Which of the following has a bond formed by overlap of $s p - {s p}^{3}$ hybrid orbitals?

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

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C H_{3} - C H = C H - C H_{3}

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

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H C \equiv C H

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Solution

The correct option is A $C H_{3} - C \equiv C - H$
Carbon forming single bond forms $s p^{3}$ hybridization due to 4 sigma bonds.
Carbon forming double bond forms $s p^{2}$ hybridisation due to 3 sigma bonds and 1 pi bond.
Carbon forming triple bond forms $s p$ hybridisation due to 2 sigma bonds and 2 pi bonds.
$C H_{3} - C \equiv C - C - H$
Only in compound 1, C1 and C2 has $s p^{3} - s p$ hybridisation while in other compounds there are no $s p^{3} - s p$ overlap.

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Similar questions

Q. Which of the following has a bond formed by overlap of

s p - s p^{3}

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A straight chain hydrocarbon has the molecular formula C8H10. The hybridisation for the carbon atoms from one end of the chain to the other are respectively $s p^{3}, s p^{2}, s p^{2}, s p^{3}, s p^{2}, s p^{2}, s p$ and $s p$ . The structural formula of the hydrocarbon would be

Q. The correct order of C - H bond length in

C H_{3} - C H_{3}, C H_{2} = C H_{2}

and

C H \equiv C H

is:

How many $σ$ and $π$ bonds are presents in each of the following molecule?

(a) $H C \equiv C - C H = C H - C H_{3}$ (b) $C H_{2} = C = C H - C H_{3}$

Q. Which amongst the following is a positional isomer of

{CH}_{3} - {CH}_{2} - CH = {CH}_{3}

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Which of the following has a bond formed by overlap of sp−sp3 hybrid orbitals?

Which of the following has a bond formed by overlap of $s p - {s p}^{3}$ hybrid orbitals?