
The molecule which has a pyramidal shape is:
A.$PC{l_3}$
B.$S{O_3}$
C.$C{O_3}^{2 - }$
D.$N{O_3}^{2 - }$
Answer
534.3k+ views
Hint: To get the answer of this type of questions we should have some knowledge for the hybridization of the compound means by using hybridization concept we can define the shape of compounds by knowing the concept of hybridization we can calculate bond order, Bond length and other properties of the compound.
Complete step by step answer:
To get the hybridization firstly we will make the electronic configuration and after this we will make an excited electronic configuration and finally we will count the number of valence electrons and by counting s, p, and d orbital we will classify the hybridization and shape.
For $PC{l_3}$: the number of valence electrons in the phosphorus is 5.
So 3 electrons of phosphorus will be shared by the 3 electrons of the chlorine.
Hence 3 bonds and 1 lone pair of electron will present in the $PC{l_3}$
And we all know that 3 bond pairs and 1 lone pair of electrons makes the pyramidal shape.
Hence, $PC{l_3}$will have a pyramidal shape.
For $S{O_3}$: the number of valence electrons in sulfur will be 6.
And valency of oxygen atom is 2
So 2 electrons of each oxygen will share the 6 electrons of sulfur
So there will be 3 double bonds in $S{O_3}$ molecule. So $S{O_3}$will not have a pyramidal shape.
For $C{O_3}^{2 - }$: the number of valence electrons in carbon is 4 and two are loosed by the atom because carbon has -2 charge on it.
So $C{O_3}^{2 - }$also does not have a pyramidal shape.
For $N{O_3}^{2 - }$:the nitrogen atom has 5 valence electrons in the outermost shell.
The electrons remaining in the nitrogen will be 3 because 2 electrons are loosen by the nitrogen to form -2 charge on it.
So it does not have a pyramidal shape.
Hence the correct option is (A).
Note:
Hybridization the process of mixing or amalgamation of atomic orbitals of nearly same energy to form a set of new equivalent orbitals is termed as hybridization, hybridization is a hypothetical concept and had been introduced by pauling and slatter in order to supplement to valence bond theory to explain the shape of molecules.
Complete step by step answer:
To get the hybridization firstly we will make the electronic configuration and after this we will make an excited electronic configuration and finally we will count the number of valence electrons and by counting s, p, and d orbital we will classify the hybridization and shape.
For $PC{l_3}$: the number of valence electrons in the phosphorus is 5.
So 3 electrons of phosphorus will be shared by the 3 electrons of the chlorine.
Hence 3 bonds and 1 lone pair of electron will present in the $PC{l_3}$
And we all know that 3 bond pairs and 1 lone pair of electrons makes the pyramidal shape.
Hence, $PC{l_3}$will have a pyramidal shape.
For $S{O_3}$: the number of valence electrons in sulfur will be 6.
And valency of oxygen atom is 2
So 2 electrons of each oxygen will share the 6 electrons of sulfur
So there will be 3 double bonds in $S{O_3}$ molecule. So $S{O_3}$will not have a pyramidal shape.
For $C{O_3}^{2 - }$: the number of valence electrons in carbon is 4 and two are loosed by the atom because carbon has -2 charge on it.
So $C{O_3}^{2 - }$also does not have a pyramidal shape.
For $N{O_3}^{2 - }$:the nitrogen atom has 5 valence electrons in the outermost shell.
The electrons remaining in the nitrogen will be 3 because 2 electrons are loosen by the nitrogen to form -2 charge on it.
So it does not have a pyramidal shape.
Hence the correct option is (A).
Note:
Hybridization the process of mixing or amalgamation of atomic orbitals of nearly same energy to form a set of new equivalent orbitals is termed as hybridization, hybridization is a hypothetical concept and had been introduced by pauling and slatter in order to supplement to valence bond theory to explain the shape of molecules.
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