1: Draw dot-cross diagrams for i) PH3, ii) CS2, iii) BF3, iv) SCl2
2: Draw a dot-cross diagram for the ion H3O+
3: Ozone and sulphur trioxide also have dative bonds. Draw a dot-cross diagram for
i) O3, ii) SO3. (Note that sulphur in SO3 has an expanded octet)
4: Predict the shapes and bond angles in:
i) sulphur dichloride (SCl2) non linear 104.5° ii) silane (SiH4) tetrahedral 109.5° iii) carbon disulphide (CS2) linear 180° iv) sulphur trioxide (SO3) trigonal planar 120° v) ozone (O3) non-linear 117.5°
5: Draw dot-cross diagrams for i) lithium oxide ii) aluminium fluoride
6: Arrange the following groups of substances in order, highest boiling point first
i) Ar, Ne, He because London forces increase with increasing number of electrons in the atoms
ii) fluoromethane, chloromethane, bromomethane because London forces increase with increasing number of electrons in the halogen atom, given that the rest of the molecule remains the same in each case
7: Consider the series of molecules CH4, CH3Cl, CH2Cl2, CHCl3 and CCl4
i) Which of these contains polar bonds ? Ans: all except CH4
ii) Which of these molecules is polar ? Ans: CH3Cl, CH2Cl2, and CHCl3
N.B. CCl4has polar bonds, but symmetrically arranged around the central carbon atom, so the dipoles cancel each other out leaving no overall dipole.
8: Explain why the ammonium ion contains polar bonds but has no overall dipole
N and H have different electronegativities so the bonds are polar. The tetrahedral shape of the ammonium ion means that the dipoles are balanced symmetrically around the ions, so there is no overall dipole.
9: Arrange the following groups of substances in order, highest boiling point first.
i) butane ii) methylpropane iii) 1-chlorobutane iv) 2-chloro methylpropane
Ans: iii) Permanent dipole-dipole, stronger London forces (linear shape)
iv) Permanent dipole-dipole, weaker London forces (less area of surface contact)
i) No permanent dipole-dipole, stronger London forces (linear shape)
ii) No permanent dipole-dipole, weaker London forces (less area of surface contact)
10: What is the main type of interaction between molecules in:
i) butanone Ans: permanent dipole-dipole (C=O is a polar bond)
ii) pentene Ans: London forces (no polar bonds)
11: Explain why pentane has a b.p. of 36°C while 2-methylbutane has a b.p. of 28°C
Ans: These are isoelectronic molecules. Pentane molecules are unbranched, less spherical – there is more area of surface contact and hence stronger London forces compared to branched 2-methylbutane.
12: Draw a diagram to illustrate the hydrogen bonding in ammonia
Your diagram should show at least TWO ammonia molecules with:
- δ+ on each H atom and δ- on each N atom
- a lone pair on each N atom
- a hydrogen bond (shaded lines) between the H of one molecule and the lone pair on the N atom of another molecule.
13: Place the following molecules into order of increasing boiling point:
Lowest: CH4 only London forces
CH2O London forces and permanent dipole-dipole
Highest: CH3OH London forces, permanent dipole-dipole and hydrogen bonding
14: Compare the intermolecular forces in ammonia, NH3, and hydrazine, N2H4, and suggest which will have the highest boiling point.
Ans: Hydrazine has a higher boiling point than ammonia. Both have hydrogen bonding (and permanent dipole-dipole, and London forces) but hydrazine can form more hydrogen bonds because it has two N atoms each with a lone pair available, while ammonia only has one.