Question

The ratio of hybrid and pure orbitals involved in the bonding of ethylene (C2H4) molecule respectively is (A) 3.2 (B) 1:3 (C) 1:1 (D) 2:3​

Answer 1

Answer:

Correct option is

C

1 : 1

In Benzene each carbon is sp

2

hybridizes. 2s and 2p orbitals of carbon participate in hybridization to form 3 sp

2

hybrid orbitals but its 1s and one 2p orbital remain unhybridized.

H has only one 1s orbital and the single orbital can not undergo hybridization, so it remains in pure form.

Thus there are 18 hybridized orbitals (3 from each carbon) and 18 pure orbitals ( 2 from each carbon and 1 from each hydrogen) in a benzene ring.

So ratio of hybridize orbitals to pure orbitals is 1:1.

Option C is correct.

hope it helps u....(✿^‿^)

Answer 2

Answer:

Correct option is

C

1 : 1

Explanation:

In Benzene each carbon is sp

2

hybridizes. 2s and 2p orbitals of carbon participate in hybridization to form 3 sp

2

hybrid orbitals but its 1s and one 2p orbital remain unhybridized.

H has only one 1s orbital and the single orbital can not undergo hybridization, so it remains in pure form.

Thus there are 18 hybridized orbitals (3 from each carbon) and 18 pure orbitals ( 2