Ionic & covalent bonding - Jak and Corinne

PHYSICAL SCIENCES LEVEL 3C

(PSC315115)

IONIC &

COVALENT BONDING

THEORY SUMMARY & REVISION QUESTIONS

Tasmanian TCE Physical Sciences Revision Guides by Jak Denny are licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International Licence.

© JAK DENNY

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INDEX:

PAGE

• INTRODUCTION 3

• SUMMARY OF BONDING TYPES 3

• IONIC BONDING 4

• STRUCTURE OF IONIC COMPOUNDS 5

• IONIC COMPOUNDS - QUESTIONS 6

• IONIC COMPOUNDS - PROPERTIES 7-8

• COVALENT COMPOUNDS 9

• COVALENT MOLECULAR SUBSTANCES 10

• ELECTRON DOT ‘LEWIS’ DIAGRAMS 11

• VAN DER WAAL’S FORCES 12

• MOLECULAR SUBSTANCES - QUESTIONS 13

• COVALENT NETWORK SUBSTANCES 14-15

• BONDING TYPES – DECISION TABLE 16

• BONDING REVISION QUESTIONS 17-18

• BONDING TYPES – A SUMMARY 19

• SAMPLE TEST ON BONDING & CRYSTALS 20-21

• BONDING & CRYSTALS TEST ANSWERS 22-23

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INTRODUCTION: When two or more atoms combine chemically, they are said to have formed CHEMICAL BONDS. The branch of chemistry that investigates this process is called chemical bonding. In our earlier unit of work we encountered METALLIC BONDING where metal atoms are bonded by the electrical attractive forces existing between positive metal ions (held in a rigid lattice) and delocalised, mobile, negatively charged electrons. These attractive forces or ‘metallic bonds’ hold the metallic structure together and give rise to the properties that are typical of metals such as malleability, electrical conductivity, strength, lustre,…….….. We shall now consider other types of chemical bonding where the atoms combining may be metallic or non-metallic elements. When atoms bond together, they do so by way of their respective electron clouds (ORBITALS) coming together and overlapping, resulting in there being a rearrangement in the locations of the outer most (VALENCE) electrons. Generally, atoms chemically combine by one of two electron change processes: i.e. • The simultaneous loss and gain of electrons (described as IONIC BONDING) • Combining atoms share electrons (described as COVALENT BONDING) As a convenient general rule, you can usually distinguish between these two main types of chemical bonding by the following: IONIC BONDING occurs when METALS chemically combine with NON-METALS e.g. Na2O, CuSO4, Al(NO3) 3, MgS, BaCl2, Fe2O3, KHCO3,………….

COVALENT BONDING occurs when NON-METALS chemically combine with NON-METALS e.g. N2, O2, Cl2, SO2, NF3, H2S, CF4, P4O10, CH3COOH, organic compounds,……

SUMMARY OF BONDING TYPES:

COMBINING ELEMENTS BONDING TYPE EXAMPLES Metals with metals Metallic bonding Cu, Fe, Sn, alloys,…..

Metals with non-metals Ionic bonding NaCl, FeSO4, K2CO3,…….... Non-metals with non-metals Covalent bonding H2O, HNO3, SBr2, I2,…….…

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IONIC BONDING: Ionic bonding occurs when metallic elements chemically combine with non-metallic elements. The metallic element LOSES electrons and becomes a positive ion (cation), whereas the non-metallic element GAINS electrons and becomes a negative ion (anion). Thus, the basis of ionic boding is the electrostatic attractive forces that occur between positive and negative ions. e.g. Consider the ionic compound sodium chloride (NaCl) Sodium (metal) atoms have the electron configuration: Na = 2)8)1 Chlorine (non-metal) atoms have the electron configuration: Cl = 2)8)7 When sodium reacts with chlorine, each sodium atom loses one electron and each chlorine atom gains one electron; i.e. Each Na atom LOSES 1e− and forms a sodium ion; Na+ = 2)8 Each Cl atom GAINS 1e− and forms a chloride ion; Cl− = 2)8)8 Note that in this process of electron loss and gain, both atoms achieve the stable energy condition associated with a completed outer electron energy level (“shell”). In this case the filled outer shell of 8 electrons is called a ‘stable-octet’ of electrons. In some cases, an outer complement of 2 or even 18 electrons is a stable state. Q1. Complete the following table by showing the electron shell arrangements. The first line has been completed for you:

ATOM CONFIGURATION

ION CONFIGURATION

ION TYPE

Al = 2)8)3 Al3+ = 2)8 cation K = K+ = O = O2− = Mg = Mg2+ = P = P3− = S = S2− = Ca = Ca2+ = F = F− = Li = Li+ = Br = 2)8)8)17 Br− = Rb = 2)8)8)18)1 Rb+ =

Q2. Complete the following statements by inserting the missing words. In the above table it can be seen that the metallic elements …………….. electrons and thus form ………ions whereas non-metallic elements ……….. electrons and thus form ……..……ions. Q3. Which of the following compounds are likely to involve ionic bonding? NiCl2, SiF4, ZnO, SeBr2, NF3, SrF2, K2Cr2O7, PI3 and gold metal (Au).

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THE STRUCTURE OF IONIC COMPOUNDS: The main features of ionically bonded compounds are given below:

• Ionic compounds are solids at room temperature. They have generally high to very high melting temperatures; e.g. NaCl(s) melts at 801oC.

• Ionic compounds consist of positively charged metal cations and negatively charged non-metal anions.

• The positive and negative ions occupy fixed positions in a regularly packed array called a ‘crystal lattice’. The only movement the ions have is vibrational motion.

• Each positive ion is surrounded by a fixed number of negative ions and similarly, each negative ion is surrounded by a fixed number of positive ions.

• The electrical charge on the positive ions cancels out the charge on the negative ions giving an overall charge of zero.

• The forces binding the ionic crystal lattice together are the strong electrical attractions existing between each ion and its neighbouring oppositely charged ions. This is the very basis of IONIC BONDING.

• Even though we write the chemical formula for an ionic compound as for example, Na2O, this does not mean that there are individual molecules of Na2O. What the formula Na2O tells us is that in the solid crystal lattice of sodium oxide there is a continuous packing of ions where the number of sodium ions is twice the number of oxide ions.

• Ionic crystals are thus described as CONTINUOUS lattices.

TWO AND THREE DIMENSIONAL REPRESENTATIONS OF SOLID NaCl

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Q4. Compare the number of positive ions with the number of negative ions in the ionic lattice comprised of the following. The first example has been completed for you.

CATIONS

ANIONS

RATIO OF CATIONS:ANIONS

IONIC FORMULA

Mg2+ Cl− 1:2 MgCl2 K+ F−

Al3+ Br− Mg2+ O2− Cr3+ S2− Na+ PO4

3− Al3+ SO4

2− Ag+ CH3COO− Li+ Cr2O7

2− Q5. Although most ionic compounds involve metal ion and non-metal ion combinations, there are some ionic compounds that do not contain any metallic ions. One important example is the polyatomic AMMONIUM ION (NH4

+). Give the chemical formula for the following ionic compounds. (i) ammonium nitrate ………………………………………………

(ii) ammonium sulfate ………………………………………………

(iii) ammonium phosphate ………………………………………………

(iv) ammonium chromate ………………………………………………

Q6. Some metals are able to form more than one type of ion. For example, nickel forms both 2+ and 3+ ions. When this occurs, chemists distinguish between the compounds using Roman numerals in the name to identify the ionic charge, e.g. Ni2+ and Cl− combination gives NiCl2 which is nickel(II) chloride Ni3+ and Cl− combination gives NiCl3 which is nickel(III) chloride Complete the following table:

CHEMICAL FORMULA CHEMICAL NAME FeSO4

Fe2(SO4) 3 copper(I) oxide copper(II) oxide

SnCl2 SnCl4

mercury(I) nitrate mercury(II) nitrate

Q7. How is a phosphide ion (P3−) (i) similar to (ii) different from a phosphorus atom (P)? Q8. How many protons and electrons are there is a hydrogencarbonate ion (HCO3

−)?

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PHYSICAL PROPERTIES OF IONIC COMPOUNDS: 1. IONIC COMPOUNDS HAVE HIGH MELTING POINTS AND HIGH BOILING POINTS. This is explained by the fact that ionic bonds involve strong electrical attractions between oppositely charged ions and thus require a large amount of thermal energy to be broken.

e.g. NaCl(s) melts at 801oC.

Q9. By considering the size of the attractive forces between ions, predict which of the two ionic compounds potassium chloride (KCl) and calcium sulfide (CaS) would have the higher melting temperature.

2. IONIC COMPOUNDS ARE HARD AND BRITTLE. This is explained by the high strength of the ionic bonds meaning the ions are strongly bound within the crystal lattice and not easily displaced. Under a shearing force however, distortion of the ionic crystal lattice causes ions of the same charge to come closer and their repulsion causes the crystal to shatter. (see diagram below)

3. IONIC COMPOUNDS DO NOT CONDUCT ELECTRICITY IN THE SOLID STATE. This is because there are no mobile charged particles to transfer electrical charge.

• There are no free electrons as all e− are localised (fixed) within the ions. • The ions are strongly held within the crystal lattice and cannot carry charge

because no movement of the ions can take place.

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4. IONIC COMPOUNDS DO CONDUCT ELECTRICITY WHEN IN AQUEOUS SOLUTION OR IN THE MOLTEN STATE (WHEN THEY HAVE BEEN MELTED) The process of dissolving an ionic compound in water or melting an ionic compound will result in the solid ionic lattice being broken apart and destroyed. The ions separate from each other and are now able to move randomly. We say the ions now possess mobility. When an electrical potential is applied, positive ions move towards the negative electrode and negative ions move towards the positive electrode; (remember, opposites attract). This movement of both types of ion in opposite directions is a net ‘movement of electrical charge’ and is called an electric current.

Q10. In the following table, identify which species would be electrical conductors and indicate the particles that carry the electrical charge if they do conduct electricity. The first two rows have been completed as examples.

CHEMICAL SPECIES TESTED

ELECTRICAL CONDUCTOR?

MOVING CHARGED PARTICLES

CuSO4(aq) yes positive & negative ions Ag(s) yes electrons KBr(s) NaI(l) Zn(l)

NaOH(s) KOH(aq)

Co(s) CaCl2(l)

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COVALENT BONDING: Covalent bonding usually occurs when non-metallic elements chemically combine with non-metallic elements. Covalent bonding results from adjacent atoms SHARING electrons. Covalent bonds are the attractive forces exerted by the positive nuclei of the adjacent atoms on the shared electrons. In this unit of study we will focus principally on the covalent bonding that occurs where combining atoms share electrons so as to achieve stable octets of electrons. The exception to this octet rule is hydrogen which achieves stability by gaining an outer complement of two electrons. Knowing which group of the Periodic Table the non-metals are from, gives a good indication of the electron sharing process. The Group number (in Roman numerals) tells us the number of outer or ‘VALENCE’ electrons the atom possesses. This then tells us how many more must be shared to bring the total up to 8. For example, nitrogen (N) is a non-metallic element in Group V (or Group 15) of the Periodic Table. Using the Roman numeral, we ascertain that each nitrogen atom has 5 outer (valence) electrons. (These are the electrons that determine the bonding possibilities for nitrogen). We then represent the 5 valence electrons as 5 DOTS or sometimes crosses around the symbol placing 4 dots individually before pairing any up. For nitrogen, this gives us the 5 valence electrons as one pair and 3 single or ‘unpaired’ electrons. It is these three unpaired valence electrons that tell us that nitrogen atoms can form 3 covalent bonds by sharing 3 electrons with other atoms.

P.TABLE GROUP

NUMBER

EXAMPLE ELEMENT

NUMBER OF VALENCE

ELECTRONS

NUMBER OF UNPAIRED

ELECTRONS

NUMBER OF COVALENT BONDS NORMALLY FORMED PER ATOM

IV carbon 4 4 4 V nitrogen 5 3 3 VI oxygen 6 2 2 VII fluorine 7 1 1 VIII neon 8 0 0 Q11. In the table below, draw in the valence electrons corresponding to that element and thus decide the likely number of covalent bonds the element can form.

S C Br P Si Cl Ar O

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COVALENT MOLECULAR SUBSTANCES: The process of combining atoms sharing electrons so as to achieve “stable octets” of electrons results in the formation of covalent molecules or covalent networks (see p. 15) Our main concern will be with the formation of covalent molecular substances. EXAMPLES: (i) Consider the compound formed between the two non-metallic elements phosphorus and chlorine. What will be the chemical formula of the compound formed? We use the ‘electron-dot’ method to find the answer.

These atoms thus combine by way of sharing unpaired valence electrons. This means that three chlorine atoms will share with one phosphorus atom as shown below:

This gives us the clue that the compound formed will be PCl3 (phosphorus trichloride). This is a covalent molecular substance in which the molecules of 4 atoms are held together by 3 single covalent bonds. Each covalent bond is indicated with a dash “―”. (ii) Where two pairs of electrons are shared between the same atoms, it results in the formation of a double covalent bond. For example, consider oxygen gas O2. Each oxygen atom has 6 valence electrons with there being 2 unpaired valence e− per oxygen atom. i.e.

(iii) Where three pairs of electrons are shared between the same atoms, it results in the formation of a triple covalent bond. For example, consider nitrogen gas N2. Each nitrogen atom has 5 valence electrons with there being 3 unpaired valence e− per nitrogen atom. i.e.

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LEWIS DIAGRAMS: Electron-dot diagrams are often referred to as LEWIS DIAGRAMS and are used to represent the sharing of valence electrons in molecular substances. A common feature of the Lewis diagrams that you will be asked to draw is that through the process of electron sharing, all atoms in the molecule achieve the stable condition of possessing the ‘stable octet’ of 8 electrons around the atom. (The exception is hydrogen, which usually achieves stable state with 2 electrons only) Q12. In the spaces provided, draw Lewis diagrams for each of the following substances. CCl4

F2 CO2 NBr3

NH3

H2 SiH4 H2O

H2O2

C2H6 PI3 SCl2

Q13. Draw the electron dot diagrams for: (i) CH3 (ii) CH4 (a) What particular feature does CH3 possess that is not possessed by CH4? (b) CH3 is highly reactive and yet CH4 is much more stable. Explain. (c) Two separate CH3 molecules are likely to join to form a single C2H6 molecule. Explain why this produces a much more stable product.

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VAN DER WAAL’S FORCES: During the formation of a covalent molecular substance like CF4, all 5 atoms present in the molecule achieve the stable octet state and have all their valence electrons paired up. Remember that unpaired valence electrons give rise the potential formation of covalent bonds! Thus, in a molecule such as CF4 with there being no unpaired e−, the bonding capacity of the molecule is fully utilised. Consider a sample of solid CF4. What bonds hold the solid together? There are (4) strong covalent bonds within the CF4 molecule but between the molecules of CF4 are weak attractive forces called VAN DER WAAL’S FORCES or DISPERSION FORCES. These forces are thought to be due to electrical effects where slightly positive and negative regions form momentarily in the molecule. Van der Waal’s forces are weak compared to covalent bonds and thus covalent molecular substances melt easily and boil easily. The van der Waal’s forces may be so weak that the substance is already a gas at room temperature; e.g. O2, CH4, CO2,……….... F Consider a molecule of CF4 اا

F − C − F اا F Within the molecule there are the 4 strong covalent bonds (these are called INTRAMOLECULAR BONDS) In a crystal of solid CF4 the forces between the molecules are weak van der Waal’s forces. (these are also referred to as DISPERSION FORCES or INTERMOLECULAR FORCES)

The dotted lines in the representation above indicate the weak van der Waal’s forces between molecules of CF4. Van der Waal’s forces vary in strength from ‘fairly weak’ to ‘very weak’ and will generally be stronger for bigger molecules. The lower the melting point, the weaker the forces!

e.g. F2 melting point = −220oC (extremely weak van der Waal’s forces) Cl2 melting point = −101oC Br2 melting point = −7oC I2 melting point = +114oC (fairly weak van der Waal’s forces)

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Q14. A chemist is given samples of two different white crystalline solid compounds to analyse. The solids are labelled “A” and “B”. She carries out the following tests in the laboratory: TEST 1: She tests the electrical conductivity of both solids. Neither conducts

electricity.

TEST 2: She attempts to dissolve them in water. Both dissolve readily.

TEST 3: She tests the electrical conductivity of the aqueous solutions. Solution B(aq) does conduct electricity but A(aq) doesn’t. TEST 4: She measures the melting temperature of either solid. Solid A melts at 87oC whereas solid B melts at 1134oC.

(i) What type of bonding is likely to be present in solid A? Explain your reasoning. (ii) What type of bonding is likely to be present in solid B? Explain your reasoning. (iii) Describe one further test that could have been used to confirm the bonding types?

Q15. Most substances that we can smell, such as perfumes, are covalent molecular substances. Explain why this is so. Q16. Any substance that is a gas or a liquid at room temperature is very likely to be a covalent molecular substance. Explain. Q17. The non-metallic element bromine exists as diatomic molecules Br2. Bromine is a liquid at room temperature. (a) Draw the electron dot (Lewis) diagram for the Br atom. (b) Draw the electron dot (Lewis) diagram for the Br2 molecule. (c) Consider the changes represented by the two equations: (i) Br2(l) → Br2(g) (ii) Br2(g) → Br(g) + Br(g)

• What type of bonds are being broken in reaction (i)?

• What type of bonds are being broken in reaction (ii)?

• Which of reactions (i) and (ii) would occur at the lower temperature? Explain.

Q18. What is the electron dot representation for an argon atom? Use this to explain why there have never been any chemical compounds formed with argon. Q19. Why are the crystal lattices of covalent molecular substances described as “discontinuous”? Q20. Explain why the chemical reactivity of nitrogen depends on the strength of covalent bonds but the boiling point of N2 does not.

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COVALENT NETWORK SUBSTANCES: In some covalent substances, rather than having relatively simple, separate molecules, the whole crystal is one ‘giant’ macromolecule held together by a continuous linkage of strong covalent bonds. This network of covalent bonds extending throughout the crystal means each atom is covalently bonded to a number of other atoms, making the substance very hard, very strong and extremely difficult to melt. These covalent structures are CONTINUOUS LATTICES. COVALENT NETWORK SUBSTANCES YOU SHOULD KNOW:

ELEMENTS COMPOUNDS C (diamond) SiO2 (silica sand) C (graphite) SiO2 (quartz) Si (silicon) Silicate minerals B (boron) SiC (carborundum)

BN (borazon) (i) SILICON DIOXIDE A familiar covalent network substance is beach sand. Beach sand is mainly silicon dioxide or ‘silica’ having the ratio* formula SiO2. In a crystal of SiO2, each Si atom is covalently bonded to four O atoms and each O atom is covalently bonded to two silicon atoms. i.e.

* NOTE: Although the chemical formula for silicon dioxide is written as SiO2, there are no individual molecules of SiO2 but instead there is a vast covalent network of atoms in which the ratio of: (silicon atoms) : (oxygen atoms) = 1 : 2

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(ii) DIAMOND & GRAPHITE Diamond and graphite are two other important covalent network substances with which you should be familiar. These are two solid forms of the element carbon and as they have different crystalline structures they are described as ALLOTROPES of carbon.

DIAMOND GRAPHITE In diamond, each carbon atom is strongly bonded (covalently) to four other carbon atoms arranged around it tetrahedrally, (see above left). The bonding is continuous in three dimensions and all valence electrons are involved in bonding. This makes diamond very hard and a non-conductor of electricity. Graphite is an unusual covalent network substance because it is soft as well as being an electrical conductor. These properties are explained by graphite having carbon atoms bonded in layers with three valence electrons used to bond each carbon atom to three other carbon atoms (see above, right). The fourth valence electron is delocalised and accounts for the electrical conductivity. Graphite is the exception, as practically all other covalent network substances very hard and are non-electrical conductors (insulators). COVALENT NETWORK SUBSTANCES: SUMMARY

PROPERTY EXPLANATION Non-electrical conductors

when solid or molten There are no ions and all electrons are localized in

covalent bonds or in the atoms Very high melting points Very strong covalent bonds extend throughout the

crystal lattice Covalent network substances

are very hard All atoms are bound into the crystal lattice by strong

covalent bonds Covalent network substances

are brittle The covalent bonding is directional and distortion by

force, breaks the covalent bonds

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BONDING TYPES – DECISION TABLE

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BONDING – GENERAL REVISION QUESTIONS Q1. Complete the following table by indicating the bond types that normally form in different combinations of elements shown. COMBINING ATOMS BONDING TYPE

(i) metal and non-metal .................................... (ii) metal and metal .................................... (iii) non-metal and non-metal .................................... Q2. Answer the following questions relating to the formation of a single covalent bond between two atoms A and B: (a) How many electrons does each atom contribute? .................................... (b) The covalent bond formed, results from the .................................... of electrons. Q3. Complete the table below PERIODIC TABLE VALENCE e- UNPAIRED e- NO. OF BONDS FORMED Group IV … … … Group V .... .... .... Group VI .... .... .... Group VII .... .... .... Group VIII … … … Q4. Draw "electron-dot" representations for the most likely compound formed between: (a) sulfur and chlorine (b) phosphorus and fluorine (c) hydrogen and nitrogen (d) carbon and oxygen Q5. In which of the following substance(s) do actual molecules exist? (a) CaO (b) NO2 (c) Na (d) C2H6 (e) NH3 (f) AsBr3 (g) BaCl2 Q6. Consider the covalent molecular substance tetrachloromethane CCl4. Give the type of bonds and general strength of bonds that occurs in this compound: TYPE OF BONDS GENERAL STRENGTH

(i) INTRAMOLECULAR BONDS .................................... ..................................................... (ii) INTERMOLECULAR BONDS .................................... .....................................................

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Q7. Consider the two substances nitrogen trifluoride(NF3) and phosphorus trichloride(PCl3). (i) Which one would have the stronger van der Waal’s (dispersion) forces? Explain briefly. (ii) Which one would have the lower melting temperature? Explain briefly. Q8. Consider the following information about two chemical substances X and Y: X: melting point = 119°C and X(l) = a non-conductor of electricity. Y: melting point = 1489°C and Y(l) = a non-conductor of electricity. What one of the 4 main crystal structure types is each solid likely to be? (i) X(s) = .................................................................. (ii) Y(s) = ................................................................. Q9. What type of bonding is likely to be present in compounds of "perfumes"? Explain briefly. Q10. The black compound used in the manufacture of emery paper and some cutting tool blades is silicon carbide (carborundum). It has the formula SiC. Carborundum is very hard and has a very high melting point. (a) Explain these properties in terms of the bonding type present. (b) What information is provided by the formula SiC? Q10. Four pure crystalline substances are investigated in the laboratory in order to determine the types of bonding involved. The results are tabulated below.

SUBSTANCE ‘A’

SUBSTANCE ‘B’

SUBSTANCE ‘C’

SUBSTANCE ‘D’

MELTING POINT

1136oC

142oC

1891oC

594oC

SOLUBILITY IN IN WATER

High

High

Insoluble

Insoluble

ELECRICAL CONDUCTIVITY OF SOLID

Non-conductor

Non-conductor

Non-conductor

Excellent conductor

ELECRICAL CONDUCTIVITY OF AQ. SOLUTION

Good conductor

Non-conductor

―

―

ELECRICAL CONDUCTIVITY WHEN MOLTEN

Good conductor

Non-conductor

Non-conductor

Excellent conductor

Which of the four substances is likely to be a: (i) metallic solid? Explain your answer. (ii) ionic solid? Explain your answer. (iii) covalent molecular solid? Explain your answer. (iv) covalent network solid? Explain your answer.

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BONDING TYPES –A SUMMARY

Observation Explanation using electrons

Relatively Dense De-localised electrons allow atoms to be close together in the metallic lattice.

Malleable and Ductile De-localised electrons allow atoms to move past each other without repelling each other

Good conductor of heat and electricity

De-localised electrons are free to move and carry heat energy (through vibration) and electrical charge

Metallic Bonding

High Melting and Boiling Points

The metal lattice is held together by the strong electrostatic attraction between the delocalised electrons and the positive ions.

High Melting and Boiling Points

The electrostatic attraction between the ions in the lattice is needs a high amount of energy to break.

Hard and Brittle When moved relative to one another within the lattice ions with like charges come close together and repel strongly. This shatters the crystal. Ionic Bonding

Conducts Electricity in Solution and when

molten.

In solution and when molten the ions are free to move. As they have a charge they will move toward an electrode with an opposite charge.

Low Melting and Boiling Points

Weak forces between molecules and strong forces within molecules due to sharing of electrons only within molecules. Covalent

Molecular Bonding Non-Conductor of

Electricity Electrons are shared and molecules have no overall charge.

High Melting and Boiling Points

Strong continuous lattice formed through sharing of electrons. These covalent bonds require a lot of energy to break. Covalent

Network Bonding Non-Conductor of

Electricity Electrons are shared (solid) or stay with their original atom (Liquid, Gas) so no overall charge is present.

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PHYSICAL SCIENCES (LEVEL 3C) BONDING & CRYSTALS TEST Total = 25 marks Answer ALL the questions in the spaces provided immediately following the question.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Q1. Elements X and Y are in period 3 of the Periodic Table; (i.e. in Na ↔Ar). They each react with bromine to form compounds XBr3 and YBr3 respectively. YBr3 dissolves in water to form an electrically conducting solution whereas XBr3 is a non-electrically conducting low boiling point liquid. (a) What type of solid is YBr3 likely to be?

............................................................................................................................................. (1 mark) (b) What type of solid is XBr3 likely to be?

............................................................................................................................................. (1 mark) (c) Which real element is likely to be X? ........................................................................... Which real element is likely to be Y? ........................................................................... (2 marks) (d) Draw an appropriate electron-dot representation of XBr3. (2 marks) (e) Explain what is occurring when an aqueous solution of YBr3 conducts electricity. Which particles are moving and in which direction? (2 marks) (f) When XBr3(l) boils, what type of bonds are being broken?

............................................................................................................................................. (1 mark) (g) If XBr3 is heated strongly enough, it decomposes into element X and Br2 gas. What type of bonds are being broken when XBr3 is decomposed by heating?

............................................................................................................................................. (1 mark) Q2. For each of the following substances, classify them in terms of the crystal type found in the solid. i.e. ionic, metallic, covalent molecular or covalent network.

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(a) gold .......................................................... (b) calcium carbonate .......................................................... (c) silicon carbide .......................................................... (d) stainless steel .......................................................... (e) silicon .......................................................... (f) water .......................................................... (g) lead(II) nitrate .......................................................... (h) quartzite .......................................................... (i) sulfur dioxide .......................................................... (j) mercury .......................................................... (4 marks) Q3. Both calcium metal (Ca(s)) and molten calcium fluoride (CaF2(l)) are good conductors of electricity. (a) What particle(s) are moving the electrical charge in: (i) calcium metal (Ca(s)) ..................................................... (ii) molten calcium fluoride (CaF2(l)) ..................................................... (2 marks) (b) Explain why calcium metal is a better conductor of electricity than molten calcium fluoride. (2 marks) 4. You are presented with an unknown solid which has the following properties: (i) the solid is a non-conductor of electricity. (ii) the solid melts at 1378°C (iii) the solid doesn't dissolve in water (iv) when melted, the resulting liquid does not conduct electricity. What is the likely bonding type present in the unknown solid? Explain briefly. (3 marks) 5. Draw electron-dot ("Lewis") diagrams for: (a) oxygen (O2) (b) carbon dioxide (CO2) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (4 marks)

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BONDING & CRYSTALS TEST - ANSWERS Answer ALL the questions in the spaces provided immediately following the question.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Q1. (a) What type of solid is YBr3 likely to be? (Ionic) (b) What type of solid is XBr3 likely to be? (Covalent molecular solid) (c) (i) Which real element is likely to be X? (Phosphorus P) (ii) Which real element is likely to be Y? (Aluminium Al) (d) Draw an appropriate electron-dot representation of XBr3. (similar to PCl3 on p.11) (e) Explain what is occurring when an aqueous solution of YBr3 conducts electricity. Which particles are moving and in which direction? ANS. YBr3 is an ionic solid with Y3+ and Br− ions in a crystal lattice. Once dissolved in water or melted, the ions are free to move independently i.e. they are ‘mobile’. Under the influence of an electric field the Y3+ ions move towards the negative electrode and the Br− ions move towards the positive electrode (opposite charges attract!) This flow of ions constitutes an ‘electric current’ and we say that the solution is conducting electricity. (f) When XBr3(l) boils, what type of bonds are being broken?

ANS. The bonds being broken are the van der Waal’s forces/dispersion forces (these forces between molecules are called ‘intermolecular forces’. (g) If XBr3 is heated strongly enough, it decomposes into element X and Br2 gas. What type of bonds are being broken when XBr3 is decomposed by heating? ANS. The bonds being broken are the strong covalent bonds (intramolecular bonds)

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Q2. (a) gold metallic (b) calcium carbonate ionic (c) silicon carbide covalent network (d) stainless steel metallic (e) silicon covalent network (f) water covalent molecular (g) lead(II) nitrate ionic (h) quartzite covalent network (i) sulfur dioxide covalent molecular (j) mercury metallic Q3. Both calcium metal (Ca(s)) and molten calcium fluoride (CaF2(l)) are good conductors of electricity. (a) What particle(s) are moving the electrical charge in: (i) calcium metal (Ca(s)) delocalised electrons

(ii) molten calcium fluoride (+) and (−) ions (b) Explain why calcium metal is a better conductor of electricity than molten calcium fluoride. ANS. The very much smaller electrons carrying the charge in Ca metal move with very little resistance but the comparatively large bulky Ca2+ and F− ions experience greater resistance as they are moving in opposite directions. 4. You are presented with an unknown solid which has the following properties: (i) the solid is a non-conductor of electricity. (ii) the solid melts at 1378°C (iii) the solid doesn't dissolve in water (iv) when melted, the resulting liquid does not conduct electricity. What is the likely bonding type present in the unknown solid? Explain briefly. ANS. Covalent network. The very strong covalent bonds extending throughout the crystal structure give rise to a very high melting and boiling point. The molten compound does not conduct electricity and this rules out the possibility of it being ionic. Q5. Draw electron-dot ("Lewis") diagrams for: (a) oxygen (O2) (b) carbon dioxide (CO2)

(see page 11)

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