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In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridisation are respectively :

  • Option 1)

    33 and 25

  • Option 2)

    33 and 75

  • Option 3)

    50 and 75

  • Option 4)

    67 and 75

 

Answers (1)

best_answer

 As we learned

 

Graphite -

C atom is sp2 hybridised and thermodynamically more stable than diamond at ordinary temperature 

- wherein

High electrical conductivity is due to presence of delocalised \pi - electron cloud.

 

 

Diamond -

C atom is sp3 hybridised and has three dimensional network structure

- wherein

It is bad conductor due to absence of valence electron 

 

 Graphite
Sp^{2} hybridization 

%p = \frac{2}{3}\times 100

=  67%

Diamond 

Sp^{3} hybridisation 
% p = \frac{3}{4}\times 100

= 75 %


Option 1)

33 and 25

Option 2)

33 and 75

Option 3)

50 and 75

Option 4)

67 and 75

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Plabita

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