Question 53: [Inorganic > Period 3]
A mixture of the oxides of two elements of the third period is dissolved in water. The solution is approximately neutral.
What could be the constituents of the mixture?
A Al2O3 and MgO
B Na2O and MgO
C Na2O and P4O10
D SO3 and P4O10
Reference: Past Exam Paper -June 2003 Paper 1 Q13
Solution 53:
Answer: C.
A cannot be answer because MgO when dissolved in water will dissolve slightly and form an alkaline solution of approximately 9 pH while on the other hand Al2O3 is insoluble therefore won’t dissolve and won’t have any effect in the mixture therefore solution left is slightly alkaline and not neutral
B cannot be answer because when Na2O dissolve in water it will form a strong alkaline solution of pH 14 and the other hand MgO will dissolve slightly forming a solution of pH 9 the solution left is strongly alkaline and nowhere near neutral.
C is the answer because when Na2O dissolve in water it will form a strong alkaline solution of pH 14 because it is a strong base whereas in the other hand P4O10 is a strong acid and when dissolved in water it forms a pH of 1 so the final solution is neutral because the strong base Na2O will neutralize the acid resulting in a neutral solution at the end.
D cannot be answer because both SO3 and P4O10 are strong acids and will each form a pH of 1 resulting in a very acidic solution at the end.
Question 14:
[Inorganic > Period 3]:
Which diagram represents the change
in ionic radius of the elements across the third period (Na to Cl)?
Reference: Past Exam Paper – November 2002 Paper 1 Q14
Solution 14:
Answer: C.
Going across the third period the first 3 elements are metals. They will
lose electrons from their valence shell to become ions. In this process they
lose a shell and thus significantly decrease the size of their radius. The rest
are non-metals and gain electrons to form ions.
When they do this the valence
shell becomes full. This is best depicted by the sudden rise in option C’s
graph; the first line representing metal cations and the second line
representing non-metal anions.
Between the metal cations the nucleus of the
ions increase in proton number, thus the effect of nuclear charge on the outer
shell is greater, ultimately causing the radius to decrease. This is reason is
the same for the non-metal anions.
The decrease in radius between cations and
anions is best shown by the negative gradient of the 2 lines in option C’s
graph.
Question 13:
[Inorganic > Period 3]:
The chloride of element Q is hydrolysed by water to form an
acidic solution and its oxide reacts with acid to form a salt.
What could be the element Q?
A magnesium
B aluminium
C silicon
D phosphorus
Reference: Past Exam Paper – November 2002 Paper 1 Q13
Solution
13:
Answer: B.
The best way forward is to focus on the second part of the question. If
the oxide reacts with an acid it must be either basic or amphoteric. This
eliminates option C because silicon(IV) oxide is a giant molecular structure
that does not react. It also eliminates option D since phosphorous can only
make acidic oxides since it is a non-metal.
Magnesium chloride is hydrolysed by
water to form an almost neutral solution, due to slight disassociation of H2O
into ions.
However Aluminium chloride disassociates in water to form
hydrochloric acid, hence option B is the “more” correct answer.
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