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In the given molecular formula, there are two carbon atoms (C) and four hydrogen atoms (H). This molecule is called ethene. Now, let's determine the structure of the ethene molecule with the help of hybridization.

To determine the hybridization of the carbon atoms, we first need to know the number of sigma (σ) bonds and valence electrons of each carbon atom. Carbon has four valence electrons and needs to form four bonds to achieve an octet configuration. Ethene has a double bond between the two carbon atoms and each carbon atom is bonded to two hydrogen atoms. Thus, each carbon atom forms three sigma (σ) bonds: one with the other carbon atom and two with hydrogen atoms. The double bond between the carbon atoms consists of one sigma (σ) bond and one pi (π) bond. Now, let's apply the formula for hybridization: \(hybridization = \frac{number\,of\,σ\,bonds + number\,of\,lone\,pair\,electrons}{2}\) For each carbon atom in ethene: \(hybridization = \frac{3\,σ\,bonds + 0\,lone\,pair\,electrons}{2} = \frac{3}{2}\) Since the value is greater than 1 and less than or equal to 2, the hybridization of each carbon atom is sp2.

In sp2 hybridization, one s and two p orbitals of the carbon atom mix to form three new sp2 hybrid orbitals. These orbitals arrange themselves in a trigonal planar geometry with bond angles of approximately 120°. The remaining p orbital, which is not involved in the hybridization, lies perpendicular to the plane of the sp2 hybrid orbitals.

Now that we know the hybridization and geometry of the carbon atoms in ethene, we can explain why all six atoms must lie in the same plane. The two carbon atoms and the four hydrogen atoms form sigma (σ) bonds using their sp2 hybrid orbitals. Since these sp2 hybrid orbitals are arranged in a trigonal planar geometry around each carbon atom, all the atoms bonded to the carbon atoms also lie in the same plane. Furthermore, the pi (π) bond between the two carbon atoms is formed by the overlap of the unhybridized p orbitals of the carbon atoms, which are perpendicular to the plane of the sp2 hybrid orbitals. This further supports the planar geometry of the ethene molecule. Thus, all six atoms in \(\mathrm{C}_{2} \mathrm{H}_{4}\) (ethene) must lie in the same plane because of the planar arrangement of the sp2 hybrid orbitals and the formation of a pi (π) bond between the carbon atoms.