Assignment: 
Questions on Topic 14.1: Further aspects of covalent bonding
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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. Explain how this can be used to explain why all the S–O bonds are the same length and strength.
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In each of the three resonance hybrid structures one of the O atoms is doubly bonded to the S atom (and contains two non-bonding pairs of electrons) and the other two O atoms are singly bonded to the S atom (and each contain three non-bonding pairs of electrons). 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).
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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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
In the preferred structure, O=C=O, with two double C=O bonds all the atoms within the molecule each have a charge of zero.
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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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 bond order of 1.5 which has a bond length between a C–O single bond and a C=O double bond.
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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Two of the p orbitals combine ‘sideways’ to form pi bonds whereas the other p orbital combines ‘head on’ forming a sigma bond.
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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For just one double bond this equates to 498 kJ (i.e. 4.98 x 10
5 J) 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= 6.63 x 10−34 (Js) x 3.00 x 108 (ms−1) / 8.27 x 10−19(J) = 2.41 x 10−7 m (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.
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.
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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.
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