Question
Question: The decreasing order of dipole moment of the following is: (I) \(C{{H}_{3}}Cl\) (II) \(C{{H}_{3...
The decreasing order of dipole moment of the following is:
(I) CH3Cl
(II) CH3Br
(III) CH3F
(A) (I)>(II)>(III)
(B) (I)>(III)>(II)
(C) (II)>(I)>(III)
(D) (II)>(III)>(I)
Solution
As we know that dipole moment arises in any system where there is a separation of charge which can arise in ionic bonds as well as in covalent bonds. It occurs due to the difference in electronegativity between two chemically bonded atoms which brings partial charges on them. Here we have to tell the order of dipole moments in
haloalkanes.
Complete answer:
Let us discuss the concept of dipole moment followed by order of dipole moment in haloalkanes:-
-Dipole moment: It arises in any system where there is a separation of charge which can arise in ionic bonds as well as in covalent bonds. It occurs due to the difference in electronegativity between two chemically bonded atoms which brings partial charges on them.
-As it the product of magnitude of charges carried by the atoms and the distance between them, therefore it mainly depends on two factors which are:-
(i) Magnitude of charges on each atom: Higher the electronegativity difference between the atoms more will be the magnitude of charge on the atoms.
(ii) Distance between the atoms: Greater the distance between them, greater will be the dipole moment.
-As we know that fluorine is the most electronegative element and the order of electronegativity decreases down the halogen group which means the order of magnitude of charge on an atom is: F>Cl>Br.
-Greater the size of atom, greater will be the bond length which means order of bond length is C−F−Thesetwofactorsopposeeachotherandtheoveralldecreasingorderofdipolemomentis:(B)(I) > (III) > (II)$
Note:
-Remember that bond dipole moment is a vector quantity because it has both magnitude and direction.
-Also in case of polyatomic molecules, dipole moment is the vector sum of the all present bond dipoles in the molecule.