Respuesta :
The correct answer is Cl₂ <CCl₄ < PCl₃ < CaCl₂ < CsCl (most polar)
Cl₂ is nonpolar or least polar as there is no differentiation in electronegativity of the two atoms of chlorine producing amongst them. CCl₄ exhibits four C-Cl bonds. The C-atom is electropositive and Cl is electronegative. Therefore, there are four dipoles in CCl₄. Though, these dipoles cancel each other because of the tetrahedral geometry of the molecule. Therefore, CCl₄ is less polar in comparison to others, however, more polar in comparison to Cl₂.
PCl₃ exhibits three dipoles because of electronegative chlorine atoms, and electropositive phosphorus atom. Though, the configuration of the molecule is trigonal pyramidal, producing it more polar in comparison to CCl₄. The CaCl₂ is ionic compound, therefore, polar in comparison to the covalent bond compounds.
CsCl is ionic compound, however, more polar in comparison to CaCl₂, as there is the higher difference in electronegativity of Cs and Cl compared with that between the Ca and Cl.
The arrangement of by increasing bond polarity, [tex]\rm \bold{ Cl_2 <CCl_4 < PCl_3 < CaCl_2 < CsCl }[/tex]
The polarity in the molecules is due to the difference in the electronegativities constituting atoms and symmetry of the molecule.
- [tex]\rm \bold{ Cl_2}[/tex] , there is no electronegative difference. hence it is nonpolar.
- [tex]\rm \bold{ CCl_4}[/tex],there is electronegative difference between C-Cl but it is a symmetrical molecule. Though polar than [tex]\rm \bold{ Cl_2}[/tex]
- [tex]\rm \bold{ PCl_3}[/tex], there is electronegative difference between P-Cl and assymetric hence polar.
- [tex]\rm \bold{ CaCl_2}[/tex], is ionic compound, therefore, polar in comparison to the covalent bond compounds.
- CsCl is ionic compound, however, more polar in comparison to [tex]\rm \bold{ CaCl_2}[/tex] , as there is the higher difference in electronegativity of Cs-Cl.
Hence, the arrangement of by increasing bond polarity[tex]\rm \bold{ Cl_2 <CCl_4 < PCl_3 < CaCl_2 < CsCl }[/tex]
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