Question

Find out incorrect order of the dipole moment among the following pairs of compound:

• A. $N{H}_3>N{F}_3$
• B. $\text{p-nitrophenol < o-nitrophenol}$
• C. $C{H}_{3}Cl>C{H}_{2}C{l}_2$
• D. $Si{F}_4<S{F}_4$

Answer 1

The correct option is B $\text{p-nitrophenol < o-nitrophenol}$


Since electronegativity of $F$ is high, thus the net dipolemoment in $N{F}_3$ is less compared to $N{H}_3$.

In p-nitrophenol, since $OH$ and $N{O}_2$ are in ${180}^o$ to each other the dipole moments gets added. Hence the net dipole moment is high in p-nitrophenol.

The net dipole moment of $C{H}_{2}C{l}_2$ is little less because of the higher electronegativity of $Cl$ atom.

$Si{F}_4$ is $s{p}^3$ hybridised and it is perfectly tetrahedral in shape. Thus the net dipole moment of $Si{F}_4$ is zero. But the shape of $S{F}_4$ is see saw. Thus the net dipole moment of $S{F}_4$ is high.