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5070/22

Chemistry 5070/22 — May/June 2012

Cambridge O-Level · Theory · worked solutions for every part, with the mark scheme

9
questions
75
marks
90
minutes

Topics Stoichiometry · Atoms, Elements and Compounds · Acids, Bases and Salts · Metals · Organic Chemistry · Electrochemistry · +5 more

Q16MAtoms, Elements and CompoundsElectrochemistryFree sample

Choose from the following particles to answer the questions below.

48Be^{8}_{4}\text{Be}

614C^{14}_{6}\text{C}

2040Ca^{40}_{20}\text{Ca}

1737Cl−^{37}_{17}\text{Cl}^{-}

1939K+^{39}_{19}\text{K}^{+}

1224Mg2+^{24}_{12}\text{Mg}^{2+}

1020Ne^{20}_{10}\text{Ne}

817O^{17}_{8}\text{O}

816O2−^{16}_{8}\text{O}^{2-}

1632S2−^{32}_{16}\text{S}^{2-}

1428Si4−^{28}_{14}\text{Si}^{4-}

Each particle can be used once, more than once or not at all.

Which particle

(a)

has only eight electrons,

______

1M
DifficultyEasy
Worked solution

Answer

817O^{17}_{8}\text{O}

Final answer

O-17

Detailed explanation

Walkthrough

The notation ZAX^{A}_{Z}\text{X} gives the mass number A on top and the proton number Z on the bottom. For a neutral atom, the number of electrons equals the proton number. 817O^{17}_{8}\text{O} has proton number 8 and no charge, so it has exactly 8 electrons. The other neutral atoms have different proton numbers, and the negative ions have more electrons than protons.

Key Takeaways

The bottom number in nuclide notation is the proton number. A neutral atom has equal numbers of protons and electrons. A negative charge means electrons have been gained.

Common Mistakes

Using the mass number as the number of electrons. Forgetting that ions have different electron numbers from their neutral atoms.

Things to Be Careful About

816O2−^{16}_{8}\text{O}^{2-} has 10 electrons, not 8, because the 2- charge means two extra electrons.

Techniques used
read proton number from nuclide notationcalculate electron number from chargecompare electron count with 8
(b)

is attracted to the cathode during electrolysis,

______

1M
DifficultyMedium-Easy
Worked solution

Answer

1939K+^{39}_{19}\text{K}^{+} or 1224Mg2+^{24}_{12}\text{Mg}^{2+}

Final answer

K+ / Mg2+

Detailed explanation

Walkthrough

During electrolysis, the cathode is the negative electrode, so it attracts positive ions (cations). Look through the bank for particles with a positive charge: 1939K+^{39}_{19}\text{K}^{+} and 1224Mg2+^{24}_{12}\text{Mg}^{2+} are the only cations. Negative ions such as Cl−\text{Cl}^{-}, O2−\text{O}^{2-}, S2−\text{S}^{2-} and Si4−\text{Si}^{4-} are attracted to the anode, not the cathode. Neutral atoms are not attracted to either electrode.

Key Takeaways

Cations are attracted to the cathode; anions are attracted to the anode. The charge on an ion is shown as a superscript after the symbol.

Common Mistakes

Choosing an anion because it is charged. The cathode attracts positive ions, not negative ones.

Things to Be Careful About

Both K+ and Mg2+ are acceptable answers; the mark scheme allows either. You do not need to give both.

Techniques used
identify cations from the charge in the notationapply the rule that cations move to the cathode during electrolysis
(c)

has only four electrons in its outer shell,

______

1M
DifficultyEasy
Worked solution

Answer

614C^{14}_{6}\text{C}

Final answer

C-14

Detailed explanation

Walkthrough

Write the electron configuration of each neutral atom in shells. Carbon has 6 electrons: 2 in the first shell and 4 in the second shell, so its outer shell contains 4 electrons. Silicon as a neutral atom would also have 4 outer electrons, but the particle in the list is Si4−\text{Si}^{4-}, which has gained 4 electrons and has a full outer shell of 8. Therefore the answer is 614C^{14}_{6}\text{C}.

Key Takeaways

Electron configurations are written shell by shell (2,8,8...). The outer shell electrons determine the group properties.

Common Mistakes

Choosing Si4- because silicon is in Group IV; the ion has a full outer shell, not 4 electrons.

Things to Be Careful About

Count the electrons after applying any charge before deciding the outer shell.

Techniques used
write electron configuration in shellscount outer-shell electrons
(d)

has only eight neutrons,

______

1M
DifficultyEasy
Worked solution

Answer

614C^{14}_{6}\text{C} or 816O2−^{16}_{8}\text{O}^{2-}

Final answer

C-14 / O2-

Detailed explanation

Walkthrough

Number of neutrons = mass number - proton number. For 614C^{14}_{6}\text{C}: 14 - 6 = 8 neutrons. For 816O2−^{16}_{8}\text{O}^{2-}: 16 - 8 = 8 neutrons. The charge does not change the number of neutrons. Other particles have different neutron numbers, e.g. 817O^{17}_{8}\text{O} has 9 neutrons.

Key Takeaways

Neutron number is the difference between mass number and proton number. Charge affects electrons only, not protons or neutrons.

Common Mistakes

Using the charge to change the neutron count. Forgetting to subtract proton number from mass number.

Things to Be Careful About

Two answers are possible here: C-14 and O2-. Either is accepted.

Techniques used
calculate neutrons as mass number minus proton numbercompare neutron count with 8
(e)

has only ten protons,

______

1M
DifficultyEasy
Worked solution

Answer

1020Ne^{20}_{10}\text{Ne}

Final answer

Ne-20

Detailed explanation

Walkthrough

The bottom number in the notation is the proton number. 1020Ne^{20}_{10}\text{Ne} has 10 protons. None of the other particles has proton number 10: Be has 4, C has 6, O has 8, Mg has 12, S has 16, Cl has 17, K has 19, Ca has 20, Si has 14.

Key Takeaways

The proton number defines the element. It is the bottom number in nuclide notation.

Common Mistakes

Reading the mass number instead of the proton number.

Things to Be Careful About

Even though Ne has 10 electrons too, the question asks for protons, so the bottom number is the key.

Techniques used
read proton number from nuclide notationcompare proton numbers with 10
(f)

has four occupied electron shells?

______

1M
DifficultyEasy
Worked solution

Answer

2040Ca^{40}_{20}\text{Ca}

Final answer

Ca-40

Detailed explanation

Walkthrough

An occupied shell is a shell that contains electrons. Calcium has 20 electrons: 2,8,8,2, so it occupies four shells. The ions K+, Cl-, S2- and Si4- all have 18 electrons, configuration 2,8,8, so they occupy three shells. The other particles have fewer shells. Therefore 2040Ca^{40}_{20}\text{Ca} is the answer.

Key Takeaways

Electron configuration shell by shell shows how many shells are occupied. 20 electrons fill four shells: 2,8,8,2.

Common Mistakes

Choosing an ion with 18 electrons because 18 seems large; those ions occupy only three shells.

Things to Be Careful About

Count occupied shells, not total electrons. Ca is the only particle with four occupied shells.

Techniques used
write electron configuration in shellscount the number of occupied shells

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