CBSE

Class 12

Chemistry

General Organic Chemistry

The number of sigma () and pi () bonds in pent-2-en-4-yne is [

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The number of sigma $(σ)$ and pi $(π)$ bonds in pent-2-en-4-yne is \[ [(2019)] \] $13 σ$ bonds and no $π$ bond $10 σ$ bonds and $3 π$ bonds 196 $61.73%$ attempts got it right Explanation: $H 5 \equiv C 4 - C 3 H = C 2 H - C 1 H_{3}$ Pent-2-en-4-yne (10 $σ$ -bonds and $3 π$ -bonds)

Name: Video answer for The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is \[ [(2019)] \] 13σ bonds and no π bond 10σ bonds and 3π bonds 196 61.73% attempts got it right Explanation: H5≡C4−C3H=C2H−C1H3 Pent-2-en-4-yne (10 σ-bonds and 3π-bonds)
Uploaded: 2024-01-17T02:17:22.000Z
Duration: 1 min
Description: The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is \[ [(2019)] \] 13σ bonds and no π bond 10σ bonds and 3π bonds 196 61.73% attempts got it right Explanation: H5≡C4−C3H=C2H−C1H3 Pent-2-en-4-yne (10 σ-bonds and 3π-bonds)

$The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is \[ [(20$

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Updated on: Jan 17, 2024

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Camila Discussed

The number of sigma

(σ)

and pi

(π)

bonds in pent-2-en-4-yne is \[ [(2019)] \]

13 σ

bonds and no

π

bond

10 σ

bonds and

3 π

bonds 196

61.73%

attempts got it right Explanation:

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

Pent-2-en-4-yne (10

σ

-bonds and

3 π

-bonds)

10 mins ago

Discuss this question LIVE

10 mins ago

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Question Text	The number of sigma $(σ)$ and pi $(π)$ bonds in pent-2-en-4-yne is \[ [(2019)] \] $13 σ$ bonds and no $π$ bond $10 σ$ bonds and $3 π$ bonds 196 $61.73%$ attempts got it right Explanation: $H 5 \equiv C 4 - C 3 H = C 2 H - C 1 H_{3}$ Pent-2-en-4-yne (10 $σ$ -bonds and $3 π$ -bonds)
Updated On	Jan 17, 2024
Topic	General Organic Chemistry
Subject	Chemistry
Class	Class 12
Answer Type	Video solution: 1
Upvotes	133
Avg. Video Duration	1 min

The number of sigma $(σ)$ and pi $(π)$ bonds in pent-2-en-4-yne is \[ [(2019)] \] $13 σ$ bonds and no $π$ bond $10 σ$ bonds and $3 π$ bonds 196 $61.73%$ attempts got it right Explanation: $H 5 \equiv C 4 - C 3 H = C 2 H - C 1 H_{3}$ Pent-2-en-4-yne (10 $σ$ -bonds and $3 π$ -bonds)

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The number of sigma (σ) and pi (π) bonds in pent-2-en-4-yne is \[ [(2019)] \] 13σ bonds and no π bond 10σ bonds and 3π bonds 196 61.73% attempts got it right Explanation: H5​≡C4​−C3​H=C2​H−C1​H3​ Pent-2-en-4-yne (10 σ-bonds and 3π-bonds)

Video solutions (1)

Practice more questions on General Organic Chemistry

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The number of sigma $(σ)$ and pi $(π)$ bonds in pent-2-en-4-yne is \[ [(2019)] \] $13 σ$ bonds and no $π$ bond $10 σ$ bonds and $3 π$ bonds 196 $61.73%$ attempts got it right Explanation: $H 5 \equiv C 4 - C 3 H = C 2 H - C 1 H_{3}$ Pent-2-en-4-yne (10 $σ$ -bonds and $3 π$ -bonds)