Answer
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Hint: We have to know that Lewis acid is a synthetic animal variety that contains a void orbital that is fit for tolerating an electron pair from a Lewis base to frame a Lewis adduct. A Lewis base, at that point, is any species that has a filled orbital containing an electron pair that is not engaged with holding yet may frame a dative bond with a Lewis acid to shape a Lewis adduct.
Complete answer:
We have to know that, any particle will consistently have an empty atomic orbital. On the off chance that not empty at any rate halfway filled antibonding MO. Also, there will be numerous particles with empty antibonding molecular orbitals.
In the event that you need to know whether a particle has an empty holding MO, electron checking and MO chart would help. However, all in all particles will have empty ABMO’s. Stringently, various holding atomic orbitals should rise to the quantity of antibonding sub-atomic orbitals. On the off chance that there is no vacant antibonding sub-atomic orbital, for example; all ABMO’s are involved, the particle won't exist.
When the boron compounds act as Lewis acids as a result of their empty orbital. Its octet is inadequate, in which boron can acknowledge an electron.
Therefore, the correct option is (D) vacant orbital.
Note:
We have to know that the boron is a synthetic component with the symbol $B$ and the nuclear number is five. Delivered completely by grandiose beam spallation and supernovae and not by heavenly nucleo-synthesis, it is a low-plenitude component in the Solar System and in the earth's outside layer.
Complete answer:
We have to know that, any particle will consistently have an empty atomic orbital. On the off chance that not empty at any rate halfway filled antibonding MO. Also, there will be numerous particles with empty antibonding molecular orbitals.
In the event that you need to know whether a particle has an empty holding MO, electron checking and MO chart would help. However, all in all particles will have empty ABMO’s. Stringently, various holding atomic orbitals should rise to the quantity of antibonding sub-atomic orbitals. On the off chance that there is no vacant antibonding sub-atomic orbital, for example; all ABMO’s are involved, the particle won't exist.
When the boron compounds act as Lewis acids as a result of their empty orbital. Its octet is inadequate, in which boron can acknowledge an electron.
Therefore, the correct option is (D) vacant orbital.
Note:
We have to know that the boron is a synthetic component with the symbol $B$ and the nuclear number is five. Delivered completely by grandiose beam spallation and supernovae and not by heavenly nucleo-synthesis, it is a low-plenitude component in the Solar System and in the earth's outside layer.