
What kind of molecule is $PC{{l}_{3}}$?
A) Ionic
B) Polar covalent
C) Non-polar covalent
D) None of these
Answer
584.4k+ views
Hint: Covalent compounds are those compounds formed by the sharing of electrons between the atoms. The atomic number of phosphorus is 15 and has 5 electrons in the valence shell and chlorine has seven electrons in the valence shell with atomic number 17.
Complete step by step solution:
-So here we have to take into consideration the electronic configurations and comparison of electronegativity of the atoms.
-Here let’s first check whether the molecule is polar or not. To check that we have to know which atom is more electronegative in the molecule.
-The value of electronegativity for F is 3.98 and for P is 2.19 Here we know that the fluorine is more electronegative than phosphorus so the electrons pair will have the affinity towards the chlorine atom forming a slight negative dipole or negative charge whereas phosphorus will yield a slight positive charge.
- So from the comparison between the electronegativity we can conclude that the molecule $PC{{l}_{3}}$ is polar.
-We know that the covalent compounds are those compounds with covalent bonds which are formed by the sharing of the electron pairs between the atoms.
-The electronic configuration of P is $\left[ Ne \right]3{{s}^{2}}3{{p}^{3}}$
The electronic configuration of Cl is $\left[ Ne \right]3{{s}^{2}}3{{p}^{5}}$
- So here the P has 5 electrons in the valence shell and requires three more electrons to complete its octet, whereas Cl requires one more electron to complete its octet.
- So both the atoms share their electrons mutually so that they obtain a stable noble gases configuration.
-So the molecule is polar and covalent also, so molecule $PC{{l}_{3}}$ is polar-covalent molecule.
So, the correct answer is “Option B”.
Note: If we speak about the hybridization for more clarification it is $s{{p}^{3}}$ hybridized with trigonal pyramidal shape with bond angle less than ${{109}^{\circ }}C$. When the P atom is in excited state an electron from 3s orbital moves to vacant 3d orbital and the vacant four orbitals i.e. one s and three p orbitals are formed in which three orbitals combined with Cl atom and one hybrid orbital will contain the lone pair.
Complete step by step solution:
-So here we have to take into consideration the electronic configurations and comparison of electronegativity of the atoms.
-Here let’s first check whether the molecule is polar or not. To check that we have to know which atom is more electronegative in the molecule.
-The value of electronegativity for F is 3.98 and for P is 2.19 Here we know that the fluorine is more electronegative than phosphorus so the electrons pair will have the affinity towards the chlorine atom forming a slight negative dipole or negative charge whereas phosphorus will yield a slight positive charge.
- So from the comparison between the electronegativity we can conclude that the molecule $PC{{l}_{3}}$ is polar.
-We know that the covalent compounds are those compounds with covalent bonds which are formed by the sharing of the electron pairs between the atoms.
-The electronic configuration of P is $\left[ Ne \right]3{{s}^{2}}3{{p}^{3}}$
The electronic configuration of Cl is $\left[ Ne \right]3{{s}^{2}}3{{p}^{5}}$
- So here the P has 5 electrons in the valence shell and requires three more electrons to complete its octet, whereas Cl requires one more electron to complete its octet.
- So both the atoms share their electrons mutually so that they obtain a stable noble gases configuration.
-So the molecule is polar and covalent also, so molecule $PC{{l}_{3}}$ is polar-covalent molecule.
So, the correct answer is “Option B”.
Note: If we speak about the hybridization for more clarification it is $s{{p}^{3}}$ hybridized with trigonal pyramidal shape with bond angle less than ${{109}^{\circ }}C$. When the P atom is in excited state an electron from 3s orbital moves to vacant 3d orbital and the vacant four orbitals i.e. one s and three p orbitals are formed in which three orbitals combined with Cl atom and one hybrid orbital will contain the lone pair.
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