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Page 1: Elements & Environments

TOPIC 2ELEMENTAL & ENVIRONMENTAL CHEMISTRY

TOPIC 2 : ELEMENTAL AND ENVIRONMENTAL CHEMISTRY

This topic deals with some underlying principles of chemistry and then considers the chemistry of the environment. The elemental chemistry focuses on the periodic table and the concept of electronegativity, which underlie most of the other topics in this curriculum statement. The environmental chemistry focuses on a small number of inorganic molecular substances and their impact on the environment.

When the chemical elements are arranged in a periodic table, similarities and trends in properties become apparent. This topic examines the properties of compounds and elements. These properties can be explained in terms of the electronegativities of the elements and their positions in the periodic table.

The effects of human beings on the environment have not always been for the better. In the last hundred years concern about these effects has extended from local to global matters. Students are often exposed to environmental concerns about life on Earth, sometimes presented in emotive language. In this topic students are exposed to factual information, and consider causes and solutions of environmental problems.

Subtopic 2.1: The Periodic Table

Key Ideas Intended Student Learning

The arrangement of electrons in any atom can be described in terms of shells and subshells.

Write, using subshell notation, the electron configuration for an atom or monatomic ion of any of the first thirty-eight elements in the periodic table.

The position of an element in the periodic table reflects its electron configuration.

Identify the s, p, d, and f block elements in the periodic table.

The periodic table is the unifying framework for the study of the chemical elements and their compounds. Elements within each group of the periodic table have similar chemical properties that can be explained in terms of their similar outer-shell electron configurations.

Predict the following properties of the s and p block elements of any of the first thirty-eight elements in the periodic table:

Metal, metalloid, or non-metal nature of the element.

Charge of the monatomic ions.

Likely oxidation state(s) of the element in its compounds (including octet expansion for phosphorus, sulfur, and chlorine).

The electronegativities of non-metallic atoms are higher than those of metals; non-metallic atoms tend to gain electrons in chemical reactions.

Find regions with elements of high, intermediate, and low electronegativity in the periodic table.

The trend from metallic to non-metallic behaviour across a period is related to the increase in electronegativity. These trends are reflected in changes in the acidic–basic character of the oxides.

Predict the acidic–basic character of the oxides of an element from the position of the element in the periodic table.

The oxides of non-metals are acidic. Their acidic character can be displayed by reaction with hydroxide ions to produce an oxyanion and, in most cases, by reaction with water to produce an oxyacid.

Write equations for the reactions of oxides of non-metals such as SiO2, CO2, SO2, SO3, and P4O10 with hydroxide ions and with water, where a reaction occurs.

The oxides of metals are basic. Their basic character can be displayed by reaction with an acid to produce a cation and, in some cases, by reaction with water to produce OH in solution.

Write equations for the reactions of oxides of metals such as MgO, Na2O, CuO, and Fe2O3 with acids and with water, where a reaction occurs.

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Key Ideas Intended Student Learning

Metalloids form amphoteric oxides. Amphoteric oxides can display basic character by reaction with hydrogen ions and acidic character by reaction with hydroxide ions.

Write equations for the reaction of an amphoteric oxide such as Al2O3, ZnO with hydrogen ions or hydroxide ions.

Small molecules are formed from elements in a small section of the periodic table. Small molecules are those either of non-metallic elements or of compounds of non-metallic elements.

Predict whether or not a compound or element is likely to be molecular, given its properties, name, elemental composition, or formula.

Atoms in a molecule are bound strongly to each other by covalent bonds. Molecules interact weakly with each other.

Compare the strengths of covalent bonds with the strengths of secondary interactions.

The strengths of secondary interactions between non-polar molecules depend on their molar mass.

Explain the higher melting-points and boiling-points of substances of large molar mass.

The shape of molecules can be explained and predicted by repulsion between pairs of bonding and non-bonding electrons.

Draw diagrams showing covalent bonds, non-bonding pairs, and shapes for two-element molecules and ions containing no more than five atoms. Examples that involve valence shell octet

expansion are limited to PO43−

tetrahedra, SO2 and SO3.

The polarity of a molecule results from the polar character of the bonds and their spatial arrangement.

Predict whether or not a molecule is polar, given its spatial arrangement.

The strengths of secondary interactions between molecules of similar molar mass depend on the polarity of the molecules.

Explain the higher melting-points and boiling-points of polar substances when compared with those of non-polar substances of similar molar mass.

Molecules containing N–H or O–H groups can form hydrogen bonds to N or O atoms in other molecules.

Describe, with the aid of diagrams, hydrogen bonding between molecules.

1. Electron Configurations (up to atomic number 38)

(a) The arrangement of electrons in any atom can be described in terms of shells and subshells. Each electron in an atom, or monotomic ion, has a certain amount of potential energy arising from the attraction between its negative charge and the positive charge of the nucleusElectrons in the atoms or monatomic ions of a particular element have energy values that are unique to that element, and these electrons are said to exist at certain energy levels.Each allowed energy level is represented by a main shell number (1,2,3 etc) and a subshell represented by the letters s, p, d, f or g. For each main shell (n), there are n subshells. The subshells for the first four shells are shown in the table.

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Main Shell Subshells

1 1s

2 2s2p

3 3s3p3d

4 4s4p4d4f

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(b) When writing electron configurations, the following principles must be observed:

in the most stable state of any atom (or ion), the electrons “occupy” the lowest available energy level. They are “allocated” to subshells in order of increasing energy as showing in the following energy sequence

1s2s2p3s3p4s3d4p5s

lower energy higher energy

for each subshell, there is a maximum number of electrons which can “occupy” that subshell as shown below

Subshell

Maximum number of electrons

s 2

p 6

d 10

f 14

A diagram to illustrate the sequence is shown below.

(c) Using subshell notationThe electron configuration is written in energy sequence order the main shell number is written, followed by the letter for the subshell and then the number of electrons (as a superscript). This is illustrated in the following examples:

sodium (11 electrons) 1s2 2s2 2p6 3s1

iron (26 electrons) 1s2 2s2 2p6 3s2 3p6 4s2 3p6

strontium (38 electrons) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2

Note that chromium and copper do not conform to the principles exactly

chromium (24 electrons) 1s2 2s2 2p6 3s2 3p6 4s1 3d5

copper (29 electrons) 1s2 2s2 2p6 3s2 3p6 4s1 3d10

(d) Using subshell notation to write electron configurations for monatomic ions

(i) For positive ions of the main group elements determine the number of electrons left. eg Ca2+ has lost 2 electrons and has 18 left and so is 1s2 2s2 2p6 3s2 3p6

(ii) For negative ions of the main group elements, determine the total number of electrons. eg the S2- has gained 2 electrons and so has 18 electrons so its configuration is 1s2 2s2 2p6 3s2

3p6.

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(iii) For positive ions of the transition metals (eg Fe2+ and Fe3+) write the electron configuration of the atom eg Fe (26 electrons) - 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Decrease the number electrons equal to the number of positive charges on the ion (ie 2 for Fe2+ and 3 for Fe3+), by removing 4s electrons first, with any further deletions being made from the 3d subshell.Thus Fe2+ (24 electrons) becomes 1s2 2s2 2p6 3s2 3p6 3d6

and Fe3+ (23 electrons) becomes 1s2 2s2 2p6 3s2 3p6 3d5

EXERCISE 1

1. Write the electron configuration of the following elements

(a) Lithium

(b) Nitrogen

(c) Magnesium

(d) Sulphur

(e) Argon

(f) Manganese

(g) Arsenic

(h) Rubidium

2. Write the electron configuration of the following ions

(a) Beryllium

(b) Oxygen

(c) Sodium

(d) Chlorine

(e) Potassium

(f) Manganese II

(g) Copper I

(h) Bromine

2. Electron configuration and the Periodic table

(a) An element’s position in the Periodic Table is determined by its electron configuration.

(i) Horizontal rows are called Periods and the number of the period in which an element is placed is equal to the highest numbered main shell that is occupied by electrons.

(ii) Vertical columns are called Groups (the main groups are normally numbered with Roman Numerals) and the number of electrons occupying its highest numbered main (or outer) shell determines the group number (for main group elements). Those with incomplete 3d subshells go into a group called the transition elements.

Examples

Phosphorus (15 electrons) - 1s2 2s2 2p6 3s2 3p3 has electrons in the 3rd shell and is in Period 3 and has 5 electrons in that main shell so is in Group V.

Bromine (35 electrons) - 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 has electrons in the 4th shell and is therefore in Period 4 and has 7 electrons in that shell so is in Group VII.

Iron (26 electrons) - 1s2 2s2 2p6 3s2 3p6 4s2 3d6 has electrons in the 4th shell so is in Period V, but has an incomplete 3d shell, so is a Transitional Element.

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(b) The following is an outline of a modern Periodic Table

GROUP NUMBERS I1

VIII18

PE

RIO

D

1II2 3 4 5 6 7 8 9 10 11 12

III13

IV14

V15

VI16

VII

17

2

3 TRANSITION ELEMENTS

4

5

6 *

7 **

* LANTHANIDES

** ACTINIDES

(i) The elements in the following groups of the Periodic Table are given collective names:

Group I elements are called the alkali metals

Group II elements are called the alkaline earth metals

Group VII elements are called the halogens

Group VIII elements are called the noble gases

(ii) For all elements in Group I, there is one highest energy electron in an s subshell.

For example: lithium (Li) 1s2 2s1

rubidium (Rb) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1

(iii) For all elements in Group II, there are two highest energy electrons in an s subshell.

For example: magnesium (Mg) 1s2 2s2 2p6 3s2

calcium (Ca) 1s2 2s2 2p6 3s2 3p6 4s2

(iv) Groups I and II form the s block of the Periodic Table.

(v) For all elements in Group III, there is one highest energy electron in a p subshell.

For example: boron (B) 1s2 2s2 2p1

gallium (Ga) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1

(vi) For all elements in Group IV, there are two highest energy electrons in a p subshell.

For example: silicon (Si) 1s2 2s2 2p6 3s2 3p2

germanium (Ge) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2

(vii) For elements in Groups V to VIII there are in turn 3 to 6 highest energy electrons in a p subshell (helium being an exception at the top of Group VIII).

(viii) Groups III to VIII form the p block of the Periodic Table.

(ix) For the Transition Elements, the highest energy electrons are in a d subshell. The following are examples of electron configurations of elements from the first row (period 4) of the transition elements:

titanium (Ti) 1s2 2s2 2p6 3s2 3p6 4s2 3d3

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cobalt (Co) 1s2 2s2 2p6 3s2 3p6 4s2 3d7

(x) The Transition Elements form the d block of the Periodic Table.

(xi) For the Lanthanides and Actinides, the highest energy electrons are in an f subshell. The following is an example of an electron configuration of an element from the lanthanides:

samarium, Sm 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6

(xii) The Lanthanides and Actinides form the f block of the Periodic Table.

(xiii) The s, p, d and f blocks of the Periodic table are summarised on the following diagram:

s-block p-block

ns1 np6

ns2 d-block np1 np2 np3 np4 np5

n=1

n=2

(n-1)d1-

10

n=3

n=4

n=5

*

n=6

**

n=7f-block

* 6f1-14

** 7f1-14

EXERCISE 2

1. Using subshell notation, write the electron configurations for the following atoms for which the atomic number is given in brackets. From the electron configuration, determine the "block" of the Periodic Table to which the element belongs.

(a) arsenic, As, (33)

(b) vanadium, V, (23)

(c) krypton, Kr, (36)

(d) cobalt, Co, (27)

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2. By inspection of a copy of the Periodic Table determine the "group" and "block" for each of the following elements.

Group No. Periodic Table block

(a) francium

(b) thorium n/a

(c) tungsten n/a

(d) thallium

Chemical properties of the elements and the Periodic Table

The elements in each Group of the s and p blocks of the Periodic Table display similar chemical properties that can be explained in terms of their similar outer-shell electron configurations.

Examples of "similar chemical properties"

The elements from Group I all react with chlorine to form a chloride of formula type MCl.

Group I elements also react with water to form an hydroxide of formula type MOH.

Each element in Group V forms a compound with hydrogen of formula type XH3.

Electron configurations of the atoms of the s and p block elements can be used as a basis for explaining and predicting their chemical properties. The connection between the electron configuration of an element and its position on the Periodic Table can be used to make predictions about the properties of an element, including its metal/metalloid/non-metal nature, the charge(s) of its monatomic ion(s) and its likely oxidation states in its compounds.

Metals Atoms of metals lose electrons in chemical reactions

Non-metals Atoms of non-metals gain or share electrons in chemical reactions

Metalloids Atoms of metalloids lose or share electrons in chemical reactions

The similarity in chemical properties of the elements within each particular Group is explained in terms of the similarity of their electron configurations. When elements react, their atoms either lose or gain electrons (to form positive or negative ions respectively) or they share electrons with those of other atoms (to form covalent bonds). The electron configurations of the resultant ions are more stable than the configurations of the atoms from which they have been formed. Similarly, when atoms share electrons they acquire more stable configurations.

The Octet Rule

For period 1 and 2 elements an electron configuration in which the outer shell is "complete" with its maximum number of electrons is more stable than a configuration with an incomplete outer shell. For example, Mg2+, 1s22s22p6, is more stable than Mg, 1s22s22p63s2.

These complete stable electron configurations are the same as for the Noble Gases of periods 1 and 2 and are often referred to as "Noble Gas configurations". For the period 2 and 3 Noble Gases, the electron configurations show 8 electrons in the outer shell and when atoms attain 8 electrons in their outer shell they are said to have conformed to the "Octet Rule".

Expansion of the Octet

Atoms of the period 3 elements from the p block often conform to the octet rule by accepting electrons to form negatively charged monatomic ions or by sharing electrons with other atoms. For example, S, 1s22s22p63s23p4, forms the S2- ion, 1s22s22p63s23p6.

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The atoms of the period 3 elements from groups V to VII can share all of their outer shell electrons and as a consequence acquire more than 8 electrons in their outer shells. The extra electrons above the octet are "accommodated" in the previously unoccupied 3d subshell. This is referred to as the "expansion of the octet".

Valence Electrons

Electrons lost or shared are those from the valence (outer) shells of atoms. Electrons gained are accepted into valence shells. These electrons are called valence or outer shell electrons.

For s block elements, the valence shell electrons are the highest energy (often called the "outer") s subshell electrons.

For each Group I element, the valence shell electron is the ns1 outer electron (where n represents the number of the Period of the element).

For each Group II element, the valence shell electrons are the ns2 outer electrons.

For p block elements, the valence shell electrons are the "outer" s and p subshell electrons.

For each Group III element, the valence shell electrons are the ns2np1 outer electrons.

For each Group IV element, the valence shell electrons are the ns2np2 outer electrons.

For each Group VII element, the valence shell electrons are the ns2np5 outer electrons.

Chemical Reactions involving the s Block Elements

Group I Elements

The atoms of the Group I elements lose their outer s1 valence electrons to form an M ion. The electron lost is gained by the other reactant. The product compounds are ionic with formulae such as MCl, M2S, M2O and MOH.

The charge on the monatomic ions of the Group I elements is always 1+.

Consequently the oxidation state of the Group I elements in their compounds is always +1.

The Group I elements are classified as metals because their atoms lose electrons in chemical reactions.

NB: As described in the box below, these generalizations do not apply to hydrogen.

The following are examples of reactions involving Group I elements:

2Na(s) + Cl2(g) 2NaCl(s)

2K(s) + S(s) K2S(s)

Note that hydrogen is not included as a member of Group I. Although its electron configuration is 1s1, its properties are quite different to those of other members of Group I. Sometimes it is not shown at the top of Group I but is given a separate "box" of its own.

In compounds with non-metals, hydrogen atoms share their one valence electron with valence electrons of the other non-metal atoms. In these compounds hydrogen has an oxidation state of +1.

In compounds of hydrogen with active metals (such as those of Groups 1 and 2) hydrogen exists as the hydride ion, H-.

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Group II Elements

The atoms of the Group II elements lose their outer 2 valence electrons to form an M2+ ion. The electrons lost are gained by the other reactant. The product compounds are ionic with formulae such as MCl2, MS, MO and M(OH)2.

The charge on the monatomic ions of the Group II elements is always 2+.

Consequently the oxidation state of the Group II elements in their compounds is always +2.

Except for beryllium (which acts as a metalloid), the Group II elements are classified as metals because their atoms lose electrons in chemical reactions.

The following are examples of reactions involving Group II elements:

2Ca(s) + O2(g) 2CaO(s)

Mg(s) + S(s) MgS(s)

Chemical Reactions involving the p Block Elements

Group III Elements

In their compounds -

Eitherthey exhibit a covalence of 3 as in the case for boron in its compounds, for example BCl3. When boron forms compounds, its atoms share the s2p1 outer shell electrons to form covalent bonds with other non metal atoms.

Covalence

The covalence of an element is equal to the number of electrons that its atoms share when forming covalent bonds with other atoms. When boron shares its 3 outer shell electrons with, for example, electrons from three chlorine atoms, then boron is exhibiting a covalence of 3.

Or they exist as triple positive ions, such as Al3+, in compounds with non metals. These ions are formed when atoms of the Group III elements lose the s2p1 outer shell electrons in electron transfer reactions.

The charge on the monatomic ions of the Group III elements is usually 3+.

The oxidation state of the Group III elements in their compounds is usually +3. In some compounds their oxidation state is -3.

The elements range from a non-metal, boron, at the top of the group, to metalloids, aluminium and gallium in the middle and metals at the bottom of the group.

The following are examples of reactions involving Group III elements:

4Al(s) + 3O2(g) 2A12O3(s)

2Ga(s) + 3Cl2(g) 2GaCl3(s)

Group IV Elements

In their compounds -

Eitherthey exhibit a covalence of 4, as in the case for carbon and silicon in their compounds, for example CCl4 and SiH4. When carbon and silicon form compounds, their atoms share the s2p2 outer shell electrons to form covalent bonds with other non-metal atoms.

Or they exist as 2+ or 4+ ions, such as Pb2+, Sn2+ or Pb4+, in compounds with non metals.4+ ions are formed when the atoms of Group IV elements lose the s2p2 outer shell electrons in electron transfer reactions.2+ ions are formed when the atoms of Group IV elements lose only the p2 electrons of the s2p2 outer shell electrons in electron transfer reactions.

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The charge on the monatomic ions of the Group IV elements is usually 2+ or 4+.

The oxidation state of the Group lV elements in their compounds is either +4, +2 or -4.

The elements range from the non-metals, carbon and silicon, at the top of the group, to metalloids for the rest of the group.

Group V Elements

In their compounds

Either they exhibit a covalence of 3, as in the case of nitrogen in all of its compounds such as NH3, and of phosphorus and arsenic in some of their compounds, for example AsCl3. In these compounds, the nitrogen, phosphorus and arsenic atoms share only the p3 electrons from the s2p3 outer shell configuration to form covalent bonds with other non-metal atoms. In sharing in this way they are conforming to the octet rule.

Or they exhibit a covalence of 5, as in the case of phosphorus and arsenic in some of their compounds, for example AsCl5 and P4O10. In these compounds, the phosphorus and arsenic atoms share all of the s2p3 electrons from the s2p3 outer shell configuration to form covalent bonds with other non metal atoms. In sharing in this way they are expanding the octet.

Or they exist as 3- ions, such as N3- or P3- in compounds with metals. 3- ions are formed when atoms of the Group V elements gain three electrons into the p subshell thereby changing the outer shell configuration from s2p3 to s2p6. The resultant ions conform to the octet rule.

The charge on the monatomic ions of the Group V elements is 3-.

The oxidation state of the Group V elements in their compounds is either +5, +3 or -3.

The elements range from the non-metals, nitrogen and phosphorus, at the top of the group, to the metalloids arsenic and antimony in the middle and the metal bismuth at the bottom of the group.

Group VI Elements

In their compounds -

Either they exhibit a covalence of 2, as in the case of oxygen in all of its compounds such as H2O, and of sulfur and selenium in some of their compounds, for example SF2. In these compounds the oxygen, sulfur and selenium atoms share only two of the p4 electrons from the s2p4 outer shell configuration to form covalent bonds with other non-metal atoms. In sharing in this way they are conforming to the octet rule.

Or they exhibit a covalence of 4, as in the case of sulfur arid selenium in some of their compounds, for example SO2 and SeF4. In these compounds, the sulfur and selenium atoms share all of the p4 electrons from the s2p4 outer shell configuration to form covalent bonds with other non-metal atoms. In sharing in this way they are expanding the octet. It must be noted that oxygen does not exhibit a covalence of 4.

Or they exhibit a covalence of 6, as in the case of sulfur and selenium in some of their compounds, for example SF6, and Se03. In these compounds, the sulfur and selenium atoms share all of the s2p4 electrons from the outer shell configuration to form covalent bonds with other non-metal atoms. In sharing in this way they are expanding the octet. It must be noted that oxygen does not exhibit a covalence of 6.

Or they exist as 2- ions, such as O2- or S2- in compounds with metals. 2- ions are formed when the atoms of Group VI elements gain two electrons into the p subshell thereby changing the outer shell configuration from s2p4 to s2p6. The resultant ions conform to the octet rule.

The charge on the monatomic ions of the Group VI elements is 2-.

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The oxidation state of the Group VI elements in their compounds is +6, +4, +2 or -2. An exception is oxygen with an oxidation number of -1 in H2O2.

The elements range from the non-metals, oxygen and sulfur, at the top of the group, to the metalloids selenium and tellurium in the middle and the metal polonium at the bottom of the group.

Group VII Elements

In their compounds -

Either they exhibit a covalence of 1, as in the compounds such as HBr and CCl4. In these compounds, the Group VII atoms share one of the p5 electrons from the s2p5 outer shell configuration to form a covalent bond with other non-metal atoms. In sharing in this way they are conforming to the octet rule.

Or they exhibit a covalence of 3, 5 or 7 and in sharing in this way they are expanding the octet. It must be noted that fluorine does not exhibit a covalence of 3, 5 or 7.

In such compounds, the atoms of the Group VII elements share electrons in the following ways:

covalence of 3 three of the p5 electrons shared

covalence of 5 all five of the p5 electrons shared

covalence of 7 all seven of the s2p5 electrons shared

or they exist as 1- ions, in compounds with metals. 1- ions are formed when the atoms of Group VII elements gain one electron into the p subshell, thereby changing the outer shell configuration from s2p5 to s2p6. The resultant ions conform to the octet rule.

The charge on the monatomic ions of the Group VII elements is 1-.

The oxidation state of the Group VII elements in their compounds is +7, +5, +3, + 1 or -1.

The elements are all non-metals.

Limitations of the Covalent Bonding Model

There are some examples of the s and p block elements exhibiting oxidation states (and covalences) which are different from those given in the summaries above. Some of the more common of these are given below

Element Oxidation state Example

Nitrogen +4 NO2

Nitrogen +2 NO

Chlorine +4 ClO2

Some anomalous oxidation states

These oxidation states cannot be explained in terms of electron configurations and shared pairs of electrons. Other bonding models, that are beyond the scope of this course, must be used.

Summary Table of Oxidation States

For the s and p block elements up to atomic number 38, the following table summarises the likely oxidation states of the elements in their compounds and the metal/metalloid/non-metal nature of the elements:

I VIII

H

Non-metal

Ox state

+1 II III IV V VI VII

Li Be B C N O F

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metal

Ox state

+1

metalloid

Ox state

+2

Non-metal

Ox state

+3, -3

Non-metal

Ox state

+4, -4

Non-metal

Ox state

+5,+3, -3

Non-metal

Ox state

-2

Non-metal

Ox state

-1

Na

metal

Ox state

+1

Mg

metal

Ox state

+2

Al

Metalloid

Ox state

+3

Si

Non-metal

Ox state

+4, -4

P

Non-metal

Ox state

+5,+3,-3

S

Non-metal

Ox state

+6, +4, +2, -

2

Cl

Non-metal

Ox state

+7, +5, +3,

+1, -1

K

metal

Ox state

+1

Ca

metal

Ox state

+2

Ga

Metalloid

Ox state

+3

Ge

Metalloid

Ox state

+4, +2

As

Metalloid

Ox state

+5, +3

Se

Non-metal

Ox state

+4, +2

Br

Non-metal

Ox state

+7, +5, +3,

+1, -1

Rb

metal

Ox state

+1

Sr

metal

Ox state

+2

EXERCISE 3

Complete the following table

Group I II III IV V VI VII

Valence electrons ns1np0

Likely charge on monatomic ion

+

Expected covalence -

Formula of chloride XCl

Formula of oxide X2O

Can it expand octet? no

Other oxidation states none

Electronegativities of the elements

The relative ability of an atom to attract electrons to itself is called its electronegativity. The higher the electronegativity, the stronger the attraction for electrons.

Metal atoms have lower electronegativity values than metalloids, which in turn have lower electronegativities than non-metals. Using the Periodic Table, two clear trends for electronegativities of the s and p block elements can be described as follows:

Group I Group VII electronegativities increase across each Period

Period I Period 4 electronegativities decrease down each Group

Oxides

An oxide is a binary compound of an element combined with oxygen eg CO2, MgO.

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Acidic oxides

Nonmetal oxides are generally acidic.

Acidic oxides

React with hydroxides to produce oxyanions

React with water to make oxyacids (if they dissolve)

Oxide Reaction with hydroxide Reaction with water

P4O10P4O10+ 12OH- 4

PO43−

+ 6H2OP4O10+ 6H2O 4H3PO4

SO2SO2+ 2OH-

SO32−

+ H2O

(sulphite)

SO2+ H2O H2SO3

(Sulphurous acid)

SO3SO3+ 2OH-

SO42−

+ H2O

(sulphate)

SO3+ H2O H2SO4

(Sulphuric acid)

CO2CO2+ 2OH-

CO32−

+ H2O

(carbonate)

CO2+ H2O H2CO3

(Carbonic acid)

SiO2SiO2+ 2OH-

SiO32−

+ H2O

(silicate)

No Reaction

EXERCISE 4

One of the oxides of chlorine is Cl2O. Its is an acidic oxide with a corresponding oxyanion,ClO- (hypochlorite) and a corresponding oxyacid, HClO (hypochlorous acid).

Write an equation for the reaction of Cl2O with

1. Hydroxide ions

2. Water

Basic oxides

Metal oxides are basic oxides

They react with

hydrogen ions to produce a cation

water to produce hydroxides (if they dissolve in water)

Oxide Reaction with hydrogen ions Reaction with water

Na2O Na2O + 2H+ 2Na+ + H2O Na2O + H2O 2Na++ 2OH-

MgO MgO + 2H+ 2Mg2+ + H2O MgO + H2O Mg2+ + 2OH-

CuO CuO + 2H+ 2Cu2+ + H2O No Reaction

Fe2O3 Fe2O3 + 2H+ 2Fe3+ + H2O No Reaction

EXERCISE 5

Barium oxide BaO and lithium oxide are both basic oxides.

1. Write an equation for the reaction of BaO with hydrogen ions.

2. Write an equation for the reaction of Li2O with water.

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Amphoteric oxides

Amphoteric oxides display basic character in reaction with acids to give cations and water.

Amphoteric oxides display acidic character in reaction with hydroxides to give oxyanions and water.

Amphoteric oxides do not react with water.

Oxide Reaction with hydrogen ions Reaction with hydroxide ions

ZnO ZnO + 2H+ 2Zn2+ + H2O ZnO + 2OH- ZnO2

2−

+ H2O

Al2O3 Al2O3 + 6H+ 2Al3+ + 3H2O Al2O3 + 2OH- 2AlO2

+ H2O

EXERCISE 6

Lead oxide, PbO, is an amphoprotic oxides.

1. Write an equation for the reaction of PbO with hydrogen ions.

2. Write an equation for the reaction of PbO with hydroxide ions.

SMALL MOLECULES

Molecules consisting of 10 or less atoms are considered small molecules eg O2, CO2, H2O, NH3.

Small molecules are formed when atoms of non-metal atoms covalently bond to each other. These are mainly found in the top right-hand corner of the periodic table and also hydrogen. Some atoms in this group also make continuous lattices eg SiO2.

Properties of small molecules

Low melting and boiling points (often gases at room temperature)

Poor conductors of electricity.

Bonding in molecular compounds and elements.

Two types of bonding operate in these compounds and elements

Intramolecular (or Primary) bonds operate between atoms within the molecule.

Intermolecular (or Secondary) bonds operate between molecules

Intramolecular Bonding is mainly covalent bonding which occur when atoms share valence electrons.

The bond arises from the electrostatic attraction between the shared electrons and the positively charged nuclei.

They are strong bonds

Bond energy is the energy required to break 1 mole of bonds between covalently bonded atoms.

Bond Bond Energy (kJ mol-1)

CC 348

CH 413

HH 436

NH 391

C=C 614

C=O 745

NºN 945

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Intermolecular (secondary) bonds are the forces of attraction that operate between molecules. These are the bonds that hold molecules together.

When molecular substances are vaporized, sufficient energy must be provided to break the secondary bonds between the molecules. The quantity of energy needed to convert 1 mole of a molecular substance from a liquid to a vapour is called the molar heat of vaporization, Hvap. The boiling point, Tb, also provides an indication of the strength of secondary bonds.

Substance Hvap (kJ mol-1) Tb (C)

H2 0.45 -253

N2 2.8 -196

CH4 8.2 -161

Br2 15 59

H2O 41 100

Note

The higher Hvap the higher Tb.

The Hvap is much smaller that the corresponding bonding energy for the intramolecular bonds.

There is a big variation in Hvap values suggesting different types of secondary bonding.

Types of secondary bonding

There are three types of secondary bonds – dispersion forces, dipole-dipole and hydrogen bonding. The type of bonding for a particular molecule depends on the polarity of the molecule.

Polarity of covalent bonds

A covalent bond exists between two atoms when they share valence electrons.

If the electronegativity of both atoms is equal, the electrons are shared equally, and the bond is non-polar. eg HH, O=O, NºN

If one element has a higher electronegativity, it gains more control of the shared electrons, and gains a slight negative charge (-). The other atom has lost partial control of its shared electrons and gains a partial positive charge (). The resultant bond is said to be polar.

examples - Hd +

−Cld−

, Hd +

−Od−

, Sd +

=Cld −

The H-O bond is one of the most polar because the difference in electronegativity between hydrogen and oxygen is greater than most other pairs of atoms involved in covalent bonding.

The shapes of molecules

The shapes of molecules can be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory). They can be determined experimentally using techniques like X-Ray diffraction.

The key idea of this theory is that pairs of valence electrons in molecules repel each other to get as far apart as possible.

To determine the shape of a molecule

Determine which is the central atom.

Work out how many groups (bonding pairs, nonbonding pairs, multiple pairs in double or triple bonds) of electrons there are

The repulsion pattern of these groups of electrons

Attach the atoms and name the shape

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SHAPES OF SOME MOLECULES AND IONS

Formula Type Diagram EXAMPLES

AB A-B

linearHCl, O2

AB2

V-shaped

H2O (105), Cl2O (111)

H2S (92),

O3 (117)

B=A=B

linearCO2

V-shaped

SO2 (120)

AB3

Trigonal planar

BCl3

CO32−

NO3−

Trigonal pyramid

NH3 (107)

PCl3 (100)

SO32−

Trigonal planar

SO3

AB4

Tetrahedral

CH4

NH4+

SO42−

AB5

Trigonal Bipyramid

PCl5

AB6

Octahedral

SF6

XeF6

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EXERCISE 7

Draw the shape and name the shape of the following molecules

PF5, SCl2, CS2, NI3, GaI3

The Polarity of Molecules

Diatomic molecules

If the bonding in a diatomic molecule is nonpolar then the molecule is nonpolar. eg Cl2, O2, N2

If the bonding in a diatomic molecule is polar then the molecule is also polar. eg HCl, HF

Polyatomic molecules

If the bonding in a polyatomic molecule is nonpolar then the molecule is nonpolar. eg P4, S8, O3

If the bonding is polar, and the charge distribution is symmetrical, then the molecules is nonpolar.

Note All all hydrocarbon molecules and hydrocarbon parts of other molecules are nonpolar.

If there is an asymmetrical charge distribution ie positive and negative ends, the the molecule is polar.

Molecule Structure Spatial arrangement of atoms

SO2

SOO

positive end

negative end

. .

- -

+

V-shaped

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CHCl3 CClCl

H

positive end

negative endCl

-

-

-

+

Tetrahedral

H2S SHH

positive end

negative end

+ -

. . V-shaped

NH3

. .N

HH

positive end

negative end

H

+

+

+-

Trigonal pyramid

EXERCISE 8

For each of the following molecules

(a) draw a structural diagram showing the covalent bonds, the nonbonding electron pairs and the polarity of the bonds using + and - symbols.

(b) Describe the shape of the molecule

(c) Indicate whether the molecule is polar or not.

1. AsH3

2. H2S

3. CCl4

Polyatomic molecules with polar and nonpolar groups

Most organic molecules contain a hydrocarbon part and a functional group which is polar. The polarity of the molecules depends on the size of the molecule and the polarity of the functional group. Such large organic molecules tend to behave as if they were nonpolar.

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Secondary bonds – London dispersion forces.

What force holds nonpolar molecules together?

The electrons in molecules are constantly moving. At any instant this produces asymmetrical charge distribution and so a temporary dipole is formed.

This can induce a dipole in a neighbouring molecule, which causes an electrostatic attraction between these temporary dipoles. These weak forces of attraction are called London Dispersion Forces.

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These forces act in all molecular substances, but they are the only forces acting in nonpolar molecules.

These disperson forces are greater in larger molecules because there are more electrons. The shape of the molecule can have some effect too, so when comparing the effect of molar mass on dispersion forces, only molecules within the same structural class should be compared.

Hydrogen Bonding

Some small molecules have much higher boiling points that suggested by their molecular mass eg

Compound Molecular mass Boiling point (C)

Methane (CH4) 16 -161

Ammonia (NH3) 17 -33

Water (H2O) 18 100

These molecules with high Boiling Points usually have H-F, O-H, or N-H bonds in them. These are capable of forming strong secondary bonds called HYDROGEN BONDS.

Eg hydrogen bonding in methanol

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other examples

Effects of hydrogen bonding

Increased melting and boiling points

Solubility in each other

Ability to absorb water

Strengthens polymers by providing cross linking

Strength of hydrogen bonds

Strongest of the dipole-dipole attractions

Weaker than covalent bonds (Bond Energy 20-30 kJ mol-1 to 300-500 kJ mol-1), but stronger than dipole-dipole bonds ( 2 kJ mol-1).

They are stronger because

Large electronegativity difference

Hydrogen has only one electron, so its proton is left unshielded

The small size of hydrogen allows other atoms to get close.

EXERCISE 9

The following molecules are all tetrahedral. List them in order of increasing Boiling Point based on the nature of their secondary bonding. CH3F, CH3NH2, SiH4, CH3OH.

4. Show the hydrogen bonding between urea molecules.

N

C

N

O

H

H

H

H

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ASSIGNMENT 2.1: PERIODIC TABLE

1. Complete the following table.

Symbol Electron configuration Group Period Block (s,p,d or f)

11Na

1s22s22p63s23p5

2 4

19K+

35Br-

26Fe3+

1s22s22p63s23p6 5 3

2. The formula of hydrogen chloride is HCl.

(a) What is the common name of HCl(aq)?

Use the formula, HCl, to work out the likely formula of the chlorides of the following elements.

(b) Sodium

(c) Lithium

(d) Potassium

5. The formula of aluminium oxide is Al2O3. What is the likely formula of gallium oxide?

6. Sodium reacts vigorously with water. The equation for the reaction is given below.

2Nasq) + 2H2O(l) 2NaOH(aq) + H2(g)

Describe the like reactions of lithium and potassium with water and write equations for these reactions

7. Complete the following table by writing the possible oxidation states of each of the elements indicated.

Group

I II III IV V VI VII VIII

Period

1

2

3

4

8. Complete the following table by writing the formulae of the chlorides of each of the elements.

Group

I II III IV V VI VII VIII

Period

1

2

3

4

9. Explain how atoms may enter into chemical combination to form:

(a) ionic bonds, and

(b) covalent bonds.

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Give one example of each of the above bonding types.

10. The table opposite shows the melting points of five substances.

(a) What is meant by electronegativity?

(i) Explain the high melting point of MgO in terms of bonding present.

(ii) How can the bond type be related to the electronegativities of the elements?

(b) Why is the Tm of MgO much higher than that of NaCl?

(c) Why is the electrical conductivity of Mg good in both solid and molten states, whereas MgO conducts only in the molten state?

(d) Both C and Si are in Group IV, yet SiO2 has a much higher Tm than CO2. Explain.

11. (a) What is a polar bond?

How is the type of bonding in the chlorides of the elements of K and P related to:

(i) Their position in the periodic table and

(ii) The number of valence electrons?

(b) CF4 is a non-polar molecule with polar bonds. Explain.

12. Explain why the Tb of CH4, C2H6, C3H8 …. increase with increase in molar mass

13. Write structural formulae for each of the following species, and clearly indicate the shape.

(a) H2O2 (b) NH4+ (c) H3O+ (d) H2CO3

(e) SO32−

(f) ClO4−

(g) HSO4− (h) SO3

14. The polarity of a molecule depends not only on the polarity of the individual bonds, but also on the geometry of the molecule. Explain why the SO2 molecule is polar while the CO2 molecule is non-polar.

15. Account for the following observations.

(a) CO2 is a gas at room temperature but SiO2 is a high melting point solid.

(b) It has been shown that chlorine (Cl2) crystals have two different chlorine-chlorine bond distances (198 pm and 360 pm). (1 pm (picometre) = 1 x 10-12 m)

16. Consider the following elements from period 3 of the Periodic Table, Na, Mg, Al, Si, P and S.

(a) Write the formula for each chloride,

(b) How is the type of bonding in the chlorides of these elements related to

(i) their position in the periodic table

(ii) the number of outer electrons?

17. A, B and C represent elements with atomic numbers 9, 19 and 34 respectively.

(a) Give the electron configuration for each.

(b) What kind of bonding would you expect between

(i) A and B,

(ii) A and C,

(iii) B and C?

(c) Draw electron dot formulae for the compounds formed in (b).

18. What is a hydrogen bond?

23

Compound Tm (K)

MgO 3073

NaCl 1074

Mg 923

CO2 216

SiO2 1973

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19. Which compound from each of these pairs exhibits the stronger intermolecular hydrogen bonding? Explain your choice.

(a) H2S and H2O

(b) HCl and HF

(c) HBr and HCl

(d) NH3 and H2O

20. Water and ammonia have much higher boiling points than expected when considering the boiling points of similar compounds from their group in the periodic table. These graphs show this. Explain, with the aid of a diagram, why this is so.

H2Te

H2SeH2S

H2O

-80-60-40

-200

204060

80100120

B. P

t (

C)

SbH3

AsH3

PH3

NH3

-100

-80

-60

-40

-20

0

B. P

t (

C)

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21. Consider the following substances:iron, iodine, diamond, sodium chloride and ice

(a) Tabulate the structure of each solid and the type of bonding in the solid using the headings

Name Description of the structure of the solid Type of bond present

Iron

Iodine

Diamond

Sodium chloride

ice

(b) Discuss the relationship between the structure and the type of bonding in each solid.

(c) How are the physical properties (hardness, melting point and conductivity) of each solid related to bonding and structure in each substance?

Subtopic 2.2: Cycles in Nature

Key Ideas Intended Student Learning

The presence (aerobic conditions) or absence (anaerobic conditions) of oxygen affects the products of the decomposition of the organic compounds derived from living organisms.

State, for aerobic and anaerobic conditions, the products of the decomposition of organic matter containing carbon, nitrogen, phosphorus, or sulfur.

Photosynthesis and respiration are important processes in the cycles of carbon and oxygen.

Describe and write equations for the processes of photosynthesis and aerobic respiration involving glucose.

Nitrogen may be converted into compounds by biological processes such as fixation or by reaction with oxygen during lightning discharges and at high temperatures such as occur in engines and furnaces.

Describe and write equations for the formation of oxides of nitrogen by the reaction of nitrogen and oxygen at high temperatures.

Nitrogen compounds are important in the chemistry of life processes.

Describe how the nitrogen cycle operates by natural processes (e.g. lightning, nitrogen-fixing bacteria, decay) and industrial processes (e.g. fertiliser manufacture, combustion engines).

Plants require substantial amounts of nitrogen and phosphorus, which they obtain from the soil.

Explain why fertilisers need to contain nutrients in soluble form.

Many small molecules (methane, ammonia and carbon dioxide) are produced and consumed in the atmosphere and biosphere by chemical and biological reactions that occur in interconnected cycles.

One common feature of all these cycles is the decomposition of organic compounds derived from living organisms.

The presence (aerobic conditions) or absence (anaerobic conditions) of oxygen affects the products of the decomposition of the organic compounds derived from living organisms.

Organic Matter Aerobic decomposition Anaerobic decomposition

Carbon Carbon dioxide (CO2) Methane (CH4)

Nitrogen Nitrates (NO3−

) Ammonia (NH3)

Phosporus Phosphates (PO43−

) Phosphine (PH3)

Sulphur Sulphates (SO42−

) Hydrogen sulphide (H2S)

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Carbon and Oxygen Cycles

Photosynthesis is the photochemical process by which green plants convert carbon dioxide and water into glucose and oxygen in the leaves.

6CO2 + 6H2O + solar energy C6H12O6 + 6O2

This glucose is the primary energy source of all food chains and the building block of many complex molecules. It is also the key to maintaining oxygen levels in the atmosphere at 20%.

Aerobic Respiration is the process whereby plants and animals use glucose to make energy. It is basically the reverse of photosynthesis.

C6H12O6 + 6O2 6CO2 + 6H2O + energy

The carbon dioxide is returned to the atmosphere and maintains the carbon dioxide level in the atmosphere.

The nitrogen Cycle

In the atmosphere nitrogen occurs as diatomic molecules, NN. The triple bond is strong and so nitrogen is unreactive. The nitrogen is maintained at a constant 80% in the atmosphere by a number of natural processes.

Natural Nitrogen Fixation

There are a nitrogen fixing bacteria in nodules on the roots of legume plants like clover and peas and some blue green algae. They contain the enzyme nitrogenase which helps catalyse the conversion of nitrogen gas from the atmosphere into ammonia or ammonium compounds.

N2 f⃗ixation NH3 /NH4+

Fixation can also be caused by lightning discharges and forest fires. At these high temperatures, the energy released can be used to break nitrogen-nitrogen bonds so it can react with oxygen

N2 + O2 2NO (nitric oxide)

2NO + O2 NO2 (nitrogen dioxide)

The nitrogen dioxide then dissolves in water forming nitrate ions which are then deposited when it rains.

2NO2 + H2O H+ + NO3−

+ HNO2

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Artificial nitrogen fixation

Nearly 50% of all nitrogen fixation results from human activities. The large scale industrial production of ammonia by the Haber Process, the large scale combustion of fuels and increased cultivation of leguminous plants all remove large quantities of nitrogen from the atmosphere.

The Haber Process uses high temperature to combine nitrogen and hydrogen.

N2 + 3H2 2NH3

Burning of fuels and industrial furnaces consume nitrogen like in lightning strikes.

Industrial Chemical Conversion of Fixed Nitrogen

The chemical industry uses the Haber process to fix nitrogen as ammonia which is used as a base for the fertiliser industry. The main fertilizer compounds are ammonium nitrate, ammonium sulphate, ammonium phosphate and urea. Once these fertilizers are added to the soil they enter reactions like nitrification and denitrification.

The continuous chemical interchange between nitrogen in the atmosphere and the biosphere is called the NITROGEN CYCLE.

Plant Fertilizers

Plants require significant amounts of the essential elements nitrogen (N) and phosphorus (P) for growth.

Fertilizers provide soils with water soluble compounds of nitrogen, phosphorus and potassium in areas where the soils lack sufficient nutrient needs for agricultural needs. Plants can only absorb these nutrients when they are in aqueous form.

ASSIGNMENT 2.2: Cycles in Nature

1. Most organic compounds contain carbon, nitrogen, phosphorus and sulfur. When the organism dies, bacteria will cause decomposition and enable these elements to be recycled. Decomposition occurs under different conditions depending on the availability of free oxygen.

(a) What are aerobic conditions?

(b) What are anaerobic conditions?

(c) What decomposition products are formed under aerobic conditions from

(i) Carbon

(ii) Sulfur

(iii) Phosphorus

(iv) Nitrogen

(d) What decomposition products are formed under anaerobic conditions from

(i) Carbon

(ii) Sulphur

(iii) Phosphorus

(iv) Nitrogen

2. Photosynthesis is an important process in the maintenance of oxygen levels in the atmosphere.

(a) Write a balanced equation for photosynthesis.

(b) What enzyme present in green plants is essential for this process?

(c) Give two reasons for the importance of photosynthesis for living things.

3. All living things use the process of respiration to make energy.

(a) Write an equation for aerobic respiration.

(b) Why is it called “aerobic”?27

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4. Nitrogen can be removed from the atmosphere in a number of processes.

(a) What general name is given to this process?

(b) Describe one process whereby plants remove nitrogen from the atmosphere.

(c) Why doesn’t nitrogen normally react with oxygen in the atmosphere?

(d) Under what conditions will nitrogen and oxygen react? Give at least three examples.

(e) Write equations for the reactions of nitrogen and oxygen under the conditions outlined in the previous question. Label all reactants and products on the equation,

5. One important use of nitrogen and phosphorus is in the manufacture of fertilizers.

(a) Name the chemicals present in common fertilizers.

(b) Why are they normally in soluble compounds?

Subtopic 2.3: Greenhouse Effect

Key Ideas Intended Student Learning

Some gases in the atmosphere, called ‘greenhouse gases’, serve as insulation to maintain the temperature of the Earth’s atmosphere. This is known as the ‘natural greenhouse effect’.

Describe the action of the common greenhouse gases, carbon dioxide and methane, that serve to maintain a steady temperature in the Earth’s atmosphere.

Human activity that affects the concentration of greenhouse gases has the potential to disrupt the thermal balance of the atmosphere. This is known as the ‘enhanced greenhouse effect’.

Describe the predicted effects of human activity on the temperature of the Earth’s atmosphere.

All objects above absolute zero radiate electromagnetic radiation. The higher the temperature, the higher the energy (and shorter the wavelength) of the radiation

Of the solar radiation coming from the sun, slightly less than half actually warms the earth’s surface. The rest is reflected by clouds and oceans or absorbed by the atmosphere. The average temperature of the surface is 15C. The surface radiates heat back towards space, but certain gases in the lower atmosphere can absorb certain wavelengths. They can do this by storing energy in the molecules by stretching or bending (note only polar molecules can do this).

These gases, called ‘greenhouse gases’, serve as insulation to maintain the temperature of the Earth’s atmosphere. This is known as the ‘natural greenhouse effect’.

Some of the common greenhouse gases are carbon dioxide and methane and help to maintain a steady temperature in the Earth’s atmosphere, but water, ozone and halocarbons also contribute

Human activity that affects the concentration of greenhouse gases has the potential to disrupt the thermal balance of the atmosphere by increasing the concentration of green house gases. This is known as the ‘enhanced greenhouse effect’.

Greenhouse Gas Main sources Human activity

Carbon dioxide Burning fossil fuels Transport, power destruction of rain forests

Methane Anaerobic breakdown of plant material

Rice growing, cattle and sheep farming

Nitrous oxide Dinitrification of nitrates by microbes

Use of fertilizers

ASSIGNMENT 2.3: GREENHOUSE EFFECT.

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1. It has been suggested that an increase in the atmospheric concentration of certain gases will lead to global warming. Such warming is referred to as the Greenhouse Effect. However, other factors make it uncertain whether the increases in concentration of these gases will in fact lead to global warming.

(a) Explain briefly why this phenomenon is referred to as the Greenhouse Effect.

(b) Describe the means by which this effect is achieved.

(c) Name the 'greenhouse' gases.

2. Carbon dioxide is an important gas in the atmosphere even though it amounts to less than 1% of the air. It is continually being formed and consumed. However, there is concern that an increase in carbon dioxide concentration could have serious implications.

(a) Name the natural process that removes carbon dioxide from the atmosphere.

(b) Write a balanced equation for the natural process that removes carbon dioxide from the atmosphere.

(c) Describe two possible effects of an increase in carbon dioxide concentration in the atmosphere.

3. The natural cycle for carbon is shown below

(a) State two effects the burning of fossil fuels has had on the carbon cycle.

(b) There is concern that human activity threatens to lead to a warming of the Earth's atmosphere.

(i) State the name given to this effect.

(ii) Explain three ways in which human activity may have this effect.

(iii) Describe two possible consequences of this global warming.

4. South Australians contribute nearly 30 million tonnes of greenhouse gases to the atmosphere per year - equivalent to about 20 tonnes for every person in the State, with the main sources being electricity (27%), agriculture (21%) and transport (19%). South Australia's emissions account for only 7.4% of Australia's emissions, but per capita are close to the national average and among the highest in the world.South Australia: Reducing the Greenhouse Effect

(a) Suggest ways in which the South Australian community can reduce greenhouse gas emissions.

(b) What would happen to the earth's climate if there were no greenhouse gases in the troposphere?

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Subtopic 2.4: Acid Rain

Key Ideas Intended Student Learning

pH is a measure of the concentration of hydrogen ions where this is taken as the concentration of hydrogen ions relative to the standard 1 mol L1 , that is

pH = log10

(concentration of H+ ionsmol L– 1 )

commonly used as pH = log [H+].

Calculate the concentration of H+ of solutions, given their pH, and vice versa.

Rain containing dissolved carbon dioxide is acidic. Write equations to show how carbon dioxide produces acidic rain.

Rainfall with a pH of less than 5.6, known as ‘acid rain’, is formed when oxides of nitrogen and sulfur dissolve in water in the atmosphere.

Describe and write equations for the formation of acid rain.

The low pH of acid rain is due to the presence of sulfuric and nitric acids.

Calculate the pH of solutions of strong bases and strong monoprotic acids.

Rain is normally acid because it contains dissolved carbon dioxide which makes carbonic acid

CO2 + H2O H2CO3

The pH of such rain is normally no less than 5.6

pH is a measure of the concentration of hydrogen ions where this is taken as the concentration of hydrogen ions relative to the standard 1 mol L-1, that is

pH = -log10 (concentration of H+ ions (mol L-1))

commonly written as pH = -log [H+]

therefore note [H+] = log-1(-pH) or 10-pH

Rainfall with a pH of less than 5.6, known as ‘acid rain’, is formed when oxides of nitrogen and sulfur dissolve in water in the atmosphere and make acids.

Oxides of nitrogen

Oxides of nitrogen are formed when nitrogen and oxygen combine at high temperatures in industrial furnaces, internal combustion engines and jet engines to make nitric oxide (NO) when combines with oxygen to make nitrogen dioxide.

N2 + O2 2NO (nitric oxide)

2NO + O2 NO2 (nitrogen dioxide)

Oxides of sulfur

When compounds of sulphur burn they make sulfur dioxide, which combines with excess oxygen to make sulfur trioxide.

Compounds of sulphur + O2 SO2

2SO2 + O2 2SO3

Acid rain

These oxides then dissolve in rain water to make acids

NO2 + H2O HNO2 (nitrous) + HNO3 (nitric)

SO2 + H2O H2SO3 (sulfurous)

SO3 + H2O H2SO4 (sulfuric)

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It is the nitric and sulphuric acids that are the main causes of acid rain.

Effects of acid rain

Plants

Destroys foliage

Low pH can leach essential metal ions from the soil eg Mg, K

Can mobilise potentially poisonous cations eg Al, Pb, Zn

Animals

Mobilized toxic metal cations can get into the water supply

Affect fish especially eggs and fry

Structures

Attacks marble (calcium carbonate) buildings

CaCO3 + H2SO4 CaSO4 + CO2 + H2O

Corrodes metals especially iron

Calculating the pH of Strong Acids And Bases

Strong monoprotic acids fully ionize in water thus the [H+] is equal to the concentration of the acid.

The pH of strong bases is calculated from the expression

pH + pOH = 14

where pOH = -log10[OH-]

ASSIGNMENT 2.4: ACID RAIN.

1. Calculate the pH of the following solutions

(a) 0.10 mol L-1 H+

(b) 0.00000300 mol L-1 H+

(c) 2.00 mol L-1 H+

(d) 0.200 mol L-1 OH-

(e) 0.00000100 mol L-1 OH-

(f) 1.50 mol L-1 OH-

(g) 0.0500 mol L-1 HCl

(h) 0.00000300 mol L-1 HNO3

(i) 2.0 mol L-1 HCl

(j) 0.0500 mol L-1 NaOH

(k) 0.00300 mol L-1 Ca(OH)2

(l) 2.0 mol L-1 KOH

2. Calculate the hydrogen ion concentration in the following solutions

(a) pH = 5

(b) pH = 3.5

3. Calculate the hydroxide ion concentration in the following solutions.

(a) pH = 13

(b) pH = 10.5

4. The presence of sulfuric acid in waters draining from some mining sites is an environmental concern. Iron pyrite, FeS2, is a common component in waste heaps at mining sites in Australia. In the presence of certain bacteria that use sulfur-containing compounds as an energy source, the following redox reaction occurs:

2FeS2(s) + 2H2O(l) + 7O2(g) 2FeSO4(aq) + 2H2SO4(aq)

(a) To raise its pH, drainage water can be treated with crushed limestone (calcium carbonate). Write an ionic equation for the reaction between the limestone and the acid (H+) in the water.

(b) An alternative treatment involves introducing sulfate - reducing bacteria to the drainage water. One of the reactions that then occurs is shown by the following equation:

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2CH3CH(OH)COOH(aq) + SO42−

2CH3COOH(aq) + 2CO2(g) + H2S(g) + 2 OH(aq )

(i) Give a reason for the resulting increase in pH.

(ii) The equation above suggests that these bacteria are anaerobic decomposers. Give a reason to support this statement.

5. Acid rain contains sulfuric acid and nitric acid produced by human and industrial activity.Rain is considered to be acidic when its pH falls below 5.6. Rain falling in the urban area of Sydney (Australia) has an average pH of 4.5.

(a) Calculate the concentration of nitric acid in a rain sample of pH 4.5 if all the hydrogen ions in the sample are due to the presence of nitric acid.

(b) One effect of acid rain is that it attacks materials such as limestone, causing erosion of buildings and statues.

(i) Write the equation for the reaction between nitric acid and limestone.

(ii) Acid rain has several other damaging or harmful effects. Briefly describe two of them.

(c) Sulfur dioxide also contributes to acid rain. The concentration of sulfur dioxide in air can be monitored by pumping air through a dilute solution of hydrogen peroxide to form sulfuric acid. The net reaction that would occur is shown by the equation below:

SO2 + H2O2 H2SO4

By means of oxidation numbers, show that this is a reaction in which oxidation has occurred.

(iii) The concentration of the resulting sulfuric acid solution can be determined by titration with a solution of a base of known concentration. Suggest a suitable base for the titration and write the equation for the reaction that would occur during the titration.

6. All rain is acidic, but not all rain is called 'acid rain'.

(a) Explain the term 'acid rain'.

(b) Explain with an equation why rain is naturally acidic.

(c) (i) Outline briefly how oxides of sulfur are released into the atmosphere during the production of metals and the generation of electricity from fossil fuels.

Describe briefly, with the aid of equations, how the oxides of sulfur lead to the formation of 'acid rain'.

(d) (i) Explain briefly how oxides of nitrogen are released into the atmosphere.

(ii) Describe briefly, with the aid of equations, how oxides of nitrogen lead to the formation of 'acid rain'.

(e) (i) Biologists have noted that rain water in remote areas of Scandinavia is remarkably acidic. Explain how this is possible, since there is little population or industry in the region.

(ii) Explain why oxides of nitrogen and sulfur are readily spread in the atmosphere while other air pollutants such as soot remain near to the source.

(f) Explain, with the aid of an equation what happens when 'acid rain' comes in contact with steel structures such as statues or bridges.

(g) Outline briefly some of the long term consequences of 'acid rain' on the natural environment.

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Subtopic 2.5: Photochemical Smog

Key Ideas Intended Student Learning

Nitrogen oxides are formed in high-temperature engines and furnaces.

Write equations for the formation of nitrogen oxides NO and NO2.

Nitrogen oxides lead to the formation of ozone in the troposphere.

Describe and write equations showing the role of nitrogen oxides in the formation of ozone in the troposphere.

Nitrogen oxides and ozone in the troposphere are pollutants.

Explain the terms ‘primary pollutants’ and ‘secondary pollutants’ with reference to the harmful effects of nitrogen oxides and ozone in the troposphere.

It is possible to reduce the quantities of nitrogen oxides generated by cars.

Describe the effect of catalytic converters in reducing the quantities of nitrogen oxides generated by cars.

Photochemical SMOG is a form of pollution formed in the lower atmosphere over industrial cities. It appears as a pale brown haze and is an irritant to the human respiratory system.

The motor car is the major cause of photochemical smog. It produces nitric oxide and hydrocarbons which are called the primary pollutants. Intense sunlight initiates reactions involving these chemicals. The chemicals produced are called secondary pollutants.

Primary pollutants

Nitric oxide is formed within the combustion chamber of motor vehicles engines.

N2 + O2 2NO (nitric oxide)

Hydrocarbons are released into the atmosphere as unburnt petrol from cars.

Secondary pollutants

Nitric oxide can be converted to nitrogen dioxide, a secondary pollutant, by a number of different pathways. It is nitrogen dioxide that gives photochemical smog its brown colour.

It can react directly with oxygen, but this is quite slow.

2NO + O2 NO2 (nitrogen dioxide)

Faster reactions involve other oxidants and free radicals

2NO + O3 NO2 + O2

NO + HOO (peroxyl) NO2 + HO

Nitrogen dioxide is quite reactive and absorbs light (h is used to represent the energy involved) and dissociates into nitric oxide and atomic oxygen.

NO2 + h NO + O

The atomic oxygen then reacts with molecular oxygen to produce another secondary pollutant ozone. This reaction occurs faster if there is a stabilising molecule (M) like N2 or another O2.

O + O2 M⃗ O3

The level of ozone is constantly monitored as an indicator of the pollution level.

Hydrocarbons react with various other molecules and free radicals to produce numerous secondary pollutants including aldehydes and peroxyacyl nitrates (collectively called PAN).

Harmful Effects

Ozone and nitrogen oxides contribute to the Greenhouse effect.

Ozone can cause perishing of the rubber.

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Ozone above 0.15 ppm can cause respiratory problems.

Ozone reduces the ability of plants to photosynthesise.

Nitrogen dioxide also contributes to acid rain.

Reduction of photochemical smog

Catalytic converters have been fitted to all cars manufactured after 1986 to reduce pollution by up to 90%.

The catalytic converter is made up of a very thin layer of platinum, palladium, rhodium or iridium coated onto a honeycomb structure of large surface area.

The catalysts convert carbon monoxide to carbon dioxide.

2CO + O2 CO2

The catalysts convert unburnt hydrocarbons to carbon dioxide and water.

2C8H18 + 25O2 16CO2 + 18H2O

The catalysts converts nitric oxide to nitrogen by reaction like the one below.

2CO + 2NO CO2 + 2N2

ASSIGNMENT 2.6 : Photochemical Smog

1. Explain the difference between a primary and a secondary pollutant.

2. The oxides of nitrogen have a role in the formation of ozone in the troposphere.

(a) Explain with the aid of appropriate equations how oxides of nitrogen contribute to the formation of ozone in the troposphere.

(b) Explain why is ozone in the troposphere is considered to be a pollutant.

(c) Define pollution.

3. The nitric oxide released into the atmosphere in the exhaust gases from vehicles begins a sequence of reactions that eventually results in the formation of photochemical smog. The following equations show two reactions that occur:Reaction 1; 2NO(g) + O2(g) 2NO2(g)

Reaction 2: NO2(g) NO(g) + O2(g)

(a) Reaction 2 is a photochemical reaction. State the meaning of the term 'photochemical reaction'.

(b) Describe the formation of photochemical smog that follows from the above reactions.

4. During a low-level fly-past over the Adelaide Grand Prix circuit the exhaust gases emitted from an F111 aircraft were orange in colour. This orange colour may be attributed to an oxide of nitrogen.

(a) Name this oxide of nitrogen and give its chemical formula.

(b) Explain, with the help of equations, its formation in this case.

(c) Provide, with equations, two reasons why oxides of nitrogen are undesirable in the atmosphere.

5. Major pollutants in the exhaust gases of petrol-powered cars are carbon monoxide, unburned hydrocarbons, and oxides of nitrogen (nitric oxide and nitrogen dioxide). Over the past twenty years catalytic converters have been developed to remove these compounds from exhaust emissions.Some of the reactions that take place in these converters are:

2NO + 2CO N2 + 2CO2 (I)

2NO2 + 4CO N2 + 4CO2 (II)

2CO + O2 2CO2 (III)

(a) Explain with the aid of equations how nitric oxide and nitrogen dioxide come to be present in the exhaust gases from a petrol engine.

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(c) Identify the reducing agent in reaction (I) and explain your choice in terms of oxidation numbers.

6. The concentrations of four pollutants in the air of a large city were measured every four hours over a 24 hour period. The pollutants monitored were hydrocarbons, nitric oxide, nitrogen dioxide and ozone. The results are presented in the following graph:

(a) Of the four pollutants monitored name the primary pollutants.

(b) Explain why the concentration of hydrocarbons increases in the time period from 4 am to 9 am.

(c) Explain, with the aid of an equation, why the concentration of nitrogen dioxide increases, while the concentration of nitric oxide decreases, in the time period from 7 am to 9 am.

(d) Explain, with the aid of equations, why the concentration of nitrogen dioxide decreases, while the concentration of ozone increases, in the time period from 9 am to 12 noon.

Subtopic 2.6: Water Treatment

Key Ideas Intended Student Learning

Suspended matter is removed from water by flocculation followed by sedimentation or filtration.

Describe the use of aluminium ions in the removal of suspended matter from water.

Hypochlorous acid, chlorine, and hypochlorites are used for water purification.

State that hypochlorous acid, chlorine, and hypochlorites kill bacteria by their oxidising action.

Chlorine is used for water purification. Explain the effect of pH on the equilibrium between chlorine, water, and hydrochloric acid and hypochlorous acid.

A potable water supply is one that is drinkable. It must be free of toxic chemicals and disease carrying bacteria and have no undesirable odours, colours or tastes. Water used in swimming pools need similar properties.

Suspended matter

Water contains many suspended solids. Some of these are clay particles which don’t settle because they have a negative charge (like charges repel).

Suspended material can be forced to settle by the addition of aluminium ions (in alum). This causes the ions to clump together or flocculate and settle where they can be removed by sedimentation or filtration.

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Bacteria

Bacteria can be killed by hypochlorous acid (HOCl), chlorine (Cl2) and hypochlorites (OCl-). The are all oxidising agents and kill bacteria by oxidising them.

Chlorine reacts with water to produce the following equilibrium.

Cl2 + 2H2O HOCl + H3O+ + Cl-

Hypochlorous acid is the most effective bactericide because its small molecular size and lack of overall charge enable to penetrate the nonpolar walls of bacteria.

The amount of hypochlorous acid in water is pH dependent. If the pH increases, the hydronium ion is removed and the equilibrium shifts to the right to replace them producing more hypochlorous acid in the process. At ph 7, most of the chlorine as been converted to hypochlorous acid.

Hypochlorous acid is a weak acid and only partially ionises in water.

HOCl + H2O H3O+ + OCl-

Hypochlorous acid and the hypochlorite ions are referred to as ‘free available chlorine’ and are the oxidising agents that are most effective in killing bacteria.

Chlorine is too dangerous and corrosive to use in the back yard swimming pool, so sodium (NaClO) or calcium (Ca(ClO)2) hypochlorite are used to chlorinate pools, where they establish this equilibrium.

OCl- + H2O OH- + HOCl

The relative concentrations of the HOCl and OCl- ions are determined by pH

pH % of Cl as OCl-

% of Cl as HOCl

pH % of Cl as OCl-

% of Cl as HOCl

6.0 3.5 96.5 7.5 50.0 50.0

6.5 10.0 90.0 8.0 78.5 21.5

7.0 27.5 72.5 8.5 90.0 10.0

ASSIGNMENT 2.6: Water Treatment

Chlorine dissolves in water, according to the following equation:

Cl2 + H2O HC1 + HOC1

(a) State whether a solution of chlorine in water would have a pH greater than 7, equal to 7, or less than 7.

(b) When HOC1 acts as an acid it produces the hypochlorite ion. Write the formula of the hypochlorite ion.

(c) The hypochlorite ion is found in many household cleaners. State its function.

(d) Sodium hydroxide is added to a solution of chlorine in water. State, and give an explanation for, the effect this will have on the amount of Cl2 in the solution.

7. Chlorine, in the forms of chlorine gas, sodium hypochlorite or calcium hypochlorite, is used to disinfect water.

(a) Give the formulae of chlorine gas, sodium hypochlorite and calcium hypochlorite and state the oxidation number of chlorine in each.

(b) Explain why calcium hypochlorite is commonly used in home pools in preference to gaseous chlorine.

(c) If sodium or calcium hypochlorite are added to water the pH will be greater than 7. Write a balanced equation to show why this is so.

(d) In a pool the pH was 6.4. Determine the concentration of H3O+ in this pool.

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8. Swimming pool water must be tested for pH during active use. This is to maintain the pH at approximately 7.8. Pool-testing kits contain methyl red as an indicator and a colour chart for colour comparison to indicate the pH when the methyl red is mixed with pool water.

(a) Explain, giving an equation, why the pH of pool water must be tested after calcium hypochlorite is added.

(b) Name a suitable chemical that could be used to lower the pH if it is found to be too high.

(c) Explain why, when using the testing kit, it is important to compare the pH quickly after mixing the methyl red with pool water.

(d) Calculate the concentration (in mol L-1) of hydroxide ions in pool water when the pH is 7.8.

9. Some calcium hypochlorite was dissolved in a large volume of water producing a solution with a pH of 9.3.

(a) Explain, with the aid of an equation, why the pH of the solution is greater than 7.

(b) Calculate the concentration of hydroxide ions in the solution and hence, using the equation from (a) above, calculate the concentration of the hypochlorous acid.

(c) Explain how the concentration of the hypochlorous acid could be increased.

ANSWERS

SOLUTIONS 2.1 : Periodic Table

1. Complete the following table.

Symbol Electron configuration Group Period Block (s,p,d or f)

11Na 1s22s22p63s1 1 3 s

17Cl 1s22s22p63s23p5 5 3 p

20Ca 1s22s22p63s23p64s2 2 4 s

19K+ 1s22s22p63s23p64s0 1 4 s

35Br- 1s22s22p63s23p64s23d104p6 7 p

26Fe3+ 1s22s22p63s23p64s23d104p6 - 4 d

15P3- 1s22s22p63s23p6 5 3 p

2. (a) Hydrochloric acid

(b) NaCl

(c) LiCl

(d) KCl

3. Ga2O3.

4. 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) (less vigourous)2K(s) + 2H2O(l) 2KOH(aq) + H2(g) (more vigourous)

5.

Group

I II III IV V VI VII VIII

P

e

r

i

o

d

1 +1 0

2 +1 +2 3 4 3,+5 -2 -1 0

3 +1 +2 +3 4 3, +5 2,+4,+6

1,+3,+5,+7

0

4 +1 +2

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6. Complete the following table by writing the formulae of the chlorides of each of the elements.

Group

I II III IV V VI VII VIII

Period

1 HCl -

2 LiCl BeCl2 BCl3 CCl4 NCl3 OCl2 FCl -

3 NaCl MgCl2 AlCl3 SiCl4 PCl3 SCl2 Cl2 -

4 KCl CaCl2

7. (a) Ionic bonds form when metals transfer outer electrons to non-metals, giving both a stable configuration. The bond forms because of the strong electrostatic forces between + and – charged ions. eg. NaCl

(b) Covalent bonds form when non-metal atoms share electrons, so forming a stable configuration. An electrostatic force of attraction between a nucleus and the shared electrons holds these atoms together. eg. H2O

8. (a) The measure of an atom’s power to attract electrons to itself in a bond.

(b) Ionic bonding present because of the high electronegativity difference between Mg atoms and O atoms.

(c) Mg2+ O2- is much stronger than Na+Cl- (F a q1.q2) therefore in theory MgO is 4 times stronger than Na+Cl-, if the ionic radii are similar (they are).

(d) Metal conductivity possible in solid and liquid; ionic conductivity only possible when ions mobile, i.e. as a melt or in solution.

(e) Structural differences: SiO2(s) continuous covalent, CO2 molecular lattice. Why? Because of atom size and electronegativity. Think of a volatile SiO2! What would this have done to our planet?

9. (a) A polar bond is a covalent bond in which the electrons are shared unequally. This results in minute

charges on the atoms, eg.d + H−Od −

(b) (i) The wider the separation of the elements in the periodic table the greater the ionic character of the bonds between them. eg. Cl2 (pure) covalent non-polar, PCl3 polar covalent, KCl ionic

(ii) The number of valence electrons determines the charge on the ion or the number of covalent bonds that can be formed.

(c) The symmetrical nature of this molecule means that the polarity of its 4 C-F bonds is cancelled out.

10. Increasing molar mass implies increase size of molecules with more atoms, therefore increased dispersion forces between molecules increases the Tb.

11. Structural formulae should be determined from electron dot formulae.

12. Symmetry determines. SO2 is bent, hence polar as has polar bonds and is assymetrical, but CO2 has polar bonds but is symmetrical molecule, therefore is non-polar.

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13. (a) See above.

(b) There are two bonds present. The smaller 198 (electric dipole moment) is for the Cl-Cl covalent bond, the larger is the distance between Cl2 molecules, i.e. for the secondary bonds.

14. (a) NaCl, MgCl2, AlCl3, PCl3, PCl5, SCl2

(b) (i) From left to right goes from ionic to polar covalent

The outer electrons increase 3s1 3s23p4. Thus electrons available for bonding increase from 1 6.

15. (a) A is F 1s2 2s2 2p5, B is K 1s2 2s2 2p6 3s2 3p6 4s1, C is Se 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4

(b) (i) AB is ionic

CA2 is polar covalent

(i) B2C is ionic

(c) diagrams required

16. A hydrogen bond is a secondary bond that forms between small highly electronegative atoms (N, O, F) and hydrogen so forming slight electrostatic forces of attraction between molecules.

17. (a) H2O, O more electronegative than S

(b) HF, F more electronegative than Cl.

(c) no hydrogen bonding, Cl too big

(d) H2O, O more electronegative than N

18. Hydrogen bonding increases the forces of attraction between H2O and NH3 molecules.(diags)

19. (a)

Name Structure Bonding

Iron Metallic lattice Metallic

Iodine Molecular lattice Covalent in molecule, Dispersion between molecules

Diamond Giant atomic covalent lattice Covalent

Sodium chloride Ionic lattice Ionic

Ice Molecular lattice Polar covalent in molecule, hydrogen bonds between molecules

Iron. The close packing of ions in the metallic lattice is due to ‘cementing’ action of mobile electrons. Iodine. Induced polarities between the iodine molecules holds them together in the solid state at room temp. Diamond. The continuous covalent bonding gives a 3D lattice.Sodium chloride. The regular arrangement of ions produces cubic crystals.Ice. The spacious tetrahedral arrangements of the water molecules is due to H bonding.

(a) Iron is dense, hard, high MPt and readily conducts in the solid state. Explained by its structure and bonding.Iodine is a low MPt (sublimes) soft solid, non-conductor. It has weak dispersion forces and no free electrons.Diamond is very hard, high Mpt, non-conductor. Covalent bonding is very strong throughout the lattice.Sodium chloride is crystalline, high MPt, conducts in solution and when molten. Ions are regular in space, bonding is throughout the lattice. Ions free to move in solution and when molten.Ice is low MPt, lighter than water and low conductivity. Hydrogen bonds are weak. No ions or electrons are free to move.

SOLUTIONS 2.2: Cycles in Nature

1. (a) Aerobic conditions are those where free or molecular oxygen (O2) is available.

(b) Anaerobic conditions are those where there is no free oxygen. Often called stagnant.

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(c) (i) Carbon gives carbon dioxide (CO2) or carbonates (CO32−

) in aerobic conditions

(ii) Sulfur gives sulfates (SO42−

)

(iii) Phosphorus(PO43−

)

(iv) Nitrogen (NO3−

)

(d) (i) Carbon gives methane (CH4) in anaerobic conditions

(i) Sulphur gives hydrogen sulfide (H2S)

(ii) Phosphorus gives phosphine (PH3)

(e) Nitrogen gives ammonia (NH3)

2. (a) 6CO2+6H2 O⃗ sunlight/chlorophyll C6 H12O6+6O2

(a) Chlorophyll

(b) Makes free oxygen, makes food from raw materials (Start of all food chains)

3. (a) C6 H12O6 + 6 O2 → 6 CO2 + 6 H2 O

Needs free oxygen

4. (a) nitrogen fixation

(b) Bacteria in the roots of leguminous plants take nitrogen from the air.

(c) Too stable

(d) Extreme temperatures (at least 800C) – lightning, volcanoes, high temperature furnaces, car engines.

(e)N2 + O2⃗ high temperature 2 NO

(nitric oxide)2NO + O2 2NO2

2NO2 + H2O HNO2 (nitrous acid) + HNO3

5. (a) Nitrates, sulfates and phosphates

(b) The plants need water soluble compounds

SOLUTIONS 2.3 : Greenhouse Effect1. (a) Heat coming into the atmosphere from the sun is trapped by gases in the atmosphere in much the same

way that infrared radiation is trapped by the glass in a greenhouse.

(b) Short wavelength infrared radiation from the sun passes through the atmosphere and is absorbed on the surface. As the surface warms up, it radiates infrared radiation back into the atmosphere at a longer wavelength which is absorbed by greenhouse gases which warms up the air (by about 30C). The heat is stored in the greenhouse gases because they start to vibrate faster when hit by the radiation.

(c) Common greenhouse gases include water, carbon dioxide, methane and CFC’s

2. (a) Photosynthesis

6CO2 + 6H2O s⃗unlight /chlorophyll 6C6H12O6 + 6O2

(a) This global warming trend can cause a significant global climate changes. Human society is highly dependent on the Earth's climate. Climate patterns and human adaptations determine the availability of food, fresh water, and other resources for sustaining life. The social and economic characteristics of society have also been shaped largely by adapting to the seasonal and year to year patterns of temperature and rainfall. Some potential effects associated with climate change are listed below.

Water Resources-The quality and quantity of drinking water, water availability for irrigation, industrial use, and electricity generation, and the health of fisheries may be significantly affected by changes in precipitation and increased evaporation. Increased rainfall may cause more frequent flooding. Climate change would likely add stress to major river basins worldwide. Probable cause of the El Niño effect.

Coastal Resource - A estimated 50 cm rise in sea level by the year 2100, could inundate low lying areas

Health - Heat-stress mortality could increase due to higher temperatures over longer periods. Changing patterns of precipitation and temperature may produce new breeding sites for pests, shifting the range of infectious diseases.

Agriculture - Impacts of Climate change in developing countries could be significant.

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Forests - Higher temperatures and precipitation changes could increase forest susceptibility to fire, disease, and insect damage.

Energy and Transportation - Warmer temperatures increase cooling demand but decrease heating requirements. Fewer disruptions of winter transportation may occur, but water transport may be affected by increased flooding or lowered river levels.

3. Burning fossil fuels can cause the speed up of photosynthesis by increasing the concentration of carbon dioxide in both forests and in the ocean where lot of carbon dioxide dissolves.

4. (a) Enhanced greenhouse effect

(b) Burning fossil fuels releasing carbon dioxide.Increased agricultural production releases more methane into the atmosphere from herd of sheep and cattle and rice growingCutting down forests removes a carbon sink

(c) See answer above

5. (a) Reduce the use of cars which release carbon dioxide.Use less electricity, because fossil fuels are used to make electricity.

(b) If there were no greenhouse gases the earth would be about 30C cooler

SOLUTIONS 2.4: Acid Rain.

(a) pH = -log[H+] = -log0.1 = 1

5.5

(c) -0.3

(d) pH=14– pOH = 14 – (-log(0.2) = 13.3

(e) 8

(f) 14.2

(g) 1.3

(h) 5.5

(i) -0.3

(j) 12.7

(k) 11.8 (nb [OH] = 2 x .003)

(l) 14.3

6. (a) [H+] = log-1(-pH) = log-1(-5) = 1.0 x 10-5 mol L-1

(a) 3.2 x 10-4 mol L-1

7. (a) [OH] = log-1(14 – pH) = log-1 -(14-13) = 1 x 10-1 mol L-1

3.2 x 10-4 mol L-1

8. (a) CaCO3 + 2H+ Ca2+ + CO2 + H2O

(i) [OH] increasing.

(i) No free oxygen

9. (a) 3.2 x 10-5 mol L-1

(i) HNO3 + CaCO3 Ca(NO3)2 + CO2 + H2O

(i) Acid rain attacks metals, affects plants by leaching valuable metals like calcium and potassium and freeing up poisonous ones like aluminium, kills young fish.

(b) (i) S+4

O2+H2O-1

2→H2 S+6

O−2

4

(i) Sodium hydroxide (but sodium carbonate would also do)2NaOH + H2SO4 Na2SO4 + 2H2O

10. (a) Acid rain is rain with a pH less than 5.6

(b) CO2 + H2O 2H+ + CO32−

(c) (i) Compounds (eg metal sulfide ores and coal and oil) containing sulfur produce sulfur dioxide when burnt.

(i) Sulfur dioxide reacts with oxygen to make sulfur trioxide2SO2 + O2 2SO3

when sulfur dioxide or sulfur trioxide dissolve in water, they make acidsSO2 + H2O H2SO3 (sulfurous acid)SO3 + H2O H2SO4 (sulfuric acid)

(d) (i) Oxides of nitrogen are released into the air when oxygen and nitrogen combine at high temperatures to make nitric oxide.N2 + O2 2NO

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The nitric oxide then combines with oxygen to make nitrogen dioxide.2NO + O2 2NO2

(i) When nitrogen oxides (NOx)dissolve in rain water, they make acids.NO2 + H2O HNO2 (nitrous acid) + HNO3 (nitric acid)

(e) (i) The acidic gases (NOx and SOX) are carried in by prevailing winds from industrialised areas.

(f) Oxides of nitrogen and sulfur are gases whereas soot is a solid that fall back to the ground.

(g) Acid corrodes metals eg iron|Fe + 2H+ Fe2+ + H2

11. Effects of acid rain

(a) Corrode metal and marble structures

(b) Kills plants by mobilising poisonous metals (Al, Cd ) and leaching valuable ones (K, Ca)

(c) Kills fish sprats and roe in waterways

SOLUTIONS 2.6: Photochemical SmogPrimary Pollutants - Air pollutants that enter the atmosphere directly.

Secondary pollutants Atmospheric pollutants that are created chemically in the atmosphere when primary pollutants and other components of the air react.

12. (a) Nitogen dioxide decomposes when exposed to ultraviolet radiation and releases an oxygen radicalNO2 + hv → NO + OThe oxygen radical then reacts with oxygen to make ozoneO + O2 → O3

Ozone is a pollutant because it causes respiratory problems and accelerates deteriation of rubber amongst other problem.

(a) Pollution is when a harmful chemical is present in concentrations higher than normal.

13. (a) A photochemical reaction is a reaction that requires light.

(b) Motor vehicles produce exhaust gases containing oxides of nitrogen such as nitrogen dioxide (NO2) and nitric oxide (NO).At the high temperatures of the car's combustion chamber (cylinder), nitrogen and oxygen from the air react to form nitric oxide (NO):

N2(g) + O2(g) 2NO(g)

Some of the nitric oxide (NO) reacts with oxygen to form nitrogen dioxide (NO2):

2NO(g) + O2(g) 2NO2(g)

The mixture of nitric oxide (NO) and nitrogen dioxide (NO2) is sometimes referred to as NOx.(c) When the nitrogen dioxide (NO2) concentration is well above clean air levels and there is plenty of

sunlight, then an oxygen atom splits off from the nitrogen dioxide molecule:

NO2(g)s⃗unlight NO(g)+O(g)

This oxygen atom (O) can react with oxygen molecules (O2) in the air to form ozone (O3):

O + O2 O3

Nitric oxide can remove ozone by reacting with it to form nitrogen dioxide (NO2) and oxygen (O2):

NO(g) + O3(g) NO2(g) + O2(g)

When the ratio of NO2 to NO is greater than 3, the formation of ozone is the dominant reaction. If the

ratio is less than 0.3, then the nitric oxide reaction destroys the ozone at about the same rate as it is

formed, keeping the ozone concentration below harmful levels.

The reaction of hydrocarbons (unburnt petrol) with nitric oxide and oxygen produce nitrogen dioxide

also in the presence of sunlight, increasing the ratio of nitrogen dioxide to nitric oxide.

Peroxyacetylnitrate (PAN) Production

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Nitrogen dioxide (NO2), oxygen (O2) and hydrocarbons (unburnt petrol) react in the presence of sunlight

to produce peroxyacetylnitrate (CH3CO-OO-NO2):

NO2(g) + O2(g) + hydrocarbons s⃗unlight CH3CO-OO-NO2(g)

14. (a) Nitrogen dioxide (NO2)

In the engine, high temperatures allow nitrogen and oxygen to combine to make nitric oxideN2 + O2 2NOThe nitric oxide reacts with oxygen in the air becoming nitrogen dioxide.2NO + O2 2NO2

(a) Oxides of nitrogen contribute to acid rain (NO2 + H2O → HNO2 + HNO3)Contribute to the formation of ozone and photochemical smog (see above)

15. (a) In the engine, high temperatures allow nitrogen and oxygen to combine to make nitric oxideN2 + O2 2NOThe nitric oxide reacts with oxygen in the air becoming nitrogen dioxide.2NO + O2 2NO2.

A catalyst speeds up (in this case) or slows down a reaction without being used up itself.

(a) 2 N+2

O+2 C+2

O →N0

2+2 C+4

O2

C is oxidized (+2→+4), it must be the reducing agent that reduces nitrogen (+2→0)Alternatively, CO combines with oxygen, and therefore is oxidized, making it the reducer.

16. (a) Hydrocarbons and nitric oxide.

Lots of people driving their cars to work etc in the morning releasing car fumes.

(a) Nitric oxide decreases as it reacts with oxygen in the air.(2NO + O2 2NO2)

(b) Nitrogen dioxide then decomposes when the sun comes out making ozone.

NO2 + hv → NO + O

The oxygen radical then reacts with oxygen to make ozone

O + O2 → O3

SOLUTIONS 2.6: Water Treatment(a) pH would be less than 7 because the equilibrium produces HCl and HOCl (both acidic).

Cl2 + H2O HC1 + HOC1

(c) OCl-

(d) OCl- is an oxidizing agent that can be used to kill bacteria and remove stains by oxidation.

(e) Sodium hydroxide consumes HCl and HOCl and so the equilibrium moves to the right, reducing the amount of chlorine.

17. (a) Chlorine (Cl0

2 ), sodium hypochlorite (NaOCl+1

), calcium hypochlorite (Ca (O Cl+1

)2 ).

Calcium hypochlorite is more convenient (a solid) and safer than gaseous chlorine.

(a) OCl- + H2O HOCl + OH- . Since this reaction produces the hydroxide ion, the pH rises above 7.

(b) pH = 6.4

[ H3O+ ]=log−1 (−pH )=log−1 (−6 . 4 )=4 . 0 x10−7 mol L−1

18. (a) Swimming pool water must be tested for pH during active use. This is to maintain the pH at approximately 7.8. Pool-testing kits contain methyl red as an indicator and a colour chart for colour comparison to indicate the pH when the methyl red is mixed with pool water.When calcium hypochlorite is added to the pool, it ionizes to make the solution basic.OCl- + H2O HOCl + OH-

Hydrochloric acid

(a) Hypochlorite is a bleach, and will decolourise the indicator if left too long

(b) pH = 7.8

[OH− ]=log−1(−pOH )=log−1 (−(14−7 .8 ))=6 .3 x10−7mol L−1

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19. (a) When calcium hypochlorite is added to the pool, it ionizes to make the solution basic.OCl- + H2O HOCl + OH-

pH = 9.3

[OH− ]=log−1(−pOH )=log−1 (−(14−9 .3 ))=2 .0 x10−5 mol L−1

Since the mole ratio is 1:1, the concentration of hypochlorous acid is also 2.0 x 10-5 mol L-1.

(a) To increase the concentration of the hypochlorous acid the equilivbrium needs to be forced to the right either by adding extra calcium hypochlorite or acid to remove the hydroxide which pulls the equilibrium to the right.

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