SULPHUR AND ITS COMPOUNDS at a glance

 

 SULPHUR

CONTENT

•             General Properties of Sulphur Group.

•             Electronic Structure of Sulphur Group.

•             Extraction of Sulphur.

•             Allotropes of Sulphur.

•             Uses of Sulphur

GENERAL PROPERTIES OF GROUP VI ELEMENTS

Group VI elements include Oxygen, Sulphur, Selenium, Tellurium and Polonium.  

1.   All the elements are solid except oxygen which is a gas at room temperature

2.  Metallic property increases down the group. Oxygen and sulphur are non-metal; selenium and tellurium are metalloid, while polonium is a metal.

3. Oxygen and sulphur exhibit allotropy.

4.   They have six electrons in their outermost shell. Hence their oxidation number is -2; though sulphur can exhibit -4 and -6 states in some compounds.

5.  Electronegativity decreases down the group. Thus, oxygen is a good oxidizing agent.

ELECTRONIC STRUCRURE OF SULPHUR GROUP

 The electronic configurations of the members of the group are shown below: 

Oxygen = 8: - 1s² 2s² 2p⁴

Sulphur = 16: - 1s² 2s² 2p⁶ 3s² 3p⁴

Selenium= 34: - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴

Tellurium = 52: - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁴

Polonium = 84: - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 5d¹⁰ 5f¹⁴ 6s² 6p⁴

SULPHUR

Sulphur is found in group VI period II in the Periodic Table. It occurs as a free element found in underground deposits. It is also  found in the combined state as metallic  Sulphide and as tetraoxosulphate (IV).

EXTRACTION OF SULPHUR

Sulphur is extracted from underground deposits by the Frasch process. In this process: Three concentric pipes are is driven down a hole drilled to the Sulphur bed.  Super  heated water at 170^oC and 10atm is forced down one of the pipes to melt the solid Sulphur, hot compressed air is then forced down the second pipe as a result the molten Sulphur is then forced out through the third pipe by the compressed air. The molten Sulphur is continuously pumped out into large tanks where it is allowed to solidify. Sulphur obtained is  about 99.5% pure.

ALLOTROPES OF SULPHUR

There are two main crystalline allotropes of Sulphur they are: -

1   Rhombic Sulphur: This is a bright yellow octahedral crystalline solid made up of S₈ molecules. It is stable below 96^oC.

2 Monoclinic Sulphur: This is a needle like crystalline molecule consisting of S8. It has an amber colour and is stable  at temperatures above 96^oC. It easily reverts to Rhombic below 96^oC. The transition temperature between Rhombic and Monoclinic is 96^oC.

Comparison of the Physical Properties of Rhombic and Monoclinic Sulphur

	Rhombic sulphur	Monoclinic sulphur
Colour	Bright yellow	Amber
Shape	Octahedral	Needle-shaped
Density	2.80 g/cm3	1.98g/cm3
Melting point	1130C	119oC
Stability	Stable below 96oC	Stable above 96oC

There are other non crysatlline allotropes of Sulphur such as 

1.  Amorphous sulphur: formed when sulphur is dissolved in carbon(IV) sulphide and the solution is allowed to evaporate.

2.  Plastic sulphur: formed when molten Sulphur is suddenly poured into cold water:00

     

PHYSICAL PROPERTIES

1.   Sulphur is a yellow solid.

2.    It melts at 119^oC and boils at 444^oC

3.   It is insoluble in water but soluble in carbon (IV) sulphide and tolune

3.    It is a poor – conductor of heat and electricity.

CHEMICAL PROPERTIES

1.  It reacts directly with metals to form sulphide (S^2-)

   Fe_(s) + S_(s) → FeS_(s)

2    It reacts with hydrogen to form hydrogen sulphide;

       H_2(g) + S_(s) → H₂S_(g)

 3.   It reacts with excess oxygen to form sulphur (IV) oxide

  O_2(g) + S_(s) →SO_2(g)

 4.  It reacts with coke (carbon) to form carbon (II) sulphide 

     C(s) + S(s) → CS₂   

USES

1.  It used in manufacturing tetraoxosulphate (IV) acid

2.  It used in vulcanization of rubber

3.   It used as germicides

4.   It used in manufacturing bleaching agent

COMPOUNDS OF SULPHUR

1.    HYDROGEN SULPHIDE, H2S

Hydrogen sulphide is found in volcanic gases, sulphur springs, coal gas and natural gas.

LABORATORY PREPARATION

Hydrogen sulphide is prepared in the laboratory by the action of dilute acids on metallic sulphide like Iron (II) sulphide 

2HCl_(aq)+ FeS_(s) → FeCl_2(aq) + H₂S_(g)

The apparatus used for regular supply of hydrogen sulphide in the laboratory is Kipp’s apparatus.

                 

PHYSICAL PROPERTIES

1.  Hydrogen sulphide is a colourless gas

2. It smells  like that of rotten egg.

2.  It is a poisonous gas 

3.  It is about 1.18 times denser than air.

4.  It is moderately soluble in water to form very weak acidic solution.

5.  It burns with pale blue flame.

CHEMICAL PROPERTIES

1. As an acid it reacts with alkali to form a normal salt and water    

   2NaOH_(aq) + H₂S_(g) → Na₂S_(aq) +2H₂O_(l)

2. It reacts with excess oxygen to form sulphur (VI) oxide but forms deposit of sulphur with limited oxygen

 2H₂S_(g) + 3O_2(g) → 2H₂O_(l) + 2SO_2(g)

 2H₂S_(g) + O_2(g) → 2H_2(l) + 2S_(s)

3.   As a reducing agent, it reacts with many oxidizing agents such as acidified KMnO₄, acidified K₂Cr₂O₇, chlorine gas, FeCl₂, SO₂, H₂SO₄ and HNO₃

TEST FOR HYDROGEN SULPHIDE

Moistened a piece of filter paper with lead (II) trioxonitrate (V) solution and dropped it into a gas jar of the unknown gas. If the paper turns black, then the gas is H₂S.

SULPHUR (IV) OXIDE, SO₂

LABORATORY PREPARATION

Sulphur (IV) oxide is prepared in the laboratory by heating sodium or potassium trioxosulphate (IV) with tetraoxosulphate (VI) acid or hydrochloric acid.

         Na₂SO_3(aq)+2HCl_(aq) →   2NaCl_(aq)+ H₂O(l) + SO_2(g)

Physical Properties 

1.            Sulphur (IV) oxide is a colourless poisonous gas.

2.            It smell like that of burning matches.

2. It is very soluble in water.

3.  It is about 2.5 times denser than air.

Chemical Properties

1.  As an acid: - it reacts with alkali to form normal salt and water only.

    2NaOH_(aq)+ SO_2(g) → Na₂SO_3(aq) + H₂O_(l)

2.   As reducing agent: - Sulphur (IV) oxide reduces oxidizing agents such as acidified KMnO₄; acidified K₂Cr₂O₇; FeCl₃, HNO₃, chlorine gas. It decolorizes acidified purple KMnO₄ and turns acidified orange K₂Cr₂O₇ to green.

3.   As an oxidizing agent: -Sulphur (IV) oxide reacts as oxidizing agent in the presence of strong reducing agent such as hydrogen sulphide.

2H₂S_(g) + SO_2(g) → 2H₂O_(l) + 3S_(s)

C_(s)+ SO_2(g)→CO_2(g)+ S_(s)

4.  As a bleaching agent: - It reacts as bleaching agent decolourising dye by its bleaching action. The bleaching action is similar to that of chlorine in that there must be water. But, while chlorine bleaches by oxidation sulphur IV oxide bleaches by reduction.

USES

1.It is used in manufacture of tetraoxosulphate (VI) acid.

2. It is used as a germicide and a fumigant especially for destroying termites.

3.  It is used as bleaching agent for straw, silt and wood.

4. It is used as preservative in some liquid e.g orange juice.

5.  Liquid sulphur (IV) oxide is used as refrigerant.

Test for SO2

1. Bubbled the unknown gas through solution of either acidified potassium heptaoxodichromate (VI) or potassium tetraoxomanganate (VII). If it changes the colour of acidified K₂Cr₂O₇ from orange to green or it changes the colour of acidified KMnO₄ purple to colourless, then the gas is SO₂ 

SULPHUR (VI) OXIDE, SO₃

Sulphur (VI) oxide is prepared by reacting sulphur (IV) oxide and oxygen at 450^oC and 1 atm pressure in the presence of vanadium (V) oxide  or platinized asbestos as catalyst

     2SO_2(g) + O_2(g) →2SO_3(g)

PHYSICAL PROPERTIES OF SO₃

1.  It is a white needle-like crystal at room temperature.

2.    It has a low boiling point and vapourizes on gentle heating.

3 It dissolves readily in water to give tetraoxosulphate (VI) acid.

TRIOXOSULPHATE IV ACID, H₂SO₃

Trioxosulphate (IV) acid is a dibasic acid with a molecular formula H₂SO₃

Laboratory Prepaaration  OF H₂SO₃

It is prepared in the lab by the action of dilute hydrochloric acid on heated sodium trioxosulphate (IV) to produce sulphur (IV) oxide, which is then dissolved in water.

    Na₂SO_3(s)+2HCl_(aq)→2NaCl_(aq) + H₂O_(l) + SO_2(g)

    H₂O_(l)+ SO_2(g) → H₂SO_3(aq)

Sulphur (IV) Oxide is the acid anhydride of trioxosulphate (IV) acid.

PHYSICAL PROPERTIES OF H₂SO₃

1. It is colourless liquid.

2.  It is readily soluble in water.

3. It has an irritating and choking smell.

Chemical Properties oF H₂SO₃

1. It reacts with alkalis to form salt and water.

     2NaOH_(aq) + H₂SO_3(aq)→ Na₂SO_3(aq) + 2H₂O_(l)

2.     It is oxidized in air to tetraoxosulphate (VI) acid

 2H₂SO_3(aq) + O_2(g) → 2H₂SO_4(aq)

3.    As a reducing agent: - It reduces oxidizing agent such as acidified KMnO₄ and acidified K₂Cr₂O₇

4    It bleaches dyes in the presence of water.

Test for SO₃^2-

Add a little amount of barium chloride solution to a solution of the unknown substance. If a white precipitate is formed which is soluble in dilute hydrochloric acid then the presence of a trioxosulphate (IV) ion is confirmed.

USES OF H₂SO₃

1.   It is used for bleaching straw and other fabrics.

2.  It is used as a germicide.

TETRAOXOSULPHATE VI ACID, H₂SO₄

Tetraoxosulphate VI acid is one of the most important chemical compounds known. It is used in almost every manufacturing process; hence it is mostly prepared industrially.

INDUSTRIAL PREPARATION OF H₂SO₄

It is manufactured industrially by Contact process. The following equations summarizes the steps involved in the Contact process.

1.   Sulphur is burnt in dry air to obtain sulphur (IV) oxide, SO₂

    S_(s) + O_2(g) → SO₂(g)

2.The Sulphur (IV) oxide produced is combined with excess oxygen in the presence of vanadium V oxide (V₂O₅ ) as catalyst at a temperature of 450^oC to yield sulphur (VI) oxide.

 SO₂(g) + O_2(g) → 2SO_3(g) + heat

3. The sulphur (VI) oxide is then dissolved in concentrated H₂SO₄ to produce a thick oily liquid called Oleum.

H₂SO_4(aq) + SO_3(g)→ H₂S₂O_7(aq)

4.    The Oleum is then combined with one mole of water to produce about 98% tetraoxosulphate (VI) acid.

 H₂O_(l)+ H₂S₂O_7(aq)  2H₂SO_4(aq)

NOTE: Dissolving Sulphur (VI) oxide in water directly is highly exothermic and will cause the acid to vaporize, producing a mist of droplets of the concentrated acid which can spread and cause acid burns.

PHYSICAL PROPERTIES

1.  It is a colourless, viscous liquid with density of 1.84gcm^-3

2.  It is corrosive and cause burns when in contact with the skin.

3. It is highly soluble in water evolving large amount of heat.

CHEMICAL PROEPERTIES

1.  As an acid, 

i.  It reacts with metal higher than hydrogen in the electrochemical series to liberate hydrogen ga Mg_(s)+ H₂SO_4(aq)→ MgSO_4(aq)+ H_2(g)

ii.  It reacts with bases to form salts and water ZnO_(s)+H₂SO_4(aq) →ZnSO_4(aq)+H₂O_(l)

iii.  It reacts with alkali to form normal and acidic salt

H₂SO_4(aq)+KOH_(aq)→NaHSO_4(aq)+ H₂O_(l)

 H₂SO_4(aq)+ KOH_(aq)→ Na₂ SO_4(aq) + 2H₂O_(l)

iv.. It reacts with trioxocarbonate (IV) to liberate carbon (IV) oxide

H₂SO_4(aq)+CuCO_3(aq)→CuSO_4(aq)+ H₂O_(l) + CO_2(g)

2.  As oxidizing agent: - Concentrated H₂SO₄ oxidize metals and non –metals to yield the corresponding tetraoxosulphate VI and itself reduced to SO2.  It oxidises hydrogen sulphide to Sulphur.

 Cu_(s)+ 2H₂SO_4(aq)→ CuSO_4(aq)+ 2H₂O_(l)+SO_2(g)

C_(s) + 2H₂SO_4(aq)→2H₂O_(l) + CO_2(g)+ 2SO_2(g)

H₂SO_4(aq)+ H₂S_(g)→S_(s)+H₂O_(l)+ SO_2(g)

3. As a dehydrating agent: - Concentrated tetraoxosulphate (VI) acid also behaves as a dehydrating agent, removing components of water from compounds like sugar and ethanedioic acid

         C₁₂H₂₂O_11(s)→12C_(s) +11H₂O_(l)

  ^{sugar                    charcoal}

  COOH

 |            + conc.                        COOH    H2SO4 → CO2 + CO + H2O            

 4.  Concentrated tetraoxosulphate (VI) displaces volatile acids from their salts 

 NaCl_(s)+H₂SO_4(aq)→NaHSO_4(aq) + HCl_(g)

Test for SO₄^2-

 Add Barium chloride solution to a solution of the unknown salt. If a white precipitate is formed which is insoluble in excess dilute hydrochloric acid, then the presence of a tetraoxosulphate (VI) ion is confirmed.

USES OF H₂SO₄

1.     It is used as an electrolyte in lead  accumulator.               

2.   It is used in production of fertilizers e.g ammonium tetraoxosulphate (VI).

3. It is used in purification of crude oil.

 4.  It is used as drying agent for many gases except NH₃ and H₂S gas.

5    It is used to clean metals before electroplating.

6.   It is used in the production of fibres

7.  It is used in the manufacture of synthetic detergents

TETRAOXOSULPHATE (VI) SALTS: - These are salts formed when metals, bases/ alkalis and trioxocarbonate IV reacts with H2SO4

USES OF TETRAOXOSULPHATE (VI) SALTS

1.  Ammonium tetraoxosulphate (VI) (NH4)2SO4 used as fertilizers

2.  Sodium tetraoxosulphate (VI) (Na2SO4) is used in paper manufacture and as a purgative

3.  Calcium tetraoxosulphate (VI) (CaSO4) is mined as gypsum and when heated forms plaster of Paris       (POP) used to set broken bones.

4. Aluminum tetraoxosulphate (VI) (Al2(SO4)3) is used to coagulate precipitate in purification of water

5. Iron II tetraoxosulphate (VI) is used to treat anemia.

OBJECTIVE QUESTIONS 

1. The acid anhydride of tetraoxosulphate (VI) acid is 

a. SO2 

b. SO3 

c. SO4 

d. SO

2. Which of the following compounds gives a white precipitate with acidified barium chloride solution? a. K2SO4 

b. NaNO3 

c. KCl 

d. CaCO3

3

4. Which of the following is used as catalyst in the Contact process? 

a. V2O5 

b. Platinum 

c. Fe3O2 

d. Nickel

5.Whatbis the effect of using vanadium  (V) oxide as a catalyst in the reaction represented by the following equation?  2SO2  + O2   SO3   H= -xkjmol-1 

A. Decreases the value of H

B. Increases the collision rate of reactant particles 

C. Shift equilibrium position to the right.

D. Reduces the time for attainment of equilibrium.

6. What is the colour of tetraoxosulphate VI acid?

 a. Colourless 

b. White 

c. Blue 

d. Pale white

7. In which of the following is the oxidation number of Sulphur equal to -2

A. S8

B. H2S

C.SO2

D.SO32-

8. 

9.            

10. Why do we acidify the solution used for testing for the presence of S042-

a. To prevent the precipitation of any other ion that may be present in the solution. 

b. To acidify the test solution. 

c. To increase the rate of the reaction 

d. The acid acts as catalyst.

THEORY QUESTIONS 

1.

(I). Name the process represented by the chart 

ii). Identify reactant X and product Y.

iii). What are the operating temperature and pressure at stage II

iv). Mention the stage which requires a catalyst and state the catalyst used. 

v) give the reason why the SO3 produced in stage II is not dissolved directly in water.

2.(a). State two physical properties of hydrogen sulphide 

(ii). Name the laboratory equipment used for intermittent production of hydrogen sulphide?

b. What property of hydrogen sulphide is illustrated in the reaction represented by the following equation? 

H2S + 2NaOH ---> Na2S + H2O 

EQUILLIBRIUM at a glance

EQUILLIBRIUM is said to be established in a reversible reaction when the rate of forward reaction is equally to the rate of backward reaction.

 We can describe equilibrium here as a state where there is no observable change.

In this Chapter we will be looking at Equilibrium in Chemistry and try to explain it as best as we can

Types of Equilibrium

Equilibrium can be Static  and  Dynamic

I. Static equilibrium: - In static equilibrium there is basically no movement within the system and the substances involved are at the same level. An example of a static equilibrium is when two children on a seesaw are suspended in space.

You will observe that both children suspended in space have the same potential energy. But any slight disturbance and the equilibrium will be lost.

Ii. Dynamic equilibrium: - In dynamic equilibrium, there is actually movements, but the changes are not observable. For example, a child going up an escalator that is actually descending at the same speed as the child, the child would seem to remain at a particular spot even though he is moving up.

Dynamic equilibrium can be grouped into two

I. Physical equilibrium and 

Ii. Chemical equilibrium

I. Dynamic physical equilibrium: -In dynamic equilibrium no new substance is formed (just like a physical change). For example, when sodium chloride is dissolved in water to give a saturated solution, the dissolved crystals will crystallize out of the solution just as undissolved crystals are dissolving into the solution, with these forward and backward reactions occurring at the same rate, we say the saturated solution is in equilibrium. 

NaCl(s) ⇌ NaCl(aq)

II. Dynamic Chemical Equilibrium: - In this equilibrium system a new substance is formed (just like a chemical change).

An example is the reaction between Nitrogen and Hydrogen to yield Ammonia

N2(g) + 3H2(g) ⇌ 2NH3(g)

In dynamic equilibrium a new substance is formed.

Properties Of a System in Equilibrium

If a system is in equilibrium, then it will possess the following properties

I. The reaction must be a reversible reaction 

Ii. The rate of forward reaction must be equal to the rate of backward reaction 

III. ∆G must be equal to zero. (delta G is the Gibbs free energy)

IV. The equilibrium can be approached from either side of the reaction.

V. The reaction must occur in a closed system.

Factors affecting a system in equilibrium position of a reversible reaction

I. Temperature 

II. Concentration 

III. Pressure

IV. Catalyst

 

Le Chatelier's Principle: See laws and principles

The effect in equilibrium position brought about by a change in any of the factors mentioned above was predicted and may be explained by Le Chatelier's principle 

i. How Temperature Affects equilibrium position of a reversible reaction: - Given a reaction where the forward reaction is exothermic, then the backward reaction will be endothermic.  For such a reaction, an increase in temperature will shift the position of Equilibrium to the left, that is, it will favor the backward reaction on the other hand, a decrease in temperature will favor the forward reaction since it does not require much heat.

How Concentration Affects the equilibrium position of a reversible reaction: - When any one of the reactants in a reversible reaction is increased this will cause the equilibrium position to shift to the right, favouring the forward reaction.   This is because an equilibrium has already been established and adding more reactants will alter that equilibrium, and so according to Le Chatelier's principle the reactants will be quickly used up and be converted to product. 

If the products are removed from the system as they produced, this can also cause the equilibrium to shift to the right.

For example, let us consider the equilibrium reaction of the Haber process, that the reaction between H2 and N2 the produce NH3 

 

                    N2(g) + 3H2(g) ⇌ 2NH3(g) 

An increase in the concentration of the N2 or H2 will cause the equilibrium position to shift to the right, favouring the forward reaction.

How Pressure Affects the equilibrium position of a reversible reaction: - For Pressure to affect the equilibrium position of a reversible reaction, at least one of the reactants or products must be a gas and the total number of mols (vol.) of the reactants must be different from the total number of mols (vol.) of the product. 

For example, considering the two reactions 

  1. Fe(s) + 4H2O(g) ⇌ Fe2O3 + 4H2(g) 

                  4vol.                      4vol.

      

         2.  N2O4(g) ⇌ 2NO2(g)

              1vol.        2vol.

Pressure cannot affect the position of equilibrium of equation 1 because the volumes of both the gaseous reactant and gaseous product are the same (i.e. 4vol.)

However, in equation 2. the total volume of reactants is different from the volume of product, so an increase in pressure for such a reaction will lead to a decrease in volume (Boyle's law) hence causing the equilibrium to shift the left (the position that has a small volume) favouring the backward reaction. Similarly, a decrease in pressure will result to an increase volume and this will cause the equilibrium to shift to the right (area with larger volume) favouring the forward reaction.

Equilibrium Constant 

The equilibrium constant for a reaction in equilibrium, is the product of the molar concentrations (or pressures) of the products divided by the product of the reactants raised to the power of the coefficient of each reactant.

For a reaction 

            aA +bB ⇌ cC + dD

The equilibrium constant for the reaction will be 

                Kc = [C]^c[D]^d
                        [A]^a[B]^b

         

OBJECTIVE QUESTIONS

1. what will happen if more heat is applied to the following system in equilibrium

   

     X_2(g) + 3Y_2(g) ⇌ 2XY_3(g); ∆H= -xkjmol^-1

   

A. The yield of XY₃ will increase.

B. More of will XY₃ decompose 

C. more of X₂ will react 

D. The forward reaction will go to completion

2. What is the expression for the equilibrium constant (Kc) for the following reaction?  N_2(g) + O_2(g) ⇌ 2NO_(g)

A.      [NO]²
      [N₂] + [O₂]  

 

b.       [2NO]²
         [N₂]]O₂]

 

c.     [N2] + [O]
          2[NO]²

 

d.       [NO]²

        [N₂][O₂]

3. A reaction represented by the equation below  

     A_2(g) + B_(2(g) ⇌ 2AB_(g); ∆H = +X kjmol^-1             

which of the following statement about the system is correct

a. The forward reaction is exothermic 

b. The reaction goes to completion at equilibrium 

c. Pressure has no effect on the equilibrium mixture

d. At equilibrium increase in temperature favours the reverse reaction.

 

4. What will happen if more heat is applied to the following system at equilibrium?

                X_2(g) + 3Y_(2(g) ⇌ 2XY_(g); ∆H = -X kjmol^-1

a). The yield will increase of XY³ will increase.

b).  More of XY₃ will decompose 

c). . More of X₂ will react 

d).  The forward reaction will go to completion

5. The position of equilibrium ina reversible reaction is affected by 

a). Particle size of the reactants 

b). Change in concentration of the reactants 

c). Change in the size of the reaction vessel 

d). Vigorous stirring of the reaction mixture.

THEORY

1(a). When few drops of aqueous KSCN are added to a solution of iron (III) salt the following equilibrium is s_{The equilibrium mixture has a pale colour}

                                          ^{Yellow      colourless         deep red} 

(i). Explain what will happen if more KSCN were added to the equilibrium mixture 

(ii). Which of the ions in the equilibrium mixture forms an insoluble hydroxide with NAOH (aq)?   Write an equation for the reaction

(iii).  State two changes observed on adding NaOH to the equilibrium mixture.

(b)i. 

 

ATOMIC NUMBER and ATOMIC MASS & RELATIVE ATOMIC MASS at a glance

CONSTITUENTS OF AN ATOM

The atom as explained earlier is made up of three sub-particles. 

i. Protons, 

ii. Electrons and 

iii. Neutrons. 

the Protons are positively charged. 

the Electrons are negatively charged    while 

the Neutrons have no charge.

S/n         sub-atomic particles        charge                mass

1.             Protons                           + (positive)            1  (unit)

2.             Electrons                         - (negative)            0.00005

3.             Neutrons                          0 (neutral)               1 (unit)                                   

Atomic number: - Of an element is the number of protons in the nucleus of the atom of the element.

atomic numbers are whole numbers and NO two elements in the world have the same atomic number. 

Mass number or atomic mass: - of an element is the sum of the protons and neutrons in the nucleus of its atom.

that is,

Atomic Mass = Number of protons + Number of neutrons

In chemistry an element X can be represented as                A

                                                                                              _ZX

where             A = Atomic mass or mass number

                      Z = Atomic number

Example      40                  mass no = 40 (20 protons + 20 neutrons)

                   20 Ca           atomic no = 20 (number of protons)

     Elements     Num. of  P        Num. of  E         Num. of  N

        40
1.    20 Ca                  20                    20                    20

   

     14
2.   7 N                       7                       7                    14

     

       39
3.    19 K                   19                   19                    20

 Examples 

1.            State the constituents of an atom.

2.            What is the number of protons, neutrons and electrons in the following elements

            11               12               23                  32
(a)        5B      (b)    6C    (c)   11Na     (d)   16S 

ISOTOPES

Isotopy is the occurrence of atoms of elements having the same atomic number but different mass numbers. This is due to the difference in the number of neutrons present in the atoms. The atoms that exhibit isotopy are called ISOTOPES.

Examples of atoms that exhibit isotopy are chlorine 35_Cl and 37_Cl

Carbon-  ¹² _C        ¹³ _C and ¹⁴ _C6

                              

Potassium – ³⁹K₁₉   and   ⁴¹K₁₉             

Oxygen -  ¹⁶O₁₆    and ¹⁸O₁₆

                                             

EVALUATION

1.            Define isotopy.

2.            Write the isotopes of chlorine.

                                             

CALCULATION OF RELATIVE ATOMIC MASS

RELATIVE ABUNDANCE:  this is the amount that a particular isotope occurs in nature, and it is expressed in percentage.

For example, Chlorine always occurs in two isotopic mixtures 35_Cl and 37_Cl with percentages of 75% and 25% respectively. This percentages are known as the relative abundance of each isotope

The following is an example of calculation of relative atomic mass of an element from percentage abundance of its isotopes.

1. X is an element which exists as an isotopic mixture containing 90% of ³⁹X₁₉ and 10% of ⁴¹X₁₉

a.    How many neutrons are present in the isotope ⁴¹X                                                                               

b.   Calculate the mean relative atomic mass of X

Solution

a.            Neutrons in  ⁴¹X_{19       :   mass number - atomic number} 

                                                         =          41-19 = 22

b.         R.A.M   =    90    x 39    +          10   x 41
                               100                          100

                                =          90  x   39  +   41  x  10
                                                          100

                                                         =  3920    =  39.20
                                                              100

1.         How many neutrons are present in the isotope ³⁷Cl₁₇ ?

2.         A given quantity of chlorine contains 75% ³⁵Cl₁₇, and 25% ³⁷Cl₁₇, determine the relative atomic mass of chlorine.

 

Example 2.

1. The following are more examples on calculations of relative atomic masses of elements.
2. An element Y exist in two isotopic forms ³⁹Y₁₈ and ⁴⁰Y₁₈ in the ratio 3:2 respectively. What is the relative atomic mass of the element?

SOLUTION                            

R.A.M of Y        =          3          x          39        +          2          x          40

                                    5                      1                     5                       1

                        =         0.6        x         39         +         0.4        x          40

                        =        23.4       +       16

                        =       39.4  

3.          An element with relative atomic mass 16.2 contains two isotopes 16P8 with relative abundance 90% and mP8 with relative abundance 10%. What is the value of m?

SOLUTION

            16.2 =  90 x 16 + 10 x m
                                  100

             16.2 = 9 x 16  +   m
                          10           10

            16.2 = 144   +   m
                        10         10

            16.2 = 144  +  m
                              10

            16.2 x 10 = 144 + m

            162   = 144 + m

            162 – 144 = m

                  18 =   m

The value of m is 18

OBJECTIVE QUESTIONS 

1.    The atomic number of an element is precisely.

(a) the number of protons in the atom

 (b) the number of electrons in the atom 

(c) the number of neutrons in the atom

(d)  the number of atoms in an element

2.    An atom can be defined more accurately as 

(a) the smallest indivisible parts of an element that can take part in a chemical reaction

 (b) the smallest part of an element that can take part in a chemical reaction

 (c) a combination of protons, neutrons

(d)  the smallest particle of a substance that can take part in a chemical reaction

3.      The mass number is 

(a) proton number + neutron number 

(b) electron number + proton number  

(c) neutron number + electron number

(d) atomic number + electron number

4.  Calculate the relative atomic mass of an element having two isotopes ¹⁰⁷ Ag and ¹⁰⁹Ag in the ratio 1:1  

(a)106             

(b)107         

(c)108     

(d) 109

5.   An element X has two isotopes ^18.8X and ^15.8X in the proportion of 1:9 respectively. Find the relative atomic mass of X           

 (a) 16.1    

(b) 13.6 

(c) 16.8 

 (d)  17

THEORY

1.            Consider the atoms represented below:

                                                                  ^q_X                      ^r_X
                                                        p                      p

a.            State the relationship between the two atoms.

b.            What is the difference between them?

c.             Give two examples of other elements which exhibit the phenomenon illustrated.  (waec)

2.            State the number of electrons, protons and neutrons present in the following atoms/ions

a)    Ca         b) S^2-         c) Al³⁺         d) P

b).            If an element R has isotopes 60% ¹²R₆ and 40% ^xR₆ and the relative atomic mass of R is 12.4, find x.

 3.(a)        Define the term isotopy.

(b)        Determine the number of electrons, protons and neutrons in each of the following: ³⁹K₁₉, ^63.5Cu₂₉

c)         If an element R has isotopes 60% of ¹²R₆ and 40%  ^xR₆ and the relative atomic mass is 12.4, find x.

     

4.        How many neutrons are present in the isotope ³⁷Cl₁₇ ?

5.        A given quantity of chlorine contains 75% ³⁵Cl₁₇, and 25% ³⁷Cl₁₇, determine the relative atomic mass of chlorine.

 

QUANTUM NUMBERS at a glance

 

Rules guiding the arrangement of electrons in an atom

When electrons are arranged or filled into the atoms of elements, certain RULES are considered   and obeyed. These rules are the Aufbau's Principle, Pauli's exclusion principle and Hund’s rule of maximum multiplicity.

PAULI EXCLUSION PRINCIPLE 

states that NO two electrons in the same atom have the sets of the four quantum numbers {n, l, m and s in an atom}.

AUFBAU PRINCIPLE states that "when electrons go into atoms they fill orbitals of lower energy first before filling orbitals of higher energy and each orbital may hold up to two electrons.

 

HUND’S RULE OF MAXIMUM MULTIPLICITY state that " When electrons fill degenerate orbitals they go in singly first before pairing up occurs.

 Degenerate Orbitals are orbitals that are at the same energy level. example of degenerate orbitals is the P-orbital, the d-orbital or the f-orbital

Examples of degenerate orbitals are the P-orbitals,  the d-orbitals  and the f-orbitals 

QUANTUM NUMBERS

The quantum numbers are a set of numbers that describes the position of an electron in an atom. 

Studies have showed  that the energy of an electron may be characterized by four quantum numbers. These quantum numbers help to locate the position of electrons in an atom

1. The principal quantum number represented by n with integral values of 1,2,3,4 e.t.c.

This quantum number describes the shell ( k, l, m....)

2. The subsidiary or Azimuthal quantum number represented by l with integral values ranging from 0 to (n-1).

This quantum number describes the sub-shells ( s,p,d,f,)

3.  The magnetic quantum number represented by m with integral values ranging from –l ,0, +l.

4. The spin quantum number represented by s with integral values – ¹/₂ and + ¹/₂.

Element   At. Numb.  Elect. Conf.

 H .            1;             1s¹

He              2;         1s²

Li.               3;         1s² 2s¹

Be              4;          1s² 2s²

B =.           5;          1s² 2s² 2p¹

C =            6;          1s² 2s² 2p²

N =.          7 ;         1s²2s²2p3      

O=            8 ;       1s² 2s² 2p⁴

F=.           9;         1s² 2s² 2p⁵

Ne=        10;         1s² 2s² 2p⁶

Na=.         11;        1s² 2s² 2p⁶ 3s¹

Mg=          12;     1s² 2s² 2p⁶ 3s²

Al=.        13;   1s² 2s² 2p⁶ 3s² 3p¹

Si =       14;    1s² 2s² 2p⁶ 3s² 3p²

P=.       15;     1s² 2s² 2p⁶3s²³3p³

S =.       16;   1s² 2² 2p⁶3s² 3p⁴

Cl =.    17;     1s² 2s² 2p⁶ 3s² 3p⁵

Ar =.   18      1s² 2s² 2p⁶ 3s²3p⁶

K=.    19  1s² 2s² 2p⁶ 3s²3p⁶ 4s¹

Ca =.  20  1s² 2s² 2p⁶ 3s²3p⁶ 4s²

OBJECTIV QUESTION

1.  Which of the following orbitals is spherical in shape?

     (a) s

      (b) p

      (c) d 

      (d) f

  

2.     Which of the following shells have a maximum of eight electrons?

             (a)  K

             (b) L 

             (c) M 

             (d) N

3.     1s² 2s² 2p⁶ 3p¹ is the electronic configuration of

              (a)  potassium

               (b) calcium 

               (c) sodium

              (d) aluminum.

4  . “Two electrons in an atom cannot have the same set for all four quantum numbers”. This statement is

              (a)  Aufbau principle 

               (b) Pauli exclusion 

               (c) Hund’s rule

              (d) Rutherford’s model.

5.    Which of the quantum number is represented by L?

              (a)  principal quantum number

              (b) subsidiary quantum number                  

              (c) magnetic quantum number

              (d) spin quantum.

6.    Pauli exclusion principles related 

a). quantity of electrons in the valence shell

b). filling the orbitals with lower energy first 

c). the filling of degenerate orbitals 

d). quantum numbers of electrons.

7. Atomic orbital is 

a). the circular path through which electrons which electrons revolve round the nucleus 

b). a region around the nucleus where electrons are most likely to be found 

c). the path around the nucleus through which electrons move 

d). the path around the nucleus through which protons move.

THEORY QUESTIONS 

  1(a)(i)what are the quantum numbers 

   (ii). The models below represent the filling of orbitals in an atom

   

State which rule(s) is/are violated or obeyed by each model

(iii).     State the following principle (a) Pauli exclusion principle. (b) Aufbau principle

(b). Write the electronic configuration of 

(i) Oxygen   (ii) Calcium (iii) Fluoride ion (Cl^-) (iv) Potassium ion  (K⁺)    (v). Aluminum ion (Al³⁺

2.a(i). List the quantum numbers that are assigned to an electron in an atom.

(ii) what is the maximum number of electrons that can occupy the 3d orbital?

3.(a)State 

I.  Pauli's exclusion principle 

ii. Hund's rule of maximum multiplicity

iii. Write the electronic configuration of of each of the following ions of copper I. Cu+ 

II. Cu2+.       [29Cu]

THE HALOGENS at a glance

 

HALOGENS

Halogens (salt formers) are found in group VII of the periodic table. The halogens  are the most reactive nonmetals. They have seven electrons in their outermost shells and so ionize to form univalent negative ions. They exist as diatomic molecules. They are coloured and They form electrovalent compounds with metals.

 In the group are chlorine, fluorine, bromine, iodine and astatine.

ELECTRONIC CONFIGURATION OF HALOGENS 

The halogens have one electron short of the noble gas structure in their  electronic configuration (i.e.  they contains seven electrons in their outermost shells), and the readiness to complete the octet arrangement by receiving an electron makes the halogens very reactive.

The electronic configurations of the halogens are shown below: 

   Fluorine = 9: 1s² 2s² 2p⁵

   Chlorine = 17: 1s² 2s² 2p⁶ 3s² 3p⁵

  Bromine = 35: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵

 Iodine   = 53: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁵

PHYSICAL PROPERTIES OF THE HALOGENS

1.  They are univalent, and readily accept one electron from other atoms to form ionic compounds (especially from metals e.g Na & K). They also share electrons with themselves or with non-metals to form covalent compounds.

2.   They exist as diatomic molecules.

3.  Fluorine and chlorine are gases, bromine is a liquid and iodine a solid.

4.   The halogens are coloured, with typical penetrating odour. The colours deepen down the group. Fluorine is pale-yellow, chlorine is greenish- yellow, bromine is red and iodine is violet.

5.   They are volatile substances. Their volatility decreases down the group.

6.   All the halogens except fluorine, dissolve to some extent in water, fluorine reacts with water to give oxygen and hydrogen fluoride.

CHEMICAL PROPERTIES OF THE HALOGENS

The halogens are very reactive elements. Their reactivity decreases down the group. Fluorine is the most reactive halogen. They are also strongly electronegative. Their Electronegativity decreases down the group.

1. As oxidizing agents. Halogens are strong oxidizing agent. They readily accept electrons. The oxidizing power decreases down a group.

2.    Reaction with metals: Halogens react with metals to form ionic compounds.

 2Na_(s)  +  F_2(g) →  2NaF_(s)

3.  Reaction with hydrogen: Fluorine explodes with hydrogen even in the dark, chlorine reacts slowly in the dark but explode in bright sunlight, bromine reacts with hydrogen in the presence of platinum catalyst, while iodine reacts partially with hydrogen on heating. Example

              H_2(g) + Cl_2(g) → 2HCl_(g)

Stability of the hydrogen halides decreases down the group. Hydrogen fluoride is a liquid with a boiling point of 19^OC. The other hydrogen halides are gases.        

4.    Reaction with water: Fluorine reacts vigorously with water to give off oxygen gas. 

Chlorine reacts very slowly with water to give a mixture of hydrochloric acid and oxochlorate (I) acid 

             Cl2(g) + H2O → HOCl + HCl

which later decomposes to give hydrochloric acid and oxygen gas.

            HOCl(aq) → HCl(aq) + O2(g)

 The oxygen gas given off by the oxochlorate (I) acid is responsible for the bleaching action of moist chlorine gas and chlorine water.

             H₂O_(g)  +  Cl_2(g)  → HCl_(aq)  +  HOCl_(aq)

CHLORINE

Chlorine is the most important member in the halogen family. It does not occur as free element in nature because it is too reactive. It is usually found in combined state as chlorides.

LABORATORY PREPARATION OF CHLORINE

1.  By the oxidation of concentrated HCl with strong oxidizing agent such as MnO₂ or KMnO₄

             MnO_2(s)+ 4HCl_(aq)→MnCl_2(aq)+ 2H₂O_(l) + Cl_2(g)

                                    

2.         By heating concentrated H₂SO₄ with a mixture of NaCl and MnO₂

                2NaCl_(s) + MnO_2(s) + 2H₂SO_4(aq) →Na₂SO_4(aq) + MnSO_4(aq) + H₂O_(l)+ Cl_2(g)

INDUSTRIAL PREPARATION

Chlorine is manufactured industrially by the electrolysis of brine and molten NaCl, MgCl₂ or CaCl₂

PHYSICAL PROPERTIES

1.   Chlorine is a greenish-yellow gas with unpleasant chocking smell.

2. It is a poisonous gas.

3. It is about 2.5 times denser than air.

4. It is liquefied under a pressure of about 6atm.

5. It is moderately soluble in water.

CHEMICAL PROPERTIES

1.   It is very reactive and forms electrovalent compound with metals and a single covalent bond compounds with non-metals.

2Na_(s)  + Cl_2(g)  →  2NaCl_(s)

Cl_2(g)  +   H_2(g) → 2HCl_(g)

2.  It displaces other halogens from solution of their acids and salts

Cl_2(g)  +   NaI_(aq)  →  2NaCl_(aq)    +    I_2(g)

3. It combines directly with other elements except oxygen, nitrogen carbon and the noble gases; to form chlorides

Ca_(s)   +  Cl_2(g)   → CaCl_2(s)

4. It displaces hydrogen from its compounds due to its strong affinity for hydrogen 

C₁₀H_12(l) + 8Cl_{2(g) →}10C_(s) + 16HCl_(g)

5. It is a powerful oxidizing agent: it oxidizes green Fe²⁺ to yellow or brown Fe³⁺

2FeCl_2(aq) + Cl₂ →2FeCl_3(aq)

6.  It is a bleaching agent:  The bleaching action of chlorine is due to its ability to react with water to form oxochlorate (I) acid which decomposes to release oxygen which in turn oxidizes the dye to form a colourless compound.

H₂O_{(l) +} Cl_2(g)  → HCl_{(aq) +}    HOCl_(aq)

HOCl_(aq)     →   HCl_(aq)  +  [O]

Dye   +   [O] →  [Dye + O]

^{Colored                    Colourless}            

7.  It reacts with hot concentrated NaOH solution to give a mixture of sodiumtrioxochlorate (V) and sodium chloride.

6NaOH  + 3Cl_2(g)    →  NaClO_3(aq)   + 5NaCl_(aq)  +  H₂O_(l)

^{hot concentrated                   Sodium trioxochlorate (V)}

With cold dilute solution of NaOH, a pale yellowish mixture of sodiumoxochlorate (I) and sodium chloride is formed.

2NaOH_(aq)  +  Cl_2(g)  →  NaOCl_(aq)  +  NaCl_(aq)  +  H₂O_(l)

^{cold dilute                             sodium oxochlorate(I)}           

8. It reacts with CaOH solutions to produce bleaching powder

Ca(OH)_2(aq)  +  Cl_2(g) →  CaOCl₂.H₂O_(s)

                                          ^{Bleaching powder}

TEST FOR CHLORINE

1.  It turns wet blue litmus paper pink and then bleaches it.

2.  It turns damped starch-iodide paper blue black.  

Chlorine turns starch-iodide paper blue black because it displaces iodine from the iodide. The iodine liberated then turns the starch blue.

USES OF CHLORINE

1. It is a powerful germicide (due to its oxidizing nature).

2.  It is used as a bleaching agent for cotton, wool, pulp etc.

3.  It is used in the manufacture of polyvinyl chloride (PVC) and synthetic rubber.

4. It is used in the manufacture of organic compounds like  trichloromethan (CHCl_3),  and   tetrachloromethane (CCl₄

5. It is used in producing KClO₃, for making matches and fireworks.

6.  It is used for making NaClO_3, a weed killer.

7.  It is used for making acidified NaClO solution a domestic antiseptic.

COMPOUNDS OF CHLORINE

HYDROGEN CHLORIDE

Hydrogen chloride exists as a gas at room temperature. It dissolves readily in water to form hydrochloric acid. It occurs in traces in the air as industrial by-product and is considered as an air pollutant; but it can be easily washed down as acid rain since it is very soluble in water.

LABORATORY PREPARATION

The gas is prepared by the action of hot concentrated H₂SO₄ on any soluble chloride. Example    2NaCl_(s)  +  H₂SO_4(aq)  →  Na₂SO_4(aq)  +  2HCl_(g)

Note: NaHSO₄ is first formed at a lower temperature and later at higher temperature HCl gas is formed. The gas is dried by passing it through concentrated H₂SO₄ in another flask and collected.

INDUSTRIAL PREPARATION

Pure HCl gas can be produced in large scale by direct combination of hydrogen and chloride gas obtained from the electrolysis of brine.

     H_2(g)+ Cl_2(g)→2HCl_(g)

PHYSICAL PROPERTIES

1.  Pure HCl gas is colourless and has sharp irritating smell

2. It turns damp blue litmus paper red

 3.  It is about 1.25times denser than air

4 It is very soluble in water, hydrochloric acid

5. It dissolved readily in non-polar solvent like chloroform and toluene. 

when HCl is dissolved in nonpolar solvents, the solution does not conduct electricity and has no acidic properties because hydrogen chloride which is a covalent molecule does not ionize when it dissolves in non-polar solvents. But it dissolves in water and ionizes. The ions formed in aqueous solution are responsible for the acidic property and conductivity of its aqueous solution. 

6.  It forms misty fumes in moist air because it dissolves in the moisture to form tiny droplets of HCl acid

CHEMICAL PROPERTIES

1.  Combustion: - Hydrogen does not support combustion 

2.  It combines directly with NH₃   producing dense white fumes of ammonium chloride

 HCl_(g) + NH_3(g) → NH₄Cl_(s)

3. It reacts with electropositive metals to form their respective chloride displacing hydrogen gas 

Zn_(s) +2HCl_(g) → ZnCl_2(s) + H_2(g)

TEST FOR HYDROGEN CHLORIDE

1.  Place a gas rod that has been dipped in ammonia solution over the gas jar containing the unknown gas, if a dense white fumes forms on the glass rod, then the gas is hydrogen chloride gas.

2.   Few drops of silver trioxonitrate (V) is added to the gas jar containing the unknown gas and shaken. If white precipitate of silver chloride is observed, then the gas is hydrogen chloride gas

CHLORIDES

Chlorides are normal salts formed when metallic ion replace the hydrogen ion in hydrochloric acid. Soluble Chlorides are prepared by neutralization reaction while insoluble are prepared by double decomposition method.

All Chlorides are soluble in water with except, AgCl, HgCl2, PbCl2

PROPERTIES

1.  Chlorides are stable to heat, that is, they are not decomposed by heat.  

They are recovered from solution by evaporation to dryness and sometimes by crystallization.

2. They react with hot concentrated tetraoxosulphate (VI) acid to produce hydrogen chloride gas.

2NaCl_(s)+H₂SO_4(aq) →Na₂SO_4(aq)+2HCl_(g)

and in the presence of a strong oxidizing agent, chlorine is produced.

ZnCl_2(s) + KMnO_4(s)+ 2H₂SO_4(aq) → ZnSO_4(aq)+ K₂SO_4(aq) + 2MnO_2(aq) + 2H₂O_(l)+Cl_2(g)

TEST FOR CHLORIDES 

Add some Few drops of AgNO_3(aq) to a solution of the sample in a test tube, if it forms   a white precipitate, now acidify the solution by adding dilute trioxonitrate acid if the white precipitate remains but readily dissolves in excess NH_3(aq) solution then a chloride ion is present.

OBJECTIVE QUESTIONS

1. The bleaching action of chlorine is through the process 

a. Hydration 

b. Hydrolysis 

c. Reduction 

d. Oxidation 

2. When chlorine is passed through a sample of water, the pH of the water sample would be 

a. <7

b. =7

c. >7

d. 0

3. Halogens generally react with metals to form 

a. Alkalis.

b. Acids.

c. Bases.

d. Salts.

4. Potassium chloride solid does not conduct electricity because 

a. It is a covalent compound 

b. Stong cohesive forces make its ions immobile 

c. Strong cohesive forces make its molecules immobile 

d. Each of Potassium and chlorine ions has a noble gas structure.

5. Chlorine water is used as a bleaching agent because it is 

a. An acidic solution 

b. An alkaline solution 

c. An oxidizing agent 

d. A reducing agent 

6. Which of the following halogens is a liquid at room temperature 

a. Iodine

b. Chlorine 

c. Bromine

d. Fluorine.

7. Chlorine, bromine and iodine belong to the same group and 

a. Are gaseous at room temperature 

b. Form whit precipitate with AgNO3(aq)

c. React violently with hydrogen without heating.

d. React with alkali 

8. Which of the following chlorides is insoluble in water? 

a. AgCl

b. KCl

c. NH4Cl

d. ZnCl2

9. Which of the following statements about chlorine and iodine at room temperature is correct 

a. Chlorine is a gas and Iodine as solid 

b. Chlorine is a liquid and iodine is a gas.

c. Chlorine and iodine are gases 

d. Chlorine is solid and iodine is a liquid.

10. 

THEORY QUESTIONS

1. Draw and label a diagram to illustrate the preparation and collection of a dry sample of chlorine gas in the laboratory. 

1b. state two use of chlorine 

2a. Write the equation for reaction between chlorine gas and            

  i. Concentrated NaOH

   Ii. Dilut NaOH

b. Write the electronic configuration of the following atoms/ions: Cl, F^-, Br.

C.  Give three physical properties of the halogens

2a.   Explain one laboratory preparation of dry chlorine gas.

b. Name the method of collection of chlorine gas and explain why it can be collected by the method. 

3a.   Mention three physical properties of chlorine.  

b.  Using balanced equations, state THREE chemical properties 

C. Explain why hydrogen chloride in toluene does not conduct electricity but its aqueous solution  conducts  electricity.

4a.   Describe a test for a soluble chloride.

bGive three uses of chlorine gas.

C.  State TWO physical and TWO chemical properties of hydrogen chloride gas

5a.  An unknown gas is colourless, has an irritating smell, fumes in moist air and turns blue litmus paper red; describe how you will confirm the gas to be hydrogen chloride gas.

b. A solid chloride E which sublimed on heating reacted with an alkali F to give a choking gas G. G turned moist red litmus paper blue.  Identify E,F and G.