Question

Arrange the following molecules in the decreasing order of their dipole moment:
A. $CB{r}_4$
B. $CHB{r}_3$
C. $C{H}_{2}B{r}_2$
D. $C{H}_{3}Br$

Answer 1

Hint As the number of hydrogens increase, the polarity also increases. The compounds which have symmetrical shape have 0 or less dipole moment as opposite dipole moment vectors cancel out each other.

Complete step by step solution:
In order to answer our question, we need to learn about the dipole moment. Now, dipole moment is used to measure the polarity of a compound. When there are positive and negative charges present at a distance, then dipole moment occurs. Dipole moment is dependent upon the nature of charges, the magnitude of charges as well as the distance between both the charges. The dipole moment is represented by the symbol $\mu$ . Mathematically, we can write:
$$\mu =q\times d$$
Where $\mu$ is the dipole moment, q is the magnitude of the charged particle and d is the distance between the charges. The value of the dipole moment is always a vector quantity. More the dipole moment of a compound, more is its polarity. And more the polarity of the compound, the more stable it is. Now, let us come to the question and study the compounds. We can see that the number of bromines is gradually decreasing and hydrogen is occupying its space. So, let us analyse the dipole moment by looking at the structures of the 4 molecules:

So, we can say that in $CB{r}_4$ the dipole moment becomes 0 due to its symmetrical figure, as the dipoles cancel out each other. Hence it has the lowest dipole. The dipole of $C{H}_{2}B{r}_2$ has more dipole moment than it as two bromines carry negative charge. Then as the hydrogens increase, the polarity increases. So, after that, $CHB{r}_3$ , and followed by $C{H}_{3}Br$ , has the highest dipole moment among the four.

NOTE: Dipole moment can be also considered similar to electric dipole and since it is a vector, it follows the vector law of addition too.