Question

$C{ H }_{ 2 }\quad =\quad C\quad =\quad O$
The hybridisation state of the middle carbon atom is:
A.$sp$
B.$s{p}^{2}$
C.$s{p}^{3}$
D.none

Answer 1

Hint:
Hybridization is the notion that newly hybridized orbitals are formed from the fusion of atomic
orbitals. This, in turn, influences the molecular bonding and geometry properties of the compound.

Complete step by step answer:
 The energy of the hybridized orbitals is lower than the energy compared to their separated, unhybridized counterparts. Due to this more stable compounds are formed.
The hybridization also has a greater role in determining the shape or the molecular geometry of the compound. the shapes can be linear, angular or V or bent, tetrahedral, seesaw, square pyramidal, and many more.
Hybridization is an expansion of valence bond theory.
The hybridization also depends on the types of the bond present between the atoms, like a single bond, double bond, or, triple bond. The sigma and the pi bonds are then identified in these bonds.
In a single bond, 1 sigma bond is present. In a double bond, there is 1 sigma and 1 pi bond. And in a triple bond, there is 1 sigma bond and 2 pi bonds.
In the case of carbon compounds, if only a single bond is present then that means it has only sigma bonds. Therefore, the hybridization is $s{p}^{3}$.
If one double bond and 2 single bonds are present then that means it has 3 sigma bonds and one pi bond. Therefore, the hybridization is $s{p}^{2}$.
If one triple bond and one single bond are present or two double bonds are present then that means it has 2 sigma bonds and 2 pi bonds. Therefore, the hybridization is $sp$.
In the given case, $C{ H }_{ 2 }\quad =\quad C\quad =\quad O$, The middle carbon atom has 2 double bonds. This shows that the hybridization of this carbon is $sp$. And hence, option (a) is the correct option.

Note:
The given facts are only applicable when we need to calculate the hybridization of carbon atoms. This would not work for other atoms.