Unit 4 worksheets Further aspects of covalent bonding
1. The bonding in sulfur trioxide is sometimes explained by sulfur expanding its octet to six pairs of
outer electrons (by using available d orbitals) to give three S=O double bonds.
It is also sometimes explained by keeping to the ‘octet rule’ where only eight outer electrons around the
central sulfur atom are involved. Draw the resonance hybrids to illustrate this model of bonding and
explain why all the S–O bonds are the same length and strength.
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2. The following is a Lewis structure for carbon dioxide:
Show that it has a formal charge of zero and explain why it is not the preferred structure.
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3. Explain why the two C–O bond lengths in propanoic acid, C2H5COOH, are different and yet the two
C–O bond lengths in the propanoate anion, C2H5COO–, are the same length.
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4. Explain why p orbitals on one atom can form two pi bonds and one sigma bond when they combine
with the p orbitals on another atom.
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5. Ultraviolet light can break the oxygen to oxygen bond in both oxygen, O2, and ozone, O3, molecules.
The O=O bond energy is 498 kJ mol-1. Calculate the wavelength of light required to break this bond.
Suggest why the light required to break the O–O bond in oxygen is of a higher frequency than the light
required to break the O–O bond in ozone.
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6. When there is a large amount of delocalised electrons in a molecule the molecule tends to be
coloured. The two structures below are for the indicator phenolphthalein in (a) acid solution and
(b) in alkali solution. Explain why phenolphthalein becomes coloured in an alkaline solution
(a) in acid solution
(b) in alkali solution
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Answers
1.
The S–O bonds are the same length and strength as the real structure lies between these three
extreme resonance structures with identical S–O bonds with an average of one and a third bonds
each (a bond order of 4/3).
2.
C = 4 – 0 – (½ x 8) = 0,
O = 6 – 2 – (½ x 6) = +1 (O with triple bond)
O = 6 – 6 – (½ x 2) = – 1(O with single bond) so overall charge = 0
However in the preferred structure with two double bonds all the atoms within the molecule each have
a charge of zero.
3. In propanoic acid, C2H5COOH, there is a C–O single bond and a C=O double bond. The C–O single
bond is longer than the C=O double bond. In the propanoate ion delocalisation occurs so that the
extra electron is shared between the two C–O bonds giving two identical bonds with an
average 1.5 bond which has a bond length between a C–O single bond and a C=O double bond.
4. Two of the p orbitals combine ‘sideways’ to form pi bonds whereas the other p orbital combines
‘head on’ forming a sigma bond.
5. For just one double bond this equates to 498 divided by Avogadro’s constant = 8.27 x 10-19 J. The
wavelength of light that corresponds to this enthalpy value (E) is calculated by combining
the expressions E = hν and c = λν to give λ =
λ= 6.63 x 10-34 (Js) x 3.00 x 108 (ms-1) / 8.27 x 10-19 (J) = 241 nm
The double O=O bond in oxygen is stronger than the 1.5 O-O bond in ozone (the average of the two
resonance hybrid structures) so more energy (higher frequency) is required to break the O-O bond in
oxygen.
6. The hybridization of the central carbon atom changes from sp3 to sp2. This provides a single electron
in a p orbital so that the electrons can delocalise throughout the whole anion via conjugated π bonds
whereas in the undissociated molecule the delocalisation is limited mainly just to the separate aromatic
rings.