        (i) At the beginning (ii) observation at the end of the experiment     (a) (i) Identify substance M..................................................................................         (ii) State two reasons for the suitability of substance M for this experiment     (b) Write the equation for the reaction of substance M and the active part of air
    (c) (i) Using the letters Y and X write an expression for the percentage of the active part of air         (ii) The expression in (c)(i) above gives lower value than the expected. Explain      (d) (i) Explain the observation made when litmus paper is dipped into the beaker at the end of the               experiment           (ii) Name the active part of air ................................................................................................            (iii) Suggest another method that can be used to determine the active part of air   15. A piece of phosphorous was burnt in excess air. The product obtained was shaken with a small  amount of hot water to make a solution     i) Write an equation for the burning of phosphorus in excess air     ii) The solution obtained in (b) above as found to have pH of 2.  Give reasons for this                     observation 16.


        (ii) State two reasons for the suitability of substance M for this experiment     (b) Write the equation for the reaction of substance M and the active part of air
    (c) (i) Using the letters Y and X write an expression for the percentage of the active part of air         (ii) The expression in (c)(i) above gives lower value than the expected. Explain      (d) (i) Explain the observation made when litmus paper is dipped into the beaker at the end of the               experiment           (ii) Name the active part of air ................................................................................................            (iii) Suggest another method that can be used to determine the active part of air   15. A piece of phosphorous was burnt in excess air. The product obtained was shaken with a small  amount of hot water to make a solution     i) Write an equation for the burning of phosphorus in excess air     ii) The solution obtained in (b) above as found to have pH of 2.  Give reasons for this                     observation 16. Study the set-up below and answer the questions that follow:- (a) State two observations that would be made after one week. Explain (b) Write the equation of the reaction taking place in the test-tube 17.


(iii) Suggest another method that can be used to determine the active part of air   15. A piece of phosphorous was burnt in excess air. The product obtained was shaken with a small  amount of hot water to make a solution     i) Write an equation for the burning of phosphorus in excess air     ii) The solution obtained in (b) above as found to have pH of 2.  Give reasons for this                     observation 16. Study the set-up below and answer the questions that follow:- (a) State two observations that would be made after one week. Explain (b) Write the equation of the reaction taking place in the test-tube 17. Fe3O4 and FeO are oxides of iron which can be produced in the laboratory (a) Write chemical equation for the reaction which can be used to produce each of the oxides (b) Wire an ionic equation for the reaction between the oxide, Fe3O4 and a dilute acid.


A piece of phosphorous was burnt in excess air. The product obtained was shaken with a small  amount of hot water to make a solution     i) Write an equation for the burning of phosphorus in excess air     ii) The solution obtained in (b) above as found to have pH of 2.  Give reasons for this                     observation 16. Study the set-up below and answer the questions that follow:- (a) State two observations that would be made after one week. Explain (b) Write the equation of the reaction taking place in the test-tube 17. Fe3O4 and FeO are oxides of iron which can be produced in the laboratory (a) Write chemical equation for the reaction which can be used to produce each of the oxides (b) Wire an ionic equation for the reaction between the oxide, Fe3O4 and a dilute acid. 18.Below is a list of oxides.


Study the set-up below and answer the questions that follow:- (a) State two observations that would be made after one week. Explain (b) Write the equation of the reaction taking place in the test-tube 17. Fe3O4 and FeO are oxides of iron which can be produced in the laboratory (a) Write chemical equation for the reaction which can be used to produce each of the oxides (b) Wire an ionic equation for the reaction between the oxide, Fe3O4 and a dilute acid. 18.Below is a list of oxides. MgO, N2O, K2O, CaO ans Al2O3 Select:- a) A neutral oxide. b) A highly water soluble basic oxide. c) An oxide which can react with both sodium hydroxide solution and dilute hydrochloric acid. 19.


b) A highly water soluble basic oxide. c) An oxide which can react with both sodium hydroxide solution and dilute hydrochloric acid. 19. The diagram below shows students set-up for the preparation and collection of oxygen gas
(a) Name substance X used (b) Write an equation to show the reaction of sodium peroxide with the substance named in 1(a) 5. Water and hydrogen 1. (a) Hydrogen can reduce coppers Oxide but not alluminium oxide. Explain (b) When water reacts with potassium metal the hydrogen produced ignites explosively                    on the surface of water.                    (i) What causes this ignition?                    (ii) Write an equation to show how this ignition occurs 2.


(a) Hydrogen can reduce coppers Oxide but not alluminium oxide. Explain (b) When water reacts with potassium metal the hydrogen produced ignites explosively                    on the surface of water.                    (i) What causes this ignition?                    (ii) Write an equation to show how this ignition occurs 2. In an experiment, dry hydrogen gas was passed over hot copper (II) oxide in a combustion             tube as  shown in the diagram below:-
(a) Complete the diagram to show how the other product, substance R could be collected                  in the laboratory. (b) Describe how copper could be obtained from the mixture containing copper (II) oxide
3.     The setup below was used to investigate the reaction between metals and water.


                   (ii) Write an equation to show how this ignition occurs 2. In an experiment, dry hydrogen gas was passed over hot copper (II) oxide in a combustion             tube as  shown in the diagram below:-
(a) Complete the diagram to show how the other product, substance R could be collected                  in the laboratory. (b) Describe how copper could be obtained from the mixture containing copper (II) oxide
3.     The setup below was used to investigate the reaction between metals and water.     (a) Identify solid X and state its purpose         Solid X ………………..………………………………………………………………………..          Purpose ………………………………………………………………………………………..     (b) Write a chemical equation for the reaction that produces the flame. 4. Gas P was passed over heated magnesium ribbon and hydrogen gas was collected as shown in the diagram below:
(i) Name gas P ...............................................................................................................


    (b) Write a chemical equation for the reaction that produces the flame. 4. Gas P was passed over heated magnesium ribbon and hydrogen gas was collected as shown in the diagram below:
(i) Name gas P ............................................................................................................... (ii) Write an equation of the reaction that takes place in the combustion tube (iii) State one precaution necessary at the end of this experiment 5.       When hydrogen is burnt and the product cooled, the following results are obtained as shown              in the diagram below:           (a) Write the equation for the formation of liquid Y           (b) Give a chemical test for liquid Y
6. Jane set-up the experiment as shown below to collect a gas. The wet sand was heated before heating Zinc granules (a) Complete the diagram for the laboratory preparation of the gas (b) Why was it necessary to heat wet sand before heating Zinc granules? 7.


(ii) Write an equation of the reaction that takes place in the combustion tube (iii) State one precaution necessary at the end of this experiment 5.       When hydrogen is burnt and the product cooled, the following results are obtained as shown              in the diagram below:           (a) Write the equation for the formation of liquid Y           (b) Give a chemical test for liquid Y
6. Jane set-up the experiment as shown below to collect a gas. The wet sand was heated before heating Zinc granules (a) Complete the diagram for the laboratory preparation of the gas (b) Why was it necessary to heat wet sand before heating Zinc granules? 7.            (a) Between N and M which part should be heated first? Explain (b) Write a chemical equation for the reaction occurring in the combustion tube. 8.


7.            (a) Between N and M which part should be heated first? Explain (b) Write a chemical equation for the reaction occurring in the combustion tube. 8. The set-up below was used to investigate electrolysis of a certain molten compound;- (a) Complete the circuit by drawing the cell in the gap left in the diagram (b) Write half-cell equation to show what happens at the cathode (c) Using an arrow show the direction of electron flow in the diagram above
9.Hydrogen can be prepared by reacting zinc with dilute hydrochloric acid. a) Write an equation for the reaction.             b) Name an appropriate drying agent for hydrogen gas. c) Explain why copper metal cannot be used to prepare hydrogen gas. d) Hydrogen burns in oxygen to form an oxide.    (i) Write an equation for the reaction.


c) Explain why copper metal cannot be used to prepare hydrogen gas. d) Hydrogen burns in oxygen to form an oxide.    (i) Write an equation for the reaction.    (ii) State two precautions that must be taken before the combustion begins and at the end of         the combustion. e) Give two uses of hydrogen gas. f) When zinc is heated to redness in a current of steam, hydrogen gas is obtained. Write an     equation for the reaction. g) Element Q reacts with dilute acids but not with cold water. Element R does not react with     dilute acids. Elements S displaces element P from its oxide. P reacts with cold water. Arrange     the four elements in order of their reactivity, starting with the most reactive. h) Explain how hydrogen is used in the manufacture of margarine. 10. a) The set-up below  is used to investigate the properties  of  hydrogen.


h) Explain how hydrogen is used in the manufacture of margarine. 10. a) The set-up below  is used to investigate the properties  of  hydrogen.           i) On the diagram, indicate what should be done for the reaction to occur           ii) Hydrogen gas is allowed to pass through the tube for some time before it is lit. Explain           iii) Write an equation for the reaction that occurs in the combustion tube          iv) When the reaction is complete, hydrogen gas is passed through the apparatus until they                cool down .   Explain          v) What property of hydrogen is being investigated?         vi) What observation confirms the property stated in (v) above?         vii) Why is zinc oxide not used to investigate this property of hydrogen gas? 11.


        vi) What observation confirms the property stated in (v) above?

        vii) Why is zinc oxide not used to investigate this property of hydrogen gas?

11.

The set up below was used to collect gas K, produced by the reaction between water and               calcium metal.

(a) Name gas K……………………………………………………………..

(b) At the end of the experiment, the solution in the beaker was found to be a weak base.

Explain       why the solution is a weak base


ATOMIC STRUCTURE AND THE PERIODIC TABLE Preface The structure of the atom is extensively covered here. The periodic table is the arrangement of atoms of elements based on their atomic structure. Emphasis on the trends across and down the periodic table of atoms is important for the teacher facilitator. This work should be covered before “Chemical bonding and structure”. The two complements each other in understanding periodicity. Candidate-user preparing for any secondary level chemistry from members of A.ATOMIC STRUCTURE The atom is the smallest particle of an element that take part in a chemical reaction.The atom is made up of three subatomic particle: (i)Protons (ii)Electrons (iii)Neutrons (i)Protons 1.The proton is positively charged 2.Is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons in a atom of an element is its Atomic number (ii)Electrons 1.The Electrons is negatively charged 2.Is found in fixed regions surrounding the centre of an atom called energy levels/orbitals. 3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element
2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements.


Candidate-user preparing for any secondary level chemistry from members of A.ATOMIC STRUCTURE The atom is the smallest particle of an element that take part in a chemical reaction.The atom is made up of three subatomic particle: (i)Protons (ii)Electrons (iii)Neutrons (i)Protons 1.The proton is positively charged 2.Is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons in a atom of an element is its Atomic number (ii)Electrons 1.The Electrons is negatively charged 2.Is found in fixed regions surrounding the centre of an atom called energy levels/orbitals. 3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element
2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements. Element Symbol Protons Electrons Neutrons Atomic number Mass number Hydrogen H 1 1 0 1 1 Helium He 2 2 2 2 4 Lithium Li 3 3 4 3 7 Beryllium Be 4 4 5 4 9 Boron B 5 5 6 5 11 Carbon C 6 6 6 6 12 Nitrogen N 7 7 7 7 14 Oxygen O 8 8 8 8 16 Fluorine F 9 9 10 9 19 Neon Ne 10 10 10 10 20
3 Sodium Na 11 11 12 11 23 Magnesium Mg 12 12 12 12 24 Aluminium Al 13 13 14 13 27 Silicon Si 14 14 14 14 28 Phosphorus P 15 15 16 15 31 Sulphur S 16 16 16 16 32 Chlorine Cl 17 17 18 17 35 Argon Ar 18 18 22 18 40 Potassium K 19 19 20 19 39 Calcium Ca 20 20 20 20 40 Most atoms of elements exist as isotopes.


3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element
2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements. Element Symbol Protons Electrons Neutrons Atomic number Mass number Hydrogen H 1 1 0 1 1 Helium He 2 2 2 2 4 Lithium Li 3 3 4 3 7 Beryllium Be 4 4 5 4 9 Boron B 5 5 6 5 11 Carbon C 6 6 6 6 12 Nitrogen N 7 7 7 7 14 Oxygen O 8 8 8 8 16 Fluorine F 9 9 10 9 19 Neon Ne 10 10 10 10 20
3 Sodium Na 11 11 12 11 23 Magnesium Mg 12 12 12 12 24 Aluminium Al 13 13 14 13 27 Silicon Si 14 14 14 14 28 Phosphorus P 15 15 16 15 31 Sulphur S 16 16 16 16 32 Chlorine Cl 17 17 18 17 35 Argon Ar 18 18 22 18 40 Potassium K 19 19 20 19 39 Calcium Ca 20 20 20 20 40 Most atoms of elements exist as isotopes. Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number. By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e.


Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number. By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e. mass number atomic number m n X symbol of element Below is the conventional method of writing the 1st twenty elements showing the mass numbers and atomic numbers; 2He 4Be 6C 147N 168O 10Ne 12Mg 14Si 16S 18Ar 20C The table below shows some common natural isotopes of some elements Element Isotopes Protons Electrons Neutrons Atomic number Mass number Hydrogen 1H(deuterium) 31H(Tritium) 1 1 1 1 1 1 0 2 3 1 1 1 1 2 17Cl 3717Cl 17 17 17 17 18 20 17 17 35 19K 4019K 19 19 19 19 20 21 19 19 39 40
4 4119K 19 19 22 19 8O 188O 8 8 8 8 8 10 8 8 16 92U 23892U 92 92 92 92 143 146 92 92 235 10Ne 10Ne 10 10 10 10 10 10 12 10 11 10 10 10 22 20 21 The mass of an average atom is very small (10-22 g).Masses of atoms are therefore expressed in relation to a chosen element. The atom recommended is 12C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) .


i.e. mass number atomic number m n X symbol of element Below is the conventional method of writing the 1st twenty elements showing the mass numbers and atomic numbers; 2He 4Be 6C 147N 168O 10Ne 12Mg 14Si 16S 18Ar 20C The table below shows some common natural isotopes of some elements Element Isotopes Protons Electrons Neutrons Atomic number Mass number Hydrogen 1H(deuterium) 31H(Tritium) 1 1 1 1 1 1 0 2 3 1 1 1 1 2 17Cl 3717Cl 17 17 17 17 18 20 17 17 35 19K 4019K 19 19 19 19 20 21 19 19 39 40
4 4119K 19 19 22 19 8O 188O 8 8 8 8 8 10 8 8 16 92U 23892U 92 92 92 92 143 146 92 92 235 10Ne 10Ne 10 10 10 10 10 10 12 10 11 10 10 10 22 20 21 The mass of an average atom is very small (10-22 g).Masses of atoms are therefore expressed in relation to a chosen element. The atom recommended is 12C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) . All other atoms are compared to the mass of 12C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to 1/12 an atom of 12C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e; RAM = mass of atom of an element 1/12 of one atom of 12C isotope Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability. Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples.


All other atoms are compared to the mass of 12C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to 1/12 an atom of 12C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e; RAM = mass of atom of an element 1/12 of one atom of 12C isotope Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability. Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples. a) Chlorine occurs as 75% 3517Cl and 25% 3717Cl isotopes. Calculate the relative atomic mass of Chlorine. Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as;
5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope.


Calculate the relative atomic mass of Chlorine. Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as;
5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope.


Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as;
5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope. d) Calculate the relative atomic mass of Argon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, NB The relative atomic mass is a measure of the masses of atoms.


b) Calculate the relative atomic mass of potassium given that it exist as;
5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope. d) Calculate the relative atomic mass of Argon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, NB The relative atomic mass is a measure of the masses of atoms. The higher the relative atomic mass, the heavier the atom. Electrons are found in energy levels/orbital. An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus.


Electrons are found in energy levels/orbital. An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons
6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure.


An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons
6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:
7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)
8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels
9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels
10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus
11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2
12 B.PERIODIC TABLE There are over 100 elements so far discovered.


By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons
6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:
7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)
8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels
9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels
10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus
11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2
12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table.


Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons
6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:
7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)
8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels
9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels
10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus
11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2
12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers.


By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]:
7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x)
8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels
9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels
10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus
11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2
12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers. This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev.


Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers. This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev. The horizontal arrangement forms period. Atoms in the same period have the same the same number of energy levels in their electronic structure. i.e. The number of energy levels in the electronic configuration of an element determine the period to which the element is in the periodic table. e.g. Which period of the periodic table are the following isotopes/elements/atoms? a) 126C Electron structure 2:4 => 2 energy levels used thus Period 2 b) 2311Na Electron structure 2:8:1 => 3 energy levels used thus Period 3 c) 3919K Electron structure 2:8:8:1 => 4 energy levels used thus Period 4 d) 11H Electron structure 1: => 1 energy level used thus Period 1 The vertical arrangement of elements forms a group. Atoms in the same have the same the same group have the same number of outer energy level electrons as per their electronic structure. i.e. The number of electrons in the outer energy level an element determine the group to which the element is ,in the periodic table. 13 a) 126C Electron structure 2:4 => 4 electrons in outer energy level thus Group IV b) 2311C Electron structure 2:8:1 => 1 electron in outer energy level thus Group I c) 3919K Electron structure 2:8:8:1=>1 electron in outer energy level thus Group I d) 11H Electron structure 1: => 1 electron in outer energy level thus Group I By convention; (i)Periods are named using English numerals 1,2,3,4,… (ii)Groups are named using Roman numerals I,II,III,IV,… There are eighteen groups in a standard periodic table. There are seven periods in a standard periodic table. When an atom has maximum number of electrons in its outer energy level, it is said to be stable. When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable. All stable atoms are in group 8/18 of the periodic table. All other elements are unstable.


When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable. All stable atoms are in group 8/18 of the periodic table. All other elements are unstable. THE STANDARD PERIODIC TABLE OF ELEMENTS
14 All unstable atoms/isotopes try to be stable through chemical reactions. A chemical reaction involves gaining or losing outer electrons (electron transfer) .When electron transfer take place, an ion is formed. An ion is formed when an unstable atom gain or donate electrons in its outer energy level inorder to be stable. Whether an atom gain or donate electrons depend on the relative energy required to donate or gain extra electrons i.e. Examples 1. 199 F has electronic structure/configuration 2:7. It can donate the seven outer electrons to have stable electronic structure/configuration 2:. It can gain one extra electron to have stable electronic structure/configuration 2:8. Gaining requires less energy, and thus Fluorine reacts by gaining one extra electrons. 2. 2313 Al has electronic structure/configuration 2:8:3 It can donate the three outer electrons to have stable electronic structure/configuration 2:8. It can gain five extra electrons to have stable electronic structure/configuration 2:8:8. Donating requires less energy, and thus Aluminium reacts by donating its three outer electrons. Elements with less than four electrons in the outer energy level donates /lose the outer electrons to be stable and form a positively charged ion called cation. A cation therefore has more protons(positive charge) than electrons(negative charge) Generally metals usually form cation Elements with more than four electrons in the outer energy level gain /acquire extra electrons in the outer energy level to be stable and form a negatively charged ion called anion. An anion therefore has less protons(positive charge) than electrons(negative charge) Generally non metals usually form anion. Except Hydrogen The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost. Examples of ion formation 1.11H H -> H+ + e (atom) (monovalent cation) (electrons donated/lost) Electronic configuration 1: (No electrons remains)
15 2. 2713 Al Al -> Al3+ + 3e (atom) (trivalent cation) (3 electrons donated/lost) Electron 2:8:3 2:8 structure (unstable) (stable) 3.


Except Hydrogen The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost. Examples of ion formation 1.11H H -> H+ + e (atom) (monovalent cation) (electrons donated/lost) Electronic configuration 1: (No electrons remains)
15 2. 2713 Al Al -> Al3+ + 3e (atom) (trivalent cation) (3 electrons donated/lost) Electron 2:8:3 2:8 structure (unstable) (stable) 3. 2311 Na Na -> Na+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:1 2:8 structure (unstable) (stable) 4. 2412Mg Mg -> Mg2+ + 2e (atom) (cation) ( 2 electrons donated/lost) Electron 2:8:1 2:8 structure (unstable) (stable) 5. 168O O + 2e -> O2- (atom) ( 2 electrons gained/acquired) (anion) Electron 2:6 2:8 structure (unstable) (stable) 6. 147N N + 3e -> N3- (atom) ( 3 electrons gained/acquired) (anion) Electron 2:5 2:8 structure (unstable) (stable) 7. 3115P P + 3e -> P3- (atom) ( 3 electrons gained/acquired) (anion) Electron 2:5 2:8 structure (unstable) (stable) 8. 199F F + e -> F- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:7 2:8 structure (unstable) (stable) 9. 3517Cl
16 Cl + e -> Cl- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:8:7 2:8:8 structure (unstable) (stable) 3. 3919 K K -> K+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:8:1 2:8:8 structure (unstable) (stable) When an element donate/loses its outer electrons ,the process is called oxidation.


199F F + e -> F- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:7 2:8 structure (unstable) (stable) 9. 3517Cl
16 Cl + e -> Cl- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:8:7 2:8:8 structure (unstable) (stable) 3. 3919 K K -> K+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:8:1 2:8:8 structure (unstable) (stable) When an element donate/loses its outer electrons ,the process is called oxidation. When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state. Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4
17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound.


When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state. Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4
17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions.


Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4
17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae.


Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae. Elements combine together to form a compound depending on their combining power. The combining power of atoms in an element is called Valency.Valency of an element is equal to the number of: (i)hydrogen atoms that an atom of element can combine with or displace. (ii)electrons gained /acquired in outer energy level by non metals to be stable/attain duplet/octet. (iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet. (iv)charges carried by ions/cations/ions
18 Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency. Table showing the valency of common radicals.


(iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet. (iv)charges carried by ions/cations/ions
18 Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency. Table showing the valency of common radicals. Radical name Chemical formulae Combining power / Valency Ammonium NH4 + 1 Hydroxide OH- 1 Nitrate(V) NO3 - 1 Hydrogen carbonate HCO3- 1 Hydrogen sulphate(VI) HSO4- 1 Hydrogen sulphate(IV) HSO3- 1 Manganate(VII) MnO4- 1 Chromate(VI) CrO42- 2 Dichromate(VI) Cr2O72- 2 Sulphate(VI) SO42- 2 Sulphate(IV) SO32- 2 Carbonate(IV) CO32- 2 Phosphate(V) PO42- 3 Table showing the valency of some common metal and non metals Element/metal Valency Element/non metal Valency Hydrogen 1 Florine 1 Lithium 1 Chlorine 1 Beryllium 2 Bromine 1 Boron 3 Iodine 1 Sodium 1 Carbon 4 Magnesium 2 Nitrogen 3 Aluminium 3 Oxygen 2 Potassium 1 Phosphorus 3 Calcium 2 Zinc 2 Barium 2 Mercury 2 Iron 2 and 3 Copper 1 and 2 Manganese 2 and 4 Lead 2 and 4
19 From the valency of elements , the chemical formular of a compound can be derived using the following procedure: (i)Identify the elements and radicals making the compound (ii)Write the symbol/formular of the elements making the compound starting with the metallic element (iii)Assign the valency of each element /radical as superscript. (iv)Interchange/exchange the valencies of each element as subscript. (v)Divide by the smallest/lowest valency to derive the smallest whole number ratios Ignore a valency of 1. This is the chemical formula.


(iv)Interchange/exchange the valencies of each element as subscript. (v)Divide by the smallest/lowest valency to derive the smallest whole number ratios Ignore a valency of 1. This is the chemical formula. Practice examples Write the chemical formula of (a)Aluminium oxide Elements making compound Aluminium Oxygen Symbol of elements/radicals in compound Al O Assign valencies as superscript Al3 O2 Exchange/Interchange the valencies as subscript Al2 O3 Divide by smallest valency to get whole number - - Chemical formula of Aluminium oxide is thus: Al2 O3 This means:2atoms of Aluminium combine with 3 atoms of Oxygen (b)Sodium oxide Elements making compound Sodium Oxygen Symbol of elements/radicals in compound Na O Assign valencies as superscript Na1 O2 Exchange/Interchange the valencies as subscript Na2 O1 Divide by smallest valency to get whole number - - Chemical formula of Sodium oxide is thus: Na2 O This means:2atoms of Sodium combine with 1 atom of Oxygen (c)Calcium oxide Elements making compound Calcium Oxygen Symbol of elements/radicals in compound Ca O Assign valencies as superscript Ca2 O2 Exchange/Interchange the valencies as subscript Ca2 O2 Divide by two to get smallest whole number ratio Ca1 O1
20 Chemical formula of Calcium oxide is thus: CaO This means:1 atom of calcium combine with 1 atom of Oxygen. (d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen.


Practice examples Write the chemical formula of (a)Aluminium oxide Elements making compound Aluminium Oxygen Symbol of elements/radicals in compound Al O Assign valencies as superscript Al3 O2 Exchange/Interchange the valencies as subscript Al2 O3 Divide by smallest valency to get whole number - - Chemical formula of Aluminium oxide is thus: Al2 O3 This means:2atoms of Aluminium combine with 3 atoms of Oxygen (b)Sodium oxide Elements making compound Sodium Oxygen Symbol of elements/radicals in compound Na O Assign valencies as superscript Na1 O2 Exchange/Interchange the valencies as subscript Na2 O1 Divide by smallest valency to get whole number - - Chemical formula of Sodium oxide is thus: Na2 O This means:2atoms of Sodium combine with 1 atom of Oxygen (c)Calcium oxide Elements making compound Calcium Oxygen Symbol of elements/radicals in compound Ca O Assign valencies as superscript Ca2 O2 Exchange/Interchange the valencies as subscript Ca2 O2 Divide by two to get smallest whole number ratio Ca1 O1
20 Chemical formula of Calcium oxide is thus: CaO This means:1 atom of calcium combine with 1 atom of Oxygen. (d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen. (e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen.


(d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen. (e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen. (f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1
21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical.


(e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen. (f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1
21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical.


(f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1
21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical.


(g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical.


(h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical.


(i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical.


(j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - -
22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical. (l)Magnesium hydrogen sulphate(VI) Elements making compound Magnesium Hydrogen sulphate(VI) Symbol of elements/radicals in compound Mg HSO4 Assign valencies as superscript Mg 2 HSO4 1 Exchange/Interchange the valencies as subscript Mg1 HSO4 2 Divide by two to get smallest whole number ratio - - Chemical formula of Magnesium hydrogen sulphate(VI) is thus: Mg(HSO4)2 This means:1 atom of Magnesium combine with 2 hydrogen sulphate(VI) radical. Compounds are formed from chemical reactions. A chemical reaction is formed when atoms of the reactants break free to bond again and form products. A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:
23 1. Write the word equation 2.


A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations:
23 1. Write the word equation 2. Write the correct chemical formula for each of the reactants and products 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. This is called balancing. Do not change the chemical formula of the products/reactants. 5. Assign in brackets, the physical state/state symbols of the reactants and products after each chemical formula as: (i) (s) for solids (ii) (l) for liquids (iii) (g) for gas (iv) (aq) for aqueous/dissolved in water to make a solution. Practice examples Write a balanced chemical equation for the following (a) Hydrogen gas is prepared from reacting Zinc granules with dilute hydrochloric acid. Procedure 1. Write the word equation Zinc + Hydrochloric acid -> Zinc chloride + hydrogen gas 2. Write the correct chemical formula for each of the reactants and products Zn + HCl -> ZnCl2 + H2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of Zn on the reactant side is equal to product side One atom of H in HCl on the reactant side is not equal to two atoms in H2 on product side. One atom of Cl in HCl on the reactant side is not equal to two atoms in ZnCl2 on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “2” to get “2” Hydrogen and “2” Chlorine on product and reactant side. Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g)
24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1.


Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g)
24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1. Write the word equation Hydrogen peroxide -> Water + oxygen gas 2. Write the correct chemical formula for each of the reactants and products H2O2 -> H2O + O2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of H on the reactant side is equal to product side Two atom of O in H2O2 on the reactant side is not equal to three atoms (one in H2O and two in O2) on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply H2O2 by “2” to get “4” Hydrogen and “4” Oxygen on reactants Multiply H2O by “2” to get “4” Hydrogen and “2” Oxygen on product side When the “2” Oxygen in O2 and the“2” in H2O are added on product side they are equal to the“4” Oxygen on reactants side. 2H2O2 -> 2H2O + O2 5. Assign in brackets, the physical state/state symbols . 2H2O2(aq) -> 2H2O(l) + O2(g) (c) Chlorine gas is prepared from Potassium manganate(VII) reacting with hydrochloric acid to form potassium chloride solution, manganese(II) chloride solution,water and chlorine gas. Procedure 1. Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.


Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 25 Number of atoms of K and Mn on the reactant side is equal to product side Two atom of H in H2O on the product side is not equal to one atom on reactant side. Four atom of O in KMnO4 is not equal to one in H2O One atom of Cl in HCl on reactant side is not equal to three (one in H2O and two in Cl2) 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “16” to get “16” Hydrogen and “16” Chlorine on reactants Multiply KMnO4 by “2” to get “2” Potassium and “2” manganese, “2 x4 =8” Oxygen on reactant side. Balance the product side to get: 2 KMnO4 +16 HCl -> 2 KCl + 2 MnCl2 +8 H2O + 5 Cl2 5. Assign in brackets, the physical state/state symbols . 2KMnO4(s) +16 HCl(aq)-> 2 KCl (aq) + 2MnCl2(aq)+8 H2O(l)+5 Cl2(g) (d)Carbon(IV)oxide gas is prepared from Calcium carbonate reacting with hydrochloric acid to form calcium chloride solution, water and carbon(IV)oxide gas. Procedure 1. Write the word equation Calcium carbonate + Hydrochloric acid -> calcium chloride solution+ water +carbon(IV)oxide 2. Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4.


Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. 5. Assign in brackets, the physical state/state symbols . CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) (d)Sodium hydroxide solution neutralizes hydrochloric acid to form salt and water. 26 NaOH(aq) + HCl(aq) -> NaCl (aq) + H2O(l) (e)Sodium reacts with water to form sodium hydroxide and hydrogen gas. 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) (f)Calcium reacts withwater to form calcium hydroxide and hydrogen gas Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g) (g)Copper(II)Oxide solid reacts with dilute hydrochloric acid to form copper(II)chloride and water. CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l) (h)Hydrogen sulphide reacts with Oxygen to form sulphur(IV)Oxide and water. 2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water.


2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water. C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l) (l)Ethyne(C2H2) gas burns in air to form Carbon(IV)Oxide and water. 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l)
27 C.PERIODICITY OF CHEMICAL FAMILES/DOWN THE GROUP. The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table.i.e The number of occupied energy levels determine the Period and the valence electrons determine the Group. Elements in the same group have similar physical and chemical properties. The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level.


Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M+ The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels). Atomic and ionic radius The distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron/s is called atomic radius. Atomic radius is measured in nanometers(n).The higher /bigger the atomic radius the bigger /larger the atomic size. The distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron/s is called ionic radius. Ionic radius is also
28 measured in nanometers(n).The higher /bigger the ionic radius the bigger /larger the size of the ion. Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g. The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels). Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level.


Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level. Table showing the atomic and ionic radius of some alkali metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Lithium Li 3 0.133 0.060 Sodium Na 11 0.157 0.095 Potassium K 19 0.203 0.133 The atomic radius of sodium is 0.157nM .The ionic radius of Na+ is 0.095nM. This is because sodium reacts by donating/losing the outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity The ease of donating/losing electrons is called electropositivity. All alkali metals are electropositive. Electropositivity increase as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and can be lost/ donated easily/with ease. Francium is the most electropositive element in the periodic table because it has the highest/biggest atomic radius. Ionization energy The minimum amount of energy required to remove an electron from an atom of element in its gaseous state is called 1st ionization energy. The SI unit of ionization energy is kilojoules per mole/kJmole-1 .Ionization energy depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer electrons/energy level and thus the lower the ionization energy. For alkali metals the 1st ionization energy decrease down the group as
29 the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium.


e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium. Physical properties Soft/Easy to cut: Alkali metals are soft and easy to cut with a knife. The softness and ease of cutting increase down the group from Lithium to Francium. This is because an increase in atomic radius, decreases the strength of metallic bond and the packing of the metallic structure Appearance: Alkali metals have a shiny grey metallic luster when freshly cut. The surface rapidly/quickly tarnishes on exposure to air. This is because the metal surface rapidly/quickly reacts with elements of air/oxygen. Melting and boiling points: Alkali metals have a relatively low melting/boiling point than common metals like Iron. This is because alkali metals use only one delocalized electron to form a weak metallic bond/structure. Electrical/thermal conductivity: Alkali metals are good thermal and electrical conductors. Metals conduct using the outer mobile delocalized electrons. The delocalized electrons move randomly within the metallic structure. Summary of some physical properties of the 1st three alkali metals Alkali metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conductivity 1st ionization energy Lithium Silvery white Not easy 180 1330 Good 520 Sodium Shiny grey Easy 98 890 Good 496 Potassium Shiny grey Very easy 64 774 Good 419 Chemical properties (i)Reaction with air/oxygen On exposure to air, alkali metals reacts with the elements in the air. Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate.


Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate. 2NaOH(aq) + CO2(g) -> Na2CO3(g) + H2O(l) (ii)Burning in air/oxygen Lithium burns in air with a crimson/deep red flame to form Lithium oxide 4Li (s) + O2(g) -> 2Li2O(s) Sodium burns in air with a yellow flame to form sodium oxide 4Na (s) + O2(g) -> 2Na2O(s) Sodium burns in oxygen with a yellow flame to form sodium peroxide 2Na (s) + O2(g) -> Na2O2 (s) Potassium burns in air with a lilac/purple flame to form potassium oxide 4K (s) + O2(g) -> 2K2O (s) (iii) Reaction with water: Experiment Measure 500 cm3 of water into a beaker. Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium.


Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium. Caution: Keep a distance Observations Alkali metal Observations Comparative speed/rate of the reaction Lithium -Metal floats in water -rapid effervescence/fizzing/bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Moderately vigorous Sodium -Metal floats in water -very rapid effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn Very vigorous
31 phenolphthalein indicator pink -pH of solution = 12/13/14 Potassium -Metal floats in water -explosive effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Explosive/burst into flames Explanation Alkali metals are less dense than water. They therefore float in water.They react with water to form a strongly alkaline solution of their hydroxides and producing hydrogen gas. The rate of this reaction increase down the group. i.e. Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions.


Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions. Chemical equations 2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g) 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g) 2Rb(s) + 2H2O(l) -> 2RbOH(aq) + H2(g) 2Cs(s) + 2H2O(l) -> 2CsOH(aq) + H2(g) 2Fr(s) + 2H2O(l) -> 2FrOH(aq) + H2(g) Reactivity increase down the group (iv) Reaction with chlorine: Experiment Cut about 0.5g of sodium into a deflagrating spoon with a lid cover. Introduce it on a Bunsen flame until it catches fire. Quickly and carefully lower it into a gas jar containing dry chlorine to cover the gas jar. Repeat with about 0.5g of Lithium. Caution: This experiment should be done in fume chamber because chlorine is poisonous /toxic. Observation Sodium metal continues to burn with a yellow flame forming white solid/fumes. 32 Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus.