Strongest negative inductive effect group between trimethylammonium, ammonium and dimethylsulfonium groups Which of the following group exerts the strongest -I effect?
  
  
  $\ce{-N(CH3)3+}$
  $\ce{-NH3+}$
  $\ce{-S(CH3)2+}$   
  $\ce{-F}$
  


My idea is that positive species exerts more -I effect than neutral species. thus, 4) is neglected. Moreover, as nitrogen and sulfur have almost the same electronegativity 3) has two methyl groups and 2) none so 2) should exert more -I effect. However, the answer key says it to 1). Why is this?
The inductive effect can be quantitatively measured by the Hammett equation

$$\sigma(\ce{X}) = \mathrm{p}K_\mathrm{a}(\ce{H}) - \mathrm{p}K_\mathrm{a}(\ce{X})$$

where $\mathrm{p}K_\mathrm{a}(\ce{X})$ refers to the $\mathrm{p}K_\mathrm{a}$ of $\ce{X}$-substituted benzoic acid and $\mathrm{p}K_\mathrm{a}(\ce{H})$ to the $\mathrm{p}K_\mathrm{a}$ of unsubstituted benzoic acid. A more electron-withdrawing group will lead to a more acidic carboxylic acid group, a lower $\mathrm{p}K_\mathrm{a}(\ce{X})$, and hence a more positive $\sigma$ coefficient.

Values for a wide range of substituents in the meta and para positions have been tabulated in Table I of Ref. 1. Also included are Swain and Lupton's $F$ ($F$ for field) parameters which are designed to remove the influence of resonance effects on the original Hammett coefficients $\sigma$ (given in the same table; see Ref. 2 for original discussion).

$$\begin{array}{ccccc} \hline
\text{Entry number} &amp; \text{Substituent} &amp; \sigma_\mathrm{m} &amp; \sigma_\mathrm{p} &amp; F \\ \hline
\text{main text, p 167} &amp; \ce{NMe3+} &amp; 0.88 &amp; 0.82 &amp; 0.89 \\
58 &amp; \ce{NH3+} &amp; 0.86 &amp; 0.60 &amp; 0.92 \\
241 &amp; \ce{SMe2+} &amp; 1.00 &amp; 0.90 &amp; 0.98 \\
15 &amp; \ce{F} &amp; 0.34 &amp; 0.06 &amp; 0.45 \\ \hline
\end{array}$$

So, perhaps the answer should be $\ce{SMe2+}$. I don't have a good explanation for why, given that sulfur is less electronegative than nitrogen. I also do not think that the difference between $\ce{NMe3+}$ and $\ce{NH3+}$ can solely be attributed to the inductive effect of the methyl groups, as it is quite possible that there are solvation or hydrogen bonding effects that affect the $\mathrm{p}K_\mathrm{a}$ values. The bottom line is that I don't think this can be rationalised as simply as the question seems to want.

References


Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. Chem. Rev. 1991, 91 (2), 165–195. DOI: 10.1021/cr00002a004. (there is a non-paywall copy here)
Gardner Swain, C.; Lupton, E. C. Field and resonance components of substituent effects. J. Am. Chem. Soc. 1968, 90 (16), 4328–4337. DOI: 10.1021/ja01018a024.