Answer
Verified
64.8k+ views
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'.
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'.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main