Question

In the compound ${C^1}{H_2} = {C^2} = {C^3}H = {C^4}{H_3}$, the hybridization of first, second carbon atom is:
A)$s{p^3} - sp$
B) $s{p^2} - s{p^3}$
C) $s{p^2} - sp$
D) $s{p^2} - s{p^2}$

Answer 1

Hint:We know that,
Hybridization:
Hybridization is the idea that atomic orbitals combine to form new hybridized orbitals which in turn, influences molecular geometry and bonding properties.
We know that the electrons which are present at the outermost shell of an atom are called valence electrons and valency of an electron is the number of electrons in which atom accepts or donate to form a bond.

Complete step by step answer:
Given,
The molecular formula of the compound is,
${C^1}{H_2} = {C^2} = {C^3}H = {C^4}{H_3}$
In this compound each of two electrons of the central atom forms collinear $sp$ sigma bonds to the terminal $s{p^2}$ hybridized carbons. The two remaining electrons of the central carbon occupy p orbitals and form pi bonds through an overlap of these p orbitals and the p orbitals of the terminal carbons.
Thus the First carbon is $s{p^2}$ hybridized and second carbon is \[sp\] hybridized.
Therefore option C is correct.

Note: The \[sp\] Hybridization:
The \[sp\] hybridization is practical when one s and one p orbital in the same shell of an atom mix to form two new equivalent orbitals. The new orbitals formed are called \[sp\] hybridized orbitals. It forms linear molecules with an angle of \[180^\circ \]. This hybridization is also called diagonal hybridization. Each \[sp\]hybridized orbital has an equal amount of s and p character, i.e., \[50\% \] s and p character.
The \[s{p^2}\] Hybridization:
This hybridization is possible when one s and two p orbitals of the same shell of an atom mix to form three equivalent orbits. The new orbitals formed are called \[s{p^2}\] hybrid orbitals. It is also called trigonal hybridization. Each of the hybrid orbitals formed has \[33.33\% \] s character and \[66.66\% \] ‘p’ character.