In \(\overset{1}{\mathrm{C}}\mathrm{H_2} = \overset{2}{\mathrm{C}} = \overset{3}{\mathrm{C}}\mathrm{H} - \overset{4}{\mathrm{C}}\mathrm{H_3}\) molecule, the hybridization of carbon 1, 2, 3 and 4 respectively, are :
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- \(sp^2,\ sp,\ sp^2,\ sp^3\)
- \(sp^2,\ sp^2,\ sp^2,\ sp^3\)
- \(sp^3,\ sp,\ sp^3,\ sp^3\)
- \(sp^2,\ sp^3,\ sp^2,\ sp^3\)
The Correct Option is A
Solution and Explanation
C-1 (\(\mathrm{CH_2}\)): 2 H plus 1 C gives 3 sigma bonds plus one pi bond, so \(sp^2\).
C-2: two double bonds, 2 sigma bonds, so \(sp\).
C-3 (\(\mathrm{CH}\)): 1 H, C-2 and C-4 give 3 sigma bonds with one pi bond, so \(sp^2\).
C-4 (\(\mathrm{CH_3}\)): 4 sigma bonds, so \(sp^3\).
The sequence is \(sp^2, sp, sp^2, sp^3\).
Answer: (1) \(sp^2,\ sp,\ sp^2,\ sp^3\)
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