The ratio of pure and hybrid orbitals of valence shell in $H_{2} C = C H - C H = C H_{2}$ is:

7 : 12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14 : 13

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 : 5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 : 6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $5 : 6$
Every double bonded carbon has $3$ $s p^{2}$ orbitals and one pure p orbital. And every hydrogen has one pure s orbital, so the number of hybrid orbitals are $4 \times 3 = 12$ , and pure orbitals are $1 \times 4 + 6 \times 1 = 10$ .
So, their ratio is $10 : 12$ or $5 : 6$ .

Suggest Corrections

Similar questions

C H_{3} 6 - C H 5 = C H 4 - C H_{2} 3 - C 2 \equiv C H 1

C H_{3} - C H = C H - C H_{2} - C \equiv C H

7 C H_{3} - 6 C H_{2} - 5 C | H C H_{3} - 4 C H = 3 C H - 2 C H_{2} - 1 C H O

C H_{2} = C H - C H_{2} - C H_{2} - C \equiv C H

C_{2} - C_{3}

6 C H_{2} = 5 C H - 4 C H_{2} - 3 C H_{2} - 2 C \equiv 1 C H

C_{2} - C_{3}

Join BYJU'S Learning Program