Question

ratio of pure orbitals to hybridised orbitals in ethylene​

Answer 1

Answer:

In ethane, the carbon atoms use sp3 hybrid orbitals for the formation of sigma bonds. The four bonds around each C atom point toward the vertices of a regular tetrahedron, and the ideal bond angles are 109.5°. The simplest compound is methane, CH4, which is the first member of the alkane family. The next few members are ethane CH3CH3, propane, CH3CH2CH3, butane, CH3CH2CH2CH3, etc..

H H

\ /

H--C---C--H

/ \

H H

Hence pure orbitals only exist in the beginning. After bonding all become hybrid orbitals. All hybrid orbitals are the same. Hence for each carbon in ethane there 4 hybrid orbitals no pure orbital.

Explanation:

Answer 2

Answer:

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 from each carbon and 1 from each hydrogen) in a benzene ring.  

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

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