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I was going through the general properties of carbonyl compounds and then found this (in the pic) data, where it says that dipole moment of acetone (if we assume $\ce{R = -CH3}$) is more than acetaldehyde which in turn is greater than that of formaldehyde. But I think, as $\ce{-CH3}$ is a $+I$ effecting group, so it will satiate the residing positive pole on carbon slightly and hence will make the $\ce{C=O}$ bond less polar and hence the sequence should be reversed! Please anyone make me correct where am I mistaking with the concept?

Picture source

EDIT :- I got my answer of the above confusion. Thanks to Physicsapproval. But with that I came to face another confusion. In the next part of the text (see the paragraph just below the pic in the Picture source) I found that "the larger the dipole moment the greater the polar character of the carbonyl group". So that means, $\ce{C=O}$ bond in acetone is more polar and least in the case of formaldehyde! Now in the next part it is mentioned that the bond energies of $\ce{C=O}$ bond in the mentioned compounds are: $$ \ce{H2C=O} \, ; 170~\mathrm{kcal/mol}$$ $$\ce{RHC=O} ~ ; 175~\mathrm{kcal/mol}$$ $$\ce{R2C=O}~; 180~\mathrm{kcal/mol}$$ Now I know that the more polar the bond is, the less is its bond energy. So, in this case why the trend is reversed, i.e., $\ce{C=O}$ bond of acetone being more polar has the highest bond energy compared to other two?

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    $\begingroup$ I think that is because -R group behave as electron donating like group, so the two vectors would further add up to make the dipole moment even greater. $\endgroup$
    – Physicsapproval
    May 29, 2017 at 14:13
  • $\begingroup$ Yes, thank you @Physicsapproval. I just forgot to consider that dipole moment is a vector quantity! $\endgroup$
    – chail10
    May 29, 2017 at 15:11

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the more polar the bond is, the less is its bond energy

I'm not sure where you are getting this from.

Take Acid halides as an example series. As the bond get more polar ( HI (0.4 D) < HBr (0.7 D) < HCl (0.9 D) < HF (1.9 D) ) the bond energies follow accordingly ( HI (295 kJ/mol) < HBr (362 kJ/mol) < HCl (428 kJ/mol) < HF (565 kJ/mol) )

As the charge separation gets larger (and therfore more polar) there is a stronger will to not have that large charge separation, as nature wants to naturally come to an equilibrium - and thus the bonds are more energetic.

Does this clear up your confusion?

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  • $\begingroup$ Thanks for your response @ChildishJack. As far as I got your point that you want to say, the more the charge separation, the more the bond tends to be ionic and hence the more is its bond energy. Yes I agree and your data too substantiate that logic. But then how would you describe the fact that ClOH is an acid and NaOH is a base, where In case of ClOH, Cl–O bond polarity < O–H bond polarity, so O–H bond dissociates easily and hence it's an acid and in case of NaOH, Na–O bond polarity > O–H bond polarity, so the Na–O bond dissociates easily and hence it's a base. Aren't both data contradict? $\endgroup$
    – chail10
    May 30, 2017 at 1:06
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    $\begingroup$ Contradict what? Your statement about the correlation between bond polarity and bond energy isn't always true. The opposite statement isn't always true either. There is simply no such rule, one way or another. There is nothing to contradict. $\endgroup$
    – Ivan Neretin
    May 31, 2017 at 19:55
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    $\begingroup$ Also, electrolytic dissociation is only distantly related to bond energy. $\endgroup$
    – Ivan Neretin
    May 31, 2017 at 20:01

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