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Ratio of hybrid and unhybrid orbitals taking part in bond formation in ethylene molecule

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1:1

2:3

3:4

1:2

answer is A.

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Detailed Solution

No of sp² hybrid orbitals = 2 x Number of hybridized orbitals on each carbon

= 2 x 3

= 6

No of pure orbitals = number of hydrogen atoms + 2 [number of pi bonds]

= 4 + 2 x 1

= 6

Therefore ratio of H.O and P.O = 1: 1

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