H Chemistry: Periodic Trends & Intermolecular Forces
NAME ___________________
Multiple Choice
1. Arrange the following in increasing order for atomic radius: Mo, Ru, Rb, I, Sr, Xe, Sn
a. Mo, Ru, Rb, Sr, Xe, Sn
c. Rb, Sr, Mo, Ru, Sn, I, Xe
e. Rb, Sr, Mo, Ru, Sn, I
b. I, Sn, Ru, Mo, Sr, Rb
d. Xe, I, Sn, Ru, Mo, Sr, Rb
Option d
2. Arrange the following in increasing order for electron affinity: Mo, Ru, Rb, I, Sr, Xe, Sn
a. Mo, Ru, Rb, Sr, Xe, Sn
c. Rb, Sr, Mo, Ru, Sn, I, Xe
e. Rb, Sr, Mo, Ru, Sn, I
b. I, Sn, Ru, Mo, Sr, Rb
d. Xe, I, Sn, Ru, Mo, Sr, Rb
Option e
3. Arrange the following in increasing order for atomic radius: Pu, W, Hf, Ra, O, Cu, Si, P
a. Ra, Pu, Hf, W, Cu, Si, P, O
c. Ra, Hf, W, Pu, Cu, Si, P, O
e. Ra, Hf, W, Cu, Si, P, O
b. O, P, Si, Cu, Pu, W, Hf, Ra
d. O, P, Si, Cu, W, Hf, Pu, Ra
Option d
4. Arrange the following in increasing order for ionization energy: Pu, W, Hf, Ra, O, Cu, Si, P
a. Ra, Pu, Hf, W, Cu, Si, P, O
c. Ra, Hf, W, Pu, Cu, Si, P, O
e. Ra, Hf, W, Cu, Si, P, O
b. O, P, Si, Cu, Pu, W, Hf, Ra
d. O, P, Si, Cu, W, Hf, Pu, Ra
Option a
5. Although they are not polar like water, some large, non-polar molecules have high viscosities
and high boiling points because they experience large:
a. ion-dipole forces
c. London-dispersion forces
e. dipole-dipole forces
Option c
b. ion-ion forces
d. hydrogen bonding
6. The bond between hydrogen and oxygen in a water molecule is classified as
a. ionic and nonpolar
c. covalent and nonpolar
e. ionic only
Option d
b. ionic and polar
d. covalent and polar
f. covalent only
7. Arrange the following in increasing order of atomic radius: K+1, Cl-1, Ar, Ca+2, S-2
a. K+1, Cl-1, Ar, Ca+2, S-2
c. Ca+2, K+1, Ar, Cl-1, S-2
e. S-2, Cl-1, Ar, K+1, Ca+2
b. Ar, K+1, Cl-1, Ca+2, S-2
d. K+1, Cl-1, Ca+2, S-2, Ar
Option c
8. Arrange the following in increasing order of ionization energy: X+1, X–1, X, X+2, X–2
a. X+1, X-1, X, X+2, X-2
c. X+2, X+1, X, X-1, X-2
e. X-2, X-1, X, X+1, X+2
b. X, X+1, X-1, X+2, X-2
d. X+1, X-1, X+2, X-2, X
Option e
9. If for a 3rd energy level element, IE1 = 150, IE2 = 450, IE3 = 900, and IE4 =7000, the element is:
a. Na
c. Al
e. None of the above
b. Mg
d. Si
Option a
10. On the Ionization Energy vs Atomic Number graph, there are minor drops in the graphs
between Column 2 and Column 13 and between Column 15 and Column 16 because:
a.
b.
c.
d.
e.
the effective nuclear charge decreases between these columns
losing an e– to obtain a filled or half-filled orbital state is easier than removing an e– from it
there are extra energy levels moving the valence e– further away from the nucleus
the number of core e– ‘s increases between these columns
there are no minor drops in the Ionization Energy vs Atomic Number graph
Option b
11. Compared to the boiling point of H2S, the boiling point of H2O is relatively high. Which type of
force/bond causes this difference?
a. Covalent Bonding
c. Dipole-Dipole Force
e. London Dispersion Force
Option d
b. Ion-Dipole Force
d. Hydrogen Bonding
f. Ionic Bonding
12. When you wash an oily pan with just hot water, the pan stays oily. By using some soap, you
are then able to clean the pan completely. Which statement below best explains why this
happens?
a.
b.
c.
d.
e.
Soap is more polar than water so that it has a stronger effect on oil
Soap has a polar end and a non-polar end, so it can interact with both water and oil.
Oil is strongly polar so it cannot interact with non-polar molecules, such as water.
Water has low surface tension so it cannot effectively move oil molecules.
Soap provides no help in cleaning oily pans.
Option b
SHORT ANSWER
13. Fill in the table below and then place the elements in their proper relative position on the
periodic table (the four-square box.)
Element
A
IE1
5.1391
IE2
47.286
IE3
71.620
IE4
98.91
# of Valence e—
1
B
6.1132
11.87
50.91
67.27
2
C
4.3407
31.63
45.81
60.91
1
D
7.6462
15.035
80.144
109.27
2
A
B
C
D
14. Explain why a paperclip can float on water but not float on CH3CH2CH2CH2CH2CH3 (C6H14)?
Density of C6H14 = 0.7785 g/ml
While water’s density = 1.0 g/ml
So, C6H14 is less denser than water and hence wont be able to bear the paper clip,. So water which
is enough dense to bear the weight of paper clip would be able to make it float.
15. For each of the following groups of substances, circle the appropriate substance. Justify your
answer using principles of intermolecular forces. You must discuss each chemical for full
credit.
a. Highest boiling point: HCl, Ar, F2
HCl since it exhibits hydrogen-bonding the strongest force of attraction
b. Highest freezing point: H2O, LiF, HF
H2O since it exhibits hydrogen-bonding the strongest force of attraction
c. Lowest boiling point: Cl2, Br2, I2
Cl2 being most covalent
Consider the picture below for questions #16 - #20:
16. What is the primary type of bonds/forces that exists
H2O
H2O
H2O
between the water particles? Hydrogen bonding
H2O
C6H14
Na+
H2O
H2O
H2O
H2O
C6H14
H2O
H2O
H2O
H2O
H2O
Na+
H2O
Cl–
Cl–
18. What is the primary type of bonds/forces that exists
between the water and CH3CH2CH2CH2CH2CH3
particles? Dipole- induced dipole interactions
H2O
H2O
H2O
17. What is the primary type of bonds/forces that exists
between the water and Na+ or Cl– particles?
Ionic bond
C6H14
19. What is the primary type of bonds/forces that exists
within the CH3CH2CH2CH2CH2CH3 particle?
London-dispersion forces
20. What will eventually happen to the chemicals in this
container?
NaCl dissolves in water and forms NaCl solution,
while C6H14 isa immiscible in water and form an
oily layer. So we have an Aqueous layer and an
Organic layer.
21. For each of the following:
i.
ii.
iii.
iv.
Draw the best Lewis structure.
Determine the molecular shape and the bond angles.
Determine the overall polarity (don’t do this for polyatomic ions.)
What intermolecular force would be most dominant between itself and water?
Lewis structure of TeF4
F
F
F
Te
F
Lewis structure of SF4
F
F
S
F
F
Molecular geometry:
Tellurium / Sulfur has a total of six valence electrons, out of which two form a
"lone pair."Remaining 4 electrons combine with 4 fluorine atoms which has one
valance electron of each to form 4 shared pairs. One of the three equatorial
positions is occupied by a nonbonding lone pair of electrons. Ignoring this lone
pair, SF4 has a see-saw molecular geometry with sulphur at the centre.
Bond angles:
The bond angles will be less than 120° in the equatorial plane and less than 90°
between the axial and equatorial positions (greater repulsions between bonding
and lone pair electrons than between bonding electron pairs).
Both TeF4 and SF4 have distroted trigonal bipyramidal geometry (known as seesaw – molecular geometry ), so it is polar.
Intermolecular force would be most dominant between itself and water:
Hydrogen bonding between Hydrogen of water and fluorine atoms of SF4 or TeF4
a. TeF4
b. SF4