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The ratio of the number of sp, sp^(2), ...

The ratio of the number of sp, `sp^(2)`, and `sp^(3)` orbitals in the compound is
`CH_(3)-CH=C=CH-C-=C-CH_(3)`

A

`1:1:1`

B

`2:2:1`

C

`3:2:1`

D

`3:3:4`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the ratio of the number of sp, sp², and sp³ orbitals in the compound CH₃-CH=CH-CH≡C-CH₃, we will analyze the hybridization of each carbon atom in the compound step by step. ### Step 1: Identify the Structure The compound can be represented as follows: - CH₃ (C1) - CH (C2) = CH (C3) - C≡C (C4 and C5) - CH (C6) - CH₃ (C7) ### Step 2: Analyze Each Carbon Atom 1. **C1 (CH₃)**: - Forms 4 sigma bonds (3 with H and 1 with C2). - **Hybridization**: sp³ (steric number = 4). 2. **C2 (CH)**: - Forms 3 sigma bonds (1 with C1, 1 with C3, and 1 with H). - **Hybridization**: sp² (steric number = 3). 3. **C3 (CH)**: - Forms 3 sigma bonds (1 with C2, 1 with C4, and 1 with H). - **Hybridization**: sp² (steric number = 3). 4. **C4 (C≡C)**: - Forms 2 sigma bonds (1 with C3 and 1 with C5) and has 2 pi bonds (due to the triple bond). - **Hybridization**: sp (steric number = 2). 5. **C5 (C≡C)**: - Similar to C4, it forms 2 sigma bonds (1 with C4 and 1 with C6) and has 2 pi bonds. - **Hybridization**: sp (steric number = 2). 6. **C6 (CH)**: - Forms 3 sigma bonds (1 with C5, 1 with C7, and 1 with H). - **Hybridization**: sp² (steric number = 3). 7. **C7 (CH₃)**: - Forms 4 sigma bonds (3 with H and 1 with C6). - **Hybridization**: sp³ (steric number = 4). ### Step 3: Count the Number of Each Hybridization Type - **sp³**: C1 (1) + C7 (1) = 2 - **sp²**: C2 (1) + C3 (1) + C6 (1) = 3 - **sp**: C4 (1) + C5 (1) = 2 ### Step 4: Calculate the Total Orbitals - **Total sp³ orbitals**: 2 carbon atoms × 4 orbitals = 8 orbitals - **Total sp² orbitals**: 3 carbon atoms × 3 orbitals = 9 orbitals - **Total sp orbitals**: 2 carbon atoms × 2 orbitals = 4 orbitals ### Step 5: Determine the Ratio The ratio of the number of sp, sp², and sp³ orbitals is: - sp : sp² : sp³ = 4 : 9 : 8 ### Final Answer The ratio of the number of sp, sp², and sp³ orbitals in the compound CH₃-CH=CH-CH≡C-CH₃ is **4 : 9 : 8**. ---
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