Molecular Dipoles – Why CO2 Is Nonpolar but CO Is Polar Assertion (A): Carbon dioxide (CO2) has no resultant dipole moment, whereas carbon monoxide (CO) has a finite dipole moment. Reason (R): The structure of CO2 is linear and symmetric, represented as O = C = O.

Difficulty: Easy

Correct Answer: Both A and R are true and R is correct explanation of A

Explanation:


Introduction / Context:
Molecular polarity depends on both the bond dipoles and molecular geometry. Even when individual bonds are polar, the vector sum can cancel, yielding a nonpolar molecule. This distinction is essential in understanding solubility, infrared activity, and dielectric behavior of gases and solids.


Given Data / Assumptions:

  • CO2 geometry: linear (O–C–O with 180° bond angle).
  • CO geometry: diatomic, inherently polar if bond is polar.
  • Bond dipoles: C=O bonds are polar due to electronegativity difference.


Concept / Approach:

In CO2, two equal C=O bond dipoles act in opposite directions along the same line. Their vector sum is zero, so the net molecular dipole moment vanishes. By contrast, CO is a heteronuclear diatomic molecule; its single bond dipole cannot be canceled by symmetry, so the molecule exhibits a finite dipole moment, albeit modest and directionally nuanced due to molecular orbital effects.


Step-by-Step Solution:

Represent CO2 as O = C = O with equal bond moments μ along one axis.Vector addition: μ (to the left) + (−μ) (to the right) = 0 ⇒ nonpolar.For CO, only one bond dipole exists ⇒ net dipole ≠ 0.


Verification / Alternative check:

Spectroscopic data show CO2 has no permanent dipole and is IR-inactive in pure rotational spectrum; CO has a permanent dipole and shows microwave rotational spectra, confirming polarity differences.


Why Other Options Are Wrong:

  • R false: The linear symmetric structure of CO2 is well established; it correctly explains A.
  • Other combinations do not align with molecular geometry and measured dipoles.


Common Pitfalls:

Assuming any molecule with polar bonds must be polar; ignoring geometry and symmetry operations that cancel dipole vectors.


Final Answer:

Both A and R are true and R is correct explanation of A

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