Question

In the organic compound $C{H_2} = CH - C{H_2} - C{H_2} - C \equiv CH$, the pair of hybridized orbitals involved in the formation of ${C_2} - {C_3}$ is:

Answer 1

Hint: We have to know that the formation of a similar number of orbitals containing the same properties of different kinds of orbitals (s and p) of carbon atom is known as hybridization and orbitals produced are called hybridized orbitals.

Complete step by step answer:
We have to know that sigma bonds are formed by hybrid orbitals of carbon form and those orbitals which do not participate in hybridization to form pi (p) bonds.
A sigma bond is always a single bond. One sigma and one pi bond are found in double bonds. One sigma and two pi bonds are found in triple bonds. Based on the number of sigma bonds in an atom of carbon, we can determine its hybridization.
In $ - C - C$ bond, the type of hybridization on the carbon atom is $s{p^3}$.
In $ > C = C < $ bond, the type of hybridization on the carbon atom is $s{p^2}$.
In $ - C \equiv C - $ bond, the type of hybridization on the carbon atom is $sp$.
Let us now number the carbon atoms in the given structure.
The given structure is,
$C{H_2} = CH - C{H_2} - C{H_2} - C \equiv CH$
We can give the numbered structures as,
$\mathop C\limits^1 {H_2} = \mathop C\limits^2 H - \mathop C\limits^3 {H_2} - \mathop C\limits^4 {H_2} - \mathop C\limits^5 \equiv \mathop C\limits^6 H$
We know that, when double and triples are found at equivalent positions, we have to give preference to the double bond when we numbered the carbon chain.
We can give the hybridization on each carbon as,
The hybridization on carbon one is $s{p^2}$.
The hybridization on carbon two is $s{p^2}$.
The hybridization on carbon three is $s{p^3}$.
The hybridization on carbon four is $s{p^3}$.
The hybridization on carbon five is $sp$.
The hybridization on carbon six is $sp$.
$
  \mathop C\limits^1 {H_2} = \mathop C\limits^2 H - \mathop C\limits^3 {H_2} - \mathop C\limits^4 {H_2} - \mathop C\limits^5 \equiv \mathop C\limits^6 H \\
  {\text{s}}{{\text{p}}^2}{\text{ s}}{{\text{p}}^2}{\text{ s}}{{\text{p}}^3}{\text{ s}}{{\text{p}}^3}{\text{ sp sp}} \\
 $
We can see that the pair of hybridized orbitals involved in the formation of ${C_2} - {C_3}$ bond is $s{p^2} - s{p^3}$.
In the compound $C{H_2} = CH - C{H_2} - C{H_2} - C \equiv CH$, $s{p^2} - s{p^3}$ is the pair of hybridized orbitals that is involved in the formation of ${C_2} - {C_3}$ bond.

Note:
We know that the geometry of the molecule depends on the hybridization.
If a compound contains single bonds, then it is $s{p^3}$ hybridized. Example for $s{p^3}$ hybridized molecule is methane. Generally, alkanes come under $s{p^3}$ hybridization. Molecules that have $s{p^3}$ hybridization will have tetrahedral geometry.
If a compound contains double bonds, then it is $s{p^2}$ hybridized. Example for $s{p^2}$ hybridized molecule is ethene. Generally, alkenes come under $s{p^2}$ hybridization. Molecules that have $s{p^2}$ hybridization will have trigonal planar geometry.
If a compound contains triple bonds, then it is $sp$ hybridized. Example for $sp$ hybridized molecule is ethyne. Generally, alkynes come under $sp$ hybridization.