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Chemical Bonding and Molecular Structure question

2019 · Shift 1 · Q12
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Chemical Bonding and Molecular Structure question

2019 · Shift 1 · Q12

JEE AdvancedChemistryChemical Bonding and Molecular StructureMultiple correct+4 / −1
Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules posses permanent dipole moment at room temperature.
  1. A
    SO2SO_2SO2​, C6H5ClC_6H_5ClC6​H5​Cl, H2SeH_2SeH2​Se, BrF5BrF_5BrF5​
  2. B
    BeCl2BeCl_2BeCl2​, CO2CO_2CO2​, BCl3BCl_3BCl3​, CHCl3CHCl_3CHCl3​
  3. C
    NO2NO_2NO2​, NH3NH_3NH3​, POCl3POCl_3POCl3​, CH3ClCH_3ClCH3​Cl
  4. D
    BF3BF_3BF3​, O3O_3O3​, SF6SF_6SF6​, XeF6XeF_6XeF6​
View written solutionFree

Correct answer: A, C

  1. Principle used

A molecule has a permanent dipole moment if the vector sum of all bond dipoles is non-zero.

So we must check, in each option, whether all four molecules are polar at room temperature.


  1. Option A: SO2SO_2SO2​, C6H5ClC_6H_5ClC6​H5​Cl, H2SeH_2SeH2​Se, BrF5BrF_5BrF5​
  • SO2SO_2SO2​: bent shape due to lone pair on S. μ≠0\mu \neq 0μ=0 So it is polar.

  • C6H5ClC_6H_5ClC6​H5​Cl (chlorobenzene): the C−ClC-ClC−Cl bond gives a net molecular dipole; molecule is not symmetrically cancelling to zero. Hence polar.

  • H2SeH_2SeH2​Se: like H2O/H2SH_2O/H_2SH2​O/H2​S, it has a bent structure. Therefore bond dipoles do not cancel. Polar.

  • BrF5BrF_5BrF5​: square pyramidal geometry. This shape is unsymmetrical, so dipoles do not cancel. Polar.

Thus, all four are polar.

✅ Option A is correct.


  1. Option B: BeCl2BeCl_2BeCl2​, CO2CO_2CO2​, BCl3BCl_3BCl3​, CHCl3CHCl_3CHCl3​
  • BeCl2BeCl_2BeCl2​: linear Cl−Be−ClCl-Be-ClCl−Be−Cl Equal bond dipoles cancel. Non-polar.

  • CO2CO_2CO2​: linear, symmetric. Non-polar.

  • BCl3BCl_3BCl3​: trigonal planar, symmetric. Non-polar.

  • CHCl3CHCl_3CHCl3​: tetrahedral, unsymmetrical. Polar.

Since not all four are polar, this option is not correct.

❌ Option B is incorrect.


  1. Option C: NO2NO_2NO2​, NH3NH_3NH3​, POCl3POCl_3POCl3​, CH3ClCH_3ClCH3​Cl
  • NO2NO_2NO2​: bent molecule. Hence polar.

  • NH3NH_3NH3​: trigonal pyramidal due to lone pair on N. Polar.

  • POCl3POCl_3POCl3​: tetrahedral arrangement around P with one P=OP=OP=O and three P−ClP-ClP−Cl bonds; substituents are not identical, so dipoles do not cancel. Polar.

  • CH3ClCH_3ClCH3​Cl: tetrahedral but unsymmetrical because one Cl and three H. Polar.

Thus, all four are polar.

✅ Option C is correct.


  1. Option D: BF3BF_3BF3​, O3O_3O3​, SF6SF_6SF6​, XeF6XeF_6XeF6​
  • BF3BF_3BF3​: trigonal planar and symmetric. Non-polar.

  • O3O_3O3​: bent, polar.

  • SF6SF_6SF6​: octahedral and symmetric. Non-polar.

  • XeF6XeF_6XeF6​: distorted octahedral / monocapped octahedron, generally polar.

Since BF3BF_3BF3​ and SF6SF_6SF6​ are non-polar, not all four are polar.

❌ Option D is incorrect.


  1. Final conclusion

The options in which all four molecules possess permanent dipole moment at room temperature are: A, C\boxed{A,\ C}A, C​


  1. Comparison with stored correct answer

Stored correct answer: A, C

My derived answer: A, C

They match.

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