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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.

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 ${180^ \circ }$ .
${C_2}{H_2}\left( {{{180}^ \circ }} \right) > B{F_3} > \left( {{{120}^ \circ }} \right) > NH_4^ + \left( {{{109}^ \circ }28'} \right) > {H_2}O\left( {104 \cdot {5^ \circ }} \right)$ .
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.