[AdiDex]

I've heard something that deuterium is more electropositive than protrium (Hydrogen)
So deuterium shows +I (Inductive ) effect..??
Can anybody give the reason...??

[Enthalpy]

Any difference of affinity for electrons is extremely small. Its heavier nucleus moves the center of mass among nucleus and electron, which changes the energy levels. Magnetic interaction would be even smaller.

Deuterium does differ a bit chemicall from hydrogen, more so than isomers of other elements, but that's essentially because the heavier deuterium is better localized than protium, and this permits stronger bonds with other atoms, especially by hydrogen bonds.

[AdiDex]

I asked to my teacher..
They said it is related about nuclear physics nd binding energy..
Nd it is clear that he will not gonna to tell me the answer....i have to find it...
So its the clue that it is about  nucleons specially neutrons..
Nd i haven't know anything about  nuclear chemistry..
Nd srry for poor english..

[phth]

The physics behind it is that two nuclei of the chemical bond vibrate modeled like 2 balls on a spring.  Force=-μ*dx²/dt². If reduced mass (μ=1/m₁+1/m₂) goes  , then the forces required to break the bond go  . The difference is the largest between deuterium and hydrogen.  This is observable when the loss of Hyodrgen or Deuterium is the rate determining step. Rate enhancements are  1-7 times larger for hydrogen.

[Irlanur]

Proton transfer reactions also often have a high tunneling probability. This probability is highly dependent on the mass.

[Enthalpy]

If reduced mass (μ=1/m₁+1/m₂) goes  , then the forces required to break the bond go  .

This defines the resonance frequency, not the force nor the energy. Anyway, if you find differences like sqrt(2)-1 Rydberg between deuterium and protium, the explanation is wrong.

[Enthalpy]

Proton transfer reactions also often have a high tunneling probability. This probability is highly dependent on the mass.

Sure, but this influences dynamic processes rather than electropositiveness that is observable on equilibria. Or?

[Enthalpy]

Its heavier nucleus moves the center of mass among nucleus and electron, which changes the energy levels.

The difference in optical spectra between atomic protium and deuterium is known to result from this, and equals 0.025% or 3.4meV, for the ground state of atomic hydrogen
http://cat.middlebury.edu/~PHManual/atomic.html (they forgot a zero there)

Whether the better localization of deuterium is more or less important than 3.4meV in a molecular bond is unclear to me.

[AdiDex]

I think by using  reduced mass explanation we can just explain that the 1st shell of deuterium is smaller than the hydrogen...
But can't explain electro positiveness of deuterium..
So please help me out....
And if you got the answer then it would be better if you  explain it because i'm beginner in chemistry..

[Borek]

I think by using  reduced mass explanation we can just explain that the 1st shell of deuterium is smaller than the hydrogen...
But can't explain electro positiveness of deuterium.

You can't treat these things as if they were unrelated. Due to the different nucleus mass deuterium reacts slightly differently than protium (for example google kinetic isotopic effect). If it reacts differently, its measurable properties (including those used to estimate its electronegativity) will differ - in the end electronegativity will be different as well.

It has nothing to do with the nuclear physics and binding energy suggested by your teacher.

[Enthalpy]

I think by using  reduced mass explanation we can just explain that the 1st shell of deuterium is smaller [BIGGER] than the hydrogen... But can't explain electro positiveness of deuterium.

The electron being farther away from the proton is less attracted.

[AdiDex]

I think by using  reduced mass explanation we can just explain that the 1st shell of deuterium is smaller [BIGGER] than the hydrogen... But can't explain electro positiveness of deuterium.

The electron being farther away from the proton is less attracted.

Deuterium's electron has greater reduced mass ..so acoording to bohr model
Radius is inversly proportional to mass of electron .
So greater mass lesser radius....

[AdiDex]

Deuterium is one of only four stable nuclides with an odd number of protons and odd number of neutrons. (2H, 6Li, 10B, 14N; also, the long-lived radioactive nuclides 40K, 50V, 138La, 180mTa occur naturally.) Most odd-odd nuclei are unstable with respect to beta decay, because the decay products are even-even, and are therefore more strongly bound, due to nuclear pairing effects. Deuterium, however, benefits from having its proton and neutron coupled to a spin-1 state, which gives a stronger nuclear attraction; the corresponding spin-1 state does not exist in the two-neutron or two-proton system, due to the Pauli exclusion principle which would require one or the other identical particle with the same spin to have some other different quantum number, such as orbital angular momentum. But orbital angular momentum of either particle gives a lower binding energy for the system, primarily due to increasing distance of the particles in the steep gradient of the nuclear force. In both cases, this causes the diproton and dineutron nucleus to be unstable.

I got it in Wikipedia  the link is http://en.m.wikipedia.org/wiki/Deuterium#Deuteron_mass_and_radius
Now can somebody explain me why deuterium is more stable...
How to apply Pauli's exlusion principle....

[rwiew]

It is worth noting that the relationship between electronegativity, however defined, and inductive effects here is more complicated that you might think. As people said above, the 1s level of the deuterium atom lies lower in energy than the 1s level of the protium atom - this you can explain using the reduced mass argument. So is it more favourable to add another electron to an H or D atom? You've got the binding energy (negative, released on binding of electron - this more favourable in D) and the added electron - electron repulsion (positive, costs - you would expect this to be a higher cost to D as he 1s orbital will be slightly shrunk compared to H), experimental data shows the electron affinity of D to be higher than that of H, so the binding energy must win out (http://scitation.aip.org/content/aip/journal/jcp/106/11/10.1063/1.473500). Ionisation energy (easiness to remove the existing electron, "electropositivity" measure if you like) must also be included in the calculation of total electronegativity - I'm using the Mulliken definition (http://en.wikipedia.org/wiki/Electronegativity#Mulliken_electronegativity), so from higher electron affinity and higher ionisation energy (because lower 1s level, (http://physics.nist.gov/PhysRefData/HDEL/energies.html) for D we get a higher electronegativity for D (i.e. it is not more electropositive, but less electropositive than H).

This however does not mean that it will show a -I effect, actually to the contrary - deuterated carboxylic acids are less acidic than the H-versions. This is due to the electronegativity differences being only slight and secondary isotope effects playing a role (these are a bit more complicated than the usually known primary effects, in a nutshell you need to be concerned with the vibrational levels differences of the whole molecule due to the isotopic change) - this JACS paper http://pubs.acs.org/doi/abs/10.1021/ja069103t is a good take on the subject. This of course you could call D showing +I effects, but shouldn't because it does not result from induction.

[Enthalpy]

I got it in Wikipedia  the link is http://en.m.wikipedia.org/wiki/Deuterium#Deuteron_mass_and_radius
Now can somebody explain me why deuterium is more stable...
How to apply Pauli's exlusion principle....

Mamma mia! You mix up the atom and the nucleus.