Group 4 element (carbon, silicon, germanium, tin, and lead) (A-level inorganic chemistry)

Group 4 Elements, carbon, silicon, germanium, tin and lead

Members

Elements                     symbol

Carbon                        C

Silicon                          Si

Germanium                Ge

Tin                               Sn

Lead                            Pb

Physical properties

• Carbon and silicon are nonmetals.
• Germanium is a metalloid
• Tin and lead are metal

Electron configuration

The general electron configuration of group 4 elements is ns²np²

Oxidation states

The common oxidation state of group IV elements is +2 and +4

Stability of oxidation states

Oxidation state +2 become more stable down the group

Oxidation state +4 becomes less stable down the group.

For instance, carbon burns in oxygen to form carbon (IV) oxide (CO₂) whereas lead form Lead (II) oxide (PbO).

Secondly, CCl₄ is stable whereas PbCl₄  easily decomposes on heating giving lead (II) chloride and chlorine

PbCl₄ (g) (heat) →PbCl₂ (s) + Cl₂ (g)

Reasons for variation of stability of oxidation states

Down the group IV; the +2 oxidation state becomes more stable due to the inert pair effect i.e. the two s-electron do not participate in the bonding due to the increase in the energy difference between ns and np electrons.

Properties of the compound with +2 and +4  oxidation states

Compounds in oxidation state +4 are predominantly covalent because formation  M⁴⁺ requires alot energy whereas compound in oxidation state +2 are ionic

This explains the fact that PbCl₄ is soluble in organic solvents like ethanol whereas PbCl₂ does not.

Note that

1. covalent compounds are soluble in organic solvents whereas ionic compounds do not.
2. Covalent compounds have low melting and boiling points whereas ionic compounds have high melting and boiling point.
3. Covalent compounds do not conduct electricity whereas ionic compounds do. On this basis compounds in the oxidation state, +4 show properties of covalent compounds because they are predominately covalent.

Chemical properties

Reaction of group (iv) elements with water

Carbon does not react with cold water but reacts with steam to form water gas

            C(s) + H₂O(g)   →   CO(g) + H₂(g)

Silicon, germanium, and Tin do not react with water up to 100⁰C because they are protected by thin layer of oxide. But they react with steam at very high temperatures to form dioxides

Ge(s) + 2H₂O(g)   →   GeO₂(s) +2H₂(g)

Lead reacts with soft water (water that contains oxygen) to form lead II hydroxide.

Pb(s) + 2H₂O(g) + O₂(g)  → Pb(OH)₂(s)

Reaction of group (IV) elements with acids

Dilute nonoxidizing acid like HCl.

C, Si, and Ge do not react

Sn liberates hydrogen

Sn(s) + 2H⁺(aq)       Sn²⁺(aq) + H₂(g)

Pb lead does not react with dilute hydrochloric acid and sulphuric acid due to the formation of insoluble salts.

Nitric acid

With concentrated nitric acid.

C, Si, Ge and Sn react to form dioxide

C(s) + 4HNO₃(aq)→ CO₂(g) +4NO₂(g) + 2H₂O(l)

Pb reacts to form lead II nitrate and nitrogen dioxide.

Pb(s) + 4HNO₃(aq)  →  Pb(NO₃)₂(aq) + 2NO₂(aq)+ 2H₂O(l)

With 50% nitric acid.

Pb reacts to form lead II nitrate and nitrogen monoxide.

3Pb(s) + 8HNO₃(aq)  → 3Pb(NO₃)₂(aq) + 2NO(aq)+ 4H₂O(l)

With sulphuric acid

Dilute sulphuric acid

C, Si, and Ge do not react.

Sn and Pb reacts dilute concentrated sulphuric acid liberating hydrogen.

 Pb (s) + H₂SO₄(aq) →  PbSO₄(s) + H₂(g)

With  hot Concentrated sulphuric acid

C, Si and Ge  react to form dioxides

C(s) + 2H₂SO₄(aq)   →   CO₂(g) + 2SO₂(g) +2H₂O(l)

Sn and Pb reacts hot concentrated sulphuric acid liberating sulphur dioxide .

Pb(s) + 2H₂SO₄(aq) →   PbSO₄(s) + SO₂(g) +2H₂O(l)

Sn(s) + 2H₂SO₄(aq) →   SnSO₄(s) + SO₂(g) +2H₂O(l)

With acetic or ethanoic acid

C, Si Ge and Sn do not react.

Pb reacts ethanoic acid in the presence of air giving lead II ethanoate and water

Pb (s) + 4CH₃COOH(aq) + O₂(g)   →       Pb(CH₃COO-)₂(aq) +2H₂O(l)

Reaction of silicon with hydrofluooric acid or aqueous hydrogen fluoride

Silicon is dissolved by hydrofluoric acid, HF, to form fluorosilicic acid, a reaction apparently driven by the stability of the Si(IV) fluoride complex [SiF₆]^2-.

Si(s) + 6HF(aq) →  2H₂(g) +H₂SiF₆ (aq) (fluorosilicic acid)

Reaction group (IV) elements  with sodium hydroxide

C does not react.

Si Ge and Sn react liberating hydrogen and silicate (IV), germates (IV) and stannates (IV) respectively

Si (s) + 2NaOH(aq) + H₂O(l)  →   Na₂SiO₃(aq) + 2H₂(l)

Lead reacts to form plumbates (II).

Pb(s) + 2NaOH(aq)  →   Na₂PbO₂(aq) + 2H₂(g)

Unique characteristics of carbon

1. Forms gaseous oxides, the rest form solid oxides

Note that

(a) Carbon dioxide is gas because its molecules are bonded by weak van der Waal forces.

(b) Silicon dioxide is a solid of high melting point because each silicon atom is bonded to four oxygen atom by strong covalent bonds.

2. Catenation: carbon forms chains, rings, and branches with single, double and triple bonds while others do not

3. Forms compounds mainly with oxidation state (IV); other elements form compounds with oxidation states +2 and +4.

Carbon tetrachloride and carbon tetrahydride do not hydrolyze in water. Chlorides of other elements hydrolyze in water.

5.  Carbon form CF₄ and not CF₆^-2

Reason for the uniqueness of carbon

•  Has a very small atomic radius
•  Has high electronegativity
• Lacks easily accessible d-orbital in its valence shell.

Allotropes of carbon

Allotropy is the existence of two or more different physical forms in the same state of a chemical element.

An allotrope is any of the different physical forms in the same state into which a chemical element can exist.

Carbon exists in two main allotropes, i.e. graphite and diamond.

Graphite

In graphite, each carbon atom is covalently bonded to 3 carbon atoms to form a layer of hexagons. Each layer is bonded to another by weak van der Waal forces.

Properties of graphite as a result of its structure

1. It has open structure and low density.
2. It is slippery and used as a lubricant.
3. Has unbonded π-electron that is free to move about making graphite a good conductor of electricity and heat

Diamond

Structure of diamond

Each carbon atom is bonded tetrahedrally to four carbon atoms to form a 3D compact structure by strong covalent bonds. As a result, the diamond has a high density, melting, and boiling point. It is the hardest substance known.

Diamond is used as an ornament, and to drill and cut other substances.

Differences between diamond and graphite

1. The density of graphite (2.3 gcm^-3) that of the diamond (3.5 g cm^-3)
2. Diamond is very hard while graphite is soft
3. Graphite is slippery while diamond is not
4. Graphite conducts electricity while diamond does not.

Experiment to show that graphite and diamond are allotropes of carbon

When equal masses of graphite and diamond are burned in air, they give an equal volume of carbon dioxide.

Compounds of carbon

1. Hydrides:

they include alkanes, alkenes, alkyne and aromatic compounds studied in organic chemistry

 2. Halides of carbon

Carbon forms all the four tetrahalides: CF₄, CCl₄, CBr₄ and CI₄

Comments

(a) The melting points of halides increase with molecular. All tetrahalides are volatile liquids

(b) The stability of the halides decreases from CF₄ > CCl₄ > CBr₄ > CI₄ due to a decrease in the bond strengths as the electronegativity of the halogen decrease.

(c) CCl₄ is a good solvent, being very stable and heavier than air; it is used in fire extinguisher to protect the burning substance from oxygen.

(d) CCl₄ is not hydrolyzed by water due to a lack of accessible d-orbital to accommodate the lone pair of an electron from water and weaken the C-X bond.

3. Oxides f carbon

Carbon monoxide (CO)

Preparation

By dehydration of methanoic acid with concentrated sulphuric acid.

HCO₂H(l)   +  Conc. H₂SO₄(l)   → CO(g) + H₂O(l)

Uses of carbon monoxide

It reduces most metal oxides to metals.

     Fe₂O₃ (s) + 3CO (g)  → 2Fe (s) + 3CO₂ (g)

       NiO (s) + CO (g) → Ni (s) + CO₂ (g)

 

This property is employed industrially in the extraction of iron and nickel.

 It is used to purify nickel: it forms a volatile carbonyl that decomposes to give a pure metal

It is used preparation of carbonyl chloride that is used in the production of polyurethane form of plastics and pesticides.

       CO (g) + Cl₂ (g)   → COCl₂ (g)

                                   [carbonyl chloride]

It reacts with fused NaOH under pressure to give sodium methanoate.  Methanoic acid is produced from this salt by the addition of dilute HCl.

NaOH (l) + CO (g)  → HCOONa (s)

then          HCOO- (aq) + H+ (aq)  →    HCOOH (l)

Methanoic acid is used to produce insecticides and for dyeing, tanning, and electroplating.

Carbon dioxide (CO₂)

Preparation

By decomposition of calcium carbonate

CaCO₃(s) (heat)  →  CaO(s) + CO₂(g)

Limestone              quick lime

Uses of carbon dioxide

1. Flavors fizzy drinks.
2. It is heavy and stable so it is used in fire extinguishers to displace oxygen from burning objects.

Carbonic acid

Preparation

By reacting carbon dioxide and water. There is, however, equilibrium in the solution of carbonic acid, hydrogen ions, hydrogen carbonate, and carbonate ions.

 CO₂ (g) + H₂O (l)   ↔     H₂CO₃ (aq)                   

H⁺(aq) + HCO₃^– (aq)  ↔  2H⁺ (aq) + CO₃^2- (aq)

Structure of carbonic acid

Has 3 resonance hybrid whose bond lengths are intermediate between a C=O and C-O bonds

Carbonates and hydrocarbonate

Both react with acids to liberate carbon dioxide

CO₃^2- (aq)+ 2H⁺ (aq)   →    H₂O (l) + CO₂ (g)

HCO₃^– (aq)+ H⁺ (aq)    →    H₂O (l) + CO₂ (g)

CO₃^2- forms white ppt with Mg²⁺ whereas HCO₃^– does not.

Silicon

Existence

Exists as silica, SiO_2, or sand used for building or making glass.

Chemical properties of SiO₂

Reaction with NaOH

   SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l)

   SiO₂(s) + 2^–OH (aq) → SiO₃^2- (aq) + H₂O (l).

Chemical properties of SiO₂

(a) Reaction with NaOH

SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l)

   SiO₂(s) + 2^–OH (aq) → SiO₃^2- (aq) + H₂O (l).

(b) Reaction with HF

SiO₂(s) + 4HF(g) → SiF₄(aq) + H₂O(l)

SiF₄(aq) + 2HF(aq) → H₂SiF₆(aq)

Reaction of silicon with hydrogen fluoride or hydrofluoric acid

Silicon does  dissolves in hydrofluoric acid, HF, to form fluorosilicic acid, a reaction apparently driven by the stability of the Si(IV) fluoride complex [SiF₆]^2-.

Si(s) + 6HF(aq) →  2H₂(g) +H₂SiF₆ (aq) (fluorosilicic acid)

Germanium, tin, and lead

Points to note

a. They form compounds with oxidation states +2 and + 4. The stability of compounds with oxidation states +2 increases down the group due to ‘inert pair effect’ while those of oxidation states +4 decreases down the group.

b. Compounds with oxidation state +2 are reducing. For instance, tin II ions reduce iron III ions to iron II ions.

Sn²⁺(aq) + Fe³⁺(aq) → Sn⁴⁺(aq) + Fe²⁺(aq)

c. Compounds with oxidation state +4 are oxidizing. For instance, lead IV oxide ions oxidize HCl ions to chlorine

PbO₂ (s)+ 4HCl (aq) →  PbCl₂(s)  +  2H₂O (l)  +  Cl₂ (g)

d. Compounds with oxidation state +4 are predominantly ionic whereas those in oxidation state +4 are predominantly covalent. This explains the fact that PbCl₂  is soluble in water but insoluble in ethanol whereas PbCl₄ is soluble in ethanol

Chemical properties

The principal reactions of germanium and tin are similar, both elements exhibiting oxidation state of 4

The principal reactions of lead are summarized below

Lead IV halide

Lead forms ionic Pb⁴⁺(F^–)_4, and covalent liquid PbCl_4.

The tetra-bromide and tetra-iodide do not exist, presumably because bromine and iodine are not sufficiently strong oxidizing agents to convert Pb²⁺ to Pb⁴⁺.

Lead IV chloride

Preparation

By reacting lead IV oxide with cold concentrated hydrochloric acid.

PbO₂ (s) + 4 HCl (aq) (conc. and cold)  → PbCl₄(aq) +2H₂O(l)

Effect of Heat on Lead (IV) chloride

Decomposes on heating liberating chlorine

PbCl₄ (g) (heat) →  PbCl₂ (s) + Cl₂ (g)

Hydrolysis of lead (IV) chloride

Lead tetrachloride is readily hydrolyzed by water (white fumes of HCl and brown precipitate of PbO₂ are observed)

PbCl₄ (s) + 2H₂O (l)    →    PbO₂ (s)  +  4HCl (aq)

Reaction of lead (IV) chloride with sodium hydroxide

Lead(IV) chloride react with sodium hydroxide to produce lead(IV) oxide, sodium chloride and water. Sodium hydroxide – diluted solution.

PbCl₄ + 4NaOH → PbO₂ + 4NaCl + 2H₂O

The compound PbO₂ reacts with the sodium hydroxide to produce the hexahydroxoplumbate (IV) ion [Pb(OH)₆]²⁻, soluble with the water.

PbO₂ +  2NaOH + 2 H₂O → Na₂[Pb(OH)₆]

Lead II halide

Preparation

By reacting soluble salt like lead nitrate with soluble halide

Pb²⁺(aq)  + 2X^–(aq)   →   PbX₂(s)  ( X = F, Cl, Br, I).

E.g.  Pb²⁺(aq) + 2Cl^–(aq) →  PbCl₂(s)

Properties

Except for the iodide, which is yellow, they are white solids. Lead (II) chloride forms chloro-complexes, e.g., [PbCl₄]^2-, with Cl^– ions and because of this, it is soluble in concentrated hydrochloric acid.

Reaction of Lead II chloride with sodium hydroxide

Lead II chloride reacts with sodium hydroxide solution to form lead (II) hydroxide which dissolves in excess sodium hydroxide to form sodium plumbate (II)

PbCl₂ (s) + 2OH^–(aq) → Pb(OH)₂(s) + 2Cl^–(aq)

Then

Pb(OH)₂(s) + 2OH^–(aq) → Pb(OH)₄^2-(aq)

Lead IV oxide

Preparation

Lead (IV) oxide, PbO₂, is a brown solid obtained by oxidation of a soluble lead (II) salt with hot sodium chlorate (I).

Pb²⁺(aq)  + H₂O(l) + ClO^–(aq)    → PbO₂(s) + Cl^–(aq) + 2H⁺(aq)    

(ii) By action of nitric acid on trilead tetra oxide, Pb₃O₄.

Pb₃O₄ (s)  + 4HNO₃ (aq)  →      2Pb(NO₃)₂(aq) +  PbO₂(s) + 2H₂O(l)

Chemical properties of lead (IV) oxide

Decomposes on heating

2PbO₂ (s) + heat →  2PbO (s) +  O₂ (g)

Reacts with cold concentrated hydrochloric acid to form lead (IV) chloride and water

PbO₂(s) + 4HCl (aq)     →  PbCl₄(s) + 2H₂O(l)

Oxidize hot concentrated hydrochloric acid to  chlorine

PbO₂(s) + 4HCl(aq) → PbCl₂ (s) + 2H₂O(l) + Cl₂(g)

Reaction of lead (IV) oxide with sodium hydroxide

 PbO₂ reacts with the sodium hydroxide to produce the hexahydroxoplumbate (IV) ion [Pb(OH)₆]²⁻, soluble with the water.

PbO₂ +  2NaOH + 2 H₂O → Na₂[Pb(OH)₆]

Qualitative analysis of Pb²⁺

1. Forms a white precipitate with sodium hydroxide soluble in excess

Pb²⁺(aq) + 2OH^–(aq)     →    Pb(OH)₂ (s) + 2OH_–(aq) →           Pb(OH)₄^2- (aq)

• Forms white precipitate with ammonia solution insoluble in excess 

Pb²⁺(aq) + 2OH^–(aq)     →    Pb(OH)₂ (s)

• Forms yellow precipitate with potassium iodide

  Pb²⁺(aq) + 2I^–(aq)       →    PbI₂ (s)

• Forms white precipitate hydrochloric acid-soluble on warming

Pb²⁺(aq) + 2CI^–(aq)       →    PbCI₂ (s)

• Forms white precipitate with sulphate ions

Pb²⁺(aq) + SO₄^2-(aq)     →    PbSO₄ (s)

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Group 4 element (carbon, silicon, germaniun, tin, and lead)

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