Explore More
Similar Questions
Match the following molecules in List-$I$ with their respective dipole moments in List-$II$.
List-$I$ (Molecules) List-$II$ (Dipole moment) $A$. $HBr$ $I$. $1.04$ $B$. $H_2S$ $II$. $0$ $C$. $NH_3$ $III$. $0.79$ $D$. $CHCl_3$ $IV$. $0.95$ $V$. $1.47$
| List-$I$ (Molecules) | List-$II$ (Dipole moment) |
| $A$. $HBr$ | $I$. $1.04$ |
| $B$. $H_2S$ | $II$. $0$ |
| $C$. $NH_3$ | $III$. $0.79$ |
| $D$. $CHCl_3$ | $IV$. $0.95$ |
| $V$. $1.47$ |
DifficultTS EAMCET 2021
View SolutionIf the dipole moment of $Images-1$ is $1.5 \ D$,then the dipole moment of $Images-2$ will be .................. $D$.
Medium
View SolutionIdentify the option in which the molecules are arranged in the correct order of their dipole moments.
Match the compounds given in List-$I$ with their dipole moments given in List-$II$.
List-$I$ List-$II$ $(1)$ ${\rm{HBr}}$ $(A)$ $0.23$ $(2)$ ${{\rm{H}}_2}{\rm{S}}$ $(B)$ $1.04$ $(3)$ ${\rm{N}}{{\rm{F}}_3}$ $(C)$ $1.07$ $(4)$ ${\rm{CC}}{{\rm{l}}_4}$ $(D)$ $0.79$ $(5)$ ${\rm{CHC}}{{\rm{l}}_3}$ $(E)$ $0.00$ $(F)$ $0.95$
| List-$I$ | List-$II$ |
| $(1)$ ${\rm{HBr}}$ | $(A)$ $0.23$ |
| $(2)$ ${{\rm{H}}_2}{\rm{S}}$ | $(B)$ $1.04$ |
| $(3)$ ${\rm{N}}{{\rm{F}}_3}$ | $(C)$ $1.07$ |
| $(4)$ ${\rm{CC}}{{\rm{l}}_4}$ | $(D)$ $0.79$ |
| $(5)$ ${\rm{CHC}}{{\rm{l}}_3}$ | $(E)$ $0.00$ |
| $(F)$ $0.95$ |
Medium
View SolutionThe cyanide ion,$CN^{-}$ and $N_2$ are isoelectronic. But in contrast to $CN^{-}$,$N_2$ is chemically inert because of
MediumIIT 1992
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo