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प्रश्न
Explain the difference in properties of diamond and graphite on the basis of their structures.
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उत्तर १
| Diamond | Graphite |
| It has a crystalline lattice. | It has a layered structure |
| In diamond, each carbon atom is sp3 hybridised and is bonded to four other carbon atoms through a σ bond. | In graphite, each carbon atom is sp2hybridised and is bonded to three other carbon atoms through a σ bond. The fourth electron forms a π bond. |
| It is made up of tetrahedral units. | It has a planar geometry. |
| The C–C bond length in diamond is 154 pm. | The C–C bond length in graphite is 141.5 pm. |
| It has a rigid covalent bond network which is difficult to break. | It is quite soft and its layers can be separated easily. |
| It acts as an electrical insulator. | It is a good conductor of electricity. |
उत्तर २
Since diamond exists as a three-dimensional network solid, it is the hardest substance known with high density and high melting point.
Whereas in graphite, any two successive layers are held together by weak forces of attraction. This makes graphite soft.
In graphite, carbon atom is sp2 hybridized whereas in diamond, carbon atom is sp3 hybridized.
Unlike diamond, graphite is good conductor of heat and electricity.
संबंधित प्रश्न
Discuss the pattern of variation in the oxidation states of C to Pb.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon?
Rationalise the given statement and give a chemical reaction:
Lead (IV) chloride is highly unstable towards heat.
Classify the following oxide as neutral, acidic, basic or amphoteric:
SiO2
Classify the following oxide as neutral, acidic, basic or amphoteric:
CO2
Classify the following oxide as neutral, acidic, basic or amphoteric:
Al2O3
Write suitable chemical equations to show the nature of the following oxide.
B2O3
Write suitable chemical equations to show the nature of the following oxide.
SiO2
Write suitable chemical equations to show the nature of the following oxide.
PbO2
The reason for small radius of Ga compared to Al is:
(i) poor screening effect of d and f orbitals.
(ii) increase in nuclear charge.
(iii) presence of higher orbitals.
(iv) higher atomic number.
The linear shape of CO2 is due to:
(i) sp3 hybridisation of carbon.
(ii) sp hybridisation of carbon.
(iii) pπ – pπ bonding between carbon and oxygen.
(iv) sp2 hybridisation of carbon.
Explain the following:
Silicon forms \[\ce{SiF^{2-}6}\] ion whereas corresponding fluoro compound of carbon is not known.
The +1 oxidation state in group 13 and +2 oxidation state in group 14 becomes more and more stable with increasing atomic number. Explain.
Explain the following:
Carbon shows catenation property but lead does not.
The principal ore of lead is ______.
\[\ce{SiCl4 ->[H2O] (A) ->[\Delta] (B) ->[Na2CO3][heat] (C)}\]. The Compound C is ______.
Which one of the following compounds of Group–14 elements is not known?
