NDA Chemistry
Carbon Compounds — Quiz
20 Questions · No Negative Marking
Question 1
The concept of sp³, sp², and sp hybridisation of carbon explains different bonding geometries. In ethylene (CH₂=CH₂), each carbon is:
Asp³ hybridised — four equivalent tetrahedral bonds (109.5°)
Bsp² hybridised — three planar σ bonds (120°) and one p orbital for the π bond
Csp hybridised — two linear σ bonds (180°) and two p orbitals for two π bonds
Dsp³d hybridised — five bonds in a trigonal bipyramidal arrangement
Question 2
The principle of aromaticity (Hückel's rule) states that a cyclic, planar, conjugated system is aromatic if it has:
A6 π electrons only (limited to benzene)
B4n + 2 π electrons (n = 0, 1, 2, ...)
C4n π electrons (antiaromatic systems)
DAny even number of π electrons
Question 3
In nucleophilic substitution reactions (SN1 vs SN2), which conditions favour SN2 mechanism?
ATertiary alkyl halide, polar protic solvent, weak nucleophile
BPrimary alkyl halide, polar aprotic solvent, strong nucleophile
CTertiary halide with strong nucleophile in polar aprotic solvent
DSecondary halide in polar protic solvent with weak nucleophile
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