OXIDATION NUMBER at a glance

 Oxidation number (O.N) of an element is the charge on an atom of the element whether it is by itself or bonded to another atom. It indicates the number of electrons the atom has gained or lose at that moment. That is, it is the charge an element will have if electrons were transferred to or from it. It is usually zero (0) for an element in the uncombined state. It is also referred to as the oxidation state of the element

          NOTE: - The sign or charge for O.N is written before the number (–2) but it is written after the number for an ionic charge i.e O^2–

Rules for calculating oxidation number

The following rules are applied when assigning an oxidation number or calculating the oxidation number of an element thus

1.       The O.N of oxygen is always equal to -2 except in peroxides (–1)

2.   O.N of hydrogen is always equal to plus one (+1) except again in metallic hydrides (–1).

3.        The O.N for an element in the elemental (or ground) state O.N = O (zero) e.g.

          Na = O, Cl₂ = O,  O₂ = O etc.

4.   For an ion or radical the O.N is equal to the charge on it for example 

   Na+ = +1,   Cl^– = –1,     O^2–  =  -2

   CO₃^2-  = –2,  NO₃^- =-1,  SO₄^2- =-2

3        The algebraic sum of the O.N of all the atoms in a compound is equal O e.g.

    H₂O = O,  i.e O.N of H + O.N of O =

(+1 x 2) + ( -2 x1) 

          2 -2 = 0

NaOH = (+1 x1) + (-2 x 1) + (1 x 1) 

                         +1 - 2+ 1

                           +2-2=0

      

Rules for specific groups table groups

I        For group 1A elements (comprising Li,Na,K, e.t.c) their O.N = +1

II       For group 2A elements (comprising Be, Mg, Ca e.t.c) their O.N = +2 in all compounds

III     For group 3 elements (B, Al, e.t.c) their O.N = +3 especially in their binary compounds.

IV     For group 5 = -3

V     For group6 = -2 except Oxygen (O) in peroxides).

 VI   For group 7 = –1 respectively especially in their binary compounds 

 

Determining the oxidation number of an element

1   Find the O.N of the underlined elements in the following

  a). ZnCl₂        b). SO₃      c). NO₃^-         d).Ca²⁺

          Solution

  To determine the O.N of the underlined elements, we must refer to the general rules for assigning O.N to an element.

   a). ZnCl₂: The algebraic sum of the O.N of all the atoms in a compound is equal to zero, i.e. ZnCl₂  = O. 

       Since Cl is a group 7(A) element and ZnCl₂ is a binary compound then the O.N of Cl is –1, therefore, the O.N of Zn is

        (O.N of Zn) + (O.N of Cl ´ 2) = 0
                   x + (–1 ´ 2) = 0
                   x – 2 = 0
                   x = +2

 

          b)      SO₃

                    Solution

(O.N of S) + (O.N of 0 ´ 3) = 0
            x + (–2 ´ 3) = 0
             x – 6 = 0
           x = +6

                   Trioxosulphate(IV) ion

 

 

          c)        NO₃^-

                   Solution

         If we recall from the rules s for assigning O.N to an element. The O.N of a radical is equal to the charge on it, hence

                    NO₃^- =  –1
                   that is,
                   (O.N of N) + (O.N of 0 ´ 3) = –1
                   x + (–2 ´ 3) = –1
                   x – 6 = –1
                   x = +6 – 1
                    =  +5

                   Trioxonitrate (V) ion

    d)Ca²⁺ – here we refer to the rule that says the O.N of an ion is the charge on it, i.e., Ca²⁺ = +2  Calcium  = ion

 Uses of oxidation number

Oxidation number is used for in the

1   It is used in the  IUPAC (Internal Union of Pure and Applied chemistry) system of naming compounds e.g. H₂SO₄: TetraoxosulphateVI acid

2. It is used to know the oxidation state or number of an element in a compound 

 1.       Find the O.N of the following underlined elements

          a)       Na₂SO₄                                  (a = +4,  b = =2,  c = +6,  d  =  –5)

          b)      [Al(H₂O)₆]³⁺                           (a = +3,  b = –3,  c = +6,  d = –6)

          c)       K₂Cr₂O₇                                 (a = +5,  b = +4,  c = –6,  d = +6)

          d)      Mn                                            (a = +6,  b = +7,  c = +5,  d = +3)

 

2        Which species undergoes reduction in the reaction represented by the equation below?

 H₂S_(aq)+2FeCl_3(aq)→S_(s) + 2HCl +3FeCl₂

   a) Fe³⁺     (b). H₂S            (c). Cl^–      (d)  S

3. Find the oxidation numbers of the following underlined elements.

 a) K₂Cr₂O₇           

   b).KMnO₄       

  c). HNO₃

   d). S^2-       

  e). Cl^-                 

  f). Cr₂

OBJECTIVE QUESTIONS 

1. Oxidation is a reaction which involves the following except 

a. Loss of electrons

b. Increase in oxidation number 

c. Gain of oxygen 

d. addition of hydrogen

2. 

THEORY QUESTIONS 

1. State two applications of oxidation numbers 

REDOX REACTION at a glance

 

Oxidation–Reduction reaction (Redox)

 Oxidation–Reduction(redox) reactions are two opposite and complementary reactions which occur simultaneously

Redox reactions have been defined in several ways before attaining a more general and simplified definition.

1        In term of addition of oxygen:  Oxidation is defined as the addition of oxygen to a substance while reduction is defined as the removal of oxygen from a substance. E.g.

            
              Reduction              CuO + C_(s) → Cu_(s) +  CO_(g)     O.A    R.A                       
   

the example above, carbon (C) is oxidized to carbon (II) oxide (CO) while Cupper (II) oxide (CuO) is reduced to metallic Copper (Cu).

   Carbon is removing oxygen from CuO and so is the reducing agent because it causes CuO to become reduced to Cu. CuO  supplies the oxygen atom that causes carbon to become oxidized to CO and so CuO  is the oxidizing agent.

2       In terms of removal of hydrogen:  Oxidation is defined as the removal of hydrogen from a substance while reduction is defined as the addition of hydrogen to a substance                    

                    ——————
               ↓Reduction   ↓
H₂S   +   Cl₂ → S_(s) + HCl(aq)
 R A        O.A    
 ↑  Oxidation  ↑
     ——————

                     

          Similarly in this reaction, H₂S is oxidized to atomic S_(s) due to the removal of hydrogen as chlorine is reduced by gaining or addition of hydrogen. H₂S is action as the reducing agent while Cl₂ is the oxidizing agent.

3        In terms of change in the oxidation number of an element: Oxidation is defined as the increase in oxidation number of an element while reduction is the decrease in oxidation number of an element. 

4.  Definition in terms of electronegative elements: - Oxidation is the addition of an electronegative element to a substance or the removal of electropositive element from a substance while reduction is the removal of electronegative element from a substance or the addition of electropositive element to a substance 

4        In terms of electron transfer: Oxidation is defined as the loss of electrons while reduction is defined as the gain of electrons. 

  When an element loses an electron to become an ion; the O.N increase to a higher number while a gain of electrons by an element will lead to a decrease in O.N of an element. For example,

        ₂₀Ca   → 20Ca²⁺+ 2e^–
        20 protons      20 protons
   20 electrons         18 electrons
    O.N 0 (zero)        O.N (+2)

        

        ₁₇Cl    +   e^–   → 17Cl^–
     17 protons            17 protons
     17 electrons         18 electrons
     O.N 0 (zero)         O.N( –1)

          In other words, as noted from the above examples, loss of electrons means a higher O.N while gain of electrons means a decrease in O.N of an element.

                            Oxidation
               Mg + Cl   →MgCl2       
               R.A   O.A     reduction

Example of redox reactions that occurs generally around us include

1. Photosynthesis

2. Rusting of iron

  Fe(s) + nH₂O → Fe₂O₃.nH₂O_(s)

3.   Combustion

             –——————
            ↓ Oxidation      ↓
          C₄H₁₀ + O2 → CO₂ + H₂O
                              red   ↑
                            ——————

Redox reactions always involve the movement of electrons ( i.e loss and gain of electrons) For example Pb° →Pb²⁺ + 2e^–  

Pb²⁺→Pb⁴⁺ + 2e^–. O.N increased from 0 to +2 to +4 in Pb. i.e., oxidation involves loss of electrons which will lead to increase in O.N.

In contrast reduction involves the gains of electrons which will lead to a decrease in the O.N of the element for example                   S° + e → S^– + e^– → S^2–.

Some examples of redox recitations are

1        Fe_(s)+ S_(s)→ FeS

          R.A        O.A

          In the above reaction Fe losses electrons to become iron (II) ions (Fe²⁺), there is an increase in its O.N from 0 to +2 and so it is the reducing agent. Sulphur on the other hand, gains electrons from the iron, its O.N decreases from 0 to (–2) and so it is the O.N

2. Pb⁺²O^-2 + C⁺²O^-2 → Pb⁰+ CO₂

          In the above example, the O.N of Pb decreased from +2 to 0, so PbO is reduced and so PbO is the oxidizing agent.

          The O.N of C increased from +2 to +4, so CO is oxidized and so CO is the reducing agent.

3        H_2(g)+ O_2(g) → H₂O_(l) 
          R.A           O.A

          O.N of H increased from 0 to (+1) i.e. H is oxidized.

          The above reaction is a combustion reaction, and in this reaction. It is important to note that all combustion reactions are redox reaction with oxygen as the oxidizing agent.

i        AgNO_3(aq) + NaCl_(aq) → AgCl_(s) + NaNO_3(aq)

          The above reaction, is a double decomposition reaction, it is not a redox reaction as there is no change in the O.N number of all the element s involved. Another non-redox reaction is a neutralization reaction, there is no change in the O.N of element involved in the reaction.

  ii  KOH_(aq) + HCl_(aq) → NaOH_(aq) + H₂O_(l)    (neutralization reaction)

          

Oxidizing and reducing agents (in summary)

OXIDIZING AGENT 1.      Supplies Oxygen 2.      Removes Hydrogen 3.      Decreases in oxidation number 4.      Gains electrons

REDUCING AGENT Supplies hydrogen Removes Oxygen Increases in oxidation number Loss electrons

Test for oxidizing agents

To common test or reactions that are used to test for an oxidizing agent involves the action on iron (II) chloride and hydrogen sulphide.

a)       Reaction with FeCl₂

          When an oxidizing agent is added to green iron (II chloride; the iron (II) ions become oxidized to yellow or brown Fe³⁺.

          Fe²⁺ →     Fe³⁺ + e^–

          green         yellow/brown

b)      Reaction with hydrogen sulphide

          When hydrogen sulphide is bubbled through a solution of an oxidizing agent, the sulphide ions S^2– becomes oxidized to elemental sulphur; and this is seen or observed as yellow deposits sulphur,                    i.e. S^2– → S(s) + 2e^–.

Test for reducing agents

Two commonest reagents that are used to test for a reducing agent are

1        Acidified potassium tetraoxomanganate(VI) (KMnO₄) and acidified potassium heptaoxodichromate(I) (K₂Cr₂O₇).

a)   Action of potassium hyptaoxodichromate (VI) (K₂Cr₂O₇)

  When acidified potassium heptaoxodichromate (VI) (K₂Cr₂O₇) is added to a sample of a reducing agent, its colour changes from orange to green, due to the reduction of the dichromate (VI) ion  (Cr⁶⁺) (orange) to chromium (III) (Cr³⁺) ion green

     Cr⁶⁺  +  3e^– → Cr³⁺
  Orange                green

 b) Test using acidified potassium tetraoxomangane(VI) (KMnO₄)

  When acidified potassium tetraoxomanganate (VII) to a sample of reducing agent, the purple colour changes to colourless: due to the reduction of the manganate ion from (+7) which is purple to (+2) which is colourless and a more stable oxidation state.

MnO₄^- + 8H⁺ + 5e^– →Mn²⁺ + 4H₂O

                  purple                       colourless         

Mn⁷⁺ + 5e^– → Mn²

  purple              colourless

                   This reaction is reversible as the purple colour is restored when an oxidizing agent is reintroduced into the mixture.

                  Mn²⁺ + 5e →Mn⁷⁺   
               colourless        purple

THEORY QUESTIONS 

1(a)State what you will see on

i) bubbling SO2 into acidified KMnO4 solution 

ii). 

(b)(i).Write the ionic equation for the reaction between zinc powder and silver trioxonitrate (V) solution 

(ii). Which substance in bi above is I. Oxidized  II. Reduced 

SODIUM at a glance

 

Sodium:  is found in group 1 period III on the periodic table. It has an atomic number of 11 and an atomic mass of 23. 

 It does not occur as a free element in nature because it is very reactive. However, it is found mainly in the combined state as sodium chloride in sea water, and as rock salt (Halite) in underground deposits. 

It is extracted from fused sodium chloride by electrolysis using the Dawn's Cell. a little amount of CaCl2 is added to lower the melting point (from about 801 to 600)

 Chemistry of the reaction

at the cathode

At the cathode: - the sodium ions migrate to the cathode where they gain an electron each to become reduced to metallic sodium

          Na⁺_(l) + e- → Na_(s)

At the anode: - the chloride ions migrate to the anode where they loss their excess charge and become reduced to atomic chlorine

                Cl^- → Cl + e^-

the chlorine atom combines with another chlorine atom and is discharged as chlorine gas

          Cl + Cl → Cl_2(g)                                                                                      

Properties of Sodium:

i.                    Sodium is soft and can easily be cut with a knife.

ii.                 It has a of 0.968g/cm³ 

iii.               It has a silvery-white appearance.

iv.               It has a low melting point.

v.        It has a boiling point.

v.                  Sodium is a good conductor of electricity.

Chemical Properties

i)                   Reaction with air or oxygen: - Sodium metal tarnishes on exposure to air

 4Na_(s) + O_2(g) →2Na₂O_(s)

                 Na₂O_(s) + H₂O_(l) → 2NaOH (s)   

                 NaOH_(aq) + CO₂ → Na₂CO₃  

 In excess air or oxygen, it burns with a golden yellow flame to yield sodium peroxide Na₂O₂,

               2 Na_(s) + O_2(g) → Na₂O_2(s)

 In limited supply of air sodium oxide (Na₂O) is formed.

                Na_(s) + O_2(g) → 2Na₂O_(s)

 Because of its reactivity sodium is stored under paraffin oil or other organic solvents like naphtha or toluene.

ii.                  Reaction with water: - It reacts violently with cold water to yield sodium hydroxide and hydrogen with large amount of heat.

  Na_(s) + H₂O_(l) → NaOH_(aq) + H_2(g)

iii.               Reaction with acids: - It reacts explosively to form a salt and hydrogen gas  

iv.                     Na_(s) + HCl_(aq) →NaCl_(aq) + H_2(g)

                       Na_(s) H₂SO_4(aq) → Na₂SO_4(aq) + H_2(g)

 This reaction is highly dangerous and should not be carried out in the school laboratory.

Reaction with non-metals: - sodium combines directly with the following non-metals when heated to form binary compounds.  

               Na_(s) + S_(s) →Na₂S_(s)

               Na_(s) + H₂ → NaH_(s)

               Na_(s) +P_(s) → Na₃P_(s)

              Na_(s) + Cl_(g) → NaCl_(s)

Sodium does not react with carbon, boron and nitrogen

Reaction with mercury: - Sodium forms various stable mixture with mercury known as sodium amalgam of varying composition such as  NaHg, Na₂Hg, Na₃Hg etc.

Sodium amalgam reacts with water to yield hydrogen.

          Na_(s) + Hg_(l) →NaHg_(l)

v.                  Reaction with ammonia: - Sodium reacts with ammonia to form sodamide and hydrogen gas.

             Na_(s) + NH_3(g) → NaNH_2(s) + H_2(g)

 As a reducing agent: - Sodium act as a strong reducing agent. It reduces some metallic chlorides and oxides to their metals.    

 Na_(s) + BeCl_2(s) → NaCl_(s) + Be_(s)

Test for sodium ions

i).   Flame test: when sodium compounds give a bright or golden yellow flame when burnt in a non-luminous flame

Uses of Sodium:

-i). Sodium is used in the manufacture of other compounds like sodamide, sodium peroxide.

ii). Sodium alloys like  NaK(sodium-potassium alloy), are used as coolant in nuclear reactors.

iii). Sodium vapor lamps are commonly used for street lighting

iv) It is used in the manufacture of tetraethyl lead (C₂H₅)₄Pb, which is used as an antiknock agent in petrol.

v) It is used as a laboratory reagent (Lassaigne's extract).

vii). It is used for producing amalgams used as reducing agents.

viii)  Sodium used as a catalyst in the preparation of artificial rubber and also as a deoxidizer in the preparation of light alloys.

COMPOUNDS OF SODIUM

 Sodium compounds are generally white crystalline salts and are mostly soluble in water.

1.      Sodium chloride (NaCl): (table salt) it is found naturally in sea water and in underground deposits as rock salt.

Properties

- It is a white anhydrous crystalline solid

- It has a melting point of 801⁰C and a boiling point of 1420⁰C.

- The pure form is not deliquescent.

Uses

1. It is used as a food preservative.

2. It is used as an important raw material for the manufacturing of Na, NaOH, Cl₂, Na₂CO₃, NaClO₃ and other compounds.

3.  It is used for salting out soap

4. It is used in glazing earthenware

5.  It is used in regenerating water softener.

2.  Sodium hydroxide (NaOH):  It is a white crystalline solid, made into flakes or pallets

Properties

-i). It is a white crystalline solid

ii).  It is highly deliquescent 

iii) It has a melting point of 320⁰C without decomposing.

iii).  It dissolves in water to give a strong alkaline solution with the evolution of heat

Chemical properties

With acids: - NaOH produce salt and water.

2NaOH_(aq) + H₂SO_4(aq) → Na₂SO_4(aq) + 2H₂O_(l)

With acidic oxides: - It form sodium salt. E.g.

NaOH_(aq) + SO_2(g) → NaHSO_3(aq)

With ammonium salts: - When heated with an ammonium salt, ammonia gas is liberated.

NaOH_(aq) + NH₄Cl_(s)→ NaCl_(aq) + H₂O_(l) + NH_3(g)

With metals – Al and Zn are amphoteric; they combine with excess NaOH to form the  alluminate (III) and respectively with hydrogen gas. 

2Al_(s) + 2NaOH_(aq) + 6H₂O_(l) → 2NaAl(OH)_4(aq) + 3H_2(g)

                                                                        sodium aluminate (III)

Zn_(s) + 2NaOH_(aq) + 2H₂O_(l) → Na₂Zn(OH)_4(aq) + H_2(g)

                                                                        sodium zincate (II)

Therefore, Aluminium or Zinc containers should not be used to store NaOH.

As a precipitating agent: - NaOH solution is most times used to precipitate insoluble hydroxides. E.g

Zn²⁺_(aq) + 2OH^-_(aq) → Zn(OH)_2(s)

Pb²⁺_(aq) + 2OH^-_(aq) → Pb(OH)_2(s)`

Zn(OH)2, Al(HO)3, Sn(OH)2, and Pb(OH)2, are amphoteric and will react excess sodium hydroxide to form complex salts. E.g

Zn(OH)_(s) + 2NaOH_(aq) → Na₂Zn(OH)_4(aq)

With non-metals: NaOH reacts with various non-metals to form sodium salts.

Reaction with glass – High concentrations of NaOH attack glass to form sodium trioxosilicate (IV). Hence, glass stoppers are not used to cover reagent bottles containing concentrated sodium hydroxide or burette because they would become stuck. This is called etching.          

Uses of NaOH

1. it is used as a strong alkali

2. it is used as an analytical and precipitating reagent

3. it is used for absorbing CO₂

4. it is used for making soap, rayon (artificial silk), 

5. I is used for making paper 

6. it is used for making various compounds like sodium trioxochlorate (V), sodium methanoate and phosphine.

7. it is used for purification of bauxite

8. it is used petroleum refining.

9. it is used for the bleaching of cotton textiles.

3.Sodium tetraoxosulphate (IV) (Na₂SO₄₎

Properties

It occurs both in the anhydrous form called saltcake or as a decahydrate form known as Glauber’s salt which is efflìorescent.

Uses Of Na₂SO₄

i.  It is used as a purgative

ii.   In producing of sodium sulphide

Iii it is in the manufacture of wood pulp, glass, and detergents

4. Sodium bicarbonate (NaHCO₃): (baking soda): - A white it is utilized in cooking and as a leavening agent in baking.

- Sodium carbonate (Na₂CO₃): Also called soda ash

Properties

-Na₂CO₃ in form of soda ash (i.e. anhydrous Na₂CO₃) is a fine white powder, while washing soda (Na₂CO₃.10H₂O) is translucent and crystalline.

i. They both dissolve in water to form an alkaline solution by hydrolysis.

ii. Washing soda is efflorescent

iii. It does not decompose on heating

iii. t reacts with acid to liberate CO₂

Uses of Na₂CO₃

i. It is used in the industrial manufacturing of glass

ii.  It is used as a water softener

iii.  It is used in manufacturing of detergent

iv. It is used in the manufacturing NaOH, borax, waterglass, soap and paper

iv.  It is used in laboratory to standardize acids and as an analytical reagent.

vi.  it is used in glass production,

vi.                It is used as a pH regulator in various industrial processes

SOLVAY PROCESS: This is the industrial preparation of NaCO3

The raw materials are sodium chloride, ammonia gas and limestone. The reactions are as follows

1.The ammonia gas in brine (conc. sodium chloride) to give a mixture known as ammoniacal brine.     

ii. This mixture is then allowed to trickle down a Solvay tower as a stream of carbon (IV) oxide is forced up the tower. It reacts with the ammonia in the mixture to yield ammonium hydrogen trioxocarbonate (IV) (NH₄HCO₃). 

  i).  NH_3(g) + CO_2(g) +H₂O →NH₄HCO_3(aq)

The NH₄HCO₃ reacts with the sodium chloride to give sodium hydrogen trioxocarbonate (IV) (NaHCO₃). 

   ii). NH₄HCO_3(aq) + NaCl_(aq) →NaHCO_3(s) + NH₄Cl_(aq)

The sodium hydrogen trioxocarbonate (IV) is slightly insoluble in water and so precipitates out as a white sludge. 

 The NaHCO₃ is then filtered, rinsed and heated to give anhydrous sodium trioxocarbonate (IV) (soda ash), steam and carbon (IV) oxide

  

 iii). NaHCO_3(s) →Na₂CO_3(s) + H₂O_(l) + CO_2(g)

The anhydrous Na₂CO_3(s) (soda ash) is redissolved in hot water and recrystallize to give the pure hydrated compound (Na2CO₃.10H₂O) called washing soda

    iv). Na₂CO_3(s) + 10H₂O_(l) →Na₂CO₃.10H₂O_(s)

some highlights of the process

-Perforated dome-shaped baffle-plates are incorporated into the Solvay tower to slow down the flow rate of the ammoniacal brine so as to allow for proper contact between the ammoniacal brine and the carbon (IV) oxide as well as increase the surface area of reaction

-The concentrated sodium chloride also serves as a carrier for the ammonia gas.

Importance and Economics of the reaction: - 

i. All the raw materials required in the Solvay process are quite cheap and are also readily available. 

ii. Almost all the carbon (IV) oxide generated during the process from the decomposition of the NaHCO_3(s) is recycled, making the process quite economical. 

iii. The sodium chloride solution is obtained from sea water or from rock salt deposits,

iv.  the carbon (IV) is got from limestone found in rich deposits around.

     CaCO_3(s) →CaO_(s) + CO_2(g)

The calcium oxide (CaO) is then reacted with the ammonium chloride to generate ammonia gas, which is also recycled back into the, producing calcium chloride as a by-product from the process.

    CaO_(s) + NH₄Cl_(aq)→ CaCl₂ +H₂O+ NH_3(g)

 5. Sodium nitrate (NaNO₃): It is a white crystalline solid produced when sodium hydroxide reacts with trioxonitrate (V) acids.

Properties

I). it is a white crystalline solid

ii). It has a melting point of 310⁰C and decomposes on further heating.

Uses

i)  it is used primarily as a nitrogenous fertilizer

ii)  In making trioxonitrate (V) acid, potassium trioxonitrate (V) and sodium dioxoxnitrate (III).

iii). It is used in the production of explosives and glass.

Objective questions 

1.

THEORY QUESTIONS 

1. Explain with equations where appropriate the functions of the following substances in the Solvay Process (i) limestone  (ii). ammonia (iii). brine.

2. Calculate the mass of sodium trioxocarbonate (IV) produced by the complete decomposition of 16.8g of sodium hydrogen trioxocarbonate (IV). [ H=1, O=16, Na=23, S=33]

ALKYNES at a glance

 UNSATURATED HYDROCARBON (ALKYNES)

Alkynes are a homologous series of unsaturated hydrocarbons containing at least one triple bond. It has a functional group of (≡) and general molecular formular of C_nH_2n-2 where n= 1,2,3, ... n for successive members of the group. 

The first member of the alkyne family is ethyne (acetylene).

 Alkynes are named by replacing ending –ane  of the corresponding alkane with –yne.

 

NOTE: Since alkynes contain triple bonds between C≡C therefore n=1 is not visible.

When n=	General Molecular Formulae CnH2n-2	Name
2.	C2H2x2-2 = C2H2	Ethyne
3.	C3H2x3-2 = C3H4	Propyne
4.	C4H2x4-2 = C4H6	Butyne
5.	C5H2x5-2 = C5H8	Pentyne
6.	C6H2x6-2 = C6H10	Hexyne
7.	C7H2x7-2 = C7H12	Heptyne
8.	C8H2x8-2 = C8H14	Octyne
9.	C9H2x9-2 = C9H16	Nonyne
10.	C10H2x10-2 = C10H18	Decyne
11.	C11H2x11-2 = C11H20	Undacyne
12.	C12H2x12-2 = C12H22	Dodecyne
13.	C13H2x13-2 = C13H24	Tridecyne
14.	C14H2x14-2 = C14H26	Tetradecyne
15.	C15H2x15-2 = C15H28	Pentadecyne
16.	C16H2x16-2 = C16H30	Hexadecyne
17.	C17H2x17-2 = C17H32	Heptadecyne
18.	C18H2x18-2 = C18H34	Octadecyne
19.	C19H2x19-2 = C19H36	Nonadecyne
20.	C20H2x20-2 = C20H38	Icosyne/Eiocosyne

 

 MOLECULAR STRUCTURES OF ALKYNES

N	ALKYNES	STRUCTURAL FORMULAR	MOLECULAR FORMULAR
2.	C2H2 Ethyne	H-C≡C-H	HC≡CH
3.	C3H4 Propyne	H  H-C-C≡C-H       H	CH3C≡CH
4.	C4H6 Butyne	H H  H-C-C-C≡C-H      H H	CH3CH2C≡CH
5.	C5H8 Pentyne	H H H  H-C-C-C-C≡C-H      H H H	CH3(CH2)2C≡CH
6.	C6H10 Hexyne	H H H H  H-C-C-C-C-C≡C-H      H H H H	CH3(CH2)3C≡CH
7.	C7H12 Heptyne	H H H H H  H-C-C-C-C-C-C≡C-H      H H H H H	CH3(CH2)4C≡CH
8.	C8H14 Octyne	H H H H H H  H-C-C-C-C-C-C-C≡C-H       H H H H H H	CH3(CH2)5C≡CH
9.	C9H16 Nonyne	H H H H H H H  H-C-C-C-C-C-C-C-C≡C-H      H H H H H H H	CH3(CH2)6C≡CH
10.	C10H18 Decyne	H H H H H H H H  H-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H	CH3(CH2)7C≡CH

NOMENCLATURE OF ALKYNES

The nomenclature of alkynes is similar to that of alkenes in many respects as shown in the structures below. The only difference lies on the type of bonds, in alkenes (double bond) and alkynes (triple bond).

 

(i)         CH₃-CH₂-C≡CCH₃              

                    Pent-2-yne  

  (ii)        CH₃CH₂CH₂C≡CCH₃    

                       hex-2-yne

            C1H₃                                         C1H₃
                 |                                                            |   
(iii)       C2HC3≡C4C5H₃              (iv)    C2H₂-C3≡C4-C5H₂
                                      |                                                                  |
                                     CH₃                                                            C6H₃
             4-methylpent-2-yne                                 hex-3-yne

               CH₃                                                        CH₃          CH_{3
                          |}                                                                           |                 |
(v)        CH₃CHC≡CCHCH₃                (vi)       C1H₃C2-C3≡C4-C5-C6H₃
                                         |                                                  |                 |
                                  CH₃                                           CH₃          CH₃
            2,5-dimethylhex-3-yne                         2,2,5,5-tetramethylhex-3-yne

                          CH₃
                                            |            
(vii)      CH₃CH-C ≡CC-CH₂CH₃                (viii)     CH₃C≡CCH₂
                          |              |                                                               |
                    CH₃        CH₃                                                        CH₃
             2,5,5-trimethylhept-3-yne                          pent-2-yne

                 CH₃                                                                          CH₃ 
                        |                                                                                |
(ix)       CH≡CC-C=CCH—CH₂CH₂C≡CH               (x)        CH₃C-CHC≡CC≡CC≡CH 
                        |              |                                                                 |
                      CH₃      CH₂CH₃                                                      CH₂CH₃  

    6-ethyl,3,3-dimethyldec-1,6-diyne                 8-ethyl, 8-methylnon-1,3,5-triyne

                                                                Cl
                                                                                        |
(xi)       CH₃C≡CCHCH₃                        (xii)      CH₃-C-C≡CH 
                             |                                                          |
                            Cl                                                       Cl
            4-chloropent-2-yne                                         3,3-dichlorobut-1-yne

(xiii)     CH₃CHC≡CC≡CCHCH₃                (xiv)     CH₃CHC≡CCHC≡CH  
                     |                     |                                                |            |
                   Cl                   Br                                             Cl         Cl
          2-bromo, 7-chlorooct-3,5-diyne                     3,6-dichlorohept-1,3-diyne

                             H    
                              |
                         H-C-H      
                              |
              H H H      |         
              |    |     |     |
(xv)  H-C6-C5-C4-C3C2≡C1H
              |    |     |     |
             H H H       |
                               |
                          H-C-H
                               |
                              H
            3,3-dimethylhex-1

LABORATORY PREPARATION OF ETHYNES (ALKYNES)

Ethyne is prepared in the laboratory by adding cold water into calcium dicarbide (CaC₂). Much heat evolved and sand is placed beneath the flask to protect the flask from breakage. Ethyne is collected over water. The main impurity, phosphine, PH₃ is absorbed by the acidified CuSO₄ solution.

 

CaC₂  +  2H₂O → Ca(OH)₂  +  C₂H₂.

                                                 Ethyne

 

 

PHYSICAL PROPERTIES OF ETHYNE

1. It is colourless gas

2. It has sweet smell when pure

3. Almost insoluble in water

4. It is neutral to litmus

5. It is strongly exothermic

CHEMICAL PROPERTIES OF ETHYNE

 Alkynes such as ethyne also undergoes addition reaction – a reaction in which one molecule of a compound is simply added on to the alkynes at the position of the carbon – carbon triple bond (C≡C) and this is converted to carbon – carbon single bond (C-C) that is, the alkanes. Examples of addition reaction are:

1.     Reaction of ethyne with hydrogen in the presence of nickel as a catalyst

•                                       Ni
•             CH≡CH + 2H₂   →   CH₃CH₃   
•             ethyne                           ethane

2. Reaction of ethyne with bromine to produce 1,1,2,2-tetrabromoethane. The reddish brown colour of bromine is destroyed.

            CH≡CH + 2Br₂ → CHBr₂-CHBr₂

3. Reaction of ethyne with chlorine to produce hydrogen chloride

            CH≡CH + Cl₂ → 2C+ 2HCl

4. Reaction of ethyne with oxygen or combustion reaction of ethyne (alkynes) to produce carbon(iv)oxide and water

            2CH≡CH + 5O₂ →4CO₂ + 2H₂O

5. Polymerization reaction of ethyne to produce benzene.

            3C₂H_{2 →} C₆H₆ 

6. Reaction of ethyne with water in the presence of dilute H₂SO₄ and mercury as a catalyst to produce ethanal

            CH≡CH + H₂O →CH₃CHO

7. Reaction of ethyne with KMnO₄ to produce 1,2-ethan-diol (glycol)

            CH≡CH +KMnO₄ →CH₂-CH₂
                                               |          |
                                               OH   OH
                                               1,2-ethan-diol

USE OF ETHYNE

1. In oxy-acetylene flame for welding and cutting of metals

2. In oxy-acetylene torch

3. In preparation of acetic acid

4. as a starting material for making polyvinylchloride (PVC) which is used in electrical insulation and water proofing.

TESTS TO DISTINGUISHED BETWEEN ALKANES, ALKENES AND ALKYNES.

The following test can be performed to distinguished clearly the different classes of hydrocarbons, that is, the alkanes, alkenes and alkynes.

All alkanes are saturated compounds while both alkenes and alkynes are unsaturated.

TEST 1: 

To the suspected hydrocarbons, add an acidified solution of KMnO₄ or K₂Cr­₂O₇ solution. Alkanes have no effect in any of these solutions while both alkenes and alkynes decolorized. Acidified KMnO₄ solution changes from purple to colourless, while K₂Cr₂O₇ changes from orange to green.

TEST 2:  Add the solution of Ammonical copper (I) chloride to the suspected hydrocarbons. it will form a yellowish or reddish –brown precipitate with terminal alkynes (alkynes with the triple bond in front of or behind the first C- atom). Alkanes and alkenes show no reaction.

2NH₄OH_(aq)+ 2CuCl + C₂H₂→ CuC₂ + 2NH₄Cl + 2H₂O

TEST: To the suspected hydrocarbon, add solution of Ammonical silver tronitrate (v). Alkanes and alkenes have no effect, but alkynes form a yellowish precipitate.  

2NH₄OH + 2AgNO₃ + C₂H₂ → 2AgC + 2NH₄NO₃ + 2H₂O

OBJECTIVE QUESTION

1.When alkynes are hydrogenated completely, they produce compounds with the general molecular formula 

a. CnHn

b. CnH2n+2

c. CnH2n

d.CnH2n-2

2. 

3

4

5

6