Which one of the following has $s p^{3}, s p^{2}, s p$ hybrid orbitals in the ratio 6 : 3 : 2?

H_{3} C - C H = C H - C H_{2} - C \equiv C - C H_{3}

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H_{3} C - C H = C H - C H_{2} - C \equiv C H

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H_{3} C - C H_{2} - C \equiv C - C H = C H_{2}

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

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Solution

The correct option is A $H_{3} C - C H = C H - C H_{2} - C \equiv C - C H_{3}$
One $s p^{3}$ hybridisation has 4 $s p^{3}$ hybrid orbitals.one $s p^{2}$ hybridisation has 3 $s p^{2}$ hybrid orbitals.One $s p$ orbital has 2 $s p$ hybrid orbitals. For ratio $6 : 3 : 2$ it should be 12 $s p^{3}$ orbitals 6 $s p^{2}$ orbitals and 4 $s p$ orbitals. So, minimum number of carbon should be seven with one double bond and one triple bond

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Similar questions

H_{3} C - C H_{2} - C H_{3} | C = C H - C | H C_{2} H_{5} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{3}

The correct IUPAC name of the compound :

H_{3} C - C H = C H_{2} + B r_{2} (1 equiv.) C C l_{4} - -- \to X a l c . K O H - ----- \to Δ Y N a N H_{2} - ---- \to Δ Z

Z is:

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

H_{2} C = C H - C H_{2} - + C H - C H_{3}

H_{3} C - C H_{3} | C | C H_{3} - + C H_{2}

Write these carbocations in decreasing order of their stabilities.

Q. Consider the following transformations.

H - C \equiv C - C H_{2} - C H_{2} - C H_{3} \to H_{3} C - C H = C H - C H_{2} - C H_{3}

Which reaction sequence will bring about the above transformation?

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Which one of the following has sp3,sp2,sp hybrid orbitals in the ratio 6 : 3 : 2?

Which one of the following has $s p^{3}, s p^{2}, s p$ hybrid orbitals in the ratio 6 : 3 : 2?