Which one of the following is a correct set concerning molecule, hybridization, and shape
A.\[BeC{l_2},{\rm{ }}s{p^2}\], linear
B. \[BeC{l_2},{\rm{ }}s{p^2}\], triangular planar
C. \[BC{l_3},{\rm{ }}s{p^2}\], triangular planar
D.\[BC{l_3},{\rm{ }}s{p^3}\], tetrahedral
Answer
281.1k+ views
Hint: Molecules possess shapes. Small molecules i.e, molecules with a single central atom maintain shapes that can be efficiently predicted.
Beryllium chloride has sp hybridization and linear shape.
Complete step by step solution:Here in this question, we have to find out the
the correct set of molecules, hybridization, and shape.
A.\[BeC{l_2},{\rm{ }}s{p^2}\], linear
Here Be is the central atom.
It has two valence electrons. These electrons combine with the electrons of chlorine atoms and form two Be-Cl bonds. So, there is sp hybridization and linear shape.
So, it is incorrect.
B. \[BeC{l_2},{\rm{ }}s{p^2}\], triangular planar
This is also incorrect as this given molecule has a linear shape.
So, B is incorrect.
C. \[BC{l_3},{\rm{ }}s{p^2}\], triangular planar
Boron is the central atom.
It has three valence electrons. These electrons combine with the electrons of chlorine atoms and form three B-Cl bonds. So, there is {\rm{ }}s{p^2}\] hybridization and trigonal planar shape.

Image: Structure of boron trichloride.
So, C is correct.
D.\[BC{l_3},{\rm{ }}s{p^3}\], tetrahedral
This is incorrect as the given molecule has a planar shape, not a tetrahedral shape.
So, D is incorrect.
So,\[BC{l_3}\] has\[s{p^2}\] and a triangular planar shape.
So, option C is correct.
Note: Boron reacts with halogens to provide the corresponding trihalides. Boron trichloride is yielded industrially by direct chlorination of boron oxide and carbon.
In the laboratory,\[B{F_3}\] reacts with\[AlC{l_3}\] to yield\[BC{l_3}\]
through halogen exchange.
Beryllium chloride has sp hybridization and linear shape.
Complete step by step solution:Here in this question, we have to find out the
the correct set of molecules, hybridization, and shape.
A.\[BeC{l_2},{\rm{ }}s{p^2}\], linear
Here Be is the central atom.
It has two valence electrons. These electrons combine with the electrons of chlorine atoms and form two Be-Cl bonds. So, there is sp hybridization and linear shape.
So, it is incorrect.
B. \[BeC{l_2},{\rm{ }}s{p^2}\], triangular planar
This is also incorrect as this given molecule has a linear shape.
So, B is incorrect.
C. \[BC{l_3},{\rm{ }}s{p^2}\], triangular planar
Boron is the central atom.
It has three valence electrons. These electrons combine with the electrons of chlorine atoms and form three B-Cl bonds. So, there is {\rm{ }}s{p^2}\] hybridization and trigonal planar shape.

Image: Structure of boron trichloride.
So, C is correct.
D.\[BC{l_3},{\rm{ }}s{p^3}\], tetrahedral
This is incorrect as the given molecule has a planar shape, not a tetrahedral shape.
So, D is incorrect.
So,\[BC{l_3}\] has\[s{p^2}\] and a triangular planar shape.
So, option C is correct.
Note: Boron reacts with halogens to provide the corresponding trihalides. Boron trichloride is yielded industrially by direct chlorination of boron oxide and carbon.
In the laboratory,\[B{F_3}\] reacts with\[AlC{l_3}\] to yield\[BC{l_3}\]
through halogen exchange.
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