Question:
Identify the correct set of molecules with zero dipole moment.
Identify the correct set of molecules with zero dipole moment.
Show Hint
Symmetrical molecules generally have zero dipole moment because individual bond dipoles cancel each other.
Examples:
\[
\mathrm{CO_2,\ CH_4,\ BF_3,\ CCl_4}
\]
Molecules containing lone pairs often become asymmetrical and usually possess non-zero dipole moment.
Updated On: Jun 3, 2026
- \(\mathrm{CO_2,\ NH_3,\ H_2O}\)
- \(\mathrm{NH_3,\ NF_3,\ BF_3}\)
- \(\mathrm{PF_3,\ NH_3,\ CH_4}\)
- \(\mathrm{CH_4,\ BF_3,\ CO_2}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
Dipole moment measures the separation of positive and negative charges in a molecule.
A molecule has:
• Zero dipole moment when the bond dipoles cancel each other due to symmetrical geometry.
• Non-zero dipole moment when the molecular shape is unsymmetrical and dipoles do not cancel. Thus, molecular geometry plays a very important role in determining dipole moment.
Step 1: Examine the molecules in option (A). \[ \mathrm{CO_2} \] has linear geometry: \[ O=C=O \] The two bond dipoles are equal and opposite, so: \[ \mu=0 \] \[ \mathrm{NH_3} \] has trigonal pyramidal geometry due to one lone pair on nitrogen. Dipoles do not cancel completely. Hence: \[ \mu \neq 0 \] \[ \mathrm{H_2O} \] has bent geometry because of two lone pairs on oxygen. Therefore: \[ \mu \neq 0 \] Thus, option (A) is incorrect.
Step 2: Examine the molecules in option (B). \[ \mathrm{NH_3} \] is polar: \[ \mu \neq 0 \] \[ \mathrm{NF_3} \] also has trigonal pyramidal geometry and possesses a small but non-zero dipole moment. \[ \mu \neq 0 \] \[ \mathrm{BF_3} \] has trigonal planar symmetrical structure. Bond dipoles cancel completely. \[ \mu=0 \] Since all molecules do not have zero dipole moment, option (B) is incorrect.
Step 3: Examine the molecules in option (C). \[ \mathrm{PF_3} \] has trigonal pyramidal geometry and is polar. \[ \mu \neq 0 \] \[ \mathrm{NH_3} \] is also polar. \[ \mu \neq 0 \] \[ \mathrm{CH_4} \] has tetrahedral symmetrical geometry. Dipoles cancel completely. \[ \mu=0 \] Hence, option (C) is incorrect.
Step 4: Examine the molecules in option (D). \[ \mathrm{CH_4} \] has tetrahedral symmetrical geometry. \[ \mu=0 \] \[ \mathrm{BF_3} \] has trigonal planar symmetrical geometry. \[ \mu=0 \] \[ \mathrm{CO_2} \] has linear symmetrical geometry. \[ \mu=0 \] Thus, all molecules in option (D) have zero dipole moment. Therefore: \[ \boxed{\text{Option (D) is correct}} \]
• Zero dipole moment when the bond dipoles cancel each other due to symmetrical geometry.
• Non-zero dipole moment when the molecular shape is unsymmetrical and dipoles do not cancel. Thus, molecular geometry plays a very important role in determining dipole moment.
Step 1: Examine the molecules in option (A). \[ \mathrm{CO_2} \] has linear geometry: \[ O=C=O \] The two bond dipoles are equal and opposite, so: \[ \mu=0 \] \[ \mathrm{NH_3} \] has trigonal pyramidal geometry due to one lone pair on nitrogen. Dipoles do not cancel completely. Hence: \[ \mu \neq 0 \] \[ \mathrm{H_2O} \] has bent geometry because of two lone pairs on oxygen. Therefore: \[ \mu \neq 0 \] Thus, option (A) is incorrect.
Step 2: Examine the molecules in option (B). \[ \mathrm{NH_3} \] is polar: \[ \mu \neq 0 \] \[ \mathrm{NF_3} \] also has trigonal pyramidal geometry and possesses a small but non-zero dipole moment. \[ \mu \neq 0 \] \[ \mathrm{BF_3} \] has trigonal planar symmetrical structure. Bond dipoles cancel completely. \[ \mu=0 \] Since all molecules do not have zero dipole moment, option (B) is incorrect.
Step 3: Examine the molecules in option (C). \[ \mathrm{PF_3} \] has trigonal pyramidal geometry and is polar. \[ \mu \neq 0 \] \[ \mathrm{NH_3} \] is also polar. \[ \mu \neq 0 \] \[ \mathrm{CH_4} \] has tetrahedral symmetrical geometry. Dipoles cancel completely. \[ \mu=0 \] Hence, option (C) is incorrect.
Step 4: Examine the molecules in option (D). \[ \mathrm{CH_4} \] has tetrahedral symmetrical geometry. \[ \mu=0 \] \[ \mathrm{BF_3} \] has trigonal planar symmetrical geometry. \[ \mu=0 \] \[ \mathrm{CO_2} \] has linear symmetrical geometry. \[ \mu=0 \] Thus, all molecules in option (D) have zero dipole moment. Therefore: \[ \boxed{\text{Option (D) is correct}} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Molecular Polarity Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions