IB Past Test Questions Key Bonding
1. D
2. C
3. boiling points increase going down the group (from H2S to H2Se to H2Te);
Mr/number of electrons/molecular size increases down the group;
Accept electron cloud increases down the group for the second marking point.
greater dispersion/London/van der Waals’ forces;
H2O/water has a higher boiling point than expected due to the hydrogen
bonding between the molecules;
Do not accept hydrogen bonding alone.
4. Tetrahedral shape with electron lone pairs repelling the two hydrogens into a molecular bent
shape. The repulsion of the lone pairs lends a 105 bond between the H-O-H bond angle.
5. Permanent dipole exists between each Si-Cl bond, making them polar. But the symmetrical
distribution of charge gives no net dipole for the molecule.
6. A= NaI B= Na C= I2
a. A= Ionic B= Metallic C= London dispersion forces (non-polar covalent)
b.
i. Na has delocalized electrons so the availability of free electrons enables
conductivity. NaI does not have delocalized electrons but the ions are mobile in
the molten state which will enable conductivity.
ii. Weaker forces (London- covalent molecular non-polar) versus the stronger ionic
and metallic bonds allow a lower boiling point for I2.
7. C
8. A
9. B
10. H-C-O-H
a. H-C-O bond angle is 120
b. C-O-H bond angle is 105
c. Length of σ bond is shorter and weaker alone, π with σ bonds result in shorter, stronger
bonds than with σ alone.
11. B
12. C
13. C
14. Br= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
Fe3+= 1s2 2s2 2p6 3s2 3p6 3d5
15. Increase in molar mass= increase in electrons= increase in London forces= increase in boiling
point. H-bonding in H2O increases strength of bonds= highest boiling point for the hydride
group.
16. Butane (non-polar) insoluble as London dispersion forces alone and no permanent dipoles will
not bond with water. Chloroethane, propanone, and propan-1-ol are soluble in water (polar) as
the dipole-dipole interactions promote bonding with water (but propan-1-ol can also H-bond).
17. Carbonate ion has longest bond σ bond only between single bond of C-O which is longest and
weakest. Less attraction of electrons (π bonds) in multiple bonds to the protons in nucleus of
atoms.
18. :N=N=N-H and :N≡N- N-H
a. N-N-N bonds are linear in both cases with only 2 negative charge centers for the central
N and no lone pair electrons. N-N-H are both bent but the first case has only 1 lone pair
so the angle is 109 and there are two lone pairs in the second structure so the bond
angle is 105
b. N=N-H is sp2 hybridized and N-N-H is sp3 hybridized
19. B
20. D
21. C
22. O=C=O and H-S-H
a. CO2 = linear as there are 2 negative charge centers and no lone pairs. H2S= bent as has
lone pairs and repulsion of the lone pairs push the Hydrogens farther away
b. CO2 is non-polar (no net dipole) and has symmetrical shape. H2S is polar as is
asymmetrical with the lone pairs being present (show net dipole/partial charges)
23. CH3Cl = permanent dipole-dipole interactions. CH4 = London dispersion forces. CH3OH = Hbonding (extreme dipole-dipole)
24. C
25. A
26. C
27. NO2- bent molecular, trigonal planar orbital (has a lone pair); NO2+ linear with no lone pairs
28. C2H6<CH3CHO<CH3CH2OH<CH3COOH
a. Only non-polar is C2H6. H-bond and dipole-dipole doe CH3COOH; H-bonding with
CH3CH2OH, net dipole-dipole with CH3CHO
29. A
30. B
31. B
32. B
33. Electronegativity: (relative) measure of an atoms attraction for electrons;
in a covalent bond / shared pair;
a. H-N-H in NH3 is 107. Polar due to net dipole with lone pair of electrons (show net
dipole)
b. H-N-H in ammonium ion is 109.5
c. Ammonia bond different from ammonium due to the repulsion of lone pair of electrons
which will repel the Hydrogens to a smaller bond angle
34. A
35. B
36. A σ bond is formed through the axial overlap of orbitals. A π bond is formed through the
sideways overlap of parallel p orbitals. Double bonds have one of each. Triple bonds have 1 σ
and 2 π bonds.
37. OF2 is sp3 hybridized with orbital shape= tetrahedral; molecular shape= bent
a. H2CO is sp2 hybridized and is trigonal planar
b. C2H2 is sp hybridized and is linear
38. Diamond vs graphite
a. Diamond: 3D array of tetrahedral Carbons (sp3 hybridized) bonded to 4 other carbons.
Graphite is made of layer sheets of trigonal planar carbons (sp2 hybrid) in hexagonal
rings. Sheets are held together by London disp. forces
b. Diamonds are harder due to all carbons covalently bonded to only 4 other carbons and
does not conduct because of the lack of delocalized electrons. Graphite is not as hard as
diamond due to the sp2 carbons and having London disp forces between layers.
Graphite contains delocalized electrons so can conduct electricity (delocalized from the
sp2 hybrid carbons in the hexagonal rings.
c. Fullerenes are harder than graphite due to the lack of London forces and only covalent
bonding; but can also conduct unlike the diamonds due to the sp2 hybrid
carbons/delocalized electrons.
39. B
40. B
41. D
42. C
43. W3+ Y3a. XZ2
44. Covalent bonding with dipole-dipole interactions in a non-polar molecule with polar bonds
45. Find number of electron pairs/charge centres in (valence shell of)
central atom;
electron pairs/charge centres (in valence shell) of central atom repel
each other;
Any one of the following:
to positions of minimum energy/repulsion / maximum stability;
pairs forming a double or triple bond act as a single bond;
non-bonding pairs repel more than bonding pairs / OWTTE;
46. SCl2 orbital geom= tetrahedral, molecular= bent with bond angles of 105 due to two lone pairs.
C2Cl2 is linear with bond angle of 180 due to no lone pairs
a. SCl2 is polar and has 2 lone pairs/asymmetrical and therefore a net dipole (show net
dipole/partial charges). C2Cl2 is non-polar/symmetrical with no net dipole
47. A
48. D
49. A
50. PCl3 = trigonal pyramidal 107 bond angle; PCl5 = trigonal bipyramidal 120 and 90 bond angles;
POCl3 = tetrahedral 109.5 bond angle
a. PCl3 and POCl3 polar- asymmetrical with net dipole; PCl5 non-polar symmetrical with no
net dipole. Show net dipoles/partial charges
51. mixing/combining/merging of (atomic) orbitals to form new orbitals (for bonding);
Allow molecular or hybrid instead of new.
Do not allow answers such as changing shape/symmetries of atomic orbitals.
a. CH3HC=CH2
b. The σ bonds are formed through axial overlap of orbitals and π bonds are formed
through the sideways overlap of parallel p prbitals
c. C-C bonds are longer and weaker than C=C bonds
d. C1 has sp3 hybridization and C2 and C3 have sp2 hybridization
52. C
53. A
54. D
55. D
56. A
57. B