Bonding and Intermolecular Forces – Answers to Practice Questions

Simple Molecules

1: Draw dot-cross diagrams for     i) PH3,   ii) CS2,   iii) BF3,   iv) SCl2

dx_q1

2: Draw a dot-cross diagram for the ion H3O+

h3oplusdx

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)

dx_q3

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°

Ionic Compounds 

5: Draw dot-cross diagrams for   i) lithium oxide     ii) aluminium fluoride

dx_q5

Intermolecular Forces

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 CHCl

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:

  1. δ+ on each H atom and δ- on each N atom
  2. a lone pair on each N atom
  3. 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.