Which has \[sp_3\] hybridization of the central atom
A. \[PCl_3\]
B. \[SO_3\]
C. \[BF_3\]
D. \[NO_3^{-}\]
Answer
273.3k+ views
Hint: Integrating two distinct entities to obtain another completely distinct entity broadly defines hybridization.
This idea is also obeyed by atomic orbitals and relying on the number and type of orbitals involved, there are distinct varieties of hybridization.
Complete step by step solution:Hybridization is the mixing of two atomic orbitals of identical or nearly identical energies to obtain new orbitals of comparable energy called hybrid orbitals.
Central atom orbitals undergo hybridization.
When one s- and the p- orbitals of the identical atomic orbitals combine to provide new hybrid orbitals of comparable energy, \[sp_3\] hybridization is seen.
The orbitals are towards the four corners of a regular tetrahedron.
Here in this question, we have to find out which has \[sp_3\] hybridization of the central atom.
A. \[PCl_3\]
P has 5 electrons in its valence shell. Three electrons are involved in the formation of three sigma bonds.
Rest two remain as lone pairs.
So, there will be \[sp_3\] hybridization.
So, A is correct.
B. \[SO_3\]
S has six electrons in its valence shell.
Here the two oxygen atoms are doubly bonded with the sulfur atom. So, all six electrons of S are involved in the formation of three sigma bonds and three pi bonds.
As there are three bond pairs, so there will be
\[sp_2\] hybridization.
So, B is incorrect.
C. \[BF_3\]
Boron has three electrons in its valence shell. All of these electrons are involved in the formation of three-sigma bonds.
As there are three bond pairs, so there will be
\[sp_2\] hybridization.
So, C is incorrect.
D. \[NO_3^{-}\]
There are six electrons on the nitrogen atom.
It forms three bonds with each oxygen atom. It also forms an additional bond with oxygen.
So, there are three sigma bonds and one pi-bond.
As there are three bond pairs, so there will be
\[sp_2\] hybridization.
So, D is incorrect.
So, only \[PCl_3\] forms \[sp_3\] hybridization.
So, option A is correct.
Note:The nitrate (\[NO_3\]) can be utilized as an organic or inorganic ester or salt of nitric acid including the \[NO_3^{-}\] ion.
Nitrates are greatly soluble in water and have an important part in the nitrogen cycle and nitrate pollution as well.
This idea is also obeyed by atomic orbitals and relying on the number and type of orbitals involved, there are distinct varieties of hybridization.
Complete step by step solution:Hybridization is the mixing of two atomic orbitals of identical or nearly identical energies to obtain new orbitals of comparable energy called hybrid orbitals.
Central atom orbitals undergo hybridization.
When one s- and the p- orbitals of the identical atomic orbitals combine to provide new hybrid orbitals of comparable energy, \[sp_3\] hybridization is seen.
The orbitals are towards the four corners of a regular tetrahedron.
Here in this question, we have to find out which has \[sp_3\] hybridization of the central atom.
A. \[PCl_3\]
P has 5 electrons in its valence shell. Three electrons are involved in the formation of three sigma bonds.
Rest two remain as lone pairs.
So, there will be \[sp_3\] hybridization.
So, A is correct.
B. \[SO_3\]
S has six electrons in its valence shell.
Here the two oxygen atoms are doubly bonded with the sulfur atom. So, all six electrons of S are involved in the formation of three sigma bonds and three pi bonds.
As there are three bond pairs, so there will be
\[sp_2\] hybridization.
So, B is incorrect.
C. \[BF_3\]
Boron has three electrons in its valence shell. All of these electrons are involved in the formation of three-sigma bonds.
As there are three bond pairs, so there will be
\[sp_2\] hybridization.
So, C is incorrect.
D. \[NO_3^{-}\]
There are six electrons on the nitrogen atom.
It forms three bonds with each oxygen atom. It also forms an additional bond with oxygen.
So, there are three sigma bonds and one pi-bond.
As there are three bond pairs, so there will be
\[sp_2\] hybridization.
So, D is incorrect.
So, only \[PCl_3\] forms \[sp_3\] hybridization.
So, option A is correct.
Note:The nitrate (\[NO_3\]) can be utilized as an organic or inorganic ester or salt of nitric acid including the \[NO_3^{-}\] ion.
Nitrates are greatly soluble in water and have an important part in the nitrogen cycle and nitrate pollution as well.
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