
The correct order of bond angles for the following is
A) $N{{H}_{3}}>PC{{l}_{3}}>BC{{l}_{3}}$
B) $BC{{l}_{3}}>N{{H}_{3}}>PC{{l}_{3}}$
C) $BC{{l}_{3}}>PC{{l}_{3}}>N{{H}_{3}}$
D) $PC{{l}_{3}}>BC{{l}_{3}}>N{{H}_{3}}$
Answer
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Hint: The answer here includes the fact that the bond angle is based on what type of hybridisation does the molecule possess and the bond angles assigned to that particular hybridisation.
Complete step by step answer:
Let us consider the hybridisation for the given molecules.
We have come across the hybridisation concept in our chapters on the lower classes of chemistry.
- Hybridisation is the mixing of the orbital of two or more atoms to form the new orbital having different energies and shapes.
- Thus, accordingly the hybridisation of $BC{{l}_{3}}$ is $s{{p}^{2}}$ hybridisation as the boron is the central metal atom and is bonded to three chlorine atoms with no presence of any lone pair on it.
Thus, the structure of $BC{{l}_{3}}$ is trigonal planar and the bond angle for $s{{p}^{2}}$ hybridised molecule having trigonal planar geometry is ${{120}^{0}}$
- For$PC{{l}_{3}}$ , the hybridisation is $s{{p}^{3}}$ as the phosphorus is the central metal atom and three chlorine atoms attached to it and has a lone pair of electrons in it. Thus, the shape of this molecule is trigonal pyramidal.
Therefore, the bond angle for this compound according to $s{{p}^{3}}$ hybridisation and the geometry is actually ${{109}^{0}}$ but due to strong lone pair – bond pair repulsion, it reduces to ${{103}^{0}}$.
- For, $N{{H}_{3}}$ hybridisation is $s{{p}^{3}}$and has a structure of pyramidal shape with a lone pair of electrons where it has bond angle of ${{109.5}^{0}}$ but due to bond pair and lone pair repulsion it reduces to ${{107}^{0}}$
Thus the correct order of bond angles for these molecules is $BC{{l}_{3}}>N{{H}_{3}}>PC{{l}_{3}}$
So, the correct answer is “Option B”.
Note: The hybridisation concept has to be made through for assigning the bond angles which is based on geometry and make sure that you know which atoms possess lone pairs and which do not because this plays an important role in deciding the geometry of molecules.
Complete step by step answer:
Let us consider the hybridisation for the given molecules.
We have come across the hybridisation concept in our chapters on the lower classes of chemistry.
- Hybridisation is the mixing of the orbital of two or more atoms to form the new orbital having different energies and shapes.
- Thus, accordingly the hybridisation of $BC{{l}_{3}}$ is $s{{p}^{2}}$ hybridisation as the boron is the central metal atom and is bonded to three chlorine atoms with no presence of any lone pair on it.
Thus, the structure of $BC{{l}_{3}}$ is trigonal planar and the bond angle for $s{{p}^{2}}$ hybridised molecule having trigonal planar geometry is ${{120}^{0}}$
- For$PC{{l}_{3}}$ , the hybridisation is $s{{p}^{3}}$ as the phosphorus is the central metal atom and three chlorine atoms attached to it and has a lone pair of electrons in it. Thus, the shape of this molecule is trigonal pyramidal.
Therefore, the bond angle for this compound according to $s{{p}^{3}}$ hybridisation and the geometry is actually ${{109}^{0}}$ but due to strong lone pair – bond pair repulsion, it reduces to ${{103}^{0}}$.
- For, $N{{H}_{3}}$ hybridisation is $s{{p}^{3}}$and has a structure of pyramidal shape with a lone pair of electrons where it has bond angle of ${{109.5}^{0}}$ but due to bond pair and lone pair repulsion it reduces to ${{107}^{0}}$
Thus the correct order of bond angles for these molecules is $BC{{l}_{3}}>N{{H}_{3}}>PC{{l}_{3}}$
So, the correct answer is “Option B”.
Note: The hybridisation concept has to be made through for assigning the bond angles which is based on geometry and make sure that you know which atoms possess lone pairs and which do not because this plays an important role in deciding the geometry of molecules.
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