Australian School of Abu DHabi
Unit 3 Periodicity
Worksheet 2 Periodic trends
1. (a) Explain why the atomic radius of sodium (Z = 11) is bigger than the atomic radius of chlorine (Z =
17).
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(b) Explain why the atomic radius of the elements increases upon descending group 17 (F → I).
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2. Suggest how the values for the atomic radii of the noble gases given in Section 9 in the IB Chemistry
data booklet have been obtained.
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3. Suggest two reasons why the ionic radius of a sodium ion is much smaller than the atomic radius of a
sodium atom.
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4. Three negative ions are the phosphide ion, P3–, the sulfide ion, S2– and the chloride ion, Cl–.
Place these three ions in order of increasing size (smallest first) and explain your logic.
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5.Give the equations for the reaction of water with:
(a) sodium oxide, Na2O.
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(b) magnesium oxide, MgO
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(c) phosphorus(V) oxide, P4O10
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(d) sulfur(VI) oxide, SO3.
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6.Describe and explain (with a relevant equation) what will be observed when chlorine water is added
to:
(a) a solution of chloride ions.
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(b) a solution of bromide ions.
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(c) a solution of iodide ions.
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7. State what will be observed when sodium metal is placed in water and give the equation for the
reaction.
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8. Suggest one reason why caesium is more reactive than lithium when it is placed in water.
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9. State what will be observed when a piece of warm sodium metal is lowered into in a gas jar
containing chlorine gas and give the equation for the reaction.
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10. Aluminium oxide, Al2O3, has a high melting point and it reacts with both hydrochloric acid and
sodium hydroxide. What information can be deduced about the chemical nature of aluminium oxide
from this information?
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Answers
1. (a) Chlorine has 7 electrons in the third energy level and sodium only has one. The seventeen protons
in the nucleus of the chlorine atom attract the energy level more strongly than the eleven protons in the
sodium atom thus making the chlorine atom smaller.
(b) The outer electrons are further away from the nucleus as they are in successively higher energy
levels.
2. Atomic radii are normally obtained by measuring half the distance between two bonded atoms. Noble
gases (essentially) do not bond with other atoms so the radius is obtained by measuring the distance
between the atoms when the noble gas is in the frozen state.(It is known as the non-bonding atomic
radius or van der Waals' radius).
3. (i) The outer electron in the sodium atom is in the third energy level, once this is removed the outer
level of the electrons in the sodium ion is the second energy level.
(ii) There are more protons than electrons in the sodium ion which attracts the outer energy level closer
to the nucleus.
4. Cl– < S2– < P3–. They all have the same electron configuration (1s22s22p63s23p6), but the 17 protons in
the Cl– ion attract the outer electrons more closely to the nucleus than the 15 protons in the P3– ion.
5. (a) Na2O(s) + H2O(l) → 2NaOH(aq)
(b) MgO(s) + H2O(l) → Mg(OH)2(s) (magnesium hydroxide is only sparingly soluble in water)
(c) P4O10(s) + 6H2O(l) → 4H3PO4(aq)
(d) SO3(g) + H2O(l) → H2SO4(aq)
6. (a) There is no change as there is no reaction between chlorine and a solution of chloride ions.
(b) A yellow/brown solution will be formed. This is due to the formation of bromine.
Cl2(aq) + 2Br–(aq) → 2Cl–(aq) + Br2(aq)
(c) A dark brown solution will be formed. This is due to the formation of iodine. (The presence of iodine
can be confirmed by shaking with a little non-polar organic solvent such as heptane.
When the two layers are allowed to separate the non-polar organic layer will be purple).
Cl2(aq) + 2I–(aq) → 2Cl–(aq) + I2(aq)
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7. The sodium metal melts into a ball and floats on the surface. It reacts vigorously giving off heat and a
gas until eventually it all disappears. The solution remains colourless. During the reaction small yellow
sparks may be observed.
Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
8. The simplistic (Standard Level) answer is that the outer electron in caesium is further from the
nucleus than the outer electron in lithium so requires less energy to remove.
Some critical thinking will show that it is not as simple as this.
The general reaction is M(s) + 2H2O(l) → M+(aq) + 2OH– (aq) + H2(g)
The first ionization energy of caesium is lower: M(g) → M+(g) + e– but first the metal must be turned into
a gas.
Again less energy is required for caesium for the process M(s) → M(g).
The final change is the formation of the aqueous ion M+(g) → M+(aq). This is an exothermic process.
Because of the small size of the lithium ion, lithium gives out more heat than caesium in this process as
it attracts water molecules more as the density of charge is greater. Overall the total energy changes for
the two metals are very similar. It would appear that the heat change in the two reactions is similar but
caesium gives heat out much quicker. This may be because it has a lower melting point and so provides
a greater surface area for the reaction to proceed faster.
9. The sodium metal burns in the green gas producing a bright light and clouds of white smoke.
The green colour disappears (unless the chlorine is in large excess) and a white solid remains.
2Na(s) + Cl2(g) → 2NaCl(s)
10. (i) As the melting point is high the bonding is ionic.
(ii) As it reacts with both acids and bases it is an amphoteric oxide.
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