Bonding and Structure How atoms bond together The

Bonding and Structure

How atoms bond together The bonds between atoms always involve their outer electrons Inert gases have ____ outer shells of electrons and are very_____. When atoms bond together they share, pool or transfer electrons to get full outer shells, making them more ______ and less ____, like the inert gases. There are three types of strong chemical bonds: ionic, covalent and ____. Mostly we show these bonds through drawing _____ structures, these are ‘dot and cross diagrams’. metallic full reactive Lewis stable unreactive

Ionic Bonding

Contents Describe the ionic bond as the electrostatic attraction between oppositely charged ions. Describe how ions can be formed as a result of electron transfer. Deduce which ions will be formed when elements in group 1, 2 and 3 lose electrons. Deduce which ions will be formed when elements in groups 5, 6 and 7 gain electrons. State that transition elements can form more than one ion.

Ionic Bonding Ionic bonding occurs between metals and nonmetals and involves transfer of _____ from metals to non-metals. Sodium chloride has ionic bonding. Remember sodium has ___ electrons, its electron arrangement is _______. Chlorine has _____ electrons, its electron arrangement is _____. 17 drawelectrons 11 electrons. 2, 8, 7 Now these atoms 2, 8, 1 showing all the

An electron is transferred. The single outer electron of the sodium atom moves into the outer shell of the chlorine atom + Na - Cl + 2, 8 , 1 2, 8, 87 -

- + Cl Na 2, 8 + 2, 8, 8 - Each outer shell is now ____. Both sodium and chlorine have an inert gas electron structure. The two charged particles are called _____. The sodium ion is ____charged, because it has _____an electron. The chloride ion is ____charged, because it has _____an electron. The two ions are attracted to each other and to other oppositely charged ions in the sodium chloride compound by ______forces gained full electrostatic positively lost ions negatively

The number of electrons lost or gained will depend on the number of electrons in the outer shell of the atom. 2 - 2+ Mg O Mg has 2 electrons in its outer shell, it can therefore lose 2 electrons and become a 2+ ion Oxygen has 6 electrons in its outer shell, it can therefore gain 2 electrons and become a 2 - ion

+ Na O + Na 2 - Look what happens when sodium bonds with oxygen? Sodium can only donate one electron, but oxygen needs two One oxygen atom needs to react with two sodium atoms to balance the charges.

In summary Ionic bonding is the result of electrostatic attraction between oppositely charged ions. The attraction extends throughout the compound making a structure called a lattice. The formula for sodium chloride is Na. Cl, because we know that for every one sodium ion there is one chloride ion. The formula for magnesium oxide is Mg. O, because there is one magnesium for every oxide. The formula for sodium oxide however is Na 2 O because two sodium ions are needed for each oxide ion

Common ions formed Metals always form positive ions, by losing electrons Non-metals always form negative ions by gaining electrons Element Ions formed +1 +3 +2 -1 -3 -2 Group 1 Group 2 Group 3 Group 5 Group 6 Group 7 Transition elements variable

Common ions formed in the transition elements Sc +III Ti V Cr Mn Fe Co Ni Cu +I +I +II +II +III +III +IV +IV +V +V +VI +VI +VII Zn +II

Contents State the formula of common polyatomic ions formed by non metals in period 2 and 3. Describe the lattice structure of ionic compounds.

Other polyatomic complex ions Name hydroxide sulphate nitrate Formula NO 3 Mn. O 4 OH- carbonate SO 42 - phosphate Cr 2 O 72 - manganate (VII) dichromate (VI) PO 43 CO 32 -

Properties of ionic compounds – melting point Ionic compounds are always solids. They have giant structures and therefore high melting points. This is because in order for them to melt an ionic compound, energy must be supplied to break up the lattice of ions.

Properties of ionic compounds – conducting electricity Ionic compounds conduct electricity when molten or dissolved in water (aqueous) but not when solid. This is because the ions that carry the current are free to move in the liquid state but are not free in the solid state. - + + - Cathode - + - - + Anode + Cathode - - - + - + + Anode + LIQUID SOLID

Properties of ionic compounds – hardness Ionic compounds tend to be brittle and shatter easily when given a sharp blow. This is because they form a lattice of alternating positive and negative ions. A blow in the direction shown may move the ions and produce contact between ions with like charges.

Covalent Bonding

Contents Describe the covalent bond as the electrostatic attraction between a pair of electrons and positively charged nuclei. Describe how the covalent bond is formed as a result of electron sharing. Deduce the Lewis (dot and cross) structures of molecules and ions for up to four electron pairs on each atom. State and explain the relationship between the number of bonds, bond length and bond strength.

Covalent bonding Covalent bonds form between _____atoms. The atoms _____ some of their outer electrons so that each has a full outer shell of electrons. A covalent bond is a shared _____ of electrons. We can represent one pair of shared electrons in a covalent bond by a line, Cl – Cl. Many non-metal elements exist as ____ molecules, covalently bonded to each other diatomic non-metal pair share

Forming molecules A small group of covalently bonded atoms is called a molecule. For example chlorine is a gas which is made of molecules, chlorine has 17 electrons and an electron arrangement 2, 8, 7. Two chlorine atoms make a molecule. The two atoms share one pair of electrons Each atom now has a full outer shell. The molecule does not contain charged particles because no electrons have been transferred from one atom to another.

Cl Cl The Cl – Cl bond is formed from the sharing of two electrons; one from each of the chlorine atoms to form the diatomic, covalent molecule, Cl 2

Methane gas is a covalent compound of carbon and hydrogen. Carbon has 6 electrons with electron arrangement 2, 4 and hydrogen has just one electron. In order for carbon to get a full outer shell, there are four hydrogen atoms to every carbon atom.

H H Four hydrogen's are needed for each carbon atom C H Hydrogen has just 1 electron in its outer shell. H H Carbon has the electron arrangement 2, 4

Double Covalent Bonds In a double bond, four electrons are shared. For example, the two atoms in an oxygen molecule share two pairs of electrons, so that the oxygen atoms have a double bond between them. O O

Activity: Draw the triple bond between two nitrogen atoms

Properties of simple covalent molecules Substances composed of molecules are gases, liquids or solids with low melting points. (The strong covalent bonds are only between the atoms in the molecules, not between the molecules themselves). They are poor conductors of electricity (because there are no charged particles to carry the current). If they dissolve in water, the solutions are poor conductors of electricity. (Again there are no charged particles).

Dative Covalent Bonding A single covalent bond consists of a pair of electrons shared between two atoms. In most covalent bonds, each atom provides one of the electrons, but in some bonds, one atom provides both the electrons. This is called dative covalent bonding, it is also called coordinate bonding. In a dative bond; The atom that receives the electrons is electron deficient. The atom that is donating the electrons has a pair of electrons that is not being used in a bond, called a lone pair.

The ammonium ion For example, ammonia, NH 3, has a lone pair of electrons. Draw the Lewis structure of ammonia. H Lone pair Ammonium ion + H H N H H+ H N H H In the ammonium ion, NH 4+, the nitrogen uses its lone pair of electrons to form a dative bond to an H+ ion (a ‘bare’ proton with no electrons at all and therefore electron deficient).

Dative covalent bonds are represented by an arrow, . The arrow points towards the atom that is receiving the electron pair. But this is only to show the bond was made. The ammonium ion is completely symmetrical and all the bonds have exactly the same strength and length. Dative bonds have exactly the same strength and length as ordinary covalent bonds between the same pair of atoms. The ammonium ion has covalently bonded atoms, but is a charged particle. Ions like this are called complex ions

Activity: Draw Lewis structures showing all electrons for the following covalent compounds: Water, H 2 O Carbon dioxide, CO 2 Fluorine, F 2 Ethane, C 2 H 6 Ammonia, NH 3

Shapes of covalent molecules

Contents Predict the shape and bond angles for species with four, three and two negative charge centers on the central atom using the valence shell electron pair repulsion theory (VSEPR). Predict whether or not a molecule is polar from its molecular shape and bond polarities.

The shapes of covalent molecules Molecules are three dimensional and they come in many different shapes. We can predict the shape of a simple covalent molecule – for example, one consisting of a central atom surrounded by a number of other atoms – by using the ideas that: A group of electrons around an atom will repel all other electron groups. The groups of electrons will therefore take up positions as far apart as possible. This is called the Valence Shell Electron Pair Repulsion theory or VSEPR (the valence shell is another name for outer shell).

A group of electrons may be: A set of two shared in a single bond A set of four in a double bond An unshared (lone) pair The shape of a molecule depends on the number of groups of electrons that surround the central atoms. To work out the shape of any molecule you must first draw a Lewis, ‘dot and cross’, diagram.

Two groups of electrons If there are two groups of electrons around the atom, the molecule will be linear. The furthest away from each other the two groups can get from each other is 180 o. Carbon dioxide has this shape, because it has two groups of electrons. Each group has four electrons in a double bond between carbon and oxygen. 180 o O C O O = C = O

Three groups of electrons If there are three groups of electrons around the central atom, they will be 120 o apart. The molecule is flat and the shape is trigonal planar. Boron trifluoride is an example of this: F F F B B Outer electrons shown only F F 120 o F

Four groups of electrons If there are four groups of electrons, they are furthest apart when they are arranged so that they point to the four corners of a tetrahedron (a triangular based pyramid) The angles here are 109. 5 o. Methane, CH 4, is an example: H H C H Outer electrons shown only H 109. 5 o H H

Ammonium ion The ammonium ion is also shaped like a tetrahedron. It has four groups of electrons surrounding the nitrogen atom. + H o 109. 5 + H H Outer electrons shown only N H H H

Molecules with lone pairs Some molecules have unshared (lone) pairs of electrons. These are electrons that are not part of a bond. The lone pairs affect the shape of the molecule. Water and ammonia are good examples of this effect. There is an increase in the repulsion between the following groups: Bonding pair – bonding pair Lone pair – bonding pair Repulsion increases Lone pair – lone pair

Ammonia, NH 3 Ammonia has four groups of electrons and one of the groups is a lone pair. With four groups, the ammonia molecule, like the water molecule, has a shape based on a tetrahedron. In this case there are only three ‘arms’. The lone pair squeezes the three shared pairs together and the bond angles are approximately 107 o. The shape is described as pyramidal H N H H Outer electrons shown only N H H H 107 o

Water, H 2 O H 104. 5 o H Outer electrons shown only O H Look at the Lewis diagram. There are four groups of electrons around the oxygen so the shape is based on a tetrahedron. However, two of the ‘arms’ of the tetrahedron are lone pairs that are not part of a bond. This results in a ‘V’ shaped or angular molecule.

Lone pairs explained The angles of a perfect tetrahedron are all 109. 5 o but lone pairs affect these angles. The shared pairs of electrons are attracted towards the oxygen nucleus and also the hydrogen nucleus. However, lone pairs are attracted only by the oxygen nucleus and are therefore pulled closer to it than shared pairs. The lone pairs therefore repel more effectively than shared pairs, and ‘squeeze’ the hydrogen together, reducing the H-O-H angle. An approximate rule of thumb is 2 o per lone pair. The actual bond angle in water is about 104. 5 o.

AHL Shapes of molecules and ions

Contents Predict the shape and bond angles for species with five and six negative charge centres using the VSPER theory.

Shapes of five and six negative charge centres Five pairs will give a VSEPR theory can be trigonal bipyramid shape extended to cover five with bond angles of 90˚, and six pairs of 120 ˚, 180˚ e. g. electrons or negative phosphorus pentachloride charge centres. PCl 5 The electron pairs will arrange themselves Six pairs will give an around the central atom octahedral shape with so that they are as bond angles of 90˚ and mutually repulsive as 180˚ e. g. sulphur possible. hexafluoride SF 6

Activity Use the orbit molecular building system to create a trigonal bipyramid shape and an octahedral shape. Cl Cl F Cl P Cl Cl Trigonal bypyramid, PCl 5 F F S F F F Octahedral, SF 6

Expanding the Octet The presence of five or six pairs of electrons around the central atom implies that the octet has been expanded. This cannot happen with the second period elements such as nitrogen, oxygen or fluorine, but can happen with the third period elements such as phosphorus, sulphur and chlorine. This is because, in the third period, the elements have 3 d orbitals, which are close enough in energy to the 3 p orbitals that they are available to be utilised. Nitrogen, for example, forms only one chloride, NCl 3, whereas phosphorus can form two chlorides, PCl 3 and PCl 5.

Non-bonding pairs When there are non-bonding pairs of electrons then the same rules apply, in that non-bonding pairs exert a greater repulsion than bonding pairs. Consider the structure of xenon tetrafluoride (Xe. F 4). There are six pairs of electrons around the central xenon atom. Four of them are bonding pairs and two are non-bonding pairs. This could result in two possible structures. In the first structure the bond angle between the non-bonding pairs is 90˚, whereas in the second structure it is 180˚, attempt to draw these two structures. Which structure will exist? What is the name of this shape? The actual shape of xenon tetrafluoride is therefore square planar

Bond Length, Strength and Resonance Hybrids

Contents State and explain the relationship between the number of bonds, bond length and bond strength. Describe resonance hybrid structures. Draw Lewis diagrams for resonance hybrid structures

Bond length and strength The more pairs of electrons that are shared between two atoms in a bond then the _______ the bond and the _____the bond length. In almost all cases single bonds are _______ and _____ than double bonds, and triple bonds are even ____ and _____ than double bonds. shorter stronger weaker stronger shorter longer

Resonance hybrids Consider the Lewis structure for the carbonate ion, CO 32 -. The carbon atom has four outer electrons of its own, and each oxygen has six outer electrons. Together with the two extra electrons this gives a total of 24 valence electrons: that is 12 pairs. The structure shows that one of the carbon-tooxygen bonds is a double bond, whereas the other two are single bonds.

If this structure is correct we would expect one of the carbon-to-oxygen bonds to be shorter than the other two. In fact all the carbon-to-oxygen bonds have the same length, which is intermediate between a carbon-to-oxygen single bond a carbon-tooxygen double bond. This can be explained by stating that the true structure of the carbonate ion lies somewhere between the three possible extreme structures, which are called resonance hybrids.

2 - O C O O Electrons gained from another species, eg 2 Na Outer electrons shown only

The resonance hybrids of the Carbonate ion 2 - O O C O 2 - 2 - O C C O O The true structure of the carbonate ion lies somewhere in between these three extremes. O

Resonance Hybrid structures Activity: Other compounds or ions for which resonance hybrid structures can be drawn include; sulphur dioxide, SO 2; ozone, O 3; the nitrate ion, NO 3 -; the nitrite ion, NO 22 -; and benzene, C 6 H 6. Look up these structures in your text books on page 67, figure 15, and copy them into your notes. NB you may be asked to draw these structures in an exam.

MOLECULAR ORBITALS

Contents Describe σ and π bonds. Explain hybridization in terms of the mixing of atomic orbitals to form new orbitals for bonding. Identify and explain the relationships between Lewis structures, molecular shapes and types of hybridization (sp, sp 2 and sp 3). Describe the delocalization of π electrons and explain how this can account for the structures of some species.

Dual Nature of the electron Remember the dual nature of the electron - ‘wave particle duality’. They possess properties associated with both particles and waves. This similarly, two waves are exactly out of phase means that theythat show the same properties of combine waves. Sodestructively. when two waves are in phase meet, they combine constructively;

Bonding molecular orbitals. A similar situation arises when two atomic orbitals combine. When two atomic orbitals from different atoms combine constructively the electron density between the two atoms increases, resulting in a molecular orbital with a lower energy than either of the two atomic orbitals. The molecular orbital is of a lower energy, and electrons tend to fill lower energy levels first, this is known as bonding molecular orbital.

Anti-bonding orbitals Similarly, if the two atomic orbitals combine destructively the electron density between the two atoms decreases, resulting in a molecular orbital with higher energy than either of the two atomic orbitals. In this case the electrons will tend to stay in their individual atomic orbitals, and the molecular orbital formed is known as an anti-bonding orbital

Hydrogen and helium The model below explains why hydrogen forms a diatomic molecule when the two 1 s atomic orbitals from each atom combine, whereas helium is monatomic. Hydrogen 1 s Helium 2 s energy Antibonding molecular orbital Atomic orbital molecular orbital 1 s orbital Atomic orbital No overall change in With lower net energy so helium atoms the hydrogen diatomic do not form helium molecule forms. molecules. Atomic orbital molecular orbital

Shapes of orbitals Do you remember the shapes of the different orbitals? Try and draw them, then click on the screen to see if you were right. s px py Recall that all of them are 3 -dimensional and thus occupy a region along axes. This is particularly important in the formation of bonds. pz

Sigma σ, and Pi π bonds When covalent bonds form there is overlap of the valence electron orbitals. The overlap can happen in two way: Head-on (along the plane of axis) Side –on Any overlapping between an s orbital with another s orbital or a p orbital is head on, this forms a sigma, σ bond. Side on overlapping can happen between two like p orbitals (these are p orbitals in the same plane), this form a pi, π bond.

Sigma σ bonds Sigma bonds (σ), occur when orbitals combine head-on. axis Two s orbitals combining axis An s orbital combining with a p orbital axis Two p orbitals combining

Pi, π bonds z Pi bonds form when like p orbitals (py with py or pz with pz) align with each other side by side. z x x y y These form electron dense areas above and below the plane of the nuclei, these are the π bonds.

The head–on nature of the sigma bond means that if a p orbital of one atom bonds head on with a p orbital of another atom then no more sigma bonds can form between those two atoms because it is not possible for more than one set of p orbitals to meet head-on, thus only pi bonds can form. y axis π π σ z axis x axis

Hybridisation The molecular orbital theory has to be adapted to explain the bonding (and shape) of tetrahedral molecules such as methane. Carbon has the electron configuration 1 s 2 2 p 2. Thus it is impossible for it to retain this configuration and form four equal bonds pointing to the corners of a tetrahedron when it combines with the 1 s 1 electrons from the four hydrogen atoms. For a start it has only two unpaired electrons, so it might reasonably be expected to have a combining power of 2. In addition, the two p electrons are in orbitals that are at 90˚ to each other, and so will not form bond angles of 109. 5˚ when they combine with the s orbitals on the hydrogen atoms, so the orbitals have to merge together.

3 sp CH , exhibits Hybridisation Methane, sp hybridisation in its 3 σ σ All 4 hybrid orbitals arenow used Carbon can now Click on the But this can only Click on the in forming σ bonds form 4 equal form 4 bonds but screen totosee how screen see how produce 2 bonds screen to see is how thus the bonds, e. g. not With they are anthe electrons and we know in the orbitals hybridisation is hydrogen to equal in energy. promoted to a 3 as arrange methane, CHit 4 is combine to form called sp form CH 4 higher energyin a the carbon themselves hybrid orbitals made up of 1 s and level during carbon atom forms 4 equal 3 p bonding bonds 4 bonding but what does this mean? σ σ Click on the screen to see how carbon can form 4 equal bonds with hydrogen

2 sp Hybridisation σ Click on on the screen to to see the howfilling the of three of the electrons orbitals arrange leaving one p electron. themselves in a carbon atom σ Ethene, C 2 H 4, exhibits sp 2 hybridisation in its bonding but what does this mean? σ π on the When bonding forms σ bonds can Only. During 3 ethene σ bonds are only 3 Click to see form, an(remember electron isa the σ bond screen only forms formed using 3 of when promoted orbitals overlap and the axis. hybrid This hybrid orbitals 1 s andalonghow orbitals and leaves the a phybrid orbital on each of the carbon 2 p thus this is called electrons atoms orbitals. The lobes are overlap above and sp 2 hybridisation below formed the plane to form a πarrange bond themselves

sp Hybridisation σ Click on the screen to see what happens in ethyne. σ Ethyne, C 2 H 2, exhibits sp hybridisation in its bonding but what does this mean? π π What do you think will happen in ethyne, when only 2 σ bonds can form on each carbon (one with hydrogen and one with carbon), but breaking the C – C bond requires more energy than just one σ bond.

Shapes of molecules and ions. We have seen that the shapes and bond angles of simple molecules and ions can be determined using VSEPR theory. They can also be determined using hybridisation theory. If the hybridisation is known, then the shape and bond angles can be predicted; similarly, if the shape and bond angles are known, the type of hybridisation can be deduced. Hybridisation can also be used to explain shapes of molecules with 5 and 6 charge centres. This involves mixing s, p and d orbitals to form, for example, d 2 sp 3 hybrid orbitals for octahedral shapes.

Hybridisation, shapes and bond angles. Shape Examples Bond angle Type of hybridisation Linear N 2, C 2 H 2, HCN 180˚ sp Trigonal planar C 2 H 4, BF 3, SO 3 120˚ sp 2 Tetrahedral CH 4, NH 4+, BF 4 - 109. 5˚ sp 3 Bent (parent shape trigonal planar) SO 2, CH 3 COCH 3 ~120˚ sp 2 Trigonal pyramid (parent shape tetrahedral) NH 3, PCl 3 107˚ sp 3 Bent (parent shape tetrahedral) H 2 O 105˚ sp 3

Delocalisation of the pi electrons and resonance hybrids Earlier we used the idea of resonance structures to explain certain structures e. g. the carbonate ion. Another example of a resonance structure is the benzene molecule. For many years chemists had a problem in determining the structural formula of benzene, which was known to have the molecular structure, C 6 H 6. This problem was solved by Friedrich Kekule, who first showed that benzene could be written in a cyclic or ring form using alternate double and single bonds between the carbon atoms. Later this was described using two resonance structures. Can you draw the two resonance structures of benzene.

Resonance hybridisation theory v molecular orbital theory. Resonance hybridisation theory can explain many of the properties of benzene, and in particular the fact that all the carbon-to-carbon bond lengths are equal and lie between a C-C single bond and C=C double bond, and the fact that benzene is more stable energetically and less reactive chemically expected. The Molecular orbital theory can also explain the structure of benzene and account for even more of its properties.

Molecular orbital theory and benzene Each atom in benzene is sp 2 hybridised so that it forms three sigma (σ) bonds, one to a hydrogen atom and one each to the adjacent carbon atoms to form a planar hexagonal ring. The remaining electron on each carbon atom is in a p orbital vertically above and below the plane of the ring. Instead of pairs of p orbitals combining together to give alternate double and single bonds, all six p orbitals combine to form one ring – shaped molecular orbital containing six delocalised pi electrons.

Pi delocalisation in benzene The pi electrons above and below the plane become delocalised to form a ring of delocalised electrons This is more stable by about 150 k. Jmol-1 this is known as the delocalisation enthalpy or resonance energy. Like the resonance hybrid model it explains why benzene does not readily undergo addition reactions, because this extra amount of energy would need to be put in to break down the delocalised pi bond. But unlike the resonance hybrid model it explains why benzene does react with electrophiles. These are attracted to the delocalised pi electrons above and below the plane of the ring to bring about substitution reactions.

Ozone The delocalisation of pi electrons can always be used as an alternative explanation to resonance hybrids. For example, in ozone the two resonance hybrids are formed depending on which two out of the three oxygen atoms combine to form the double bond. Using molecular orbital theory the p atomic orbitals on all three of the oxygen atoms combine together to form a “banana” shaped molecular orbital containing two delocalised pi electrons.

Activity: Other examples of the delocalisation of pi electrons include; the nitrate ion, NO 3 -; the nitrite ion, NO 2 -; the carbonate ion, CO 32 -; and the ethanoate ion, CH 3 COO-. Draw the two resonance structures of the ethanoate ion, and then draw a diagram to show the p orbitals on the carbon atom and two oxygen atoms can combine to form delocalised pi bonding.

Electronegativity and the nature of bonds

Contents Predict whether a compound of two elements would be ionic from the position of the elements in the periodic table or from their electronegativity values. Predict whether a compound of two elements would be covalent from the position of the elements in the periodic table or from their electronegativity values. Predict the relative polarity of bonds from electronegativity values. Compare and explain the properties of substances resulting from different types of bonding.

The nature of the bond. We can predict whether a bond between two elements will be ionic or covalent by looking at the ______ in electronegativity values. Elements in groups 1, 2 and 3 tend to have ______ electronegativity values. Elements in Groups 5, 6 and 7 tend to have ______ electronegativity values. The _____ the difference in electronegativity values, then the more likely it is that the bond will be _____. Generally, the difference in electronegativity value between the two elements needs to be about 1. 8 or ____ than for ionic bonding to occur. greater difference low ionic high greater

Copy out the table and fill in the data to determine the type of bonding present in these structures Mg Electronegativity Difference Formula Type of bonding Cl 1. 2 3. 0 Be Cl 1. 5 3. 0 Al O 1. 5 3. 5 Al Cl 1. 5 3. 0 1. 8 1. 5 2. 0 Mg. Cl 2 Be. Cl 2 Al 2 O 3 Al 2 Cl 6 ionic covalent 1. 5

Exceptions to the rule Activity: Look up and record neatly in your notes the electronegativity values for lead and bromine. Calculate the difference between these two values. What type of bonding would you expect this compound to display, explain your answer? Do the same for boron and fluorine.

Lead bromide is ionic and yet the difference in electronegativity values is only 1. 0, whereas boron trifluoride is not ionic and yet the difference in electronegativity values is 2. 0. Many ionic compounds show some covalent character, and many covalent compounds show some ionic character, so often it is not a case of ionic or covalent but somewhere in between. Generally if a compound conducts electricity when molten or in aqueous solution then ionic bonding predominates and if a compound is a poor conductor of electricity when molten or in aqueous solution then covalent bonding predominates

Polarity of bonds When a single covalent bond is formed between two atoms of the same element, for example a chlorine molecule, Cl 2, the electron pair will on average be shared equally between the two atoms, and the bond will be non-polar. Cl Cl Electron pair shared equally

However, if two different atoms are bonded together covalently, for example a hydrogen chloride molecule, HCl, then the nuclei of the different atoms will exert different attractive forces on the electron pair, and it will not be shared equally. The atom that attracts the electron pair more strongly will then be slightly negatively charged, δ -, compared with the other atom, which will be slightly positively charged, δ+. + δ δ This results in a polar bond H Electron pair pulled towards the chlorine Cl

If the polar molecule is placed between two electrically charged plates, then it is said to have a dipole moment, because the δ- end of the molecule will be attracted to the positive plate and the δ+ end of the molecule will be attracted to the negative plate. The bigger the difference in electronegativities, then the more polar the bond and the greater the dipole moment. Calculate the difference in electronegativity between Carbon and Hydrogen. Now try and explain why methane CH 4 is not a polar molecule.

Intermolecular forces

Contents Describe the types of intermolecular forces (attractions between molecules that have temporary dipoles, permanent dipoles or hydrogen bonding) and explain how they arise from the structural features of molecules. Describe and explain how intermolecular forces affect the boiling points of substances.

Types of intermolecular forces. So far you have learnt about the strong intramolecular forces which hold atoms to each other to form molecules, these affect the energy in chemical reactions, but there are other forces that control the physical properties of the molecules. These are intermolecular forces, there are three main types of intermolecular force: Dipole – dipole van der Waals Hydrogen bonding

Polarity of bonds To understand dipole – dipole forces you need to understand how some molecules are polar. When a single covalent bond is formed between two atoms of the same element, e. g. a chlorine molecule, Cl 2, the electron pair, on average, will be shared equally between the two atoms, and the bond will be non-polar. However, if two different atoms are bonded together covalently, e. g. a hydrogen chloride molecule, HCl, then the nuclei of the different atoms will exert different attractive forces on the electron pair, and it will not be shared equally.

The atom that attracts the electron pair more strongly (thus more electronegative), will pull the electron pair toward itself making it slightly negative, this is denoted by the symbol, δ Thus the less electronegative atom will become slightly positive, δ+. The molecule is polar, if it is placed between two electrically charged plates, the δ- end of the molecule will be attracted to the positive plate and the δ+ end will be attracted to the negative plate and it is said to have a dipole moment. The bigger the difference in electronegativities, then the more polar the bond and the greater the dipole moment.

Polar molecules The polarity of a molecule depends on the polarity of the bonds in the molecule and the shape of the molecule. There is a degree of polarity between Carbon and oxygen making the bond polar; δ+ δC O This molecule has polar bonds and is asymmetrical so the molecule is polar. But carbon dioxide is not polar. . . can you suggest why. δδ+ δO C O

Distinguishing between polar and non-polar molecules We can distinguish between polar and non-polar molecules by: Comparing boiling points; molecules with higher boiling points but with similar molecular mass and number of electrons will be polar. Comparing the effect of a charged rod on a stream of different liquids; a charged rod will only have an effect on polar molecules. Microwave radiation will only effect a polar liquid; a microwave oven works by causing the polar water molecules to line up with the microwave radiation.

Activity: Using chapter 10 of your course companion find out the displayed formulae for the following compounds and decide if you think they are polar molecules. Cyclohexane Dichlorohexane Ethanol Hexane Propanone Water Watch the video clip on the following slide for the answers

Electrostatic deflection of different covalent molecules showing their polarity. Click on the box to start movie A piece of perspex was charged by rubbing it with nylon. Different covalent liquids were allowed to run out of a burette. Their polarity was measured by the amount the liquid was deflected when the charged perspex was brought close to it.

van der Waals forces For non-polar molecules there are no permanent electrostatic forces of attraction between them. The forces that do exist are called van der Waals’ forces. The stronger the van der Waals’ forces, then the higher is the boiling point, as more energy is required to overcome the attraction between the molecules and separate them.

Factors affecting the strength of van der Waals’ Look up and compare the boiling points of the following non-polar molecules: Fluorine Methane, CH 4 Chlorine Ethane, C 2 H 6 Bromine Propane, C 3 H 8 Iodine Butane, C 4 H 10 Now compare the structures of the halogens and the alkanes separately. Can you work out the possible factors affecting the strength of the van der Waals’ ?

Hydrogen Bonding When hydrogen is bonded directly to one of the small, highly electronegative atoms fluorine, oxygen or nitrogen, then the polarity of the covalent bond is very high. In addition, as the electron pair is drawn away from the hydrogen atom, all that remains is the proton in the nucleus, as there are no outer electrons. The negative electronegative atom of another molecule is thus attracted by a very strong dipole attraction. . This type of very strong dipole-dipole attraction is given its own name – hydrogen bonding.

Strength of hydrogen-bonding The strength of hydrogen bonding can be demonstrated by the hydrides of the elements of groups 5, 6 and 7. Each of the first members of their respective series – ammonia, NH 3, water, H 2 O and hydrogen fluoride, HF – have a much higher boiling point than the other members of the group. This is particularly noticeable with water, which is a liquid at room temperature with a boiling point of 100 o. C at atmospheric pressure compared with all the other group 6 hydrides, which are gases at room temperature and atmospheric pressure.

Hydrogen bonding for Group IV, V, VI and VII hydrides. Activity: Using the boiling points for the hydrides opposite, plot a graph of boiling point (on the y axis) against period number (on the x axis). Describe and explain the shape of the graph in terms of types of intermolecular forces.

Hydrogen bonding for Group IV, V, VI and VII hydrides.

The unique properties of water Water is in fact an almost unique liquid, because when it freezes it expands, nearly all other liquids contract in volume when they freeze. The structure of ice is very open. Each oxygen atom is bonded to four hydrogen atoms in a giant tetrahedral arrangement. Two of these bonds are the strong covalent O-H bonds in the water molecule; the other two are weaker and longer hydrogen bonds between the 2δ- charge on each oxygen atom and the δ+ charge on each of the two hydrogen atoms from other water molecules. When ice melts, the molecules can move closer together, water has its maximum density at 4 o. C

Properties of giant covalent structures

Contents Describe and compare the structure and bonding in the three allotropes of carbon (diamond, graphite and C-60 Fullerene) Describe the structure of and bonding in silicon and silicon dioxide.

Giant covalent structures Carbon exhibits allotropy. This means that it can exist in more than one physical form. The three main allotropes of pure carbon are: Diamond Graphite Buckminster Fullerene In all three allotropes the carbon atoms are bonded covalently, but in diamond and graphite, instead of small simple molecules, the covalent bonds link across the carbon atoms to form a single large (or giant) molecule.

Diamond In diamond each carbon is bonded equally to four other carbon atoms to form a giant tetrahedral structure. All the C-C bond lengths are equal, and there is no plane of weakness through the structure, so diamond is an extremely hard substance. All the outer electrons around each carbon atom are localized to form the four bonds to other carbon atoms, so diamond does not conduct electricity, because there are no delocalized electrons.

Diamond All bonds are equal length and the bond angle is perfectly tetrahedral 109. 5 o

Graphite Each carbon atom forms strong covalent bonds to three other carbon atoms in a trigonal planar structure so that the carbon atoms link up to form hexagonal rings. The C – C bonds in the ring are in fact stronger and shorter than the C – C bond in diamond. The forces of attraction between the layers are very weak, because they are formed by delocalised electrons that can move between the layers. This means that graphite is one of the few non-metals to be a good conductor of electricity This also means that the layers can easily slide over each other, so graphite feels waxy to the touch and is a good lubricant.

Graphite Three bonds are equal length and in fact shorter than those in diamond, their bond angles are trigonal planar 120 o The red lines represent forces of attaction between the layers, these are delocalised electrons that are free to move

Buckminster Fullerene In 1996 Robert Curl (1933 - ), Harold Kroto (1939 - ) and Richard Smalley (1943 – 2005) were jointly awarded the Nobel prize in Chemistry for their discovery of fullerenes. The basic fullerene is C 60 in which 60 carbon atoms are joined in a combination of hexagonal and pentagonal rings to form a sphere. Since their discovery, more than 1000 new compounds involving fullerenes have been made. Some contain metals (e. g. Lanthanum) trapped inside a fullerene cage; others consist of long tubes that can be closed or open at one end. These are called nanotubes, because they have an extremely small diameter, in the order of one nanometer (1 x 10 -9 m)

Activity: Watch the Horizon video ‘Molecules have sunglasses’, this gives an idea of the struggles throughout the discovery of the fullerenes. It is about 50 minutes long.

Metallic bonding

Contents Describe the metallic bond as the electrostatic attraction between a lattice of positive ions and delocalized electrons. Explain the electrical conductivity and malleability of metals.

Metallic bonding When atoms of metals bond together in the solid state, one or more of their valence electrons becomes detached from each atom to become delocalised. These valence electrons are no longer associated with a particular atom, but are free to move throughout the metallic structure. The bonding in metals thus consists of the attraction between these delocalized valence electrons and the remaining positive metal ions (cations). It is sometimes said that metals are made up of an array of cations in a ‘sea’ of moblie electrons.

Physical properties of metals The physical properties of metals include: Good conductors of electricity Good thermal conductors Generally high melting points but varies with the number of valence electrons and size of the cation. Malleable – the ability to be beaten into shape without breaking Ductile – it can be drawn into a wire

Conductors of electricity and heat It is because the valence electrons are no longer located on a particular atom, but are free to move throughout the structure, that metals are such excellent conductors of electricity. If there are impurities in the metal, then this can hinder the movement of electrons and increase the electrical resistance: this explains why copper needs to be refined or purified before it is used for electrical wiring. The movement of electrons through metals also enables the transmission of kinetic energy, so metals are also good conductors of heat.

Melting points The melting point of metals is related to the strength of the attractive forces holding the cations in the ‘sea’ of delocalised electrons. This depends on the number of valence electrons delocalized from each atom, the size of the cations and the way in which the cations are packed together. In general, the melting point decrease as the size of the cation increases, which explains why melting points decrease down Group 1 (alkali metals).

Malleablity and ductility Many metals are malleable and ductile. Both of these properties can be explained by the cations being able to slide past each other to arrange the overall shape of the solid. Because the electrons are delocalised this can happen without significant change in the bonding forces.

Properties related to bonding

Contents Compare and explain the properties of substances resulting from different types of bonding.

Physical properties Type of force Mp / bp Solubility in water Solubilty in organic solvent Electrical conductor van der Waals low insoluble Non conductor Dipoledipole Higher than v d W miscible Non-conductor Hydrogen bonding Higher than d-d miscible Less soluble Non-conductor Ionic high Mostly soluble insoluble Conducts when molten or in solution Metallic high insoluble Good conductor Giant covalent high insoluble Non-conductor except graphite