state electronic configuration?
a) alkali metals
b) chalcogens
c) pnictogens
d) coinage metals
e) tetrels
a) Sn (Z = 50) [Kr] 5s24d105p2 : 4 valence e–
b) Fe (Z = 26) [Ar] 4s23d6 : 8 valence e–
c) P (Z = 15) [Ne] 3s23p3 : 5 valence e–
d) Cs (Z = 55) [Kr] 6s1 : 1 valence e–
e) none of the above
5
Valence electrons are usually those with the highest principal QN n. Transition‐metals are an exception, and include the (n–1)d electrons as well. ns1(n–1)d10 is Group 11: Cu Ag and Au
6
2