PEC140 SB2 Tutorial Answers

8/12/2019 PEC140 SB2 Tutorial Answers

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PEC140 Introduction to Chemistry

STUDY BLOCK 2 Tutorial Answers

1. a) Manganese is a transition element. Examples of transition elements include Cu, Fe, Cr,etc.

b) Potassium forms a K+ion. It is in group 1 of the periodic table and thus has 1 outer shell

(valence) electron. The valence electron is readily lost to produce an ion that has oneless electron than the number of protons in its nucleus. That is K

+has one more proton

than the number of electrons, and thus the ion has an overall charge of +1. Removal ofthe valence electron gives the K

+ ion the very stable electron configuration of Ar,

corresponding to a full outer shell of electrons.

c) Oxygen can form the O2

ion. Oxygen is in main group 6 of the periodic table and, in

order to get a stable electron configuration, i.e.full outer shell of electrons, the neutral

atom will accept 2 electrons. It will thus have 2 more negative charges than positive and

the overall charge on the atom will be 2. Other elements that may form X2 ions are

those in the same group as oxygen, i.e.group 6 elements: S, Se, Te, etc.

2. a) Elements 19 (potassium, K) and 30 (zinc, Zn) belong to the sameperiod (period 4).

b) Elements 8 (oxygen, O) and 52 (tellurium, Te) belong to the same family or group(group 6).

c) Elements 24 (chromium, Cr) and 74 (tungsten, W) belong to the same family or group(in the transition metals section of the periodic table).

d) Elements 3 (lithium, Li) and 10 (neon, Ne) belong to the sameperiod.

e) Elements with similar chemical properties are expected to have the same number of

electrons in their outer shells (more on this in Study Block 5), i.e.they should occur inthe same group of the periodic table. For the examples above that means you would

expect similar chemical properties from O and Te, and for Cr and W.

3. a) Mass number,A= 19

b) Number of electrons = number of protons = atomic number (Z) = 9

c) Number of neutrons = Mass number number of protons = 19 9 = 10

4. A molecule is defined as the smallest particle of a pure substance as it occurs in nature.

Molecules can be elements or compounds. Some elements can exist as more than one type of

molecule (e.g.O2and O3) whilst some elements exist as monoatomic molecules (the group 18

elements).

5. a) MgBr2. According to the textbook (Zumdahl) type I binary ionic compounds have

ions formed from a metal (Mg) and a non-metal (Br), and the metal forms only one typeof cation, in this case Mg

2+.


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b) CoCl2. Type II binary ionic compounds contain ions formed from a metal (Co) and anon-metal (Cl), where the metal can form cations that have different charges, i.e.Co

2+

(as is the case here) or Co3+

. Type II binary compounds usually have a transition metal

as the cation.

c) N2O5. Type III binary compound as it contains only non-metal elements.

d) Na2CO3. Not a binary compound (because it has more than two elements). This

compound consists of a metal cation (Na+) and apolyatomicanion (carbonate, CO3

2).

e) Ca(OH)2. Not a binary compound. Has Ca2+

cation and the polyatomic hydroxide (OH

)anion.

f) Al(NO2)3. Not a binary compound. Has Al3+

cation and the polyatomic nitrite (NO2

)anion.

6. a) phosphate b) sulfide c) chromate

d) ammonium e) acetate f) dihydrogenphosphate

7. a) Magnesium nitride b) Calcium oxide c) Iron(III) chloride

d) Lithium carbide e) Lithium carbonate f) Magnesium sulfate

g) Sodium sulfite h) hydrogen cyanide i) Potassium permanganate

8. a) Isotopes are atoms of the same element (i.e.atoms with the same number of protons =

same atomic number) that have different numbers of neutrons. For example C12

6

, C13

6

and C146 are all isotopes of carbon. All have 6 protons but have

6, 7, and 8 neutrons, respectively.

b) Average atomic mass = average mass of all the isotopes multiplied by the natural

abundance of each isotope. Making the assumption that isotopic (atomic) mass massnumber for each isotope,

14N has a mass of 14 exactly and constitutes 99.50% of all

naturally-occurring N. Similarly15N has a mass of 15 exactly but only constitutes

0.500% of natural. To calculate the average atomic mass of N (as it occurs in nature):

{(99.50/100) 14} + {(0.500/100) 15} = 13.93 + 0.075 = 14.005

Average atomic mass = 14.01 (Answer in 2 decimal places. Because the last operation is

an addition, need to use rules for addition/subtraction when determining significant

figures, bearing in mind that the mass number (14 and 15) are pure numbers.)

9. FeO iron (II) oxide. Have Fe2+

and O2

.

Fe2O3 iron (III) oxide. Have Fe3+

and O2.

It is a characteristic of transition metalsthat they can form stable cations with different charges.

10. a) phosphoric acid b) sulfuric acid c) sulfurous acid

d) nitrous acid e) hydrobromic acid


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11. a) HClO4 b) HNO3 c) HF

12. b) MgS is magnesium sulfide, MnS manganese(II) sulfide.

d) N2O5is dinitrogen pentoxide, N2O3is dinitrogen trioxide

e) ClO2is the chloriteion; ClO3

is the chlorateion.

13. a) Isotopes refer to the same element (i.e.have the same number of protons) but have adifferent mass number (i.e.different number of neutrons).

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PEC140 SB2 Tutorial Answers