
The correct order of increasing bond angle order is:
(a) \[B{{F}_{3}}\]<\[N{{F}_{3}}\]<\[P{{F}_{3}}\]<\[Cl{{F}_{3}}\]
(b) \[Cl{{F}_{3}}\]<\[N{{F}_{3}}\]<\[P{{F}_{3}}\]<\[B{{F}_{3}}\]
(c) \[B{{F}_{3}}\]=\[N{{F}_{3}}\]<\[P{{F}_{3}}\]<\[Cl{{F}_{3}}\]
(d) \[B{{F}_{3}}\]<\[N{{F}_{3}}\]<\[P{{F}_{3}}\]<\[Cl{{F}_{3}}\]
Answer
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Hint: Bond angle is defined as the angle which has originated from the common atom and is present between two atoms. It is usually formed between the two covalent bonds. The angle of torsion for four atoms is bonded together like a chain is the angle present between the plane which is created by the first three atoms and the plane created by the last three atoms.
Complete answer:
The bond angle arises from the molecular geometry of the molecule. The three dimensional arrangement of the atoms in the molecule is known as molecular geometry. It tends to involve the shape of the molecule, the bond length, bond dissociation energy and bond angle.
The geometry of the molecule \[B{{F}_{3}}\] is trigonal planar in which the atom of fluorine lies in the same plane and exist the bond angle of \[{{120}^{o}}\]. The bond angle of the molecule \[P{{F}_{3}}\] is \[{{109}^{o}}\] because of the presence of lone pair at the top and the shape of the molecule is trigonal pyramidal. The bond angle of the molecule \[Cl{{F}_{3}}\] is \[{{90}^{o}}\] and it contains a lone pair in its molecule so it has a T shape. The bond angle for the molecule \[N{{F}_{3}}\]is\[{{107}^{o}}\] and it has a pyramidal shape.
So the bond angle of \[B{{F}_{3}}\] is the highest then comes the \[P{{F}_{3}}\] and then is \[N{{F}_{3}}\] and then the lowest is \[Cl{{F}_{3}}\].
So the correct option is, ‘(b) \[Cl{{F}_{3}}\]<\[N{{F}_{3}}\]<\[P{{F}_{3}}\]<\[B{{F}_{3}}\]’.
Note: The factors which affect the bond angle are hybridisation of the central atom. It also depends on the lone pairs of repulsion as if the number of lone pairs increases the bonded atoms will feel more repulsion and thus this will affect the bond angle. It also depends on the electronegativity so if the electronegativity of the central atom is less then it will have a lesser bond angle.
Complete answer:
The bond angle arises from the molecular geometry of the molecule. The three dimensional arrangement of the atoms in the molecule is known as molecular geometry. It tends to involve the shape of the molecule, the bond length, bond dissociation energy and bond angle.
The geometry of the molecule \[B{{F}_{3}}\] is trigonal planar in which the atom of fluorine lies in the same plane and exist the bond angle of \[{{120}^{o}}\]. The bond angle of the molecule \[P{{F}_{3}}\] is \[{{109}^{o}}\] because of the presence of lone pair at the top and the shape of the molecule is trigonal pyramidal. The bond angle of the molecule \[Cl{{F}_{3}}\] is \[{{90}^{o}}\] and it contains a lone pair in its molecule so it has a T shape. The bond angle for the molecule \[N{{F}_{3}}\]is\[{{107}^{o}}\] and it has a pyramidal shape.
So the bond angle of \[B{{F}_{3}}\] is the highest then comes the \[P{{F}_{3}}\] and then is \[N{{F}_{3}}\] and then the lowest is \[Cl{{F}_{3}}\].
So the correct option is, ‘(b) \[Cl{{F}_{3}}\]<\[N{{F}_{3}}\]<\[P{{F}_{3}}\]<\[B{{F}_{3}}\]’.
Note: The factors which affect the bond angle are hybridisation of the central atom. It also depends on the lone pairs of repulsion as if the number of lone pairs increases the bonded atoms will feel more repulsion and thus this will affect the bond angle. It also depends on the electronegativity so if the electronegativity of the central atom is less then it will have a lesser bond angle.
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