Periodicity Assignment
1. State the name of an element: (4)
a. In the same group as neon.
Helium, argon, krypton, xenon or radon
b. In the same period as sodium.
Magnesium, aluminum, silicon, phosphorus, sulfur, chlorine or
argon
c. With similar properties to nitrogen.
Phosphorus
d. That is a transition metal.
Copper, nickel, zinc, cobalt, etc
2. Give the name, symbol and ground state electron configuration of the
element: (4)
a. In group 2, period 3.
Mg 1s22s22p63s2
b. With an atomic number 1 greater than that of oxygen.
F
1s22s22p5
c. That is the noble gas in period 2.
Ne 1s22s22p6
d. That is the smallest metal in group 3.
Al
1s22s22p63s23p1
3. The electron configuration of a chemical species is 2, 8, 8.
a. State the element with the configuration. (5)
Argon
b. State an element with similar properties to the element in part a.
Neon or krypton
c. Under what circumstances could this electron configuration
belong to an element in group 6?
If it has a charge of -2
d. Under what circumstances could this electron configuration
belong to an element in group 2?
If it has a charge of +2
e. Is it possible for this electron configuration to belong to an
element in period 2?
No. Period 2 elements have only two shells of electrons.
4. Define the term electronegativity. (1)
Electronegativity is a measure of the attraction that an atom has for a
shared pair of electrons when it is covalently bonded to another atom.
5. a. State and explain the trend in electronegativity across a period.
Electronegativity increases across a period, because the nuclear
charge increases and the size of the atom decreases; thus there is
a greater attraction between electrons and the nucleus.
b. State and explain the trend in electronegativity down a group.
Electronegativity decreases down a group. As the valence shell is
further from the nucleus, there is less attraction between the
nucleus and the valence electrons.
(4)
6. Consider these elements of period 3:
Al
Cl
Mg Na P
S
Si
a. Identify each of these elements as a metal or a non-metal.
Na, Mg, Al: metals; Si: metalloid; others: non-metals
b. List these elements in order of increasing electronegativity.
Na, Mg, Al, Si, P, S, Cl
c. List these elements in order of increasing atomic radius.
Cl, S, P, Si, Al, Mg, Na
(3)
7. Compare the radii of each of the following pairs of particles and
explain the differences between: (1)
a. A sodium ion and a sodium atom.
The radius of a sodium ion, Na+, is smaller than that of a sodium
atom because a sodium ion has lost its outer-shell of electrons.
b. A chloride ion and a chloride atom.
The radius of a chlorine ion, Cl–, is greater than that of a chlorine
atom because it has gained an extra electron while retaining the same
core charge. The increased electron repulsions make the radius of the
Cl– ion greater than that of Cl.
c. A chloride ion and a sodium atom.
A chloride ion and a sodium atom have similar radii. Cl– is slightly
larger than Na. The similarity is due to Cl– being larger than Cl,
which is smaller than Na (increased core charge draws in same shell
of electrons).
8. a. List the following in order of increasing size (smallest to largest):
Cl-, Cl+, Cl
Cl+, Cl, Cl–
b. Explain why you have chosen this order.
All of these have 17 protons, but the number of electrons is 16, 17
and 18 respectively. The more electrons there are, the greater the
electrostatic repulsion between the electrons, and the larger the
size.
(2)
9. Elements exhibit a wide range of chemical and physical properties.
Many of these properties can be predicted from the position of the
element in the periodic table. Four elements are given as examples in
the table below. Copy these table and complete it by adding your
answer to parts a and b.
Electron
Electronegativity
Configuration
Sodium
1s22s22p63s1
4
2
2
6
2
Aluminum
1s 2s 2p 3s 3p1 3
Phosphorous 1s22s22p63s23p3 2
Chlorine
1s22s22p63s23p5 1
a. State the electron configuration of each element. (2)
b. Rank the elements in order of electronegativity (1 = highest). (1)
(3)
10. a. Explain why the chemical properties of barium and magnesium
are similar.
Both are group 2 metals with two valence electrons. Electron
configuration governs chemical properties.
b. Compare the physical properties of barium and magnesium and
explain the differences.
Barium is larger, hence its valence electrons are more weakly
held. This means that the melting point of barium is lower (same
charge on cations, larger size of ion) than that of magnesium, and
barium conducts electricity less easily than does magnesium.
(2)
11. For each of the following pairs of elements, compare the
electronegativities and state which element would be expected to have
the highest electronegativity: (2)
a. O or S
b. P or Al
c. B or Ne
d. N or Si
12. Explain each of the following statements concisely: (2)
a. There are eight elements in period 2. (Do not mark this question)
An element belongs to the period that matches its outer shell. In
period 2 there can only be 8 electrons in the second shell, and so there
can only be 8 elements in period 2.
b. Iodine has a lower electronegativity than fluorine.
Iodine is a larger element, hence its valence electrons are less
strongly attracted.
c. Although sulfur has a greater atomic mass than sodium, its atomic
radius is smaller than that of sodium.
Sodium and sulfur are in the same period, so the valence electrons are
in the same shell. However, sulfur has a greater nuclear charge than
sodium and so there is a greater attraction of the electrons to the
nucleus. The electrons of sulfur are pulled closer and so its radius is
smaller than that of sodium.
13. Element X is more electronegative and smaller than nitrogen. X has
valence electrons in the second shell and forms ions with a charge of 1.
Deduce the identity of X and state its name. (1)
Element X is fluorine
14. a. Define the term first ionization energy.
First ionization energy is the amount of energy required to remove
one mole of electrons from one mole of atoms in the gaseous state.
b. Using potassium as an example, state the equation for the
reaction that occurs during this ionization process.
K(g)
K+(g) + e(2)
15. a. State and explain the trend in first ionization energies across a
period.
First ionization energies increase across a period.
Electrons are being added to the same outer shell, but the nuclear
charge is increasing from left to right across the period and it is more
difficult to remove a valence electron. Also, there is a greater
attraction between the valence electrons and the nucleus.
b. State and explain the trend in first ionization energies down a
group.
First ionization energies decrease down a group.
Electrons are being added to outer shells further from the nucleus
and so the attraction between valence electrons and the nucleus is
less and it is easier to remove electrons from the outer shell the
further down the group.
(4)
16. Below is an outline of the periodic table with a number of elements
identified by letters that are not their normal symbols. Use these letters
when answering the following questions. (3)
a. Select a halogen and an alkali metal in the same period.
F(halogen) and D(alkali) are both in period 4
b. Which element has the highest electronegativity?
B
c. Which element has the lowest first ionization energy?
I
d. Which element has an electron configuration ending in 4?
C
e. Which element does not readily form bonds?
H
f. Select a member of the transition metal series.
G
17. Identify each of the following elements: (3)
a. The element with one more proton than titanium.
Vanadium
b. The noble gas that is lighter than neon.
Helium
c. The element closest in mass to iodine.
Tellurium
d. The more electronegative element of oxygen and sulfur.
Oxygen
e. The more electronegative element of magnesium and sulfur.
Sulfur
f. The larger atom of aluminum and fluorine.
Aluminum
18. The alkali metals all react with water. (4)
a. Describe what happens as each of lithium, sodium and potassium
react with water.
Lithium, sodium and potassium react with water producing a
flame and hydrogen gas.
b. State the difference in the reactivity of these alkali metals with
water.
Down the group, each alkali metal reacts more violently.
c. State the equation for each of the reactions in part a.
2Li(s) + H2O(l) → Li2O(aq) + H2(g)
2Na(s) + H2O(l) → Na2O(aq) + H2(g)
2K(s) + H2O(l) → K2O(aq) + H2(g)
d. Describe what you could do experimentally to show what the
product(s) are.
These products then react further with water to produce an alkaline
solution, forming LiOH, NaOH and KOH respectively.
Phenolphthalein can be added to the water before the reaction and the
indicator will turn pink due to the presence of the metal hydroxide .
19. Compare the melting points of each of the following pairs of
elements and explain the difference: (3)
a. Aluminum and sodium
Aluminum has a higher melting point than sodium.
Aluminum is a smaller metal atom with cations that have a charge of
+3 and 3 delocalized electrons. Sodium is a larger atom with cations
that have a charge of +1, and only 1 delocalized electron. There is a
much greater attraction between the cations and delocalized electrons
in aluminium and thus it has a higher melting point.
b. Magnesium and chloride
Magnesium has a higher melting point than fluorine.
Magnesium is a metal; chlorine is a non-metal. Magnesium has
strong metallic bonding but chlorine is a diatomic molecule with
only weak van der Waals’ forces between the molecules. Thus,
magnesium has a higher melting point than chlorine.
c. Bromine and fluorine
Bromine has a higher melting point than fluorine.
Bromine and fluorine are both non-metals and they are both
diatomic molecules with weak van der Waals’ forces between the
molecules. However, since bromine is larger than fluorine, it has
stronger van der Waals’ forces and therefore has a higher melting
point.
20. a. State the trend in reactivity of the halogens down group 7.
The halogens become less reactive down the group.
b. Identify the halide ion(s) with which chlorine will react.
Chlorine will react with bromide and iodide ions only.
c. State balanced chemical equations for the reaction(s) that occur.
Cl2(g) + 2I–(aq) → 2Cl–(aq) + I2(aq)
Cl2(g) + 2Br–(aq) → 2Cl–(aq) + Br2(aq)
(3)