Exam-style Questions: Group II and Group IV

1. Barium is a member of Group II of the periodic table.

a) Write an equation to show the reaction of barium metal with water.

(1 mark)

b) Write an equation to show the thermal decomposition
of barium nitrate

(1 mark)

c) Explain why magnesium sulphate is soluble in water but barium sulphate
is not.

(2 marks)

(Marks available: 4)

Answer

Answer outline and marking scheme for question: 1

a) Ba(s) + 2H2O(aq) = Ba(OH)2(aq) +
H2(g)

(1 mark)

b) 2Ba(NO3)2(s) = 2BaO(s) + 4NO2(g)
+ O2(g)

(1 mark)

c) There are two factors that affect the solubility in water; lattice
energy and enthalpy of hydration.

As both salts are sulphates, there is little difference in lattice enthalpy
(1)

The enthalpy of hydration is much higher for barium as it has a higher ionic
radius (lower charge density) which makes it difficult for water to hydrate.
(1) (less of an attraction between cation and polar water molecules)

(2 marks)

(Marks available: 4)

2.The following questions are about the members of Group IV of the periodic table.


a) What type of bonding is seen in silicon tetrachloride, SiCl4,

(1 mark)

b) Write an equation to show the reaction of silicon tetrachloride,
SiCl4, with water.

(1 mark)

c) Explain why silicon tetrachloride can react with water but carbon
tetrachloride does not.

(2 marks)

b) Explain why lead can form lead (II) chloride but silicon cannot form
silicon (II) chloride.

(2 marks)

(Marks available: 6)

Answer

Answer outline and marking scheme for question: 2

a) Covalent

(1 mark)

b) SiCl4(l) + 2H2O(l) = SiO2(aq)
+ 4HCl(g)

(1 mark)

c) The silicon has 3d electron orbitals available (1)

Which can accept a lone pair of electrons from the oxygen atom of water. (1)

Carbon does not have any d electron orbitals available.

(2 marks)

d) The 6s electrons in lead are drawn in to the nucleus of the atom
due to poor shielding by d and f orbital electrons. This means that they behave
like an 'inert pair' of inner electrons rather than an 'active pair' of outer
electrons. (1)

Silicon does not have any d or f electron orbitals filled therefore the 3s electron
orbitals become hybridised with the 3p electrons giving silicon a valency of
4. (1)

(2 marks)

(Marks available: 6)