(a) A hydrocarbon having the formula C(_{10})H(_{22}) was cracked to produce C(_6)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, H(_2), H(_2)S, NH(_4)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: Is(^2)2s(^2)2p(^6)3s(^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).
Explanation
(iii) because it is an alkene / has a double bond/unsaturated
(b) – electrons enter orbitals of lowest energies first (Aufbau Principle)
– (Pauli Exclusion Principle) – Two electrons in the same orbital will have different spins / no two electrons will have all four quantum numbers the same.
– (Hund’s Rule of maximum multiplicity) – electrons filling into degenerate orbitals / orbitals of same energy fill in singly before pairing up.
(c) (i) NaH: Ionic/electrostatic forces, charges (of opposite sign) holding Na(^2) and H(^-) ions together. Do not accept ionic/electrovalent bond.
H(_2): Van der Waal’s forces weak (instantaneous)dipole – dipole interactions holding molecules together
H(_2)S: Dipole – dipole interactions / H – bonding ( resulting from charge separation along S-H bond) because of higher electronegativity of sulphur)
NH(_4)Cl: Ionic/ electrostatic forces between NH(_4)(^+) and Cl(^-) holding the NH(_4)+ and Cl(^-) units
(ii) – Redox reaction occurs/NaH dissolves/gas is evolved/alkaline solution formed
– H(_2) to water no reaction occurs / H(_2) is slightly soluble
– H(_2)S to water, H(_3)O(^+) and S(_2^-) is formed/ H(_2)S dissolves / slightly acidic solution
– It dissolves/NH(_4)(^+) hydrolyses/slightly acidic solution
(iii) – NaH + H(_2)O (to) NaOH + H(_{2(g)}) – H(_2)S + 2H(_2)O (to) 2H(_3)O(^+) + S(_2^-) – NH(_4)Cl + H(_2)O (to) NH(_3) + H(_3)O(^+) + Cl(^-) OR
NH(_4) + H(_2)O (to) NH(_3) + H(_3)O(^+)
(d) (i) There are no unpaired electrons/Zero
(ii) It will be a good reducing agent
(iii) because it loses the two electrons in its outermost shell / It readily gives out electron/electron donor/electropositive