Question

In the organic compound $C{H}_2=CH-C{H}_2-C{H}_2–C\equiv CH$, the pair of hydridised orbitals involved in the formation of: $C_2-C_3$ bond is:
(a) $sp-s{p}^2$ (b) $sp-s{p}^3$ (c) $s{p}^2-s{p}^3$ (d) $s{p}^3-s{p}^3$

Answer 1

$C{\stackrel{6}{H}}_2=\stackrel{5}{C}H-C{\stackrel{4}{H}}_2-C{\stackrel{3}{H}}_2-\stackrel{2}{C}\equiv \stackrel{1}{C}H$
In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are $sp,sp,s{p}^3,s{p}^3,s{p}^2,$ and $s{p}^2$ hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of $C_2-C_3$ bond is $sp-s{p}^3$