Chemistry

Key unit Competence

To be able to compare and contrast the chemical properties of the Group 14 elements and their compounds in relation to their position in the Periodic Table.

Learning objectives

By the end of this unit, learners should be able to:

•Compare and contrast the physical properties of Group 14 elements.

•Compare the relative stabilities of the higher and lower oxidation states in oxides.

•Distinguish between the chemical reactions of the oxides and chlorides of Group 14 elements.

•Explain the trends in thermal stability of the oxide, halides and hydrides of Group 14 elements.

•Explain the variation in stability of oxidation state of +2 and +4 down the Group 14 elements.

•Define the diagonal relationship.

9.1. Physical properties of group 14 elements.

Group 14 elements are carbon (6C), silicon (14Si), germanium (32Ge), tin (50Sn) and lead (82Pb). The elements of group 14 have one important feature in common, each of them has four valence electrons. They are characterized by the electronic configuration of 4 electrons in the outermost shell (ns2np2 ). The group is also called carbon family.

Group 14 elements show the clearest trend from non-metal to metal character down the group from carbon to lead.

In group 14 of the periodic table, there is a considerably greater change in physical properties from top to bottom in the group than there is for example, in the alkali metals of group 1 on the left and the halogens of group 17 on the right.

Carbon is a typical non-metal element, silicon and germanium are semi-metals or metalloids, tin and lead are metals. However, the elements of group 14 have one important feature in common, namely each of them has four valence electrons.

In the compounds of carbon, it almost invariably completes the valence shell by forming four covalent bonds. When it forms 4 simple covalent bonds, they have a tetrahedral arrangement around the carbon atom.

Silicon and germanium also usually form four simple covalent bonds with a tetrahedral arrangement.

Tin and lead, which are larger in volume have smaller ionization energies than carbon and silicon, and they often lose just two of their valence electrons to form the Sn2+ and Pb2+ ions.

Activity 9.1

1) Write the electronic configuration of carbon, silicon and tin in terms of s,p, d and f notation.

2) Indicate 2 uses of carbon and lead in daily use.

3) Explain how carbon forms 4 bonds in its compounds, e.g in CO2 .

4) Write the chemical formula of:

a) 2 inorganic compounds of carbon.

b) 2 organic compounds of carbon.

Table 9.1: Physical properties of Group 14 elements

•Carbon, the first element of the group has two main allotropes: graphite and diamond.

•In graphite allotrope of carbon, each carbon is bonded to 3 other carbon atoms to form a hexagonal structure. The structure of graphite is made of hexagonal layers which are attracted to each other by weak Van der Waals forces such that the layers slide over each other to make the structure soft(Fig.9.1). In graphite structure, there are delocalised double bonds with mobile electrons that allow graphite to conduct electricity.

•In diamond, each carbon is covalently bonded to 4 other carbon atoms forming a giant tetrahedral structure that makes it to be very hard (Fig.9.1). In diamond, there are no mobile electrons as in graphite, hence diamond does not conduct electricity.

.As you move down the group in the carbon family, the atomic radius and ionic radius increase while the electronegativity and ionization energy decrease.

•Atomic size increases on moving down the group due to additional electronic shells

.•Density increases as you move down the group.

•Carbon is the only element in the family that can be found in pure form in nature, as diamond and graphite.

•Lead is the only element of group 14 that does not exist in various allotropes.

•Tin occurs as white, grey and rhombic tin.

•Group 14 elements have much higher melting points and boiling points than the group 13 elements.

•Melting and boiling points tend to decrease as you move down the group mainly because•inter atomic bonding between the larger atoms reduce in strength as you move down the group.

Moving down the group, there is an increase in atomic size which results in less attraction of valence electrons by the nucleus. This change results in weaker metallic bonding down the group and therefore there is a decrease in melting point, boiling point, enthalpy change of atomization and first ionization energy.

The decrease in first ionization energy from silicon to lead is relatively little compared to that from carbon to silicon because there is a large increase in nuclear charge which counterbalances the increase in atomic radius from silicon to lead.

i) The increase in metallic character down the group causes a general increase in conductivity.

arbon is typically a solid, non-metal.Carbon graphite is a non-metal but conducts electricity due to delocalized electrons in its structure.

In its compounds, carbon almost invariably completes its valence shell by forming four covalent bonds

Silicon is solid at room temperature and pressure, it is a semi-metallic element and semi-conductor of electricity which is the second most abundant element on earth, after oxygen.

Germanium is solid at room temperature and pressure which usually forms four covalent bonds with a tetrahedral arrangement. Germanium, is a hard, brittle, greyish-white and crystalline semi-metallic element which is a semi-conductor of electricity.

Tin is a silver-white solid metal that conducts electricity at room temperature and pressure. At temperatures below 13 °C, it often changes into an allotropic (distinctly different) form known as grey tin, which is an amorphous, greyish powder.

Lead, is a dense, bluish-grey solid metallic element at room temperature and pressure which conducts electricity and has a smaller ionization energy than carbon and silicon, it often loses two of its valence electrons to form the Pb2+ ions.

Special features of carbon chemistry

i. Catenation

Catenation is the ability of an element to form bonds between atoms of the same element to form long chains or ring: -C-C-C-C-C-C-C-C-C-C-C-C-C-

Carbon forms chains and rings, with single, double and triple bonds. To be able to catenate, an element such as carbon must have a valency of 2 and above.

It should also form bonds like C-C which are similar in strength to those of C to other elements, particularly C-O bonds.

Silicon forms -Si-O-Si-bonds predominantly.

ii) Multiple bonds

Carbon forms double bonds and triple bonds between carbon atoms and that bonding is formed by one Sigma bond and one π bond for double bond, one Sigma bond and two π bonds in a triple bond.

Checking-up 9.1

1. Explain the reason why diamond has a higher melting point than silicon.

2. Discuss the increase in metallic character when moving down in group 14 elements from carbon to lead.

3. Diamond and graphite are allotropes of carbon,

a) Draw their three dimensional structures.

b) With reference to their structures, compare the hardness of diamond and graphite.

c) With reference to their structures, compare their electrical conductivity and explain.

4. Germanium has the same structure as diamond. Explain the type of bonds that exist in the two elements.

5. The first element in a group in the periodic table exhibits anomalous properties compared with other members. Use carbon to illustrate this statement.

9.2. Chemical properties of Group 14 elements

Activity 9.2 (a)

1. Get a piece of charcoal and burn it. Observe and write the chemical equation that represents the change that takes place when the charcoal burns.

2. a) Put about 1 gram of carbon charcoal in a boiling tube.

b) Add 1 ml of concentrated nitric acid.

c) Heat strongly on a Bunsen burner flame using a test tube holder.

d) Observe and note the changes during heating.

e) Deduce the chemical changes that have occurred.3. Write the molecular structure of carbon dioxide, carbonate ion and carbon monoxide.

4. Describe how CO2 gas dissolves in water and state the nature of the solution formed when it is in aqueous solution.

5. Describe 2 chemical properties of amphoteric substances.

All the elements of Group 14 form four single bonds such as in tetrahalides of formula CCl4, SiCl4, GeCl4, SnCl4 with the exception of PbBr4 and PbI4 that do not exist because bromine and iodine are not sufficiently strong oxidizing agents to convert Pb to Pb4+. Germanium, tin and lead can ionize to form Ge2+, Sn2+ and Pb2+ and also are capable of forming Ge4+, Sn4+, Pb4+ respectively.

Germanium dioxide GeO2 is more stable than germanium monoxide GeO. The oxide of tin in +4 oxidationstate is slightly more stable than that in +2 state. Silicon oxide predominantly exists as SiO2. SiO is too unstable to exist at room temperature and pressure but may exist at about 2000 0C. Carbon oxides are stable in +4 state. Thus CO which is unstable reacts exothermically to form the stable compound of CO2. In lead compounds, the +2 state is more stable than +4 state. PbO2 is unstable and is a very strong oxidizing agent.

The stability of compounds with the oxidation state of +2 in group 14 elements generally increases on moving down the group from carbon to lead.

The stability of compounds with the oxidation state of +4 in group 14 elements increases on moving upward the group from lead to carbon. The bonding in tetravalent compounds is predominantly covalent.

Inert pair effect

Inert pair effect refers to the inability of the outermost s-electrons to participate in chemical bonding in Ge, Sn and Pb elements of group 14 and hence only 2 electrons of the outermost p orbital are involved.

The outermost s sub-energy level electrons are much more tightly attracted to the nucleus than the outermost p orbital. As we move down the group, the difference in energy level between s and p orbitals becomes wider.

So if we use weak oxidizing agents, only 2 electrons in p orbitals are removed. If we use a strong oxidizing agent, 2 electrons in s orbital and 2 electrons in p orbital are all removed from the shell.

Reaction of group 14 elements with chlorine:

The group 14 elements mainly form tetrachlorides in the form MCl4 but lead doesn’t directly with Chlorine.

PbCl4 is unstable molecule and is prepared from PbO2. It decomposes to make stable PbCl2

Reaction of group 14 elements with hydrogen: Carbon graphite reacts with hydrogen gas at 30 0C to 950C to produce a mixture of methane (CH4), ethane (C2H6) and propane (C3H8). The primary product is methane and the other products are produced due to free radical reaction of the methane.

Apart from carbon, other elements of group 14 do not react with hydrogen directly.

Activity 9.2 (b)

1. Put about 0.5 g of Al2O3solid into2 test tubes then add 2ml of water.

2. Pour 3 ml of 1 mol/litre HCl solution in the third test tube and add 2 drops of universal indicator.

3. Pour 3 ml of 1 mol/litre NaOH solution in the fourth test tube and add 2 drops of universal indicator.

4. Put the first portion of Al2O3 solid in the test tube containing HCl solution and shake gently.

5. Put the first portion of Al2O3 solid in the test tube containing NaOH solution and shake gently.

6. Note the observable changes in both tests.

7. Put about 0.1 g of MgO in a test tube.

8. Add 2ml of water to the MgO solid in the test tube and shake the mixture.

9. Put 2 drops of phenolphthalein indicator in MgO solution.

10. Add 5ml of 1 mol/litre HNO3 to the solution containing MgO solution.

11. Note the observable changes.

12. Interpret the acid –base character of the solutions of CaO and Al2O3

Reaction of group 14 elements with acids and bases:

Carbon does not react with dilute acids but reacts with hot, concentrated acids:

Silicon reacts with concentrated HNO3:

Silicon reacts with concentrated sodium hydroxide (base)

Tin reacts with concentrated nitric acid to form tin dioxide and nitrogen dioxide:

Cold dilute nitric acid reacts with tin to form a mixture of tin II nitrate and ammonium nitrate solution.

Tin is sufficiently metallic to form various mixtures of Sn(SO4)2 and hydrated SnO2and sulphur dioxide when it reacts with hot, concentrated sulphuric acid.

Tin reacts with strongly non-oxidising acids in solution to form tin(II) salts and hydrogen gas:

Tin reacts with concentrated or molten sodium hydroxide base to form a stannate ion SnO22- and hydrogen gas:

Lead reacts with hot concentrated sulphuric acid according to the following equation to form lead(II) sulphate and sulphur dioxide:

The reaction of lead with concentrated nitric acid forms lead(II) nitrate and nitrogen dioxide as shown in the equation:

The reaction of lead with dilute nitric acid forms lead(II) nitrate and nitrogen monoxide as shown in the equation:

The reaction of lead with boiling, concentrated hydrochloric acid forms lead(II) chloride and hydrogen gas following the equation:

Lead which is amphoteric; reacts with hot, concentrated sodium hydroxide to form trihydroxoplumbate (II) and hydrogen gas as shown by the equation:

Checking-up 9.2

1) Explain why CO2 does not react with acids but other dioxides of group 14 react with acids.

2) State the expected observation when CCl4 and SiCl4 are treated separately with water. Write the relevant balanced equations if the reaction is possible.

3) Compare the stability of PbO and PbO2 upon heating. Write the balanced equa-tions.

4) Suggest the way that can be used to show that Pb is amphoteric.

9.3 Difference between the chemical reactions of the oxides and chlorides of Group 14 elements.

Activity 9.3

1. Measure 0.5g of lead oxide or decompose the same quantity of lead nitrate crystals by heating.

2. Divide it into 5 portions and put each portion in a test tube.

3. In the first test tube, add 2mL of dilute hydrochloric acid solution in which uni-versal indicator has been dissolved.

4. In the second test tube, add 2ml sodium hydroxide solution in which phenol-phthalein indicator has been dissolved.

5. Note the observations and deduce the acid–base nature of lead oxide.

Interpretation of results of the above activity 1

The reactions that take place are:

Conclusion

Lead (II) oxide behaves as a base when it reacts with acids and behaves as an acid when it reacts with bases, thus it is an amphoteric oxide.

Reaction of CO2 with water:

Carbon dioxide reacts with water to form carbonic acid.

he reaction is reversible, but H2CO3 dissociates to produce H+ and HCO3- ; the dissociation of carbonic acid is also reversible since carbonic acid is a weak acid.

Carbon dioxide is an acidic oxide, so H2CO3 does not react with acids.

Reaction of CO2 with sodium hydroxide:

Carbon dioxide reacts with sodium hydroxide, NaOH, to form sodium carbonate salt.

Reactions of CCl4 with water, acids and bases:

Carbon tetrachloride does not react with water, acids, and alkaline solutions.

Reaction of SiO2with water: Silicon dioxide does not react with water.

Reaction of SiO2with acids: Silicon dioxide reacts with HF acid only:

This reaction is the one used to write on the glass, for example on the windscreen of vehicle.

Reaction of SiO2 with bases: Silicon dioxide reacts with hot concentrated NaOH solution

Reaction of SiCl4with water:

Silicon dioxide undergoes hydrolysis in water, this is due to the fact that Silicon atom possesses vacant d-orbitals that can partially accommodate an oxygen atom during the reaction unlike in the case of CCl4..

Reaction of SiCl4 with acids, bases and water: SiCl4(l) does not react with acids, bases and water.

Reaction of SnO2 with water: SnO2 does not react with water.

Reaction of SnO and SnO2 with acids:SnO2does not dissolve in dilute acids but it reacts with hot concentrated sulphuric acid, H2SO4

Checking-up 9.3

1. For each of the following statements, state whether it is true or false.If you consider the statement to be false, justify your answer and if it is true, give a brief explanation or write a chemical equation.

a) All the oxides of group 14 with the formula of MO are amphoteric.

b) Some carbon allotropes can conduct electricity.

2. Write the equations to show the amphoteric nature of lead (II) oxide.

3. Discuss the reactions of tin (II) oxide and lead (II) oxide with sulphuric acid.

4. Explain why tetrachlorides of Si, Sn and Pb hydrolyse in water but carbon does not. Write the equations ofthereaction for the hydrolysis of SiCl4 and SnCl4

5. Explain why PbCl2 does not hydrolyse in water but PbCl4 does.

he thermal stability of dioxides and tetrachlorides of group 14 decreases down the group from carbon to lead due to the increase in sizeof the group 14 atoms.

The increase in size of group 14 atoms down the group results in weaker bonds. Thus PbCl4 is the least stable of the chlorides of group 14 elements.

9.4.1. Stability of oxides of group 14 elements

The thermal stability of XO increases downward the group from carbon to lead.

X represents: C, Si, Sn and Pb. The stability of XO increases in the order:

CO < SiO < SnO < PbO

Silicon monoxide is obtained by heating SiO2 and Si under special conditions.

The main valency of oxides of tin, is +4, in other words SnO2 is the most stable tin oxide; the valency of +2, SnO, is subsidiary.

On exposure to air at room temperature, SnO2 is formed.

The main valency of oxides of lead, is +2, in other words, PbO is the most stable lead oxide; the valency of +4 is subsidiary. PbO is the most stable oxide of group 14 elements. PbO2 cannot be prepared from lead monoxide because it decomposes at 3000C.

9.4.2. Thermal stability of halides of group 14 elements

CCl4, SiCl4 and GeCl4 are very stable and they do not decompose even at high temperatures.

SnCl4 decomposes only when it is heated to form SnCl2 and Cl2, so SnCl4 is more stable.

PbCl4 decomposes readily without heating to form PbCl2 and Cl2.

Thermal stability of tetrahalides of group 14 halides decreases down the group from carbon to lead because the bonds in Group 14 halides become longer and weaker down the group.

Thermal stability of tetrahalides of group 14 also decreases from fluorides to iodides.

All the tetrahalides are volatile covalent compounds except SnF4 and PbF4 which have some ionic character.

PbBr4 and PbI4 do not exist since iodine and bromine are not strong oxidizing agents enough to remove all 4 valence electrons.

9.4.3. Thermal stability of hydroxides of group 14 elements

Lead (II) hydroxide Pb(OH)2 decomposes at temperatures ranging between 1000C and 1450C to produce PbO and H2O .

Tin (II) hydroxide Sn(OH)2 decomposes easily between 600C and 1200C to produce SnO and H2O.

Carbon and silicon are non-metals, so they don’t form hydroxides.

9.4.4. Chemical tests of Sn2+, Pb2+, HCO3- :

Table 9.2: Chemical tests of ions, Pb2+,Sn2+, HCO3- (aq) and CO32-(aq) :

•Mercury chloride, HgCl2 solution is reduced to white precipitate of dimercury dichloride, Hg2Cl2 which is further reduced to a grey-black deposit of Hg by Sn (II) but not with Sn(IV)

•Excess NaOH solution causes the white precipitate of SnO and SnO2 oxides to re-dissolve.

•Sodium stannate (II) reduces the colorless bismuth nitrate, Bi(NO3)3 solution to elemental bismuth (black deposit).

•Sodium stannate (IV) has no reducing properties on Bi(NO3)3

Table 9.3: Identification of ions

When heat is applied to a solid, to test the presence of carbonate or hydrogencarbonate, use calcium hydroxide solution (lime water) which turns milky in the presence of

CO2 formed by the decomposition of the sample.

CO2 can evolve from compounds containing: CO32-, HCO32-, C2O42- .

Checking-up 9.4

1.Give an explanation why carbon tetrachloride molecule is non-polar despite the fact that the electronegativity difference between carbon and chlorine is so big.

2. State what is observed when lead dioxide is warmed with concentrated HCl acid solution and write the equation for the reaction.

3. Lead hydroxide Pb(OH)2 can react with an acid and with a base. Write the ionic equation for the reaction of Pb(OH)2 with: a) An acidb) A base.

4. Explain the reason why lead tetrachloride is unstable such that it decomposes easily compared to other tetrachlorides of group 14 tetrachlorides.

5. You are given a mixture of HCO3- and SO42- ions, explain how you can show that the mixture contains HCO3- ions.

6. Give a chemical reagent test(s) you can use to confirm that a solution contains Sn2+ ions.

7. a) Give a chemical reagent test you can use to differentiate between HCO3- and CO32- ions and deduce the observable changes.

b) Write the chemical equations of reactions to illustrate the reactions that have taken place.

8.Explain the trend of thermal stability of group 14 tetrachlorides and dichlorides on descending down the group.

9.5 Variation in stability of oxidation state +2 and +4 down the Group 14 elements.

Activity 9.5

1. Write the electronic configuration of carbon and silicon.

2. Predict the stability of dichloride compounds in group 14 elements from car-bon to lead and explain

3. Discuss the influence of variation in electronegativity to the bond strength in compounds of group 12 from beryllium to barium.

Use the bond strength between MgCl2 and CaCl2 to explain the answer.

The 2 oxidation states exhibited by elements of group 14 are +2 and +4 in their compounds.

The oxidation state+4 involves loss of 4 valence electrons, s2 and p2, while the oxidation state +2 involves loss of only 2 electrons in the p2.

The stability of the +2 oxidation state increases from carbon to lead while +4 oxidation state increases from lead to carbon.

Inert pair effect refers to the inability of the outermost s-electrons to participate in chemical bonding in Ge, Sn and Pb elements of group 14 and hence only the outermost p-electrons are involved.

The electrons in s orbital are much more tightly bound to the nucleus than p-electrons. As we move down the group, the difference in energy level between s sub-shell and p sub-shell becomes wider.

So if we use weak oxidizing agents, only 2-p electrons are removed. If we use a strong oxidizing agent 2 s-electrons and 2-p electrons are all removed from the shell.

If the elements in group 14 form +2 ions, they will lose the p electrons leaving the s-electrons pair unused. For example, to form Pb2+ ions lead will lose the two 6p electrons but the 6s electrons will remain in its sub-energy level.

The inert pair effect shown in Pb2+ explains why the compounds of lead are predominantly ionic.

The compounds of group 14 with +4 oxidation state are more covalent than those with +2 oxidation state. The covalent character increases as a bigger anion is bonded to the group 14 element. For example, PbBr2 is more covalent than PbCl2. A big anion is more polarisable than a small anion.

Checking up 9.5

1. Explain the reason why carbon does not form CCl2 compound.

2. Explain the term”inert pair effect”. Give examples to illustrate.

3. Indicate which compound between PbCl2 and PbF2 is more covalent than an-other and justify.

4. Explain the reason why PbBr4 and PbI4 do not exist whereas PbCl4 exists

9.6 Uses of Group 14 elements

Activity 9.6

1. State 2 uses of carbon in daily use.

2. Diamond is considered to be very hard, predict its application.

3. Carbon forms hydrocarbons such as methane when combined with hydrogen. State where methane is obtained in Rwanda and indicate its uses.

4. Silicon is a semi-conductor. Predict its uses in electric gadgets.

5. Tin resists corrosion. Discuss where it can be used to prevent corrosion.

Group 14 elements are used in numerous applications. The wide applications are due to the fact that some of the elements in the group are non-metals, metals and metalloids.

Carbon uses:

•As a component of fuel for combustion as charcoal or coal.

•As the main component of crude oil and its derivatives used in our everyday life such: fuel, plastics, etc...

•As good chemical reducing agent used in extraction of metals (metallurgy).

.As a lubricant in moving parts of machines, to make electrodes, in lead pencils when mixed with clay. •Carbon isotope, C-14 isotope is used in archaeological dating.

•Diamond is used to make glass cutters, drilling devices and as abrasive for smoothing hard materials as precious gemstone in jewelry and ornamental objects; it is also a precious stone appreciated in jewelry.

Silicon uses:

•Silicon is used as a semi-conductor in transistors in electrical gadgets such as radios, computers, amplifiers etc.

.•Silicon in form of silicates is used in ceramics and in glass production.

•Silicon is also used in medicine to make silicone implants.

•Many rocks that we use for building our houses and other buildings are Silicates.

•Ferrosilicon alloy is used as a deoxidizer in steel manufacture.

•Silicon dioxide can be used to produce toothpastes and in semiconductors; silicon dioxide is the main component of sand, a raw material in the manufacture of glass.

Germanium uses:

•Germanium being a metalloid, is used in transistors in electrical gadgets such radios, computers, amplifiers etc..

Tin uses:

•Tin is used in plating steel sheets to resist corrosion; it is used for example to make canned tins to avoid the corrosion of the materials which are in contact with an acid medium.

Lead uses:

•Lead is used in making linings of vessels which are used in industrial production of sulphuric acid.

•Lead bricks alloyed with 4% antimony is used in radiation shielding

•Lead is used in accumulator plates of the batteries and as shielding materials against dangerous radiations such as X-rays, gamma rays etc.

Checking –up 9.6

1. Describe how diamond is used on a large scale.

2. Describe the materials that are manufactured using lead, tin and silicon as the main component.

3. Describe 3 compounds which contain:

i) Silicon

ii) Lead.

9.7 Define diagonal relationship

Activity 9.7

1. Discuss and compare the type of bonding in beryllium and aluminium elements.

2. Describe 2 chemical properties of oxides of beryllium and aluminium com-pounds.

3. Describe the solubility of lithium and magnesium in water.

4. Discuss the similarities in physical properties and chemical properties of BeCl2and AlCl3.

9.7.1.The diagonal relationship in groups1 & 2, 13 &14 elements

Diagonal relationships are similarities between pairs of elements in different groups which are adjacent to one another in the second and the third period of the periodic table.

These pairs are in Groups 1 and 2(Li/Mg), Groups 2 and 13(Be/Al) and Groups 13 and 14(B/Si). They exhibit similar properties; for example, boron and silicon are both semi-conductors, they form halides that are hydrolyzed in water and have acidic oxides.

NB: This relationship does not hold for allcertain groups; the only significant relationships are the ones mentioned above.

Such a relationship occurs because as we move across and descend in the periodic table, there are opposite effects.

On moving across a period of the periodic table from left to right, the size of the atoms decreases, and on moving downward a group, the size of the atoms increases.

If you look into those two movements in the first 3 positions of Period 2 of the periodic table, the result is: Li and Mg, Be and Al, B and Si, all couples in diagonal relationship, or diagonally adjacent resemble each other in their behavior. This resemblance, called diagonal relationship is attributed to the similarity of their size; for example the atomic radii of lithium and magnesium are similar (1.55×10-10m and 1.6×10-10m respectively).

Due to this relationship, those elements will show similar chemical properties and their compounds will have similar properties.

9.7.2. Diagonal Relationship of Li with Mg

At room temperature and pressure:

1. Li and Mg react with air to form only normal oxides, whereas Na forms a peroxide but metals below Na in the group, form superoxide.

2. The bicarbonates of Li and Mg do not exist in solid state, they exist in solution only.

3. Nitrates thermally decompose, to give oxides, nitrogen dioxide and oxygen:

4. Hydroxides of both Li and Mg decompose on heating to give oxides.

Hydroxides of both Li and Mg are weak alkalis.

NB: Hydroxides of other alkali metals are stable on heating while their nitrates

9.7.3. Diagonal Relationship of Be with Al

•Due to many similarities in their properties and reactions, aluminium and beryllium will be presented together even though they are in different groups of the periodic table.

Atomic radii: Be= 111 pm, Al= 143 pm

Ionic radii: Be2+= 31pm, Al3+ = 50 pm

Boiling point: Be = 2477oC, Al =2467oC

•Oxides: BeO and Al2O3 are both amphoteric, the rest of Group 2 oxides are only basic.

•Reaction with dilute acids: Be and Al are fairly resistant unless amalgamated or when they are very finely divided, other Group 2 metals readily react.

•Like aluminum, beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.

• Reaction with conc. nitric acid: Be and Al are rendered passive, the rest of Group 2 metals readily form salts.

•Beryllium and aluminium have an appreciable covalent character of compounds (e.g. the chlorides are predominantly covalent).

9.7.4. Diagonal relationship between Boron and Silicon

Due to its small size and similar charge/mass ratio, boron differs from other group 13 members, but it closely resembles silicon, the second element of group 14 to exhibit diagonal relationship. Some important similarities between boron and silicon are given below:

•Both boron and silicon are typical non-metals that exist as non-metallic giant covalent 3D lattices, having high melting points and boiling points, they have almost the same densities (B=2.35gml–1 S=2.34 g//ml), low atomic volumes and bad conductor of electricity. However both elements are used as semiconductors.

•Both B and Si do not form cations but only form covalent compounds.

•Both B and Si form numerous volatile hydrides which spontaneously catch fire on exposure to air and are easily hydrolyzed.

•Both B and Si form weak acids like H3BO3 and H2SiO3when they react with water.

•B and Si are non-metals and their oxides are readily soluble in alkalis.

Checking-up 9.7

1. Write the chemical equation for the reaction between sodium hydroxide and:

a) Boron oxide

b) Silicon dioxide.

2. Be and Al belong to two different groups but their chlorides present the same properties; explain the origin of that resemblance.

3.Describe physical properties ofhydrides of boron and silicon and their behaviour in water.

4. Explain the acidity of solutions of beryllium chloride and aluminium chloride.

9.8. End unit assessment

I: Fill in the following statements with a missing word:

1. The arrangement of atoms in diamond structure is called....................

2.......................is the only element of group 14 whose chloride does not hydrolyse in water.

3.....................is a semi-metallic element of group 14 whose oxide reacts with HF acid only.

4.....................is the only element of group 14 that does not exist in various allotropic forms.

5......................is the only element of group 14 whose compounds in the oxidation state of +2 is more stable than that of +4.

II. Answer the following questions:

6. Write the equations for the reaction of decomposition of:a) Lead (II) hydroxideb) Tin tetrachloride

7. Explain the amphoteric character of tinby using appropriate equations of reaction.

8. Discuss the stability of +2 oxidation state as you move down in group 14 elements.

9. Explain the reason why the melting and boiling points of group 14 elements decrease down the group.

10. Explain by using appropriate equations, how you can confirm the presence of Sn2+ ions in a solution.

11. Discuss the origin of similarities of chemical properties of boron and silicon, use appropriate equations to emphasize the explanation.

12. Given the following three chloride, CCl4, SnCl2, SnCl4. Classify them in two groups of ionic and covalent compounds and justify your answer.

13. Complete the following chemical equation if possible and justify: