Assertion: In $C{{O}_{2}}$ molecule, C-atom undergoes $s{{p}^{2}}$ hybridization.
Reason: $C{{O}_{2}}$ molecule has net dipole moment zero.
(A) Both assertion and reason are correct and reason is the correct explanation of assertion.
(B) Both assertion and reason are correct but reason is not the correct explanation of assertion.
(C) Assertion is incorrect but reason is correct.
(D) Both assertion and reason are incorrect.

Hint: Hybridization is a concept of valence bond theory. When orbitals of different energy mix and pool their energy to form new orbitals of similar shape, size and energy, then this phenomenon is known as hybridization.

Complete step by step answer:
Let’s look at the solution of the question.
> To answer this question first we will calculate the hybridization of carbon atoms in $ C{{O}_{2}}$ molecule.
The formula for hybridization is given by:

Hybridization$=\dfrac{1}{2}(V+M+A-C)$…….equation 1

Where, V= number of valence electron on the central atom
M= number of monovalent atom
A= anionic charge present on the molecule
C= cationic charge present on the molecule
Now, in a $C{{O}_{2}}$ molecule
V=4, M=0, A=0, C=0
On putting the values in equation 1, we get.
Hybridization=2
Hence, the hybridization of $C{{O}_{2}}$ molecule is sp. So, the given assertion statement is false.
> Now, we know that dipole moment is dependent on the net bond moment in a molecule. In $C{{O}_{2}}$ molecules the bond moment of both the C=O bond are equal and aligned opposite to each other. So, they cancel their effect and the net dipole moment of the $C{{O}_{2}}$ molecule is 0.
Hence, the reason is correct.

So, the answer of the given question is option (C).

Note: The geometry and shape of the molecule depends on the hybridization of the molecule. The shapes of sp, $s{{p}^{2}}$ and $s{{p}^{3}}$ are linear, triangular planar and tetrahedral respectively. Carbon is found in these hybridizations only.