Chemistry 11 (HL)
Unit 4 / IB Topics 4 and 14
Unit 4 – Chemical Bonding
Review Questions
ANSWERS – Please do not print.
MULTIPLE CHOICE QUESTIONS (from Past Papers)
1.
D
2.
B
3.
B
4.
B
5.
B
6.
D
7.
D
8.
D
9.
A
10. B
11.
A
12. B
13.
A
14. B
15.
B
16. B
17.
B
18. B
19.
D
20. B
21.
B
22. A
23.
C
24. B
25.
D
26. C
27.
C
28. B
29.
B
SHORT ANSWER QUESTIONS (from Past Papers) – SELECTED ANSWERS
30.
(i)
see your notes
(ii)
H 3O +
C 2H 4
1+
H
H
O
H
tetrahedral, 109.5º
H C
H
C H
H
trigonal planar (around each C); 120º
(iii) H3O+  polar
 the O-H bonds are polar and are arranged asymmetrically, so there IS a
net dipole moment
C2H4  nonpolar
 C-H bonds are slightly polar and are arranged symmetrically around the
carbon atoms  bond dipoles cancel out and there is no net dipole
moment
(iv) O-H is the most polar  ∆EN (1.4) is the greatest
∆EN for O-N is 0.5 and that of N-H is 0.9
Chemistry 11 (HL)
Unit 4 / IB Topics 4 and 14
31.
molecules = BF3, N2O, P4O6, CBr4
These substances are molecular covalent because they are made of non-metals only.
32.
malleable  atoms are able to slip past each other
electrical conductors  bonding electrons are delocalized and can flow within the
metal
33.
NH4+ > NH3 > NH2–
All particles have 4 negative charge centres around N.
NH4+  no lone electron pairs  bond angles of 109.5º
NH3  one lone electron pair which occupies more space and repels the other
electron pairs more  bond angles = 107º
NH2–  two lone electron pairs, two bonding pairs  2 lone pairs repel more, giving a
bond angle of 104.5º
34.
similarities  both have carbon atoms covalently bonded to other carbon atoms in a
giant network
differences
 all C atoms covalently bonded to 4 other C atoms in a 3d structure; C atoms
in graphite covalently bonded to 3 other atoms in a plane, with van der
Waals forces between the planes
 C in diamond is sp3 hybridized; C in graphite is sp2 hybridized
 only sigma bonds in diamond; sigma and pi bonds in graphite
35.
36.
Aluminum is made of fixed positive ions in sea of delocalized electrons.
It has a higher MP than sodium because aluminum forms +3 ions, so those ions will
have a stronger forces of attraction for the electrons, giving stronger metallic bonds.
Sodium ions only have a charge of +1.
F
F
Xe
F
F
6 NCCs
4 bonding e pairs,
2 lone e pairs
F
F
F
P
F
F
5 NCCs
5 bonding e pairs
0 lone e pairs
F
F
B
1-
F
F
4 NCCs
4 bonding e pairs
0 lone e pairs
square planar
trigonal bipyramidal
tetrahedral
90º
90º and 120º
109.5º
Chemistry 11 (HL)
37.
Unit 4 / IB Topics 4 and 14
For the following compounds:
Cl
P
Cl
Cl
trigonal pyramidal
107º
PCl3, PCl5, POCl3
Cl
Cl
Cl
P
O
Cl
Cl
trigonal bipyramidal
90º, 120º
polar molecule
P-Cl bonds are polar
nonpolar molecule
P-Cl bonds are polar
non-symmetrical
net dipole moment
symmetrical
no net dipole moment
Cl
P
Cl
Cl
tetrahedral
109.5º
polar molecule
P-Cl and P-O polar
bonds
nonsymmetrical
net dipole moment
38.
polar covalent bonds  ∆EN for the Si-Cl bond is 1.3
non-polar molecule  shape of molecule is symmetrical (tetrahedral), so all the bond
dipoles cancel out and there is no net dipole moment
39.
H2O has the highest BP because it is made of polar molecules able to make hydrogen
bonds, which are strong intermolecular forces.
The other three Group 6 hydrides have increasing BP as their molar mass increases.
H2S and H2Se are very weakly polar with very weak dipole-dipole forces, and H2Te is
non-polar. The strength of the van der Waals forces increases with increasing molar
mass (or total number of electrons). This is more significant than the very weak dipoledipole forces.
40.
41.
(a)
Hybridization is the blending of an s orbital with 1, 2 or 3 p orbitals to make new
orbitals that are equivalent to each other, and with characteristic shapes and
energy values that are a intermediate between the pure s and p orbitals.
(b)
sp hybridization
sigma bonds present in the C-N triple bond = 1
pi bonds present in the C-N triple bond = 2
(i)
σ bonding = end to end overlap of 2 orbitals, where the shared electron pair lies
on the axis joining the 2 nuclei of the bonded atoms
(ii)
π bonding = side by side overlap of 2 pure p orbitals, where the shared electron
pair lies above and below the axis joining the 2 nuclei of the bonded atoms
(iii)
double bonds = sharing of 2 electron pairs (or 4 electrons)
(iv)
triple bonds = sharing of 3 electron pairs (or 6 electrons)
Chemistry 11 (HL)
42.
(i)
Unit 4 / IB Topics 4 and 14
NO2– (18 e)
 3 NCCs, 1 LP  angular; 117º
ICl5 (42 e)
 6 NCCs, 1 LP  square planar; ~90º
SF4 (34 e)
 5 NCCs, 1 LP  seesaw; 90º, 120º(or 117º), (180º)
(ii)
The lone electron pair occupies more space (or repels the other electron pairs
more), so the other bond angles are slightly less than those when all 5 negative
charge centres are used for bonding.
(iii)
Each C atom has sp2 hybridization.
The sp2 hybrid orbitals form when 1 s orbital blends with 2 p orbitals to give three
equivalent sp2 hybrid orbitals. The hybrid orbitals have a trigonal planar
geometry and a bond angle of 120º.
One pure p orbital remains around the carbon atom.
(iv)
sp2 hybridization
Both nitrogen-oxygen bonds will have the same length.
(v)
O-H ∆EN = 1.2
O-N ∆EN = 0.4
N-H ∆EN = 0.8
The most polar bond is O-H because its atoms have the greatest
electronegativity difference.