Question

If the dipole moment of C-Cl bond is 1.2 D, then the resulting value of dipole moment in:

A) 1.2 and 0.6 respectively
B) 0.6 and 1.2 respectively
C) 1.2 D and more than 1.2 D respectively
D) More than 1.2 in both

Answer 1

Hint: Dipole moment is a vector quantity and has magnitude and direction and for simple molecules, dipole moment is product of charge and distance and formula for non-planar molecule (like in this question) involves angle factor as well.

Complete Step by step Solution
In the given question, the dipole moment of the Carbon-Chlorine bond is given and before proceeding directly to the calculation part of the answer first let us know about the dipole moment.
When a bond is formed, there are 2 atoms involved generally and if one of them is more electronegative than other than a partial positive and partial negative charge is created in both of the atoms which makes the bond polar in nature.
Now Dipole moment is just basically a measure of polarity of a chemical bond between 2 atoms or the molecules present. This dipole moment occurs due to the electronegativity difference between the 2 atoms involved.
Dipole moment is a vector quantity which involves magnitude as well as direction. Generally, a more electronegative atom has an arrow towards its side and this is the representation of the dipole moment.
Generally, dipole moment for a planar molecule: Dipole moment $(\mu )$ = Charge $ \times $ Distance
When the molecule is non planar (like in this case of question): Dipole moment $(\mu )$= $\sqrt {{\mu _1}^2 + \mu _2^2 + 2{\mu _1}{\mu _2}\cos \theta } $
Because dipole moment is a vector quantity, so it has direction also.
Now in the question, the molecules provided are the same in molecular formula but structure is different in which angle is different. So, in the formula, the value of $\theta $ will be different and rest all other things are the same.

So, in the first molecule given, the angle is $120^{\circ}$ between two chlorine atoms.
So Net dipole moment
$\mu = \sqrt {{{(1.2)}^2} + {{(1.2)}^2} + 2 \times 1.2 \times 1.2 \times \cos 120} = 1.2D$
Where
\[cos{\text{ }}120{\text{ }} = {\text{ }} - \dfrac{1}{2}\]
For second molecule, the net dipole moment:
$\mu = \sqrt {{{(1.2)}^2} + {{(1.2)}^2} + 2 \times 1.2 \times 1.2 \times \cos 60} = \sqrt 3 \times 1.2D = 2.07D$ , where \[\cos 60{\text{ }} = {\text{ }}\dfrac{1}{2}\]

So, the correct answer is C i.e. 1.2 D and more than 1.2 D respectively.

Note: If the two opposite atoms are there in the same molecule, then dipole moment becomes zero because they cancel each other and nullify their effect. Like in this case if it had been Para-chlorobenzene then the dipole moment would have been zero because as we discussed, it is a vector quantity and involves direction as well as magnitude.