AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements – Group 14

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14 Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 11th Lesson The p-Block Elements – Group 14

Very Short Answer Questions

Question 1.
Discuss the variation of oxidation states in the group -14 elements.
Answer:

  • The common oxidation states exhibited by group – 14 elements are +4 and +2.
  • Carbon exhibits negative oxidation states.
  • Heavier elements exhibits +2 oxidation state.
  • The tendency to show +2 oxidation state increases in the order Ge < Sn < pb.
  • pb exhibits +2 oxidation state as stable state because of inert pair effect.

Question 2.
How the following compounds behave with water
a) BCl3
b) CCl4.
Answer:
a) BCl3 reacts with water (hydrolysis) to form boric acid.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 1
b) CCl4 does not undergo hydrolysis due to lack of d-orbitals in the central atom ‘C1 and due to its highly non polar nature, CCl4 does not acts as Lewis acid.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 3.
Are BCl3 and SiCl4 electron-deficient compounds ? Explain.
Answer:

  • BCl3 and SiCl4 are electron-deficient compounds.
  • These two compounds behave as Lewis acids.
  • These compounds are electron pair acceptors.
  • The following reacts support the electron deficiency of BCl3 and SiCl4.
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 2

Question 4.
Give the hybridization of carbon in
a) CO3-2
b) diamond
c) graphite
d) fullerene
Answer:
a) In C03-2 carbon atom undergoes sp2 hybridisation.
b) In Diamond carbon atom undergoes sp3 hybridisation.
c) In Graphite carbon atom undergoes sp2 hybridisation.
d) In Fullerenes carbon atom undergoes sp2 hybridisation.

Question 5.
Why is’CO’poisonous ? [T.S. Mar. 16]
Answer:
‘CO’ gas is highly poisonous because it has the ability to form a stable complex with haemoglobin.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 3
Carboxy haemoglobin is 300 times more stable than oxyhaemoglobin.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 6.
What is allotropy ? Give the crystalline allotropes of carbon. [Mar. 13]
Answer:

  • The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
  • Crystalline allotropes of carbon are
    a) Diamond
    b) Graphite
    c) Fullerenes.

Question 7.
Classify the following oxides as neutral, acidic, basic or amphoteric,
a) CO
b) B2O3
c) SiO2
d) CO2
e) Al2O3
f) PbO2
g) Tl2O3
Answer:
a) CO is neutral oxide.
b) B2O3 is acidic oxide.
c) SiO2 is acidic oxide.
d) CO2 is acidic oxide.
e) Al2O3 is amphoteric oxide.
f) PbO2 is amphoteric oxide.
g) Tl2O3 is basic oxide.

Question 8.
Name any two manmade silicates.
Answer:
Glass and cement are man made silicates.

Question 9.
Write the outer electron configuration of group -14 elements.
Answer:
The general outer most electronic configuration of group – 14 elements is ns2np2.
1) Carbon – [He] 2s22p2
2) Silicon – [Ne] 3s23p2
3) Germanium – [Ar] 3d104s24p2
4) Tin – [Kr] 4d10 5s2 5p2
5) Lead – [Xe] 4f4 5d10 6s2 6p2

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 10.
How does graphite function as a lubricant ?
Answer:
Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

Question 11.
Graphite is a good conductor – explain.
Answer:
In graphite carbon undergoes sp2 hybridisation. Each carbon forms three a – bonds with three neighbouring carbon atoms. Fourth electron forms TC – bond and it is delocalised. Due to the presence of these moving (or) free electrons graphite acts as good conductor.

Question 12.
Explain the structure of silica.
Answer:
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 4

  • Silica is a giant molecule with 3 – dimensional structure.
  • In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
  • Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
  • Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
  • In SiO2 Silicon atom undergoes sp3 hybridisation.

Question 13.
What is ‘Synthesis gas’?
Answer:

  • Water gas is also called as synthesis gas.
  • It is a mixture of CO and H2.
  • It s prepared by passing steamover hot coke.
  • It is used for the synthesis of methanol and a number of hydrocarbons. Hence it is called synthesis gas.

Question 14.
What is producer gas?
Answer:

  • Producer gas is mixture of CO and N2.
  • It is prepared by passing air over hot coke.

Question 15.
Diamond has high melting point – Explain.
Answer:

  • In Diamond each carbon undergoes sp3 hybridisation and it is surrounded by four other carbon atoms with strong a – bonds tetrahedrally.
  • The C – C bond energy in diamond is very high and it has 3 – dimensional structure.
  • Due to these reasons diamond has high melting point.
  • It has melting point 4200 K.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 16.
Give the use of CO2 in photosynthesis.
Answer:
The process of converting the atmospheric CO2 into Carbohydrates by green plants is known as ‘photosynthesis’.
In Photosynthesis CO2 changes to carbohydrates such as glucose.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 5

Question 17.
How does CO2 increase the green house effect ?
Answer:

  • Green plants absorbs CO2 gas for photosynthesis and releases O2 gas.
  • Due to deforestation, decomposition of lime stone and burning of fossil fuels CO2 concentration is increased in atmosphere.
  • The increase of CO2 level disturbs the O2 – CO2 balance in the atmosphere and it is responsible for green house effect (or) global warming.

Question 18.
What are silicones ?
Answer:

  • Silicones are the organo Silicon polymers containing R2 SiO – repeating unit.
  • These are synthetic compounds containing Si – O – Si.
  • Linkage preparation : These are formed by the hydrolysis of chlorosilanes.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 19.
Give the uses of silicones.
Answer:
Uses of silicones :

  • These are used in surgical and cosmetic plants.
  • These are used in preparation of silicone rubbers.
  • These are used as sealoant, greases etc.
  • These are used in preparing water proof clothes and papers.
  • These are used as insulators.
  • These are used in paints and enamels.

Question 20.
What is the effect of water on tin ?
Answer:

  • Tin metal reacts with steam to form tin dioxide and dihydrogen gas.
  • In this reaction steam is decomposed.
    Sn + 2H2O AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 6 SnO2 + 2H2

Question 21.
Write an account of SiCl4.
Answer:

  • Silicon tetrachloride (SiCl4) is also called as tetra chloro silico methane.
  • SiCl4 can acts as Lewis acid due to availability of 3d orbital in ‘Si’.
  • SiCl4 undergoes hydrolysis due to presence of vacant 3d-orbital. Here water molecules forms dative bonds with empty 3d-orbitals of Siratom.

Uses :

  • SiCl4 and NH3 mixture used to produce smoke screens.
  • Ultra pure Silicon is used to make transistors.
  • SiO2 prepared from SiCl4 used in epoxypaints, resis etc..

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 22.
SiO2 is a solid while CO2 is a gas – explain.
Answer:

  • Silica (SiO2) has giant molecular structure.
  • In SiO2 ‘Si’ undergoes sp3 hybridisation.
  •  It is a 3 – dimension structure in which each ‘Si’ atom is tetrahedrally surrounded by four oxygen atoms.
  • Hence it exists as solid compound.
  • CO2 has linear structure.
  • In CO2 ‘C’ undergoes sp hybridisation.
  • In between CO2 molecule weak Vander Waal’s forces are present.
  • In CO2 molecule two double bonds are present.
  •  Hence CO2 exists as a gas.

Question 23.
Write the use of ZSM – 5.
Answer:

  1. ZSM – 5 is a zeolite.
  2. It is used to convert alcohols directly into gasoline.

Question 24.
What is the use of dry ice ?
Answer:

  1. Solid CO2 is called as dry ice.
  2. It is used as refrigirent for frozen food and ice – creams.

Question 25.
How is water gas prepared ?
Answer:
Water gas is prepared by passing superheated steam over hot coke.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 7

Question 26.
How is producer gas prepared ?
Answer:
Producer gas is prepared by passing air over white not coke.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 8

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 27.
C-C bond length in graphite is shorter than C-C bond length in diamond – explain.
Answer:

  1. In graphite each carbon undergoes sp2 hybridisation and hence bond length is 1.42 A° (or) 141.5 pm.
  2. Graphite has hexagonal layer like lattice. It is a 2-dimensional structure.
  3. In diamond each carbon undergoes sp3 – hybridisation and hence bond length is 1.54 A° (or) 154 pm.
  4. Diamond has regular tetrahedral giant polymeric structure. It is a 3-dimensional structure.

Question 28.
Diamond is used as precious stone – explain.
Answer:

  • Diamonds are used as precious stones.
  • Diamonds are clear, colourless form of pure carbon.
  • These are hardest substances occurring naturally.
  • The weight of diamond expressed in carats.
    1 carat = 200 mg.

Question 29.
Carbon never shows co-ordination number greater than four while other members of carbon family show co-ordination number as high as six – explain.
Answer:
Carbon never shows co-ordination number greater than four because of absence of d-orbitals in carbon atom.

The other members of carbon family show co-ordination number as high as six because of availability of d – orbitals.

Question 30.
Producer gas is less efficient fuel than water gas – explain.
Answer:

  1. Producer gas has calorific value 5439.2 KJ/m3
  2. Water gas has calorific value 13000 KJ/m3.
  3. Due to high calorific value of watergas, it is more efficient fuel than producer gas (or) producer gas is less efficient than watergas.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 31.
SiF6-2 is known while SiCl6-2 is not. Explain. [A.P. Mar. 16]
Answer:
SiF6-2 is known while SiCl6-2 is not because

  • Si+4 has small size so it cannot be accomodate six large chloride ions.
  • The interaction between lone pairs of Cl ion and Si+4 is not very strong.

Short Answer Questions

Question 1.
Explain the difference in properties of diamond and graphite on the basis of their structure.
Answer:
Diamond
a) Each carbon is sp3 hybridised.
b) Each carbon is bonded to 4 other carbons tetrahedrally.
c) It has a 3 dimensional structure.
d) C – C bond length is 1.54 Å and bond angle is 109° 28′.
e) Carbon atoms are firmly held with strong covalent bonds.
f) Diamond is very hard.
g) Density = 3.5 g/cc.
h) Graphite is a conductor due to the presence of free electrons.
i) It is transparent to light and X-rays. It has high refractive index (2.45).

Graphite
a) Each carbon is sp2 hybridised.
b) Each carbon is bonded to 3 other carbon atoms to form hexagonal rings. It has sheet like structure.
c) It has a 2 dimensional structure.
d) C – C bond length in hexagonal rings is 1.42 A° and bond angle is 120°.
e) The distance between two adjacent layers is 3.35 A°. These layers are held by weak Vander Waal’s forces.
f) Graphite is soft.
g) Density – 2.2g/cc.
h) Diamond is an insulator due to the absence of free electrons.
i) It has layer, lattice. The layers are slippery. Hence it is greasy.

Question 2.
Explain the following.
a) PbCl2 reacts with Cl2 to give PbCl4
b) PbCl4 is unstable to heat,
c) Lead is not known to form PbI4.
Answer:
a) PbCl2 + Cl2 → PbCl2
But PbCl4 is unstable than PbCl2. Because compounds of lead in +2 oxidation state are stable than +4 oxidation state.

b) PbCl4 is unstable to heat:

  • In PbCl4 lead exhibits +4 oxidation state.
  • he compounds of lead in +2 oxidation state are stable than +4 oxidation state. Hence PbCl4 is unstable to heat.

c) Lead is not known to form PbI4:

  • Pb – I bond formed initially during the reaction does not release enough energy to unpair the 6s electrons.
  • Lead compounds in +2 state are stable than +4 state.
    Due to inert pair effect Pb exhibits stable +2 oxidation state.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 3.
Explain the following :
a) Silicon is heated with methyl chloride at high temperature in presence of copper.
b) SiO2 is treated with HF.
c) Graphite is a Lubricant
d) Diamond is an abrasive.
Answer:
a)

  • Methyl chloride reacts with silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl3, Me2SiCl2, Me3SiCl with small amount Me4Si.
  • Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (silicone) (Me = CH3 – group)
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 9

b) SiO2 is treated with HF to form SiF4. Which on hydrolysis to form H4SiO4 and H2SiF6.
SiO2 + 4HF → SiF4 + 2H2O
SiF4 + 4H2O → H4SiO4 + 2H2SiF6

c) Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant.

d) The covalent bonds in diamond are very strong and difficult to break, Hence diamond is used as an abrasive for sharpening hard tpols, in making dyes and in the manufacturing of tungsten filaments etc.

Question 4.
What do you understand by
a) Allotropy
b) Inert pair effect
c) Catenation.
Answer:
a) Allotropy :
The phenomenon of existence of an element in different physical forms having similar (or) same chemical properties is called allotropy.
Crystalline allotropes of carbon are
a) Diamond
b) Graphite.

b) Inert pair effect : The reluctance of ‘ns’ pair of electrons to take part in bond formation is known as inert pair effect.
(or)
The occurrence of oxidation states two units less than the group oxidation states is known as inert pair effect.
Eg : Lead exhibits +2 oxidation state as stable oxidation state due to inert pair effect. (Instead of +4 state).

c) Catenation : The phenomenon of self linkage of atoms among themselves to form long chains (or) rings is called as catenation.
Carbon has highest catenation tendency due to its high bond energy (348 KJ/mole).

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 5.
If the starting material for the manufacturing of Silicons is RSiCi3. Write the structure of the product formed.
Answer:
When RSiCl3 type of compound is used for the manufacturing of silicones a cross – linked silicon is formed.
Eg : When Methyl trichloro silane (CH3SiCl3) undergoes hydrolysis to give monomethyl silane triol. This undergoes polymerisation to form a very complex cross-linked polymer (Silicone).
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 10

Question 6.
Write a short note on Zeolites.
Answer:
Zeolites : The three dimensional structure contain no metal ions. If some of the Si+4 in them is replaced by Al+3 and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si2O8]-2n form zeolites. Zeolites act as ion exchanges and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH3, CO2 and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca+2 ions from hard water and replace them by Na+2 ions.
Uses of Zeolites :
Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 7.
Write a short note on Silicates.
Answer:
Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.
The silicates can be divided into six types. They are mentioned here.

  1. Orthosilicate or Nesosilicates : Their general formula may be M211 (SiO4).
    Ex : Willemite Zn2 (SiO4).
  2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO7-6 units. Pyrosilicates are rare.
    Ex: Thorveitite Ln2 (Si2O7).
  3. Chain silicates : They have the units (SiO3)n2n-
    Ex : Spodumene LiA/ (SiO3)2.
    Amphiboles are one type chain silicates. Generally double chains are formed in them.
  4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO3)n2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
    Ex : Beryl Be3Al2 (Si6O18)
  5. Sheet silicates : When SiO4 units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si2O5)n2n-. These compounds appear in layer struc-tures. They can be cleaved.
    Ex : Kaolin Al2 (OH)4 Si2O5.
  6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO4 tetrahedron results in three dimensional lattice of formula SiO2.
    Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na8 [Al6 Si6 O24] S2), zeolites.

Question 8.
What are Silicones ? How are they obtained ?
Answer:

  • Silicones are the organo silicon polymers containing R2 SiO – repeating unit.
  • These are synthetic compounds containing Si – 0 – Si.
    Preparation : These are formed by the hydrolysis of chlorosilanes.
  • Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl3, Me2SiCl2, Me3SiCl with small amount Me4Si
  • Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH3 – group).
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 11

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 9.
Write a short note on Fullerene.
Answer:
Fullerenes :

  • Fullerenes are one type of crystalline allotropes of carbon.
  • These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
  • These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
  • C60 molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
  • C60 contains twenty 6 – membered rings and twelve 5 – membered rings.
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 12
  •  In C60 6 – membered rings can combine with 5 (or) 6 – mem-bered rings while 5 – membered rings only combine with 6 – membered rings.
  • In Fullerens each carbon undergoes sp2 hybridisation.
  • Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
  • The C – C bond length in these compound lies between single and double bond lengths.
  • Spherical fullerene are also named as bucky balls.
  • The ball shaped molecule has 60 vertices.
  • The C – C bond distances are 1,43A° and 1.38A° respectively.

Question 10.
Why SiO2 does not dissolve in water ?
Answer:
Silica (SiO2) is a non reactive compound in it’s normal state.

  1. This non reactivity is due to very high Si – O bond enthalpy.
  2. Silica is a giant molecule with 3-dimensional structure.
  3. In Silica each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms
  4. Hence SiO2 is insoluble in water.
  5. But slightly dissolves at high pressures when heated.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 11.
Why is diamond hard ?
Answer:
In diamond, each carbon undergoes sp3 hybridization. A carbon atom is bound to four carbon atoms, arranged in a tetrahedral symmetry, with single bonds. A three dimensional arrangement of the tetrahedral structures result in giant molecule. The bond energy is very high (348 kJ mol-1). It is very difficult to break the bonds. So, diamond is hard.

Question 12.
What happens when the following are heated
a) CaCO3
b) CaCO3 and SiO2
c) CaCO3 and excess of coke.
Answer:
CaCO3 up on heating gives Quick lime.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 13
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 14
Quick lime (CO) with silica gives Calcium silicate.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 15
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 16
Coke reacts with Quick lime and form Carbides.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 17

Question 13.
Why does Na2CO3 solution turn into a suspension, when saturated with CO2 gas.
Answer:
An aq. solution of Na2CO3 when saturated with CO2, gives Sodium bicarbonate (NaHCO3).
Na2CO3 + H2O + CO2 → 2NaHCO3
NaHCO3 is less soluble compared to Sodium carbonate, hence suspension is formed.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 14.
What happens when
a) CO2 is passed through slaked lime
b) CaC2 is heated with N2.
Answer:
a) Slaked lime, Ca(OH)2 is turned milky on passing CO2 with the formation of insoluble calcium
carbonate Ca(OH)2 + CO2 → CaCO3 + H2O on passing more ‘CO2‘, CaCO3 is converted into Calcium bicarbonate.
CaCO3 + H2O + CO2 → Ca(HCO3)2
b) CaC2 on heating with N2 gives calcium cyanamide.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 18

Question 15.
Write a note on the anomalous behaviour of carbon in the group -14.
Answer:
Carbon shows ariamalous behaviour in group – 14 elements. The following facts support that anamalous behaviour.
Except carbon all other elements of group -14 has available d-orbitals and can expand octet in valency shell.
Carbon occurs in free state but not the other elements of this group.

  • Maximum covalency of carbon is four but for silicon is six.
  • C – C bond energy is very high (348 kJ/Mde).
  • Carbon can form multiple bonds with C, O, S, etc.
  • Hydrocarbons are more stable thermally than silanes.

Long Answer Questions

Question 1.
What are Silicones ? How are they prepared ? Give one example. What are their uses?
Answer:

  • Silicones are the organo silicon polymers containing R2 SiO – repeating unit.
  • These are synthetic compounds containing Si – 0 – Si.
    Preparation : These are formed by the hydrolysis of chlorosilanes.
  • Methyl chloride reacts with Silicon at high temperature in presence of copper catalyst to form various types of methyl substituted chlorosilane of formula MeSiCl3, Me2SiCl2, Me3SiCl with small amount Me4Si
  • Hydrolysis of dimethyl dichloro silane followed by condensation polymerisation forms a straight chain polymer (Silicone) (Me = CH3 – group).
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 11

Uses of Silicones:

  • These are used in surgical and cosmetic plants.
  • These are used in preparation of Silicons rubbers.
  • These are used as sealoant, greases etc.
  • These are used in preparing water proof clothes and papers.
  • These are used as insulators.
  • These are used in paints and enamels.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 2.
Explain the structure of Silica. How does it react with
a) NaOH and
b) HF.
Answer:

  • Silica is a giant molecule with 3 – dimensional structure.
  • In Silica eight membered rings are formed with alternate Silicon and Oxygen atoms.
  • Each ‘Si’ is tetrahedrally surrounded by four Oxygen atoms.
  • Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
  • In SiO2 Silicon atom under goes sp3 hybridisation.
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 19

a) Silica reacts with NaOH and forms Sodium Silicate (Na2SiO3)
SiO2 + 2 NaOH → Na2SiO3 + H2O

b) SiO2 is treated with HF to form SiF4. Which on hydrolysis to form H4SiO4 and H2SiF6.
SiO2 + 4HF → SiF4 + 2H2O
SiF4 + 4H2O → H4SiO4 + 2H2SiF6

Question 3.
Write a note on the allotropy of carbon.
Answer:
The property of an element to exist in two or more physical forms due to difference in the arrangement of atoms is called Allotropy. Allotropes have more or less similar chemical properties but different physical properties.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 20
Diamond, graphite and fullerenes are the crystalline allotropes of carbon.
Structure of Diamond : In diamond, each carbon atom bonded to four carbon atoms situated tetrahydrally around it.
In diamond, each carbon atom is in sp3 hybridisation and is linked to four carbon atoms by single covalent bonds.
C – C bond distance in diamond is 1.54 A°.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 21 bond angle in diamond is 109° 28″.
Uses of Diamond :

  1. Diamonds are used as precious stones for jewellery because of their ability to reflect light.
  2. Diamonds are used for cutting glass and drilling rocks due to their remarkable hardness.

Structure of Graphite : Graphite consists of a series of layers in which hexagonal rings made up of carbon atoms.
In Graphite, each carbon atom undergo sp2 hybridisation and forms three covalent bonds with three other carbon atoms.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 22
The fourth electron present in the pure ‘p orbitaI which is unhybridised. The electron become Free Electron.
The C — C bond length in graphite is 1.42 A°.
AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 23
The distance between the two layers in graphite is 3,4 A°.
These layers are held together by Vander Waal’s forces which are weak.
Graphite is a layer lattice structure.
Uses of Graphite:

  1. Graphite is used as a lubricant.
  2. It is used in the manufacturing of lead pencils.
  3. It is used in the manufacturing of Electrodes and Refractory crucibles.

Fullerenes :

  • Fullerenes are one type of crystalline allotropes of carbon.
  • These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
  • These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon.
  • C60 molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
  • C60 contains twenty 6 – membered rings and twelve 5 – membered rings.
  • In C60 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
  •  In Fullerens each carbon undergoes sp2 hybridisation.
  • Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
  • The C – C bond length in these compound lies between single and double bond lengths.
  • Spherical fullerene are also named as bucky balls.
  • The ball shaped molecule has 60 vertices.
  • The C – C bond distances are 1.43A° and 1.38A° respectively.
  • Amorphous allotropes of carbon are coal, coke, animal charcoal wood charcoal, lamp black, carbon black, gas carbon, petroleum coke and sugar charcoal.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 4.
Write a note on
a) Silicates
b) Zeolites
c) Fullerenes.
Answer:
a) Silicates : Many building materials are silicates.
Ex : Granites, slates, bricks and cement. Ceramics and glass are also silicates. The Si – O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.

The silicates can be divided into six types. They are mentioned here.

  1. Orthosilicate or Nesosilicates : Their general formula may be M211 (SiO4).
    Ex : Willemite Zn2 (SiO4).
  2. Pyrosilicates or sorosilicates or Disilicates : These contain SiO7-6 units. Pyrosilicates are rare.
    Ex : Thorveitite Ln2 (Si2O7).
  3. Chain silicates : They have the units (SiO3)n2n-
    Ex : Spodumene LiAl (SiO3)2.
    Amphiboles are one type chain silicates. Generally double chains are formed in them.
  4. Cyclic silicates : They are silicates having ring structures. They may be formed of general formula (SiO3)n2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
    Ex : Beryl Be3Al2 (Si6O18)
  5. Sheet silicates : When SiO4 units share three corners the structure formed is an inifinite two dimensional sheet. The empirical formula (Si2O5)n2n-. These compounds appear in layer struc-tures. They can be cleaved.
    Ex : Kaolin Al2 (OH)4 Si2O5.
  6. Frame work silicates or three dimensional silicates : Sharing all the four corners of a SiO4 tetrahedron results in three dimensional lattice of formula SiO2.
    Ex : Quartz, Tridymite; Cristobalite; Feldspar and ultramarine (Na8 [Al6 Si6 O24] S2), zeolites.

b) Zeolites : The three dimensional structure contain no metal ions. If some of the Si+4 in them is replaced by Al+3 and an additional metal ion an infinite three dimensional lattice is formed. Replacements of one or two silicon atoms in [Si2O8]-2n form zeolites. Zeolites act as ion ex-changes and as molecular sieves. The structures of zeolites permit the formation of cavities of different sizes. Water molecules and a variety of other molecules like NH3, CO2 and ethanol can be trapped in these cavities. Then zeolites serve as molecular sieves. They trap Ca+2 ions from hard water and replace them by Na+2 ions.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Uses of Zeolites :

  • Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and in Isomerisation.
  • Zeolite ZSM – 5 is used to convert alcohols directly into gasoline.

c) Fullerenes :

  • Fullerenes are one type of crystalline allotropes of carbon.
  • These are formed by heating graphite in an electric arc in presence of inertgases such as Helium (or) Argon.
  • These have smooth structure without dangling bonds. Hence Fullerenes are the only pure forms of carbon. .
  • C60 molecule is called Buck minster fullerene and it’s shave was like a soccer ball.
  • C60 contains twenty 6 – membered rings and twelve 5 – mem- bered rings.
    AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 12
  • In C60. 6 – membered rings can combine with 5 (or) 6 – membered rings while 5 – membered rings only combine with 6 – membered rings.
  • In Fullerens each carbon undergoes sp2 hybridisation.
  • Fullere has aromatic nature due to delocalisation of electrons in unhybrid p-orbitals.
  • The C – C bond length in these compound lies between single and double bond lengths.
  • Spherical fullerene are also named as bucky balls.
  • The ball shaped molecule has 60 vertices.
  • The C – C bond distances are 1.43A° and 1.38A° respectively.

Solved Problems

Question 1.
Select the member(s) of group 14 that

  1. forms the most acidic dioxide
  2. is commonly found in +2 oxidation state
  3. used as semiconductor.

Solution:

  1. Carbon
  2. lead
  3. Silicon and germanium

Question 2.
[SiF6]2- is known whereas [SiCl6]2- not. Give possible reasons.
Solution:
The main reasons are :

  1. Six large chloride ions cannot be accommodated around Si4+ due to limitation of its size.
  2. Interaction between lone pair of chloride ion and Si4+ is not very strong.

AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14

Question 3.
Diamond is covalent, yet it has high melting point. Why ?
Solution:
Diamond has a three – dimensional network involving strong C – C bonds, which are very difficult to break and in turn has high melting point.

Question 4.
What are Silicones ?
Solution:
Simple Silicones consist of AP Inter 1st Year Chemistry Study Material Chapter 11 The p-Block Elements - Group 14 24 chains in which alkyl or phenyl groups occupy the remaining bonding positions on each silicon. They are hydrophobic (water repellant) in nature.