(a) A hydrocarbon having the formula \( \mathrm{C}_{10}\mathrm{H}_{22} \) was cracked to produce \( \mathrm{C}_6\mathrm{H}_{14} \) and another hydrocarbon P...
Question 1 Report
(a) A hydrocarbon having the formula \( \mathrm{C}_{10}\mathrm{H}_{22} \) was cracked to produce \( \mathrm{C}_6\mathrm{H}_{14} \) and another hydrocarbon P.
(i) Give the molecular formula of P.
(ii) Draw the structures of two isomers of P.
(ii) Give a reason why P could be polymerized.
(b) State the guiding principles which are used to explain the way electrons of the atoms of the elements are arranged in atomic orbitals.
(c) Consider each of the following substances: NaH, \( \mathrm{H}_2 \), \( \mathrm{H}_2\mathrm{S} \), \( \mathrm{NH}_4\mathrm{Cl} \).
(i) Describe the nature of the intermolecular forces holding the units or molecules together in the condensed (liquid or solid) state.
(ii) Explain briefly what happens when a sample of each of the substances is added to water.
(iii) Write the chemical equations of any reactions occurring or of any equilibria established.
(d) Element J has the following electron configuration: \( 1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^6 3\mathrm{s}^2 \).
(i) How many unpaired electrons can be found in J?
(ii) State whether J would be a good oxidizing or reducing agent.
(iii) Give a reason for the answer in (d)(ii).
Answer Details
Cracking of a Hydrocarbon
A hydrocarbon with formula C10H22 is cracked to produce C6H14 and another hydrocarbon, P.
(a)(i) Molecular formula of P
C10H22 → C6H14 + P
P = C10H22 − C6H14 = C4H8
Therefore, P is an alkene with the molecular formula C4H8.
(a)(ii) Two isomers of P
The structures shown are but-1-ene, CH2=CHCH2CH3, and 2-methylpropene, CH2=C(CH3)2.
(a)(iii) Why P can be polymerised
P can be polymerised because it is an alkene containing a carbon–carbon double bond (C=C). The double bond can open and form bonds with other P molecules, producing a long-chain polymer.
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