Question

The hybridization of carbon atoms in $C - C$ the single bond $HC \equiv C - CH = C{H_2}$ is:
A.$s{p^3}$, $s{p^3}$
B.$s{p^2}$,$s{p^3}$
C.$sp$,$s{p^2}$
D.$s{p^3}$,$sp$

Answer 1

To answer this question, you should recall the concept of Hybridization is defined as the concept of mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals. This intermixing is based on quantum mechanics.

Complete step by step solution:
The atomic orbitals of the same energy level can only take part in hybridization and both full-filled and half-filled orbitals can also take part in this process, provided they have equal energy. During the process of hybridization, the atomic orbitals of similar energy are mixed. Double bond has $s{p^2}$hybridization, single bonds have $s{p^3}$and triple bonds have ${\mathbf{sp}}$ hybridization.
$HC \equiv C - CH = C{H_2}\;$
In the above compound (from left): First carbon atom is in ${\mathbf{sp}}$ hybridization, Second carbon atom is as well in ${\mathbf{sp}}$ hybridization, Third carbon atom is in $s{p^2}$ hybridization and the Fourth carbon atom is also $s{p^2}$hybridization.

Therefore, the correct answer to this question is option C.

Note: Key Features of Hybridization:
1.Atomic orbitals with equal energies undergo hybridization and the number of hybrid orbitals formed is equal to the number of atomic orbitals mixing.
2.Filled orbitals with slightly different energies can also participate.
3.Hybridization happens only in case of bond formation and not in an isolated gaseous atom.
4.The shape of the molecule can be predicted if hybridization of the molecule is known.
You can use the formula to calculate the hybridization of complex molecules: the hybridization $(X)$ using the formula: $\dfrac{1}{2}(V + H - C + A)$ where
$V$= Number of valence electrons in the central atom
$H$= Number of surrounding monovalent atoms
$C$= Cationic charge
$A$= Anionic charge. The value of X will determine the hybridisation of the molecule. If $X$is 2 then $sp$; is 3 then $s{p^2}$ ; is 4 then $s{p^3}$; is 5 then $s{p^3}d$ ; is 6 then $s{p^3}{d^2}$ ; is 7 then $s{p^3}{d^3}$hybridization.