The correct order of dipole moment is

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.$CH_4$ is a non-polar tetrahedral molecule with $\mu =

Vedclass · Accepted Answer

$CH_4 < NF_3 < NH_3 < H_2O$

Vedclass · Answer

(A) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.$CH_4$ is a non-polar tetrahedral molecule with $\mu = 0 \ D$.In $NH_3$,the bond dipoles of $N-H$ bonds and the lone pair dipole are in the same direction,resulting in a high dipole moment $(\mu \approx 1.46 \ D)$.In $NF_3$,the bond dipoles of $N-F$ bonds and the lone pair dipole are in opposite directions,which partially cancel each other,resulting in a lower dipole moment $(\mu \approx 0.24 \ D)$.$H_2O$ has a bent geometry with two $O-H$ bonds and two lone pairs,resulting in a high dipole moment $(\mu \approx 1.85 \ D)$.Thus,the correct order is $CH_4 (0 \ D) < NF_3 (0.24 \ D) < NH_3 (1.46 \ D) < H_2O (1.85 \ D)$.

Similar Questions

The resultant gate and its Boolean expression in the given circuit is

The solution of $x \frac{dy}{dx} = y + x e^{y/x}$ with $y(1) = 0$ is

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $C$ is equal to (in $^{\circ}$)

${ }^{15}P_8 = A + 8 \cdot { }^{14}P_7 \Rightarrow A = $

The first,second and third Ionisation energies ($E_1, E_2$ and $E_3$) for an element are $7 \ eV, 12.5 \ eV$ and $42.5 \ eV$ respectively. The most stable oxidation state of the element will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module