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The non-planar molecule among the following is:
A. \[{B_2}{H_6}\]
B. \[{C_2}{H_4}\]
C. \[{C_6}{H_6}\]
D. \[BC{l_3}\]

Last updated date: 13th Jun 2024
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Hint: In order to solve this problem, we must firstly understand the Lewis structures of the compounds to identify the types of bonds and number of lone pairs present in each compound. After that, we can make the geometrical structures of the compounds, and then compare these structures.

Complete Step-by-Step answer:
To proceed with this problem, let us first draw the Lewis Structures of the given compounds:

We can observe that some of these compounds are forming double bonds within the structure while some of them have lone pairs of electrons. The property of double bonds or pi – bonds that we must know is that they exist in the form of the shape of p – orbitals. P – orbitals have the shape of a two- lobe structure that expands over a given axis. Hence, these bonds have three dimensional geometries. Also, when it comes to lone pairs, these electrons always exist outside the lattice of the bond formation. Hence, existence of a lone pair, makes the over geometry of a relatively flat molecule into a 3 – dimensional structure.
From the Lewis structures represented above, the only compound with either a pi bond or any lone pairs of electrons is \[{B_2}{H_6}\].
Hence, the non – planar molecule among the given compounds is \[{B_2}{H_6}\]

Hence, Option A is the correct option.

Note: The bridging hydrogen atoms provide one electron each. The\[\;{B_2}{H_2}\] ring is held together by four electrons which form two 3-center 2-electron bonds. This type of bond is sometimes called a 'banana bond'.