Question

The ratio of sp\(^3\), sp\(^2\) and sp hybridized carbons present in CH\(_3\)-CH = CH-CN is:

• A. 1 : 2 : 1
• B. 1 : 1 : 2
• C. 1 : 1 : 1
• D. 2 : 1 : 1

Hint:

Count sigma bonds to determine hybridization: sp\(^3\) – 4 sigma, sp\(^2\) – 3 sigma, sp – 2 sigma bonds.

Answer 1

The Correct Option is A

In CH\(_3\)-CH = CH-CN: - The CH\(_3\) carbon is sp\(^3\) hybridized. - The two CH=CH carbons are sp\(^2\) hybridized. - The CN carbon (triple bond) is sp hybridized. So, the ratio is 1 (sp\(^3\)) : 2 (sp\(^2\)) : 1 (sp).

Answer 2

The ratio of sp³, sp² and sp hybridized carbons present in CH₃–CH = CH–CN is:

Step 1: Analyze the structure of the molecule:
CH₃–CH = CH–CN is a four-carbon compound. Let us label each carbon for clarity:
- C₁: CH₃ group (methyl group)
- C₂: CH with double bond to C₃
- C₃: CH with double bond to C₂ and single bond to C₄
- C₄: CN group, where C is triple bonded to nitrogen

Structure:
CH₃–CH=CH–C≡N

Step 2: Determine the hybridization of each carbon:
- C₁ (CH₃): forms 4 sigma bonds → sp³ hybridized
- C₂ (CH): involved in one sigma bond with C₁, one sigma bond and one pi bond with C₃ → 3 sigma bonds → sp² hybridized
- C₃ (CH): involved in one sigma bond and one pi bond with C₂, and one sigma bond with C₄ → 3 sigma bonds → sp² hybridized
- C₄ (the carbon in CN): triple bonded to nitrogen → 2 regions of electron density → sp hybridized

Step 3: Count the hybridized carbons:
- sp³: 1 carbon (CH₃)
- sp²: 2 carbons (C₂ and C₃)
- sp: 1 carbon (C in CN group)

Final Answer:
\[ \boxed{1 : 2 : 1} \]