Question

The compound with lowest dipole moment is.
A. \[C{H_3}F\]
B. \[C{H_3}Cl\]
C. \[C{H_3}Br\]
D. \[CHC{l_3}\]

Answer 1

Hint: The dipole moment is created once there is an electronegativity difference among the atoms. The dipole moment determines the compound's polarity. When the moment of the dipole is in the direction of the pair of electrons, then the dipole moment is more.

Complete Step by Step Solution:
In \[C{H_3}F\] 1 Carbon atom is attached to 3 Hydrogen atoms and 1 Fluorine atom. Fluorine is highly electronegative in nature than Carbon will going to attract the electron pairs so strongly, which makes the dipole moment quite high. That is why \[C{H_3}F\] is having a dipole moment of 1.82D.
In \[C{H_3}Cl\] 1 Carbon atom is attached to 3 Hydrogen atoms and 1 Chlorine atom. As Chlorine is more electronegative than carbon, which makes the electron pairs attract. As Chlorine is an electron-withdrawing group and only one Chlorine is present in the lattice. This makes \[C{H_3}Cl\] dipole quite high.
In \[C{H_3}Br\] 1 Carbon atom is attached to 3 Hydrogen atoms and 1 Bromine atom. Bromine is less electronegative in nature and the size and charge of the Bromine are quite low with respect to the given options. It is quite evident that \[C{H_3}Br\] will have a low dipole moment.
In \[CHC{l_3}\] 1 Carbon atom is attached to 3 Chlorine atoms and 1 Hydrogen atom. Chlorine is more electronegative than carbon, which makes the electron pairs attract. But due to the presence of C-Cl and C-H, the resultant dipole moment of the 2 C-Cl bond is cancelled. This makes the dipole moment of \[CHC{l_3}\] quite low at 1.03D.
Therefore, the correct option is C.

Note: The dipole moment also changes with the number of atoms attached to the carbon atom as we can see in options B and D.