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Hint: For the formation of bond angle there must be at least three atoms and two bonds. The bond angle is the angle that is formed between three atoms across at least two bonds. This angle is different for different compounds.
Complete answer: We know the bond angle is the angle which is formed between two bonds. It depends on many factors which are stated below:
Forces of attraction or repulsion between atoms of molecules may increase or decrease the bond angle.
Hybridization also affects the bond angle. As the s character of hybrid orbital increases bond angle increases.
Electronegativity and lone pair also affect the bond angle. Increasing the lone pair and decreasing the electronegativity decreases the bond angle.
In the given question all atoms are the same but the charge is different so let’s discuss hybridization. Hybridization is the process of mixing atomic orbitals into new hybrid orbitals. $NO_2^ + $ has two bond pairs and zero lone pairs so hybridization of $NO_2^ + $is $sp$, and it will have linear geometry. It is clear that for linear geometry bond angle is ${180^ \circ }$. Therefore the bond angle of $NO_2^ + $ is${180^ \circ }$. $N{O_2}$ has two bond pairs and one lone pair, so it will have triangular planar geometry and $s{p^2}$ hybridization and $NO_2^ - $ will have two bond pairs, one lone pair, and also a negative charge. Due to this negative charge, there will be repulsions between negative charge and lone pair and thus it will decrease the bond angle. So the bond angle of $NO_2^ - $will be less than $N{O_2}$. From the given options ${180^ \circ }$ is the bond angle of $NO_2^ + $ remaining angles are ${134^ \circ }$ and ${115^ \circ }$, also bond angle of $NO_2^ - $is less than bond angle of $N{O_2}$. So the bond angle of $N{O_2}$is ${134^ \circ }$and bond angle of $NO_2^ - $is ${115^ \circ }$.
So the correct answer is option A that is ${180^ \circ },{134^ \circ },{115^ \circ }$.
Note:
Factors that affect the bond angle of a molecule are electronegativity, lone pair, hybridization, Forces of attraction, or repulsion. Lone pair- lone pair repulsions are stronger than lone pair-bond pair repulsions and these are stronger than bond pair-bond pair repulsions.
Complete answer: We know the bond angle is the angle which is formed between two bonds. It depends on many factors which are stated below:
Forces of attraction or repulsion between atoms of molecules may increase or decrease the bond angle.
Hybridization also affects the bond angle. As the s character of hybrid orbital increases bond angle increases.
Electronegativity and lone pair also affect the bond angle. Increasing the lone pair and decreasing the electronegativity decreases the bond angle.
In the given question all atoms are the same but the charge is different so let’s discuss hybridization. Hybridization is the process of mixing atomic orbitals into new hybrid orbitals. $NO_2^ + $ has two bond pairs and zero lone pairs so hybridization of $NO_2^ + $is $sp$, and it will have linear geometry. It is clear that for linear geometry bond angle is ${180^ \circ }$. Therefore the bond angle of $NO_2^ + $ is${180^ \circ }$. $N{O_2}$ has two bond pairs and one lone pair, so it will have triangular planar geometry and $s{p^2}$ hybridization and $NO_2^ - $ will have two bond pairs, one lone pair, and also a negative charge. Due to this negative charge, there will be repulsions between negative charge and lone pair and thus it will decrease the bond angle. So the bond angle of $NO_2^ - $will be less than $N{O_2}$. From the given options ${180^ \circ }$ is the bond angle of $NO_2^ + $ remaining angles are ${134^ \circ }$ and ${115^ \circ }$, also bond angle of $NO_2^ - $is less than bond angle of $N{O_2}$. So the bond angle of $N{O_2}$is ${134^ \circ }$and bond angle of $NO_2^ - $is ${115^ \circ }$.
So the correct answer is option A that is ${180^ \circ },{134^ \circ },{115^ \circ }$.
Note:
Factors that affect the bond angle of a molecule are electronegativity, lone pair, hybridization, Forces of attraction, or repulsion. Lone pair- lone pair repulsions are stronger than lone pair-bond pair repulsions and these are stronger than bond pair-bond pair repulsions.
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