CHEMISTRY NOTES FORM 2-PERIODIC TABLE AND CHEMICAL FORMULA 

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.

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?

1. ¹²₆C

Electron structure 2:4 => 2 energy levels used thus Period 2

• ²³₁₁Na

Electron structure 2:8:1 => 3 energy levels used thus Period 3

• ³⁹₁₉K

Electron structure 2:8:8:1 => 4 energy levels used thus Period 4

• 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.

1. ¹²₆C

Electron structure 2:4 => 4 electrons in outer energy level thus Group IV

• ²³₁₁C

Electron structure 2:8:1 => 1 electron in outer energy level thus Group I

• ³⁹₁₉K

Electron structure 2:8:8:1=>1 electron in outer energy level thus Group I

• 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.

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 in order 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.

•  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 fewer 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)

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                     ->                  O^2-                  

                 (atom)            ( 2 electrons gained/acquired)      (anion)

Electron    2:6                                                                     2:8

structure   (unstable)                                                     (stable)

6. ¹⁴₇N

N       +                3e                     ->                  N^3-                  

                 (atom)            ( 3 electrons gained/acquired)      (anion)

Electron    2:5                                                                     2:8

structure   (unstable)                                                     (stable)

7. ³¹₁₅P

P        +                3e                     ->                  P^3-                  

                 (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

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 donates/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	11H 21H(deuterium) 31H(Tritium)	H+ H+ H+	+1 +1 +1
Chlorine	3517Cl 3717Cl	Cl– Cl–	-1 -1
Potassium	3919K 4019K 4119K	K+ K+ K+	+1 +1 +1
Oxygen	168O 188O	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
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:

• Copper (I) means Cu⁺ as in Copper(I)oxide
• Copper (II) means Cu²⁺ as in Copper(II)oxide
• Iron (II) means Fe²⁺ as in Iron(II)sulphide

(iv)    Iron (III) means Fe³⁺ as in Iron(III)chloride

• Sulphur(VI)mean S⁶⁺ as in Iron(III)sulphate(VI)
• Sulphur(VI)mean S⁶⁺ as in sulphur(VI)oxide
• Sulphur(IV)mean S⁴⁺ as in sulphur(IV)oxide
• 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

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

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.

 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

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.

(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

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.

(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	–	–

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 chemical equations:

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

• 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)

• 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)

• 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.

      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 KMnO4by “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. 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.

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 with water 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.

                             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)