In the organic compound $C H_{2} = C H - C H_{2} - C H_{2} - C \equiv C H$ , the pair of hydridised orbitals involved in the formation of: $C_{2} - C_{3}$ bond is:
(a) $s p - s p^{2}$ (b) $s p - s p^{3}$ (c) $s p^{2} - s p^{3}$ (d) $s p^{3} - s p^{3}$

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Solution

$C {6 H}_{2} = 5 C H - C {4 H}_{2} - C {3 H}_{2} - 2 C \equiv 1 C H$
In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are $s p, s p, s p^{3}, s p^{3}, s p^{2},$ and $s p^{2}$ hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of $C_{2} - C_{3}$ bond is $s p - s p^{3}$

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Similar questions

6 C H_{2} = 5 C H - 4 C H_{2} - 3 C H_{2} - 2 C H \equiv 1 C H_{2}

C_{2} - C_{3}

In the molecule CH ≡ C -CH = $C H_{2}$ , the hybridisation of C -Cbond is

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

C_{2} - C_{3}

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

C_{2} - C_{3}

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

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