CharacteristicTrend (left to right)Reason

Atomic radiusdecreases in size from left to rightincreased attractive force (acting on the same energy shell) of the nucleus increases as the number of protons increases

Ionic radiusdecreases across the period until formation of the negative ions then there is a sudden increase followed by a steady decrease to the endIn general as above. The sudden increase on formation of negative ions is due to the new (larger) outer shell

ElectronegativityIncreasesMore electron attracting power of the larger nuclear charge as we move to the right

Metallic characterDecreases - Na, Mg, Al metals; Si metalloid; P, S, Cl, Ar non-metalsMetallic character is a measure of the ease of loss of electrons from the outer shell. This decreases with increasing nuclear charge.

Melting pointNaAl steady increaseIncreasing availability of electrons in the metallic bonding associated with greater charge density of the metal ion

Si massive increaseSi giant macromolecular structure

P large decreaseP4molecules

S small increaseS8crown shaped molecules

ClAr decreaseCl2molecules and Ar atoms

======================E-Z NOTATION FOR GEOMETRIC ISOMERISM

This page explains the E-Z system for naming geometric isomers.

Important! If you have come straight here via a search engine, you should be aware that this page follows on from anintroductory pageabout geometric isomerism. Unless you are already confident about how geometric isomers arise, and the cis-trans system for naming them, you should follow this link first. You will find links back to this current page at suitable points on that page.

The E-Z systemThe problem with the cis-trans system for naming geometric isomersConsider a simple case of geometric isomerism which we've already discussed on the previous page.

You can tell which is the cis and which the trans form just by looking at them. All you really have to remember is that trans means "across" (as in transatlantic or transcontinental) and that cis is the opposite. It is a simple and visual way of telling the two isomers apart. So why do we need another system?There are problems as compounds get more complicated. For example, could you name these two isomers using cis and trans?

Because everything attached to the carbon-carbon double bond is different, there aren't any obvious things which you can think of as being "cis" or "trans" to each other. The E-Z system gets around this problem completely - but unfortunately makes things slightly more difficult for the simple examples you usually meet in introductory courses.

How the E-Z system worksWe'll use the last two compounds as an example to explain how the system works.You look at what is attached to each end of the double bond in turn, and give the two groups a "priority" according to a set of rules which we'll explore in a minute.In the example above, at the left-hand end of the bond, it turns out that bromine has a higher priority than fluorine. And on the right-hand end, it turns out that chlorine has a higher priority than hydrogen.

If the two groups with the higher priorities are on the same side of the double bond, that is described as the (Z)- isomer. So you would write it as (Z)-name of compound. The symbol Z comes from a German word (zusammen) which means together.

Note: I'm not getting bogged down in the names of these more complex compounds. As soon as I put the proper full names in, the whole thing suddenly looks much more complicated than it really is, and you will start to focus on where the whole name comes from rather than on if it is a (Z)- or (E)- isomer.

If the two groups with the higher priorities are on opposite sides of the double bond, then this is the (E)- isomer. E comes from the German entgegen which means opposite.So the two isomers are:

Summary (E)- : the higher priority groups are on opposite sides of the double bond. (Z)- : the higher priority groups are on the same side of the double bond.

Note: Three possible suggestions for remembering this: E is for "Enemies", which are on opposite sides.You don't, of course, need a way of remembering the Z as well - it's just the other way around from E. In Z isomers, the higher priority groups are onzee zame zide. That works best if you imagine you are an American speaking with a stage German accent! This is the way I remembered it when I first came across E-Z notation. It is more visual than the other methods.It relies on the fact that the shapes of E and Z isomers are theoppositeof the shapes of the letters E and Z.In the letter E, the horizontal strokes are all on thesameside; in the E isomer, the higher priority groups are onoppositesides. In the letter Z, the horizontal strokes are onoppositesides; in the Z isomer, the groups are on thesameside.Choose whichever of these methods make most sense to you.

Rules for determining prioritiesThese are known as Cahn-Ingold-Prelog (CIP) rules after the people who developed the system.The first rule for very simple casesYou look first at the atoms attached directly to the carbon atoms at each end of the double bond - thinking about the two ends separately. The atom which has the higher atomic number is given the higher priority.Let's look at the example we've been talking about.

Just consider the first isomer - and look separately at the left-hand and then the right-hand carbon atom. Compare the atomic numbers of the attached atoms to work out the various priorities.

Notice that the atoms with the higher priorities are both on the same side of the double bond. That counts as the (Z)- isomer.The second isomer obviously still has the same atoms at each end, but this time the higher priority atoms are on opposite sides of the double bond. That's the (E)- isomer.

What about the more familiar examples like 1,2-dichloroethene or but-2-ene? Here's 1,2-dichloroethene.

Think about the priority of the two groups on the first carbon of the left-hand isomer.Chlorine has a higher atomic number than hydrogen, and so has the higher priority. That, of course, is equally true of all the other carbon atoms in these two isomers.In the first isomer, the higher priority groups are on opposite sides of the bond. That must be the (E)- isomer. The other one, with the higher priority groups on the same side, is the (Z)- isomer.

And now but-2-ene . . .This adds the slight complication that you haven't got a single atom attached to the double bond, but a group of atoms.That isn't a problem. Concentrate on the atomdirectly attachedto the double bond - in this case the carbon in the CH3group. For this simple case, you can ignore the hydrogen atoms in the CH3group entirely. However, with more complicated groups youmayhave to worry about atoms not directly attached to the double bond. We'll look at that problem in a moment.Here is one of the isomers of but-2-ene:

The CH3group has the higher priority because its carbon atom has an atomic number of 6 compared with an atomic number of 1 for the hydrogen also attached to the carbon-carbon double bond.The isomer drawn above has the two higher priority groups on opposite sides of the double bond. The compound is (E)-but-2-ene.

A minor addition to the rule to allow for isotopes of, for example, hydrogenDeuterium is an isotope of hydrogen having a relative atomic mass of 2. It still has only 1 proton, and so still has an atomic number of 1. However, it isn't the same as an atom of "ordinary" hydrogen, and so these two compounds are geometric isomers:

The hydrogen and deuterium have the same atomic number - so on that basis, they would have the same priority. In a case like that, the one with the higher relative atomic mass has the higher priority. So in these isomers, the deuterium and chlorine are the higher priority groups on each end of the double bond.That means that the left-hand isomer in the last diagram is the (E)- form, and the right-hand one the (Z)-.

Extending the rules to more complicated moleculesIf you are reading this because you are doing a course for 16 - 18 year olds such as UK A level, you may well not need to know much about this section, but it really isn't very difficult!Let's illustrate this by taking a fairly scary-looking molecule, and seeing how easy it is to find out whether it is a (Z)- or (E)- isomer by applying an extra rule.

Focus on the left-hand end of the molecule. What is attacheddirectlyto the carbon-carbon double bond?In both of the attached groups, a carbon atom is attached directly to the bond. Those two atoms obviously have the same atomic number and therefore the same priority. So that doesn't help.In this sort of case, you now look at what is attacheddirectlyto those two carbons (but without counting the carbon of the double bond) and compare the priorities of these next lot of atoms.You can do this in your head in simple cases, but it is sometimes useful to write the attached atoms down, listing them with the highest priority atom first. It makes them easier to compare. Like this . . .In the CH3group:The atoms attached to the carbon are H H H.In the CH3CH2group:The atoms attached directly to the carbon of the CH2group are C H H.In the second list, the C is written first because it has the highest atomic number.Now compare the two lists atom by atom. The first atom in each list is an H in the CH3group and a C in the CH3CH2group. The carbon has the higher priority because it has the higher atomic number. So that gives the CH3CH2group a higher priority than the CH3group.Now look at the other end of the double bond. The extra thing that this illustrates is that if you have a double bond, you count the attached atom twice. Here is the structure again.

So, again, the atoms attached directly to the carbon-carbon double bond are both carbons. We therefore need to look at what is attached to those carbons.In the CH2OH group:The atoms attached directly to the carbon are O H H.In the CHO group:The atoms attached directly to the carbon are O O H. Remember that the oxygen is counted twice because of the carbon-oxygen double bond.In both lists, the oxygens are written first because they have a higher atomic number than hydrogen.So, what is the priority of the two groups? The first atom in both lists is an oxygen - that doesn't help. Look at the next atom in both lists. In the CH2OH group, that's a hydrogen; in the CHO list, it's an oxygen.The oxygen has the higher priority - and that gives the CHO group a higher priority than the CH2OH group.The isomer is therefore a (Z)- form, because the two higher priority groups (the CH3CH2group and the CHO group) are both on the same side of the bond.That's been a fairly long-winded explanation just to make clear how it works. With a bit of practice, it takes a few seconds to work out in any but the most complex cases.

One more example to make a couple of additional minor points . . .Here's an even more complicated molecule!

Before you read on, have a go at working out the relative priorities of the two groups on the left-hand end of the double bond, and the two on the right-hand end. There's another bit of rule that I haven't specifically told you yet, but it isn't hard to guess what it might be when you start to look at the problem. If you can work this out, then you won't have any difficulty with any problem you are likely to come across at this level.Look first at the left-hand groups.In both the top and bottom groups, you have a CH2group attached directly to the carbon-carbon double bond, and the carbon in that CH2group is also attached to another carbon atom. In each case, the list will read C H H.There is no difference between the priorities of those groups, so what are you going to do about it? The answer is to move out along the chain to the next group. And if necessary, continue to do this until you have found a difference.Next along the chain at the top left of the molecule is another CH2group attached to a further carbon atom. The list for this group is again C H H.But the next group along the chain at the bottom left is a CH group attached to two more carbon atoms. Its list is therefore C C H.Comparing these lists atom by atom, leads you to the fact that the bottom group has the higher priority.Now look at the right-hand groups. Here is the molecule again:

The top right group has C H H attached to the first carbon in the chain.The bottom right one has Cl H H.The chlorine has a higher atomic number than carbon, and so the bottom right group has the higher priority of these two groups.The extra point I am trying to make with this bit of the example is that you must just focus on one bit of a chain at a time. We never get around to considering the bromine at the extreme top right of the molecule. We don't need to go out that far along the chain - you work out one link at a time until you find a difference. Anything beyond that is irrelevant.For the record, this molecule is a (Z)- isomer because the higher priority groups at each end are on the same side of the double bond.

Can you easily translate cis- and trans- into (Z)- and (E)-?You might think that for simple cases, cis- will just convert into (Z)- and trans- into (E)-.Look for example at the 1,2-dichloroethene and but-2-ene cases.

But it doesn't always work! Think about this relatively uncomplicated molecule . . .

This is clearly a cis- isomer. It has two CH3groups on the same side of the double bond. But work out the priorities on the right-hand end of the double bond.The two directly attached atoms are carbon and bromine. Bromine has the higher atomic number and so has the higher priority on that end. At the other end, the CH3group has the higher priority.That means that the two higher priority groups are on opposite sides of the double bond, and so this is an (E)- isomer - NOT a (Z)-.Never assume that you can convert directly from one of these systems into the other. The only safe thing to do is to start from scratch in each case.Does it matter that the two systems will sometimes give different results? No! The purpose of both systems is to enable you to decode a name and write a correct formula. Properly used, both systems will do this for you - although the cis-trans system will only work for very straightforward molecules.

================================================isomermelting point (C)boiling point (C)

cis-8060

trans-5048

In each case, the higher melting or boiling point is shown in red.You will notice that: the trans isomer has the higher melting point; the cis isomer has the higher boiling point.=======================================

================================The standard enthalpy of formation"The enthalpy of formation is the energy change when 1 mole of a substance is formed from its constituent elements in their standard states"Particular points to note: The elements are in their usual states under standard conditions. i.e at 25C and 1 atmosphere pressure (100 kPa). The enthalpy of formation of an element in its standard state is zero by this definition. The energy is given per mole of substance. The standard enthalpy of formation is represented by the symbol Hf.Example:Give the equation for the enthalpy of formation of sulfuric acidThe enthalpy of formation of sulfuric acid is represented by the following equation:H2(g) + S(s) + 2O2(g)H2SO4(l)Hf= -811 kJ mol-1

Example:The enthalpy of formation of calcium carbonate is represented by the following equation:Ca(s) + C(s) + 1O2(g)CaCO3(s)Hf= -1207 kJ mol-1

The sign of the enthalpy of formation of a compound gives us the relative stability of the compound being formedwith respect to its constituent elements.A negative value tells us that it is an exothermic compound (compared to its elements). ItDOES NOTtell us that the elements can be made to react directly together to form the compoundSimilarly, a positive enthalpy of formation value tells us that the compound is relatively unstable with respect to its elements, it is an endothermic compound.It DOES NOT mean that it is unstable, only that in comparison with its constituent elements its formation would require energy. Once again there is no implication regarding the possibility of direct formation from the elements.

topIonisation enthalpyThis is defined as the energy required to remove 1 mole of electrons from 1 mole of gaseous particles producing 1 mole of ions. It can be stated as 1st ionisation energy, 2nd ionisation energy etc.The 1st ionisation energy is the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to produce 1 mole of singly charged positive ions. It is always endothermic.Key points to remember are: The value is quoted per mole of species The particles must be gaseousExample: The first ionisation energy of sodiumNa(g)Na+(g)H = +496 kJ mol-1

topEnthalpy of atomisationThis is the energy needed to transform an element in its standard state into 1 mole of gaseous atoms. It is always endothermic.Example: The enthalpy of atomisation of sodiumNa(s)Na(g)H = +107 kJ mol-1

topBond dissocation enthalpyThis is energy required to break 1 mole of a specific bond in a specific compound. It is not the same as the bond energy term, which is an average measured over many compounds.Example: The first dissociation of methaneCH4(g)CH3+ HH = +412 kJ mol-1

topThe electron affinityThis is the energy change when 1 mole of electrons are captured by 1 mole of gaseous atoms forming 1 mole of negative ions. Once again the electron affinity may be quoted in terms of 1st electron affinity, 2nd electron affinity etc.This value is usually negative, but in the case of second electron affinities may be positive, as some processes are exothermic, while others are endothermic.Example: The first electron affinity of chlorineCl(g)Cl-(g)H = -349 kJ mol-1

topLattice enthalpyThe is possible to define from two points of view: Forming of 1 mole of a crystal lattice from gaseous ions (exothermic) Separating 1 mole of a crystal lattice into gaseous ions at infinite separation (endothermic).The way you define it is not important providing that you understand the consequences for the enthalpy change in terms of the exo- and endothermic processes.Example: The lattice enthalpy of sodium chlorideNa+(g) + Cl-(g)NaCl(s)H = -786 kJ mol-1Note that in this definition the enthalpy change is exothermic (negative)

topEnthalpy of hydrationThis is the energy change when 1 mole of a gaseous species dissolves in an infinite volume of water. It is caused by the solvent water molecules bonding to the dissolving species. It is always an exothermic process as bonds are being formed.Example: The hydration enthalpy of the sodium ionNa+(g) + (aq)Na+(aq)H = -405 kJ mol-1

topEnthalpy of solutionDefinitionThe energy change when 1 mole of a substance is dissolved in a solution to infinite dilution.MgSO4(s)Mg2+(aq) + SO42-(aq)

Clearly, the idea of infinite dilution is not to be taken literally. It just means that a solution is prepared by dissolving the solute to such a dilution as would produce no further energy change.Enthalpy of solution dataProcesses involvedAt face value, dissolution is a simple process, however it comprises at least two stages. 1The solute is bonded together in some way. These bonds must be broken. 2The separated solute particles must bond to the water (or solvent) molecules.MgSO4(s)Mg2+(g) + SO42-(g)Mg2+(g) + SO42-(g) + (aq)Mg2+(aq) + SO42-(aq)If an ionic substance is being dissolved then the individual ions must be dealt with separately in the second stage.===============Exothermic ReactionsExothermic and endothermic reactions cause energy level differences and therefore differences inenthalpy(H), the sum of all potential and kinetic energies. H is determined by the system, not the surrounding environment in a reaction. A system thatreleasesheat to the surroundings, an exothermic reaction, has anegativeH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system (Figure 1).C(s)+O2(g)CO2(g)(H = 393.5 kJ)H2(g)+1/2O2(g)H2O(l)(H = 285.8 kJ)The enthalpies of these reactions arelessthan zero, and are therefore exothermic reactions. ======================A system of reactants thatabsorbsheat from the surroundings in an endothermic reactionhas apositiveH, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.N2(g)+O2(g)2NO(g)(H = +180.5 kJ> 0)C(s)+2S(s)CS2(l)(H = +92.0 kJ > 0)Because the enthalpies of these reactions aregreaterthan zero, they areendothermic reactions. =====================Lewis acids and bases Lewis acid accept electron pairs. They don't have lone pairs on the central atom. eg. BF3 Lewis bases donate electron pairs. They have lone pairs on their central atom. eg. NH3

Periodictrends of the oxides have been thoroughly studied. In any given period, the bonding in oxides progresses from ionic to covalent, and their acid-base character goes from strongly basic through weakly basic, amphoteric, weakly acidic, and finally strongly acidic. In general,basicity increases down a group (e.g., in the alkaline earth oxides, BeO < MgO < CaO < SrO < BaO).Acidityincreases with increasing oxidation number of the element. For example, of the five oxides of manganese, MnO (in which manganese has an oxidation state of +2) is the least acidic and Mn2O7(which contains Mn7+) the most acidic. Oxides of thetransition metalswith oxidation numbers of +1, +2, and +3 are ionic compounds consisting of metal ions and oxide ions. Those transition metal oxides with oxidation numbers +4, +5, +6, and +7 behave as covalent compounds containing covalent metal-oxygen bonds. As a general rule, the ionic transition metal oxides are basic. That is, they will react with aqueous acids to form solutions of salts and water; for example,CoO + 2H3O+ Co2++ 3H2O.The oxides with oxidation numbers of +5, +6, and +7 are acidic and react with solutions of hydroxide to form salts and water; for example,CrO3+ 2OH- CrO42+ H2O.Those oxides with +4 oxidation numbers are generallyamphoteric(from Greekamphoteros,in both ways), meaning that these compounds can behave either as acids or as bases. Amphoteric oxides dissolve not only in acidic solutions but also in basic solutions. For example,vanadium oxide(VO2) is an amphoteric oxide, dissolving in acid to give the blue vanadylion, [VO]2+, and in base to yield the yellow-brown hypovanadate ion, [V4O9]2. Amphoterism among the main group oxides is primarily found with the metalloidal elements or their close neighbours.==========================If you place a piece of aluminum metal into a solution of barium nitrate nothing will happen. Just because someone writes an equation on paper, doesn't meant that it will actually occur in practice. Aluminum is below barium in the activity series, which tells us that aluminum metal cannot reduce barium ions. Therefore, there is no AlCl3(aq) and Ba as you might have expected. If Ba metal was produced we could not have left it simply as Ba(s). It would have reacted with water to make Ba(OH)2(aq) and H2 gas.

============== Follow up ===============

Here is perfect example of why you must specify the states of the compounds and the conditions. In solution, the nitrate ion does nothing, but when aluminum powder is combined with solid barium nitrate powder then a reaction can occur between the aluminum metal and the decomposing barium nitrate.

Individual reactions:2Ba(NO3)2(s) ---> 2BaO(s) + 4NO2(g) + O2(g)4Al(s) + 3O2(g) --> 2Al2O3(s)

Overall reaction:2Al(s) + 3Ba(NO3)2(s) --> 3BaO(s) + 6NO2(g) + Al2O3(s)========================TheOctet Ruleand Its ExceptionsThe octet rule states thatatomsbelowatomic number20 tend to combine so that they each have eightelectronsin theirvalenceshells, which gives them the same electronic configuration as a noblegas. The rule is applicable to the main-groupelements, especially carbon,nitrogen,oxygen, and thehalogens, but also tometalssuch as sodium and magnesium.Valence electronscan be counted using a Lewis electron dot diagram. In carbon dioxide, for example, each oxygen shares four electrons with the central carbon. These four electrons are counted in both the carbon octet and the oxygen octet because they are shared.

Carbon dioxideA Lewis dot diagram for carbon dioxide.Hydrogen and LithiumHowever, many atoms below atomic number 20 often formcompoundsthat do not follow the octet rule. For example, with the duet rule of the first principalenergylevel, thenoble gashelium, He, has two electrons in its outer level. Since there is no 1p subshell, 1s is followed immediately by 2s, and thus level 1 can only haveat mosttwovalence electrons. Hydrogen only needs one additional electron to attain this stable configuration, through either covalent sharing of electrons or by becoming thehydrideion(:H-), while lithium needs to lose one by combining ionically with other elements. This leads to hydrogen and lithium both having two electrons in theirvalence shellthe same electronic configuration as heliumwhen they formmoleculesby bonding to other elements.BoronandAluminumThere are also a variety of molecules in which there are too few electrons to provide an octet for every atom. Boron and aluminum, from Group III (or 13), display different bonding behavior than previously discussed. These atoms each have three valence electrons, so we would predict that these atoms want tobondcovalently in order to gain 5 electrons (through sharing) to fulfill the octet rule. However, compounds in which boron or aluminum atoms form five bonds are never observed, so we must conclude that simple predictions based on the octet rule are not reliable for Group III.Consider boron trifluoride (BF3). The bonding is relatively simple to model with aLewis structureif we allow each valence level electron in the boron atom to be shared in acovalent bondwith each fluorine atom. In this compound, the boron atom only has six valence shell electrons, but the octet rule is satisfied by the fluorine atoms.

Lewis structure of boron trifluorideEach pair of dots represents a pair of electrons. When placed between two atoms, the electrons are in a bond. A bond can be drawn as a line between two atoms, which also indicates two electrons. Notice that the central boron atom has only 6 electrons in the final Lewis diagram/structure of this molecule.We might conclude from this one example that boron atoms obey a sextet rule. However, boron will form a stable ion with hydrogen, BH4-, in which the boron atom does have a complete octet. In addition, BF3will react with ammonia (NH3), to form a stable compound, NH3BF3, for which a Lewis structure can be drawn that shows boron with a complete octet.

Boron trifluoride-ammonia complexThis covalent compound (NH3BF3) shows that boron can have an octet of electrons in its valence level.Compounds of aluminum follow similar trends. Aluminum trichloride (AlCl3), aluminum hydride (AlH3), and aluminum hydroxide (Al(OH)3) indicate a valence of three for aluminum, with six valence electrons in the bonded molecule. However, the stability of aluminum hydride ions (AlH4-) indicates that Al can also support an octet of valence shell electrons.Although the octet rule can still be of some utility in understanding thechemistryof boron and aluminum, the compounds of these elements are harder to predict than for other elements.===================================================================