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How do you arrange $ NH_4^ + $ , $ B{F_3} $ , $ {H_2}O $ , $ {C_2}{H_2} $ , in increasing order of bond angle?

Last updated date: 23rd Apr 2024
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Hint: Before we get into the question, let's review what a bond angle is. The bond angle is the average angle between the orbitals containing bonding electron pairs around the core atom in a molecule. The temperature is measured in degrees, minutes, and seconds. The bond angle is influenced by the number of lone pairs, the size and electronegativity of the core atom, and the size of the surrounding atoms.

Complete answer:
A bond angle is the angle generated by three atoms over at least two bonds. The angle between the planes generated by the first three atoms and the plane created by the last three atoms is the angle of torsion for the four atoms linked together in a chain.
Now, coming to the question, in order to arrange the following according to the increasing order of their bond angle we will determine the bond angle first, then we will arrange accordingly.
The bond angle of $ NH_4^ + $ is $ {109^ \circ }28' $ because it has a complete tetrahedral structure.
Because $ B{F_3} $ has a trigonal planar structure, its bond angle is $ {120^ \circ } $ .
 $ {H_2}O $ does not have a complete tetrahedral structure because the two $ s{p^3} $ hybridised orbitals are occupied by two lone pairs. Because there are two lone pairs, the lone pair-bond pair repulsion is significantly higher, lowering the bond angle to $ {104.5^ \circ } $ .
Because $ {C_2}{H_2} $ is a linear molecule, its bond angle is $ {180^ \circ } $ .
Therefore from the above given information we can arrange the in the following way:
 $ {C_2}{H_2}\left( {{{180}^ \circ }} \right) > B{F_3} > \left( {{{120}^ \circ }} \right) > NH_4^ + \left( {{{109}^ \circ }28'} \right) > {H_2}O\left( {104 \cdot {5^ \circ }} \right) $ .

It's important to note that when it comes to bond angles for structures with lone pairs, all we need to know is that the bond angle will be smaller than expected. For a trigonal pyramid, for example, we just need to know that the bond angle that results is less than $ {109.5^ \circ } $ . This is due to the fact that it has a tetrahedral geometry, yet the lone pairs press down on the bonding pairs, resulting in a somewhat smaller bond angle.
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