Question

Which compound does not possess linear geometry
(a) \[C{H_2} = C{H_2}\]
(b) \[HC \equiv CH\]
(c) \[BeC{l_2}\]
(d)\[C{O_2}\]

Answer 1

Hint: The planar molecules have all their atoms in the same plane and possess \[s{p^2}\]hybridization. Unlike linear molecules, planar molecules do not have free rotation around bonds. For example., Phosgene (\[COC{l_2}\]), Sulphur trioxide (\[S{O_3}\]), Ethene, etc.

Complete step by step solution:To know which molecule is not linear, first, we must look at the hybridization of the given molecules.

For the determination of the hybridization of any molecule, we can count the total number of sigma bonds and lone pairs around the atom for which we are going to predict the hybridization.

The ethene (\[C{H_2} = C{H_2}\]) molecule has a carbon-carbon double bond and it exhibits 3 sigma bonds. Therefore, the value of 3 will be equal to \[s{p^2}\] hybridization and the bond angle for it will be \[120^\circ \] . Hence, we know that the molecule with \[s{p^2}\] hybridization always has a planar structure. The Sulphur trioxide (\[S{O_3}\]), boron trifluoride (\[B{F_3}\]), etc. also belong to the class of planar molecules.
Now look at the structure of acetylene (\[HC \equiv CH\]) molecule which has a carbon-carbon triple bond and it exhibits 2 sigma bonds. Therefore, the value of 2 will be equal to \[sp\] hybridization i.e., linear geometry, and the bond angle for it will be \[180^\circ \] .
Similar to, acetylene (\[HC \equiv CH\]) molecule, the \[BeC{l_2}\]and \[C{O_2}\]also have \[sp\] hybridization and linear geometry (\[Cl - Be - Cl\]and \[O = C = O\]).

Therefore from the above explanation we can say option (a) will be the correct option:

Note: The ethene is considered as the simplest alkene and it is obtained from the cracking of long-chain hydrocarbons. \[BeC{l_2}\]is an inorganic compound and it is highly soluble in polar solvents. \[C{O_2}\]is used in a fire extinguisher and it acts as a refrigerant.