Chem 121
Spring 2015
Answers to EXAM IV
1. Strategy: To determine ml you must determine l first as ml is dependent on l.
For n = 4, l can be 0, 1, 2, 3.
For l = 0, ml = 0
For l = 1, ml = 1, 0, +1
For l = 2, ml = 2, 1, 0, +1, +2
For l = 3, ml = 3, 2, 1, 0, +1, +2, +3
Thus, the possible ml are all of the above: 3, 2, 1, 0, +1, +2, +3
For n = 4 and l = 3 we are considering the 4f orbitals, which come in a set of 7. With 2 electrons in
each orbital, that would be 2x7 = 14 electrons allowed in 4f orbitals.
2. Strategy: First fill in the arrows for the neutral Fe atom. Then remove 3 electrons starting with the
ones in the outershell (4s2).
E
_
_
_
3d
_
__ __ __
_
4p
__
4s
3p
3s
2p
2s
1s
Electron configuration: 1s2 2s22p6 3s23p63d5
3. sulfide: 1s2 2s22p6 3s23p6
4. Valence electrons of the bromine atom: 4s24p5
5. a) strontium bromide is ionic!
SrBr2 separates out as Sr2+ and two Br
Sr has two valence electrons, which are removed when it formed Sr 2+ (so it shows no dots).
Each bromine atom has 7 valence electrons.
The bromide ion has one extra, which makes 8 (4 pairs).
b) cyanide ion is CN . This is one of the polyatomic ions you should be familiar with.
Answer (no resonance
structures as there is only
one arrangement possible).
6. Consider the iodate ion, IO3 for the following questions.
a) Give its number of pairs of valence electrons: Ans. 7 + 3x6 + 1 = 26 = 13 pairs
b) Give its Lewis structure that has an octet on all atoms:
c) Fill in the formal charges (if any) on the Lewis structure you gave to part (b).
Remember not to put zero if an atom is neutral.
Formal charges are determined as follows:
Each O has 3 lp and one bond = 6 +1 =7 valence electrons.
O should have 6 electrons and so it has one extra, that gives it a formal charge of 1.
For iodine, the central atom, it has 5 valence electrons and should have 7, so it is missing 2
electrons, giving it a formal charge of 2+.
The net charge is +2 1 1 1 = 1 which matches the charge of the cyanide ion CN .
d) Based on the formal charges, is/are there better way(s) of arranging the electrons?
YES (there are too many formal charges in the figure shown above).
So, we look at the 2+ on the central atom. It tells us we have to borrow TWO pairs of
electrons from its neighbor to share. But we see there are 3 oxygen from which to borrow,
this tells us there are resonance forms. (see belo)
e) If there are resonance forms, show them all (and of course any formal charges that may be
present):
Formal charges:
O with the double bond has 6 electrons and it should have 6, so it is neutral.
O with the single bond has 7 electrons and it should have 6, so it is 1.
I has 7 electrons and it should have 7, so it is neutral.
Each resonance form has a net charge of 1 which matches the formula of iodate: IO3 .
f) All three are equally stable. (All 3 should be circled.)
7. The geometry is determined by the number of charge clouds of the central atom.
Iodine has 1 lone pair + 3 atoms attached = 4 charge clouds. The orbitals will be tetrahedral in shape
but we cannot “see” the lone pair as electrons is much smaller.
Molecular geometry: trigonal pyramidal
Bond angle: Any structure with 4 charge clouds would have bond angles of 109.5 .
8. (6 pts) a) Fill in the formal charges (if any) of the following structures and CIRCLE the most stable
structure:
+
2+
..
: N N O.. :
2..
: N O N.. :
b) These are isomers as the N and O are not in the same location.
9. (8 pts) Fill in the blanks for each of the following. You can use the bottom of the page as scratch
paper. That section of the page will not be graded.
B
B A.. B
..
..
B ..A B
B A B
molec. geometry:
bond angles
5
see saw
90 , 120 , 108
B
B
B
# of charge clouds on A:
B
..
B A
..
4
bent
109.5
4
trigonal pyramid
109.5
5
T-shape
10. The Octet Rule states that “all” atoms try to attain 8 electrons in their outershell by...
a) gaining electrons, to form anions,
b) losing electrons, to form cations,
or c) sharing electrons, to form covalent bonds.
11. An orbital is...a shaped space surrounding the nucleus of the atom where a particular electron is
found 95% of the time.
12. They are isoelectronic but K+ has less protons (a lower nuclear charge) than Ca2+ to pull the electron inwards.
Thus K+ is larger.
13. C is larger for the same reason as in question #12. It has less protons than N to pull the electrons inwards
14.
↑↓ ↑↑ ↑↓
3p
Two electrons within the same orbital cannot have the same spin. That is the Pauli Exclusion Principle.
15. Nonmetals have a higher ionization energy. You can look at it in two ways:
1) Nonmetals are on the right and higher than the metals in the periodic table. They are smaller than metals
and therefore the electrons are closer to the nuclear charge and held more tightly.
2) Nonmetals are closer to the octet configuration. It would be easier to attain the octet by adding more
electrons than to lose any.
16. The extra large increase in ionization energy will occur for the beryllium atom between the second and third
ionization energy. The first two electrons removed are from the 2s orbital. The third electron will be much
harder to remove as it would be coming from the 1s orbital, which is much closer to the nucleus.
17. Remember that the energy levels of H are not evenly spaced by become closer and closer together as we move
up in energy. Thus, levels n =4 and n = 3 are closer together than n =2 and n= 1. Thus the energy produced
when an electron drops from n=2 to n=1 is larger than an electron dropping from n=4 to n= 3.
n=4
n=3
n=2
n=4
n = 3 is a smaller
drop.
n=2
n = 1 is a larger
drop.
n=1
18. P, Sn, S, H: S is the most electronegative. (It is the smallest and therefore its nuclear charge is more exposed
and will attract electrons more.
19. The Heisenberg’s Uncertainty Principle tell us that we can not determine both the location of the electron and
the direction in which is traveling, accurately and at the same time.
20. The term “dual” in the dual nature of the electron refers to the electron behaving both as a wave as well as
matter with mass and occupying space.
21. Each line in the Emission Line Spectrum of hydrogen represents the energy released when an electron drops
from a higher energy level to a lower energy level. It does NOT represent an energy level. One cannot SEE an
energy level. We can SEE only when light is given off during a transition.
“Boiling” is a physical change as a substance goes from a liquid state to a gaseous state. The chemical structure
does not change. Therefore the gas formed is C8H18 (g).