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Why does the electron-donating inductive effect (+I) of the isotopes of hydrogen decrease in the order $\ce{T} > \ce{D} > \ce{H}$? (where T is Tritium and D is Deuterium)

Google has nothing to offer. Does it have to do anything with mass, as the order implies?

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  • $\begingroup$ Where did you see this trend referenced? $\endgroup$ – jerepierre Dec 7 '14 at 16:05
  • $\begingroup$ Solomons and fryhle $\endgroup$ – Karan Singh Dec 7 '14 at 16:30
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Yes, it has a lot to do with mass. Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see "Note" below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.

A shorter bond has less volume to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) over resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.

The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

Similar arguments can be applied to tritium and it's even shorter $\ce{C-T}$ bond should be even more inductively electron donating towards carbon than deuterium.

Note: Bohr Radius Detail

Most introductory physics texts show the radius of the $n^\text{th}$ Bohr orbit to be given by

$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 m_\mathrm{e}}$$

where $Z$ is the atom's atomic number, $k_\mathrm{c}$ is Coulomb's constant, $e$ is the electron charge, and $m_\mathrm{e}$ is the mass of the electron. However, in this derivation it is assumed that the electron orbits the nucleus and the nucleus remains stationary. Given the mass difference between the electron and nucleus, this is generally a reasonable assumption. However, in reality the nucleus does move too. It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\mathrm{e}$ with the electron's reduced mass, $\mu_\mathrm{e}$

$$\mu_\mathrm{e} = \frac{m_\mathrm{e}\times m_\text{nucleus}}{m_\mathrm{e} + m_\text{nucleus}}$$

Now the equation for the Bohr radius becomes

$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 \mu_\mathrm{e}}$$

Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus

$$r_\text{heavy} \lt r_\text{light}$$

and consequently an electron will orbit closer to a deuterium nucleus than it will orbit a protium nucleus.

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    $\begingroup$ I don't get it. If D is more electronegative than H , then it should show -I effect (as our standard is based on H) ! (because it attract electrons more strongly than H , so acting like electron withdrawing) $\endgroup$ – Help needed Apr 19 '15 at 14:54
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    $\begingroup$ The problem is that OP actually meant +I effect : March's advanced chemistry , my class notes , my teacher , textbook , various websites after searching all have written that D shows +I effect (counterintuitively) . Also your statement "Therefore, deuterium is inductively more electron donating than protium towards carbon." implies the same - electron donation means +I effect. What I don't understand is your argument that D donates more electron density due to higher electronegativity , shouldn't it be the other way around ? For eg. F is electronegative and shows strong -I effect .(not +I) $\endgroup$ – Help needed Apr 19 '15 at 16:57
  • $\begingroup$ How big is that effect in numbers (for the Protium/Deutererium/Tritium example)? $\endgroup$ – Paŭlo Ebermann Jan 13 '18 at 17:01
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    $\begingroup$ This answer is wrong. The reduced mass of the electron in deuterium and tritium differs less than $4 \cdot 10^{-4}$ from the electron mass. This can not account for the observed differences in bond lengths (which e.g. are two orders of magnitude greater in $\ce{H2O / D2O}$). $\endgroup$ – aventurin Mar 25 '18 at 22:37
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A $\ce{C-D}$ bond has a shorter distance, thus can be in a more favorable position to donate electron density towards an electron deficient center. The $\ce{D}$ has a greater mass, and would show a shorter bond distance. This is reflected in the $+ I$ effects noted in the literature.

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    $\begingroup$ This answer offers nothing in addition to the existing one. $\endgroup$ – Jan Dec 24 '17 at 15:20

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